Category Archives: Teachers’ Resources

SCHEMES OF WORK FORM ONE ENGLISH TERM 1-3

 
SCHEME  OF WORK               FORM  ONE  ENGLISH            TERM  ONE  20…………………
 
WK NO

 

L/
NO
TOPIC
 
 
SUBTOPIC
LESSON / SPECIFIC
OBJECTIVES
TEACHING/ LEARNING
ACTIVITIES
MATERIALS       /
RESOURCES
REFERENCES
1-4
 
Selection of students.
 
 
5 1 Listening and speaking. Introductions.

 

By the end of the lesson, the learner should be able to

conduct self introduction and express own interests.

Ice-breaking.

Listening.

Speaking.

Role- playing.

 

 

 

Name tags.

 

 

 

Teacher’s resources.

 
2-3 Introduction to the English course. The language skills. The learner should be able to

state the importance of the language skills.

Brain storming.

Listening.

Speaking.

Reading.

Writing.

Newspaper extracts.  

 

Teacher’s resources.

 
4 Listening. Pronunciation. The learner should be able to

demonstrate awareness that spelling of words and their pronunciation may not be related.

Listening.

Speaking.

Reading.

Writing.

Flash cards.

Charts.

Head Start English

Bk I Pg 1

TG Pg 1

 
5 Reading. Comprehension. The learner should be able to

read a passage and answer derived questions.

Reading aloud.

Listening.

Writing.

Discussion.

Supplementary readers.

Dictionary.

Head Start English

Bk I Pg 1-2

TG Pg 1

 
6 Reading. Word power. The learner should be able to

demonstrate ability to use the dictionary to build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 3-4

TG Pg 1

 

 

6 1-2 Grammar. Parts of speech. The learner should be able to

identify the parts of speech.

Reading aloud.

Listening.

Writing.

Discussion.

Extracts from readers.

Pictures.

 

Head Start English

Bk I Pg 4-6

TG Pg 2

 
3 Grammar. Nouns. The learner should be able to

identify and explain functions of the main groups of nouns.

Reading.

Listening.

Answering questions. Discussion.

 

Extracts from readers.

Charts.

 

 

Head Start English

Bk I Pg 4-6

TG Pg 2

 
6 4 Grammar. Articles. The learner should be able to

use definite and indefinite articles correctly.

Reading.

Listening.

Answering questions. Discussion.

 

Pictures.

Charts.

Extracts from readers.

Head Start English

Bk I Pg 6

TG Pg 3

 
5 Writing. Handwriting. The learner should be able to

explain the importance of writing neatly and legibly.

Listening;

Writing.

Library books;

Samples of hand-writing

Head Start English

Bk I Pg 6-7

TG Pg 3

 
6 Listening and speaking. Close shave. The learner should be able to

express views on a given subject with supporting arguments.

Reading.

Listening.

Answering questions. Discussion.

 

Supplementary readers. Head Start English

Bk I Pg 7

TG Pg 3

 

 

7 1 Speech work. Etiquette. The learner should be able to

use words and expressions that show courtesy.

Reading.

Listening.

Role-playing.

Discussion.

 

Audio tapes.

Pictures.

 

Head Start English

Bk I Pg 8

TG Pg 4

 
2 Speech work. /b/ and /d/ sounds. The learner should be able to

pronounce the sounds correctly.

Listening.

Reading.

Writing.

Flash cards.

Charts.

Head Start English

Bk I Pg 8

TG Pg 4

 
3 Reading. Comprehension. The learner should be able to

read a passage and answer derived questions.

Reading aloud.

Listening.

Writing.

Discussion.

Supplementary readers.

Dictionary.

Head Start English

Bk I Pg 9-11

TG Pg 4

 
4 Reading. Word power. The learner should be able to

demonstrate ability to use the dictionary to build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 11-12

TG Pg 5

 
5 Grammar. Abstract nouns and concrete nouns. The learner should be able to

differentiate between abstract nouns and concrete nouns.

 

Reading aloud.

Listening.

Writing.

Exercise review.

Extracts from readers.

Pictures.

Head Start English

Bk I Pg 12-13

TG Pg 5

 
6 Writing. Spelling. The learner should be able to

identify words that are pronounced the same way but have different spelling and meaning.

Listening.

Reading.

Answer questions.

Writing.

Review exercise.

 

Flash cards. Head Start English

Bk I Pg 12-13

TG Pg 6

 
8 1 Reading. Close shave. The learner should be able to

explain how tone and choice of words can be used to show respect on telephone.

Role-playing.

Reading.

Speaking.

Listening.

 

Supplementary readers. Head Start English

Bk I Pg 15

TG Pg 6

   
2 Listening & Speaking. /k/ and /g/ sounds. The learner should be able to

pronounce the sounds correctly.

Listening.

Reading.

Writing.

Flash cards.

Charts.

Head Start English

Bk I Pg 16

TG Pg 7

   
3 Reading. Comprehension. The learner should be able to

read a passage and answer derived questions.

Reading aloud.

Listening.

Writing.

Discussion.

Supplementary readers.

Dictionary.

Head Start English

Bk I Pg 16-19

TG Pg 8

   
4 Reading. Word power. The learner should be able to

demonstrate ability to use the dictionary to build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 19-20

TG Pg 8

   
5 Grammar. Countable and non-countable nouns. The learner should be able to use countable and non-countable nouns in sentences correctly. Listening.

Writing.

Discussion.

Exercise review.

Pictures.

Extracts from readers.

Head Start English

Bk I Pg 20-22

TG Pg 8

   
6 TEST              

 

9 1 Reading. Comprehension. The learner should be able to

read a passage and answer derived questions.

Reading aloud.

Listening.

Writing.

Discussion.

Supplementary readers.

Dictionary.

Head Start English

Bk I Pg 25-27

TG Pg 10

 
2 Reading. Word power. The learner should be able to

demonstrate ability to use the dictionary to build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 27-28

TG Pg 10

 
3 Listening & Speaking. /t∫/ and /dЗ/ sounds. The learner should be able to

pronounce the sounds correctly.

Listening.

Reading.

Writing.

Flash cards.

Charts.

Head Start English

Bk I Pg 25

TG Pg 9

 
9 4 Grammar. Plural nouns. The learner should be able to

use plural nouns in sentences correctly.

Listening.

Writing.

Discussion.

Exercise review.

Supplementary readers. Head Start English

Bk I Pg 28-30

TG Pg 11

 
5 Writing. Simple sentences and compound sentences. The learner should be able to

write a variety of sentences.

Listening.

Speaking.

Reading.

Writing.

Supplementary readers. Head Start English

Bk I Pg 31

TG Pg 11

 
6 Speech work. Close shave. The learner should be able to

tell similar stories.

Narrating.

Listening.

Discussing.

Readers. Head Start English

Bk I Pg 31

TG Pg 12

 

 

10 1 Reading. Forms of literature. The learner should be able to

Describe different forms of oral literature.

Listening.

Speaking.

Reading.

Writing.

Charts.

Readers.

Relevant copies of resources.  
2 Reading. Introduction to oral literature. The learner should be able to state characteristics of  oral literature. Listening.

Speaking.

Reading.

Writing.

Chart. Teacher’s notes.  
3 Listening & speaking.

 

Trickster stories. The learner should be able to

Identify features of trickster narratives.

Narrating.

Reading.

Speaking.

Writing.

Pictures. Head Start English

Bk I Pg 32

TG Pg 12-13

 
4 Listening & speaking.

 

The sounds /∫/ and /t∫/ The learner should be able to

distinguish between the two sounds.

Narrating.

Reading.

Speaking.

Writing.

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 32

TG Pg 13

 
5 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Inferring meanings of given words.

Writing.

 

 

Readers.

Charts.

Head Start English

Bk I Pg 32-34

TG Pg 14

 
6 Grammar. Articles. The learner should be able to

use definite and indefinite articles correctly.

Narrating.

Reading.

Speaking.

Writing.

Audio tapes.

Readers.

Head Start English

Bk I Pg 35-37

TG Pg 14-15

 
11 1 Writing. Paragraphs. The learner should be able to

write in paragraph form correctly.

Reading.

Speaking.

Writing.

Supplementary readers. Head Start English

Bk I Pg 37-38

TG Pg 15

 
2 Listening & speaking.

 

The sounds /f/ and /v/. The learner should be able to

pronounce the sounds /f/ and /v/correctly.

Listening.

Speaking.

Reading.

Writing.

 

Flash cards. Head Start English

Bk I Pg 39

TG Pg 16

 
3 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Inferring meanings of given words.

Writing.

 

Readers.

Charts.

Head Start English

Bk I Pg 39-41

TG Pg 16

 
4 Writing. Paragraphing. The learner should be able to

Write organized paragraphs to illustrate preparation of a given dish.

Listening.

Speaking.

Reading.

Writing.

 

Extracts from readers. Head Start English

Bk I Pg 43-44

TG Pg 17

 
5 Listening & speaking.

 

The sounds /s/ and /z/. The learner should be able to

pronounce the sounds /s/ and /z/correctly.

Listening.

Speaking.

Reading.

Writing.

 

Flash cards.

Audio tapes.

Head Start English

Bk I Pg 45

TG Pg 18

 
6 Grammar. Personal pronouns. The learner should be able to identify personal pronouns and use the correctly. Speaking.

Listening.

Reading.

Writing.

 

Pictures.

Extracts from readers.

Head Start English

Bk I Pg 48-50

TG Pg 19-20

 

 

12 1 Reading. Using the library. The learner should be able to identify sections of a library and use the library facilities properly.

 

Listening.

Speaking.

Reading.

Writing.

 

School library.

Model of library sections.

Teacher’s resources.

 

 
2 Reading. Word power. The learner should be able to

demonstrate ability to use the dictionary to build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 27-28

TG Pg 10

 
12 3 Listening & speaking.

 

The sounds /r/ and /l/. The learner should be able to

pronounce the sounds /r/ and /l/correctly.

Listening.

Speaking.

Reading.

Writing.

 

Flash cards.

Audio tapes.

Head Start English

Bk I Pg 53

TG Pg 20

 
4 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Inferring meanings of given words.

Writing.

 

Readers.

Charts.

Head Start English

Bk I Pg 61-64

TG Pg 24-25

 
5 Grammar.

 

 

Possessive pronouns. The learner should be able to identify and use possessive pronouns correctly. Listening.

Speaking.

Reading.

Writing.

 

Flash cards.

Audio tapes.

Head Start English

Bk I Pg 53,57-58

TG Pg 20, 25

 
6 Writing.

 

 

Paragraph unity. The learner should be able to link ideas into cohesive paragraphs. Narrating.

Listening.

Speaking.

Reading.

Writing.

 

Supplementary resources. Head Start English

Bk I Pg 53,57-58

TG Pg 20, 25

 
13,

14

END  OF  TERM  ONE  EXAMINATIONS  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 
SCHEME  OF WORK               FORM  ONE  ENGLISH            TERM  TWO    20…………………
 
WK NO

 

L/
NO
TOPIC
 
 
SUBTOPIC
LESSON / SPECIFIC
OBJECTIVES
TEACHING/ LEARNING
ACTIVITIES
MATERIALS       /
RESOURCES
REFERENCES
 
1 1 Listening and speaking. Riddles. The learner should be able to mention characteristics of riddles. Listening.

Speaking.

Listening.

Flash cards. Head Start English

Bk I Pg 71-72

TG Pg 30

 
2 Reading and writing. Reading

Comprehension.

The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Inferring meanings of given words.

Writing.

 

Readers.

Charts.

Head Start English

Bk I Pg 72-74

TG Pg 31

 
3 Speaking. Debating. The learner should be able to

argue for or against a given subject.

Listening.

Speaking.

Reading.

Writing.

 

Audio tapes. Head Start English

Bk I Pg 75

TG Pg 31

 
4 Grammar.

 

 

Simple present tense. The learner should be able to

construct sentences in simple present tense.

Listening.

Speaking.

Reading.

Writing.

 

Pictures.

Extracts from readers.

Head Start English

Bk I Pg 75-76

TG Pg 32

 
5 Writing. Capital letters and commas. The learner should be able to use capital letters and commas correctly. Narrating.

Listening.

Speaking.

Reading.

 

 

Extracts from readers. Head Start English

Bk I Pg 77

TG Pg 32

 
6 Speaking. Close shave. The learner should be able to tell double-meaning in a given story and tell similar stories. Listening.

Speaking.

Reading.

Role – playing.

 

Supplementary resources. Head Start English

Bk I Pg 78

TG Pg 32

 

 

 

 

 

 

 

 

 

 

 

 

2 1 Listening & speaking.

 

The sounds /a/ and /æ/. The learner should be able to

pronounce the sounds /a/ and /æ/.correctly.

Listening.

Speaking.

Reading.

Writing.

 

Flash cards.

Audio tapes.

Head Start English

Bk I Pg 79

TG Pg 33

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Inferring meanings of given words.

Writing.

 

Readers.

Charts.

Head Start English

Bk I Pg 61-64

TG Pg 24-25

 
3 Reading. Word power. The learner should be able to

demonstrate ability to use the dictionary to build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 82-83

TG Pg 33-34

 
 4

 

Grammar. Simple past tense. The learner should be able to construct sentences in the simple past tense. Competitions.

Listening.

Writing.

 

Extracts from readers. Head Start English

Bk I Pg 83-84

TG Pg 34-35

 
5 Writing. Punctuation marks. The learner should be able to use punctuation marks correctly. Listening.

Speaking.

Reading.

Writing.

 

Extracts from readers. Head Start English

Bk I Pg 85-86

TG Pg 35

 
6 Speaking. Close shave. The learner should be able to display the ability to reason and write tell similar stories. Listening.

Speaking.

Reading.

 

Extracts from readers. Head Start English

Bk I Pg 86

TG Pg 35

 

 

3 1 Reading. Short story. The learner should be able to read, interpret and analyse the plot, characters and themes in a short story. Express reading.

Dramatization.

Discussing.

Supplementary readers. Anthology of selected stories.  
2 Reading. Extensive reading. The learner should be able to deduce information from an extract. Silent reading.

Writing short notes.

Discussing.

Newspapers.

A variety of resources.

Suitable references.  
3 3 Listening & speaking.

 

The sounds /ae/ and /Л/. The learner should be able to

pronounce the sounds /ae/ and / Л /correctly.

Listening.

Speaking.

Reading.

Writing.

 

Flash cards.

Audio tapes.

Head Start English

Bk I Pg 87

TG Pg 36-38

 
4 Reading. Comprehension. The learner should be able to answer questions based on a passage.

Infer meanings of new words and phrases.

Reading.

Writing.

Review exercise.

Supplementary readers.

Dictionary.

 

Head Start English

Bk I Pg 87-88

TG Pg 38-39

 
5 Grammar. Regular and irregular verbs. The learner should be able to identify regular verbs and irregular verbs and use both correctly. Listening.

Speaking.

Reading.

Writing.

 

Extracts from readers.

Charts.

Head Start English

Bk I Pg 91-92

TG Pg 40

 
6 Writing. Diaries. The learner should be able to keep a diary for one week. Listening.

Speaking.

Reading.

Writing.

 

Extracts from readers. Head Start English

Bk I Pg 92

TG Pg 40

 

 

4 1 Listening. Poetry. The learner should be able to identify characteristics and functions of poems. Listening.

Discussing.

Samples of poems. Poetry anthology.  
2 Reading. Using a library. The learner should be able to explain how books are arranged in a library using a catalogue. Listening.

Note taking.

Discussion.

School library /

Model of a school library

Teacher’s resources.  
3 Listening and speaking. Ideophones and onomatopoeia. The learner should be able to define ideophones and onomatopoeia and use them effectively. Listening.

Speaking.

Reading.

Writing.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 94-95

TG Pg 41

 
4 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Inferring meanings of given words.

Writing.

 

Dictionary. Head Start English

Bk I Pg 95-98

TG Pg 41

 
4 5 Grammar. Regular and irregular verbs. The learner should be able to identify regular verbs and irregular verbs and use both correctly. Listening.

Speaking.

Reading.

Writing.

 

Extracts from readers.

Charts.

Head Start English

Bk I Pg 98-99

TG Pg 42

 
6 Writing. Formal letter. The learner should be able to write a formal letter in the correct format. Listening.

Reading.

Writing.

 

Real letters. Head Start English

Bk I Pg 101-2

TG Pg 43

 

 

5 1 Listening and speaking. Pronunciation of /a:/ and /З:/ Pronounce the sounds /a:/ and /З:/ correctly. Listening.

Speaking.

Reading.

Writing.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 103

TG Pg 45

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Inferring meanings of given words.

Writing.

 

Dictionary. Head Start English

Bk I Pg 103-5

TG Pg 45

 
3 Reading. Word power. The learner should be able to

demonstrate ability to use the dictionary to build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 105

TG Pg 45

 
4 Reading. Short story. The learner should be able to read, interpret and analyse the plot, characters and themes in a short story. Express reading.

Dramatization.

Discussing.

Supplementary readers. Anthology of selected stories.  
5 Writing. Spelling. The learner should be able to spell words correctly as read out. Listening.

Writing.

Flash cards.

Pictures.

Teacher’s resources.  
6 Speech work. Story telling. The learner should be able to identify various devices used in narratives. Narrating.

Listening.

Answering questions.

Audio tapes. Teacher’s resources.  

 

6 1 Writing. Addressing envelops. The learner should be able to write addresses on envelops correctly. Reading.

Writing.

Addressed envelops. Head Start English

Bk I Pg 108

TG Pg 46

 
2 Grammar. Comparatives and superlatives. The learner should be able to identify and use comparatives and superlatives. Listening.

Speaking.

Reading.

Writing.

 

Extracts from readers. Head Start English

Bk I Pg 106

TG Pg 76

 
3 Speech work. Close shave. The learner should be able to polish speech skills. Listening.

Speaking.

Reading.

Writing.

 

Extracts from readers. Head Start English

Bk I Pg 109

TG Pg 46

 
4 Oral literature. Legends. The learner should be able to identify characteristics of legends. Listening to narratives.

Group discussion.

  Teacher’s choice of resources.  
5 Reading. Reading skills. The learner should be able to practise skimming skills. Silent reading.

Group discussions.

Oral presentations.

Suitable library resources. Teacher’s choice of resources.  
6 Reading. Poetry. The learner should be able to answer questions derived from a given poem. Expressive reading.

Answering questions.

Review of exercise.

  Teacher’s choice of resources.  

 

7 1 Listening and speaking. Repetition in poetry. The learner should be able to state the use of repetition in poetry. Listening.

Speaking.

Reading.

Writing.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 111-3

TG Pg 47

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Inferring meanings of given words.

Writing.

 

Dictionary. Head Start English

Bk I Pg 103-5

TG Pg 45

 
3 Reading. Word power. The learner should be able to

demonstrate ability to use the dictionary to build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 113

TG Pg 47-8

 
7 4 Grammar. Regular and irregular adjectives. The learner should be able to form comparative and superlative forms of adjectives.

 

Listening

Oral exercise.

Writing.

Pictures.

Flash cards.

Head Start English

Bk I Pg 114

TG Pg 48

 
5 Reading. Short story. The learner should be able to read, interpret and analyse the plot, characters and themes in a short story. Expressive reading.

Dramatization.

Discussing.

Supplementary readers. Anthology of selected stories.  
6 TEST          

 

8 1 Listening and speaking. Narratives;

/u:/ and /υ/ sounds.

The learner should be able to identify the sounds /u:/ and /υ/ as used in a narrative. Listening.

Speaking.

Reading.

Writing.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 117

TG Pg 49

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Inferring meanings of given words.

Writing.

 

Supplementary readers. Head Start English

Bk I Pg 17-120

TG Pg 50

 
3 Reading. Vocabulary. The learner should be able to

build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 120-1

TG Pg 51

 
4 Grammar. Gradable and non-gradable adjectives. The learner should be able to identify and use gradable and non-gradable adjectives correctly.

 

Exposition.

Discussion.

Writing.

Pictures.

 

Head Start English

Bk I Pg 121-3

TG Pg 51-52

 
5 Writing. Informal letter:

Punctuation and appropriate language.

The learner should be able to write a formal letter with the correct punctuation and appropriate language. Listening.

Writing.

Discussion.

Samples of formal letters. Head Start English

Bk I Pg 123

TG Pg 52

 
6 Oral literature. Trickster narratives. State characteristics of trickster narratives. Narration.

Listening.

Role-play.

Writing.

Supplementary readers. Anthology of selected stories.  

 

9 1 Listening and speaking. Turn taking in a conversation:

/Л/ and /α:/ sounds.

The learner should be able to identify the sounds /u:/ and /υ/ as used in a narrative. Listening.

Speaking.

Reading.

Writing.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 117

TG Pg 49

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Inferring meanings of given words.

Writing.

 

Supplementary readers. Head Start English

Bk I Pg 17-120

TG Pg 50

 
3 Reading. Vocabulary. The learner should be able to

build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 120-1

TG Pg 51

 
4 Grammar. Adverbs of manner. The learner should be able to identify and use adverbs of manner correctly.

 

Discussion.

Reading.

Writing.

Extracts from readers. Head Start English

Bk I Pg 129-130

TG Pg 54

 
5 Writing. Informal letter. The learner should be able to write a formal letter with the correct punctuation and appropriate language. Listening.

Writing.

Discussion.

Samples of formal letters. Head Start English

Bk I Pg 131

TG Pg 55

 
6 Reading. Poetry :

Rhyme.

The learner should be able to identify rhyme in a poem. Reading.

Discussion.

Teacher’s choice of reference. Teacher’s resources.  

 

10 1 Listening and speaking. /Л/ and /ά/ sounds. The learner should be able to identify the sounds /Л/ and /ά/  as used in a text. Listening.

Speaking.

Reading.

Writing.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 133

TG Pg 55

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Inferring meanings of given words.

Writing.

 

Supplementary readers.

Dictionary.

Head Start English

Bk I Pg 133-5

TG Pg 56

 
3 Grammar. Adverbs. The learner should be able to use adverbs correctly.

 

Reading

Discussion.

Writing.

Extracts from readers. Head Start English

Bk I Pg 138

TG Pg 57

 
10 4 Writing. Creative writing. The learner should be able to write creatively on a given topic. Reading.

Listening

Sample writing.

Discussion.

 

Samples of creative texts. Head Start English

Bk I Pg 138

TG Pg 57

 
5 Reading. Poetry. The learner should be able to answer questions derived from a poem. Reading.

Listening

Discussion.

Writing.

 

Anthology of poems. Teacher’s resources.  
  Reading. Short story. The learner should be able to read, interpret and analyse the plot, characters and themes in a short story. Expressive reading.

Dramatization.

Discussing.

Supplementary readers. Anthology of selected stories.  

 

11 1 Listening and speaking. Ogre narrative. The learner should be able to answer questions based on an ogre narrative. Reading.

Listening.

Speaking.

Writing.

Discussing.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 139

TG Pg 58-59

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Writing.

Inferring meanings of given words.

 

Supplementary readers. Head Start English

Bk I Pg 140-1

TG Pg 60

 
3 Reading. Vocabulary. The learner should be able to

build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 140-1

TG Pg 60

 
4 Grammar. Plurals, adverbs and tenses. The learner should be able to use plurals, adverbs and tenses.

 

Writing.

Discussion.

 

Extracts from readers. Head Start English

Bk I Pg 142

TG Pg 60-61

 
5 Writing. Informal letter. The learner should be able to demonstrate a mastery of language and format of an informal letter. Listening.

Writing.

Discussion.

Samples of formal letters. Head Start English

Bk I Pg 143

TG Pg 62

 
6 Oral literature. Dilemma narrative. The learner should be able to state the characteristics of dilemma narratives and their uses. Narration.

Writing.

Discussion.

 

Teacher’s choice of reference. Teacher’s resources.  

 

12 1 Listening and speaking. The sound /ei/ The learner should be able to pronounce words with sound /ei/. Reading.

Listening.

Speaking.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 144

TG Pg 62

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Silent reading.

Writing.

Discussion.

Supplementary readers. Head Start English

Bk I Pg 145-6

TG Pg 63

 
3 Reading. Vocabulary. The learner should be able to

build vocabulary base.

Group work.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 147

TG Pg 63

 
4 Grammar. Co-ordinating conjunctions. The learner should be able to use co-ordinating conjunctions.

 

Reading.

Writing.

Discussion.

 

Extracts from readers. Head Start English

Bk I Pg 148-9

TG Pg 63-64

 
5 Reading. Short story. The learner should be able to read, interpret and analyse the plot, characters and themes in a short story. Expressive reading.

Dramatization.

Discussing.

Supplementary readers. Anthology of selected stories.  
6 Reading. Metaphors in poetry. The learner should be able to identify metaphors in a poem and explain their use. Expressive reading.

Discussing.

Supplementary readers. Anthology of selected stories.  
13

14

END  OF   TERM  TWO  EXAMINATIONS  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 
SCHEME  OF WORK               FORM  ONE  ENGLISH            TERM  THREE  20…………………
 
WK NO

 

L/
NO
TOPIC
 
 
SUBTOPIC
LESSON / SPECIFIC
OBJECTIVES
TEACHING/ LEARNING
ACTIVITIES
MATERIALS       /
RESOURCES
REFERENCES
 
1 1 Writing. Note making. The learner should be able to make notes on a given text. Reading.

Listening.

Note making.

Discussing.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 151

TG Pg 64

 
2 Listening and speaking. The sound /ai/. The learner should be able to pronounce words with sound /ai/. Reading.

Listening.

Speaking.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 152

TG Pg 65

 
3 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Silent reading.

Writing.

Inferring meanings of given words.

 

Supplementary readers. Head Start English

Bk I Pg 152

TG Pg 65

 
4 Reading. Vocabulary. The learner should be able to

build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 152

TG Pg 65

 
5 Grammar. Prepositions. The learner should be able to identify and use prepositions correctly.

 

Listening.

Writing.

Discussion.

 

Extracts from readers. Head Start English

Bk I Pg 155

TG Pg 66-67

 
6 Oral literature. Aetiological tales. The learner should be able to state the characteristics of  aetilogical narratives and their uses. Reading.

Writing.

Discussion.

 

Teacher’s choice of reference. Teacher’s resources.  

 

2 1 Listening and speaking. The sound /oi/ The learner should be able to pronounce words with sound /oi/. Reading.

Listening.

Speaking.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 158

TG Pg 67

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Silent reading.

Writing.

Inferring meanings of given words.

 

Supplementary readers. Head Start English

Bk I Pg 158

TG Pg 68

 
2 3 Reading. Vocabulary. The learner should be able to

build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 158

TG Pg 68

 
4 Grammar. Prepositions. The learner should be able to identify and use prepositions correctly.

 

Listening.

Writing.

Discussion.

 

Extracts from readers. Head Start English

Bk I Pg 162-4

TG Pg 68-9

 
5 Writing. Note taking and note making. The learner should be able to distinguish note taking from note making. Discussion.

Note taking.

Note making.

 

  Head Start English

Bk I Pg 164

TG Pg 69

 
6 Reading. Drama. The learner should be able to state the characteristics of  written drama. Reading.

Discussion.

 

Teacher’s choice of reference. Teacher’s resources.  

 

3 1 Listening and speaking. The sound /oo/ The learner should be able to pronounce words with sound /oo/. Reading.

Listening.

Speaking.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 165

TG Pg 70

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Writing.

Inferring meanings of given words.

 

Supplementary readers. Head Start English

Bk I Pg 165

TG Pg 70

 
3 Reading. Vocabulary. The learner should be able to

build vocabulary base.

Reading aloud.

Listening.

Writing.

Discussion.

Flash cards.

Dictionary.

Head Start English

Bk I Pg 165

TG Pg 70

 
4 Grammar. Noun phrase pre modifiers. The learner should be able to identify noun phrase pre modifiers.

 

Exposition.

Reading.

Writing.

Discussion.

 

Extracts from readers. Head Start English

Bk I Pg 168-9

TG Pg 72

 
5 Speech work. Debate. The learner should be able to argue for or against a given topic. Speaking.

Group work.

Chart showing process of debating. Head Start English

Bk I Pg 169

TG Pg 72

 
6 Reading. Drama. The learner should be able to state the characteristics of  written drama. Reading.

Discussion.

 

Teacher’s choice of play. Teacher’s resources.  

 

4 1 Writing. Note taking. The learner should be able to make notes on a given text. Reading.

Listening.

Note making.

Discussing.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 169

TG Pg 72

 
2 Reading. Irony in poetry. The learner should be able to

identify irony in poetry.

Reading aloud.

Writing.

Inferring meanings of given words.

 

Supplementary readers. Anthology of poems.  
3 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Reading aloud.

Writing.

Inferring meanings of given words.

 

Supplementary readers. Head Start English

Bk I Pg 173

TG Pg 73

 
4 Reading. Vocabulary. The learner should be able to

build vocabulary base.

Expressive reading.

Listening.

Discussion.

Anthology of poems. Head Start English

Bk I Pg 173

TG Pg 73

 
5 Listening and speaking. The sound /o/ The learner should be able to pronounce words with sound /o/. Reading.

Listening.

Speaking.

 

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 171

TG Pg 73

 
6 Oral literature. Tongue-twisters. The learner should be able to state the characteristics of tongue twisters. Listening.

Speaking.

 

Audio tapes.

Flash cards.

Teacher’s resources.  

 

5 1 Grammar. Noun phrase post modifiers. The learner should be able to identify noun phrase post- modifiers.

 

Exposition.

Reading.

Writing.

Discussion.

 

Extracts from readers. Head Start English

Bk I Pg 177

TG Pg 75

 
2 Writing. Imaginative writing. The learner should be able to write imaginatively. Expressive reading.

Discussion.

Writing.

Extracts from readers. Head Start English

Bk I Pg 177

TG Pg 75

 
3 Reading. Poetry. The learner should be able to answer questions based on a poem. Reading.

Listening.

Speaking.

 

Anthology of poems.    
5 4 Reading. Extensive reading. The learner should be able to write down information read from magazines and periodicals. Reading.

Note making.

Discussion.

  Magazines.

Journals and periodicals.

 
5-6 Reading. Play. The learner should be able to analyse plot, themes and characters in a play. Reading.

Listening.

Discussion.

  Teacher’s choice.  

 

6 1 Listening and speaking. The sound /iЭ/ The learner should be able to articulate the sound /iЭ/ Dictation.

Pronunciation drills.

Writing.

Reading.

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 178

TG Pg 75

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Silent reading.

Writing.

Exercise review.

Supplementary readers. Head Start English

Bk I Pg 178

TG Pg 76

 
3 Grammar. Declarative and negative sentences. The learner should be able to

identify  and use declarative and negative sentences.

Exposition.

Listening.

Writing.

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 169

TG Pg 72

 
4 Writing. Imaginative composition. The learner should be able to write imaginatively.

 

  Supplementary readers. Head Start English

Bk I Pg 169

TG Pg 72

 
5 Reading. Play. The learner should be able to analyse plot, themes and characters in a play. Reading.

Listening.

Discussion.

  Teacher’s choice.  
6 Oral literature. Tongue-twisters. The learner should be able to state the characteristics of tongue twisters. Listening.

Speaking.

 

Audio tapes.

Flash cards.

Teacher’s resources.  

 

 

 

 

 

 

 

 

7 1 Listening and speaking. The sound /ea/ The learner should be able to articulate the sound /ea/. Dictation.

Pronunciation drills.

Writing.

Reading.

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 185

TG Pg 78

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Silent reading.

Writing.

Exercise review.

Supplementary readers. Head Start English

Bk I Pg 185

TG Pg 78

 
3 Grammar. Interrogative clauses. The learner should be able to

identify and use interrogative clauses.

Exposition.

Listening.

Writing.

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 188

TG Pg 79

 
4 Writing. Simple verses and poems. The learner should be able to write down simple verses and poems.

 

Discussion.

Writing.

Supplementary readers. Head Start English

Bk I Pg 169

TG Pg 77

 
5 Reading. Poetry. The learner should be able to discuss the subject matter in a poem. Reading.

Dramatization.

Listening.

Speaking.

 

Anthology of poems.    
6 TEST          

 

8 1 Listening and speaking. The sound /ua/ The learner should be able to articulate the sound /ua/. Dictation.

Pronunciation drills.

Writing.

Reading.

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 192

TG Pg 81

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Silent reading.

Writing.

Exercise review.

Supplementary readers. Head Start English

Bk I Pg 192

TG Pg 81

 
3 Reading. Vocabulary. The learner should be able to

build vocabulary base.

Expressive reading.

Listening.

Discussion.

  Head Start English

Bk I Pg 194

TG Pg 81

 
4 Grammar. The ‘wh’ questions. The learner should be able to ask the ‘wh’ questions. Oral exercise.

Writing.

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 196

TG Pg 82

 
8 5 Writing. Public notices. The learner should be able to write notices.

 

Discussion.

Reading.

Writing.

Supplementary readers.

Pictures.

   
6 Reading. Intensive reading. The learner should be able to discuss the subject matter in a play. Reading.

Listening.

Speaking.

 

  Teacher’s resources.  

 

9 1 Listening and speaking. The sounds /e/ and /ei/. The learner should be able to articulate the sounds correctly. Dictation.

Pronunciation drills.

Writing.

Reading.

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 200

TG Pg 84

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Silent reading.

Writing.

Exercise review.

Supplementary readers. Head Start English

Bk I Pg 200

TG Pg 84-5

 
3 Reading. Vocabulary. The learner should be able to

build vocabulary base.

Expressive reading.

Listening.

Discussion.

Anthology of poems. Head Start English

Bk I Pg 200

TG Pg 85

 
4 Grammar. Imperative and explanative clauses. The learner should be able to

identify  and imperative and explanative clauses.

Exposition.

Listening.

Writing.

Audio tapes.

Flash cards.

Head Start English

Bk I Pg 203-4

TG Pg 85

 
5-6 Reading. Intensive reading. The learner should be able to discuss the subject matter in a play. Reading.

Listening.

Speaking.

Dramatization.

 

Supplementary readers. Teacher’s resources.  

 

10 1 Writing. Keeping clear records. The learner should be able to make an inventory  of items. Exposition.

Discussion.

Writing.

Real records. Head Start English

Bk I Pg 205

TG Pg 86

 
2 Speech work. Emerging issues. The learner should be able to express views about something. Discussion.

Writing.

Readers. Newspapers and magazines.  
3 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Silent reading.

Writing.

Exercise review.

Supplementary readers. Head Start English

Bk I Pg 205

TG Pg 86

 
10 4 Writing. Completing a story. The learner should be able to complete a given story.

 

Group discussion.

Writing.

  Teacher’s choice.  
5-6 Reading. Intensive reading. The learner should be able to discuss the subject matter in a given play. Reading.

Listening.

Speaking.

Dramatization.

 

Supplementary readers. Teacher’s resources.  

 

11 1 Listening and speaking. Oral narrative. The learner should be able to answer questions based on an oral narrative. Listening.

Writing.

Discussion.

 

Pre-recorded material. Head Start English

Bk I Pg 206

TG Pg 87

 
2 Reading. Comprehension. The learner should be able to

answer comprehension questions correctly.

Silent reading.

Writing.

Exercise review.

Supplementary readers. Head Start English

Bk I Pg 206

TG Pg 87

 
3 Grammar. Revision exercise. The learner should be able to

complete sentences correctly.

Written exercise.

Exercise review.

  Head Start English

Bk I Pg 208

TG Pg 88

 
4 Writing. A public notice. The learner should be able to write a public notice with all important features.

 

Group discussion.

Writing.

Group presentations.

  Head Start English

Bk I Pg 210

TG Pg 89

 
5 Reading. Intensive reading. The learner should be able to discuss the subject matter in a play. Reading.

Listening.

Speaking.

Dramatization.

 

Supplementary readers. Teacher’s resources.  
6 Writing. A paragraph. The learner should be able to arrange sentences into  a meaningful paragraph. Reading.

Writing.

Extracts from readers. Head Start English

Bk I Pg 210

TG Pg 89

 
12-13 END  OF   YEAR  EXAMINATIONS  

 

CRE FORM 3 END TERM EXAMS PLUS ANSWERS IN PDF

ULTIMATE ACHIEVERS ACADEMY

Name___________________________________________________AdmNo:__________

Date__________________________________________Sign______________

END OF TERM  EXAMINATION 

FORM THREE

CHRISTIAN RELIGIOUS EDUCATION

TIME 2 ½ HRS

Instructions to candidates:

  1. This paper consists 6 questions
  2. Answer any 5 questions in the answer booklet provided.
Questions 1 2 3 4 5 6
Marks            

 

QUESTIONS

  1. a) State 7 teachings from the Genesis stories of creation. (7mks)
  2. b) Outline 4 differences between the traditional African view of evil and biblical concept of sin. (8mks)
  3. c) Explain how human beings fail to carry out the responsibilities given to them by God. (5mks)
  4. a) List down the teachings of Amos on social justice and responsibility.  (7mks)
  5. b) State 7 ways God would punish Israel
  6. c) How can Christians assist the church leaders to perform their duties effectively. (6mks)
  7. a) Outline the teachings of Jeremiah on judgment and punishment (7mks)
  8. b) Identify 6 symbolic acts of Jeremiah on Judgment and punishment (6mks)
  9. c) What evidences from the book of Jeremiah show that he was a true prophet of God. (7mks)
  10. a) Describe the birth of Jesus.   Luke 2: – 7 (6mks)
  11. b) State the activities that took place when Jesus was born (8mks)
  12. c) Outline the significance of the actions that took place during the birth of Jesus (6mks)
  13. a) Outline Jesus teachings on watchfulness and readiness.  Luke 12: 35 – 38 (7mks)
  14. b) List down he Jewish practices that Jesus condemned. Luke 11: 37 – 54; 121: 1 – 3  (7mks)
  15. c) Mention ways in which Christians today observe the Sabbath (6mks)
  16. a) Explain how unity of believers is expressed in the symbol of the bride. (7mks)
  17. b) Identify 7 factors that caused disunity in the early church at Corinthians (7mks)
  18. c) Give 6 factors that have contributed to national unity in Kenya (6mks)


CHRISTIAN RELIGIOUS EDUCATION

FORM THREE MARKING SCHEME

  1. a) Teaching from the Genesis stories of creation
    1. .God is the creator of everything on earth
    2. Human being are co-creators with God
  • Man and women are equal before God
  1. Human beings are superior to the rest of creation.
  2. Marriage is blessed by God.
  3. Work and leisure are instituted by God.
  • Human being should devote one day a week to worship God.
  • The environment is God gifted for human beings to export and their own benefits.
  1. Human life is God given and sacred.

7 x 1 = 7mks

b). Difference between traditional Africa views of evil and biblical concept of sin.

  1. The biblical account emphasizes on personal nature of sin while Africa emphasizes more on the social nature and consequences of evil.
  2. In the bible sin is punished in the next life while Africa believe that punishment of sin is here on earth.
  • In the bible sin does not leave human beings doomed forever but they are offered escape route and a message of hope through Christ death and resurrection whereas African society offer a final solution to evil.
  1. In the bible people are born sinners while in African society a child is born free from evil
  2. Bible teaches that sin in caused by mans thoughts and deeds while African society believe that evil is caused by evil spirits.
  3. In the bible there is eternal punishment for sinners while traditional African communities believe that punishment is here on earth.

4  x 2 = (8mks)

  1. c) Ways in which human beings fail to carry out responsibilities given to them by God.
  2. Carrying out abortion
  3. By using contraceptives
  • Practicing celibacy/failing to marry
  1. Through deforestation
  2. Failing to prevent soil erosion
  3. Failing to worship him/filling to observe the Sabbath
  • Mining and excavation leading to open ground
  • Polluting the environment
  1. Murder
  2. Sexual immorality
  3. a) Teachings of prophet Amos on social justice and responsibility
  4. The poor were sold for silver and a a pair of sandals as payment of debt.
  5. The rich women of Samaria lived in excessive luxury at the expense of the poor.
  • The rich exploited the farmers by taking an excessive share of their harvest,
  1. The rulers lived in magnificent houses.
  2. The merchants exploited the poor by selling to them poor quality goods and using false scales.
  3. The rulers were arrogant, trusted in material possessions and indulged in wine.
  • The judges were corrupt and took bribes from the rich.
  • Excessive drinking was wide spread and the Nazirites were forced to drink.
  1. Those who spoke the truth were hated.
  2. The accused told lies in law courts
  3. There was sexual immorality and temple prostitution.
  • There was misuse of garments taken as pledges.
  • There was robbery with violence.

7  x 1 = 7mks)

  1. b) Ways through which God would punish Israel
  1. Invasion by a foreign nation
  2. Destruction of places of worship
  • Attack by epidemics
  1. God would send an eclipse
  2. They would be taken to exile
  3. God would send an earth quake to destroy their houses.
  • The rich who oppressed the poor would not enjoy their wealth.

7  x 1 = 7mks

  1. c) How Christians can assist church leader perform their duties
  2. giving financial/material help
  3. Advising counseling them on various issues
  • Encouraging them in their work
  1. Participating fully in church activities/functions
  2. Giving tithes/offerings faithfully
  3. Praying for them
  • Respecting them
  • Practicing/obeying the word of God
  1. Defending them against unfair criticism
  2. Providing training opportunities for them.

6  x  1 =  6mks)

  1. a) Teachings of Jeremiah on judgment and punishment
  2. God was going to punish the people for their sins against God.
  3. God judges people by looking into their hearts.
  • Divine judgment would be fall the people of Judah as a group
  1. God would execute judgment by means of political and historical events.
  2. Gods punishment was unavoidable.
  3. The purpose of Gods punishment is to correct the sinner.
  • God gave the people a chance to repent before he punished them
  • Gods judgment is universal
  1. Gods judgment is real.

7  x 1 = 7mks)

  1. b) Symbolic acts of Jeremiah on judgment and punishment
  2. the wearing of the linen waist cloth
  3. the wearing of the wooden ox yoke.
  • Jeremiahs personal life.
  1. Remolding of the clay by the potter.
  2. The parable of the wine jars/wineskins
  3. The vision of the two baskets of figs

6 x 1 = 6mks)

  1. c) Evidences from the book of Jeremiah that shows that he was a true prophet.
  2. What he prophesied came to pass.
  3. He received his call from God
  • He was prayerful
  1. His messages came from God.
  2. He mediated between God and the people.
  3. He condemned social evils in Israel.
  • He lived according to the covenant way of life
  • He faced stiff opposition and rejection from the people
  1. He obeyed Gods command e.g Not to marry.

 

  1. a) The birth of Jesus Luke 2 : 1 – 27
  2. in those days Caesar Augustus issued a decree that a census should be taken of the entire Roman world.
  3. Everyone went to their own town to register.
  • Joseph also went from Nazareth to Bethlehem because he belonged to the house of David.
  1. He went to register with Mary who was expecting a child.
  2. While they were there she gave birth to her first born, a son.
  3. She wrapped him in cloths and laid him in a manger.

6   x  1 +  6mks)

  1. b) Activities that took place when Jesus was born. Luke 2 : 8  – 20

 

  1. There were shepherds looking after their flocks nearby at night
  2. An angel of the Lord appeared to them.
  • The glory of the lord shone around them and they were terrified.
  1. The angle told them not to be afraid for he had brought good news to them.
  2. The told them that a savior had been born.
  3. He gave them a sign: They will find a baby wrapped in cloths and lying in a manger.
  • Suddenly a great company of heavenly hosts appeared with the angels singing
  • When the angels disappeared, the shepherds left for Bethlehem.
  1. They found Joseph, Mary and the baby as they were told.
  2. They told them what the angel had told and they were amazed.
  3. Mary treasured up all these things and pondered them in her heart.
  • The shepherds returned glorifying God.

 

8  x  1 = 8mks)

 

  1. c) significance of the actions that took place during the birth of Jesus.
  2. Jesus birth in Jerusalem fulfilled Micah’s prophecy
  3. The census helped to know payment of taxes and those to join the army
  • Jesus being born in a stable and lain in a manger shows that he was humble
  1. The angles reporting the news to the shepherds shows that Jesus is a good shepherd.
  2. The angels singing in heavens showed that Jesus is indeed the son of God.
  3. The shepherds visiting Mary and Joseph confirmed the message of the angel.

6 x 1 = 6mks)

 

 

  1. a) Jesus teachings on watchfulness and readiness Luke 12: 35 – 48
  2. believers should be dressed for his coming
  3. They should keep their lamps burning
  • They should be awake/alert.
  1. Those who are found waiting will be blessed
  2. They should be prepared at all time since they don’t know the time for his coming.
  3. Believers should continue working until Christ comes back.
  • The servants who mistreat fellow workers will be punished
  • Those who do not do what is required of them will receive a severe beating.
  1. Believers are expected to respond to God according to how much they have received from him.
  2. Those who do wrong out of ignorance will receive a lesser beating/punishment

7  x 1  = 7mks)

  1. b) Jewish practices that Jesus condemned. Luke 11: 37 – 54, 12 : 1 – 3
  2. Cleaning the outside while the inside was not holy/dirt
  3. They gave tithes according to the law but failed to practice love, mercy and justice
  • They loved the high seats in the synagogue to be noticed as righteous people
  1. Scribes and teachers of the law failed to make people understand the interpretations of the law of God
  2. They attended to their animals on Sabbath day but failed to do so to human life.
  3. They loved money a lot.
  • They prayed with pride in their hearts.
  • They were a stumbling block to those who would be sincere
  1. They loved to be greeted with respect in the market places

 

 

7  x 1 =   7mks)

 

 

  1. c) Ways in which Christians observe the Sabbath
  2. Christians attend church services
  3. They read the scriptures
  • They listen to the word of God from the preachers
  1. They give offerings and pay tithes
  2. They visit the sick and preach to them
  3. They retrain from work and spend time with their families.’
  • They help the less fortunate in the society

 

6  x  1 =  6mks)

 

 

  1. a) How unity of believers is expressed in the symbol of the bride.
  2. The bride represents Christians
  3. The bride groom represents Jesus.
  • Christians as the bride should be pure.
  1. Gods relationship with his people is perfect and lasts eternally.
  2. Christians should be committed to Christ as the bride is committed to the groom
  3. The church is loved by Christ as the bride is loved by the bride groom.
  • Christians will be taken to a new home where they will live with Christ forever.

 

7  x 1 = 7mks)

  1. b) Factors that caused disunity in the early church
  2. Eating of the meat offered to the idols
  3. Dispute over church leadership
  • Misuse of spiritual gifts
  1. The abuse of the Lord’s supper
  2. Misunderstanding of the covering of the head during worship
  3. Rampant cases of sexual immorality
  • Disputes over marriage and divorce
  • Setting disputes in civil courts.
  1. Misunderstanding of the resurrection of the body/dead

 

  1. c) Factors that have contributed to national unity in Kenya
  2. The constitution
  3. The national assembly
  • The national anthem
  1. The presidency
  2. Religion
  3. National language
  • Education
  • Common currency
  1. Sports. 6  x  1 =  6mks)

Moi Girls High School Eldoret KCSE 2025 Joint Exam Papers & Answers

Moi Girls High School Eldoret KCSE 2025 Joint Exam Papers & Answers

MOI GIRLS PHYC PP3 CONF.pdf
MOI GIRLS PHYC PP3 CONF.docx
MOI GIRLS PHYC PP3 2025 PREDICTIONS.pdf
MOI GIRLS CHEM PP3 CONF.pdf
MOI GIRLS CHEM PP3 2025 PREDICTIONS.pdf
MOI GIRLS CHEM PP3 2025 PREDICTIONS.docx
MOI GIRLS BIO PP3 CONF.pdf
MOI GIRLS BIO PP3 CONF.docx
MOI GIRLS BIO PP3 2025 PREDICTIONS.pdf
MOI GIRLS CRE PP1 2025 PREDICTIONS.pdf
MOI GIRLS CRE PP1 2025 PREDICTIONS.docx
MOI GIRLS HOME SCI PP3 2025 PREDICTIONS.pdf
MOI GIRLS HOME SCI PP3 2025 PREDICTIONS.docx
MOI GIRLS HOME SCI PP2 2025 PREDICTIONS.pdf
MOI GIRLS HOME SCI PP2 2025 PREDICTIONS.docx
MOI GIRLS HOME SCI PP1 2025 PREDICTIONS.pdf
MOI GIRLS HOME SCI PP1 2025 PREDICTIONS.docx
MOI GIRLS COMP PP2 2025 PREDICTIONS.pdf
MOI GIRLS COMP PP1 2025 PREDICTIONS.pdf
MOI GIRLS COMP PP1 2025 PREDICTIONS.docx
MOI GIRLS MATH PP2 2025 PREDICTIONS.pdf
MOI GIRLS MATH PP2 2025 PREDICTIONS.docx
MOI GIRLS MATH PP1 2025 PREDICTIONS.pdf
MOI GIRLS BIO PP2 2025 PREDICTIONS.pdf
MOI GIRLS BIO PP1 2025 PREDICTIONS.pdf
MOI GIRLS PHYC PP2 2025 PREDICTIONS.pdf
MOI GIRLS PHYC PP2 2025 PREDICTIONS.docx
MOI GIRLS PHYC PP1 2025 PREDICTIONS.pdf
MOI GIRLS CHEM PP2 2025 PREDICTIONS.pdf
MOI GIRLS CHEM PP2 2025 PREDICTIONS.docx
MOI GIRLS CHEM PP1 2025 PREDICTIONS.pdf
MOI GIRLS BST PP2 2025 PREDICTIONS.pdf
MOI GIRLS BST PP2 2025 PREDICTIONS.docx
MOI GIRLS BST PP1 2025 PREDICTIONS.pdf
MOI GIRLS AGRI PP2 2025 PREDICTIONS.pdf
MOI GIRLS AGRI PP2 2025 PREDICTIONS.docx
MOI GIRLS AGRI PP1 2025 PREDICTIONS.pdf
MOI GIRLS AGRI PP1 2025 PREDICTIONS.docx
MOI GIRLS KISW PP3 2025 PREDICTIONS.pdf
MOI GIRLS KISW PP3 2025 PREDICTIONS.docx
MOI GIRLS KISW PP2 2025 PREDICTIONS.pdf
MOI GIRLS KISW PP1 2025 PREDICTIONS.pdf
MOI GIRLS KISW PP1 2025 PREDICTIONS.docx
MOI GIRLS ENG PP3 2025 PREDICTIONS.pdf
MOI GIRLS ENG PP3 2025 PREDICTIONS.docx
MOI GIRLS ENG PP2 2025 PREDICTIONS.pdf
MOI GIRLS ENG PP1 2025 PREDICTIONS.pdf
MOI GIRLS ENG PP1 2025 PREDICTIONS.docx
MOI GIRLS GEO PP2 2025 PREDICTIONS.pdf
MOI GIRLS GEO PP1 2025 PREDICTIONS.pdf

POETRY LATEST NOTES- UPDATED FREE

INTRODUCTION TO POETRY

Definition of Poetry

Ø  poetry has no one set definition because it can mean so many things to different people. The following are some common definitions:

·       it is the art of writing thoughts, ideas, and dreams into imaginative language which may contain verse, pause, meter, repetition, and/or rhyme.

·       writing that formulates a concentrated imaginative awareness of experience in language chosen and arranged to create a specific emotional response through meaning, sound,and rhythm

  • A poet is the one who writes poetry.
  • A poem is a written expression of emotion or ideas in an arrangement of words or verse, most often rhythmically.

 

Different Types of Poetry

  • There are over 50 types of poetry.
  • Poetry is categorized by:
  1. the number of lines in the poem, for example, sonnet
  2. the words in the poem,
  3. whether it rhymes or not, and
  4. what it is about, for example, love poetry, death poetry, etc
  • Below are the common types of poetry:
  • Haikus
  • Sonnets
  • Name poems
  • Free verse poems

Haikus

  • The haiku originated from Japan,.
  • It’s the shortest type of poem and, often, the most difficult to understand.
  • Haiku poems consist of 3 lines.
  • The first and last lines of a Haiku have 5 syllables and the middle line has 7 syllables.
  • The lines rarely rhyme.

Look at the example below:

Easter Chocolate Haiku

by Kaitlyn Guenther

Easter bunny hides
Easter eggs are out of sight
Kids look everywhere

 

Free verse poems

  • A free verse is the loosest type of poem.
  • It can consist of as many lines as the writer wants.
  • It can either rhyme or not, and it does not require any fixed metrical pattern.

Look at the example below:

From Marriage
Marianne Moore

This institution,

perhaps one should say enterprise
out of respect for which
one says one need not change one’s mind
about a thing one has believed in,
requiring public promises
of one’s intention
to fulfill a private obligation:
I wonder what Adam and Eve
think of it by this time,
this fire-gilt steel
alive with goldenness;
how bright it shows—

Sonnets

  • A sonnet is best described as a lyric poem that consists of fourteen lines.
  • Sonnet’s have at least one or two conventional rhyme schemes.

An example of a sonnet is the poem below:

From Visions
Francesco Petrarch

Being one day at my window all alone,

So manie strange things happened me to see,
As much as it grieveth me to thinke thereon.
At my right hand a hynde appear’d to mee,
So faire as mote the greatest god delite;
Two eager dogs did her pursue in chace.
Of which the one was blacke, the other white:
With deadly force so in their cruell race

They pincht the haunches of that gentle beast,

That at the last, and in short time, I spide,
Under a rocke, where she alas, opprest,
Fell to the ground, and there untimely dide.
Cruell death vanquishing so noble beautie
Oft makes me wayle so hard a desire.

Name poems

  • They are popular among children and are often used in schools.
  • The name of the person becomes the poem.
  • Each letter in the name is the first letter in the line of the poem.

Study DUNCAN poem below:

Drew naughty cucumbers frenetically
Umbilical yet chubby
Notified earls generously
Cavorted willfully
Apologized selfishly
Napped frankly but courageously

Note:There is a very wide definition of what constitutes poetry, and although some types of poetry can be grouped together in specific styles, creativity is the key to poetry and a new poet can choose to write in any style he wants, even if it doesn’t fit into one of the recognized types.

PERSONA

PERSONA

  • A persona is a character taken on by a poet to speak in a poem.
  • Dramatic monologues create a persona; however, as a poem using a persona need not tell a reader anything about the situation of the speaker, the narrative, or the person that the poem is spoken to, a persona-poem need not be a dramatic monologue.

A Freedom Song

Atieno washes dishes,
Atieno plucks the chicken,
Atieno gets up early,
Beds her sacks down in the kitchen,
Atieno eight years old,
Atieno yo.

Since she is my sister’s child
Atieno needs no pay.
While she works my wife can sit
Sewing every sunny day:
With her earnings I support
Atieno yo.

Atieno’ sly and jealous,
Bad example to the kids
Since she minds them, like a schoolgirl
Wants their dresses, shoes and beads,
Atieno ten years old,
Atieno yo.

Now my wife has gone to study
Atieno is less free.
Don’t I keep her, school my own ones,
Pay the party, union fee,
All for progress! Aren’t you grateful
Atieno yo?

Visitors need much attention,
All the more when I work night.
That girl spends too long at market.
Who will teach her what is right?
Atieno rising fourteen,
Atieno yo.

Atieno’s had a baby
So we know that she is bad.
Fifty fifty it may live
And repeat the life she had
Ending in post-partum bleeding,
Atieno yo.

Atieno’s soon replaced;
Meat and sugar more than all
She ate in such a narrow life
Were lavished at her funeral.
Atieno’s gone to glory,
Atineo yo.

Marjorie Oludhe-Macgoye

The persona in the poem is Atieno’s uncle. He says; “since she’s my sister’s child/Atieno needs no pay.

 

SUBJECT-MATTER

  • A poem’s subject matter is what the poem is about.
  • To understand subject matter, you need to answer the questions below.
  1. What event, situation, or experience does the poem describe or record?
  2. Who is the speaker? Is the poet speaking in the role of another person, an animal, a thing?
  3. To whom is the speaking talking?
  4. What is the time setting – hour of day, season, era?
  5. What is the place setting?

In the poem “Freedom Song”, can you discuss what the poem is about?

The poem is about a young girl called Atieno who is mistreated by her own uncle. She works without pay and ends in death due to post partum bleeding.

 

Now read the poem below.

 

THE GRACEFUL GIRRAFE CANNOT BECOME A MONKEY

Okot P B’TEK (Uganda)

 

My husband tells me

I have no ideas

Of modern beauty.

He says

I have stuck

To old-fashioned hair styles.

 

He says

I am stupid and very backward,

That my hair style

Makes him sick

Because I am dirty.

 

It is true

I cannot do my hair

As white women do.

 

Listen,

My father comes from Payira,

My mother is a woman of Koc!

I am a true Acoli

I am not a half-caste

I am not a slave girl;

My father was not brought home

By the spear

My mother was not exchanged

For a basket of millet.

 

 

Ask me what beauty is

To the Acoli

And I will tell you;

I will show it to you

If you give me a chance!

 

You once saw me,

You saw my hair style

And you admired it,

And the boys loved it

At the arena

Boys surrounded me

And fought for me.

 

My mother taught me

Acoli hair fashions;

Which fits the kind

Of hair of the Acoli,

And the occasion.

 

Listen,

Ostrich plumes differ

From chicken feathers,

A monkey’s tail

Is different from that of a giraffe,

The crocodile’s skin

Is not like the guinea fowl’s,

And the hippo is naked, and hairless.

 

The hair of the Acoli

Is different from that of the Arabs;

The Indians’ hair

Resembles the tail of a horse;

It is like sisal strings

And needs to be cut

With scissors.

It is black,

And is different from that of a white woman.

 

A white woman’s hair

Is soft like silk;

It is light

And brownish like

That of a brown monkey,

And is very different from mine.

A black woman’s hair

Is thick and curly;

It is true

Ring-worm sometimes eat up

A little girl’s hair

And this is terrible;

But when hot porridge

Is put on the head

And the dance is held

Under the sausage-fruit tree

And the youths have sung

 

You, Ring worm

Who is eating Duka’s hair

Here is your porridge,

 

Then the girl’s hair

Begins to grow again

And the girl is pleased.

What is this poem about?

 

BUILDING THE NATION

“Today I did my share in building the nation.
I drove a Permanent Secretary to an important, urgent function
In fact, to a luncheon at the Vic.

The menu reflected its importance
Cold bell beer with small talk,
Then fried chicken with niceties
Wine to fill the hollowness of the laughs
Ice-cream to cover the stereotype jokes
Coffee to keep the PS awake on the return journey.

I drove the Permanent Secretary back.
He yawned many times in back of the car
Then to keep awake, he suddenly asked,
Did you have any lunch friend?
I replied looking straight ahead
And secretly smiling at his belated concern
That I had not, but was slimming!

Upon which he said with a seriousness
That amused more than annoyed me,
Mwananchi, I too had none!
I attended to matters of state.
Highly delicate diplomatic duties you know,
And friend, it goes against my grain,
Causes me stomach ulcers and wind.

Ah, he continued, yawning again,
The pains we suffer in building the nation! So the PS had ulcers too!
My ulcers I think are equally painful
Only they are caused by hunger,
Not sumptuous lunches!

So two nation builders
Arrived home this evening
With terrible stomach pains
The result of building the nation-in different ways!”

(a) Who is the persona in this poem?

The persona is the senior government officer’s driver who drives him to an expensive hotel.

(b) What is the subject matter of the poem?

The poem is about the government officer who attends an official function at a hotel. The meals he takes here are expensive. The driver looks at him as he eats..

 

 

 

THEMATIC MESSAGE

  • Thematic message is the lesson the poet is attempting to express in the poem.
  • Here we ask ourselves questions like: What seems to be the poet’s purpose in writing this – what message, ideas, issues, themes, (etc.) are communicated?

Thematic Message in “THE GRACEFUL GIRRAFE CANNOT BECOME A MONKEY”

 

Africans should be proud of their identity. No matter what they do Africans will remain Africans.

 

MESSAGE in Building the Nation

  • If we wish to do well in nation building we must as public servants respect everything that belongs to the state; money, property, working hours etc.
  • If the high class is not careful with nation building one day the oppressed may revolt.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

STYLISTIC DEVICES (RHETORICAL DEVICES, FIGURES OF SPEECH)

Stylistic devices

  • In this section, we will learn some of the most important stylistic devices (also called rhetorical devices or figures of speech)
  • Stylistic devices make poem:
  1. more interesting and lively; and
  2. help you to get and keep reader’s / listener’s attention.
  • Some of these stylistic devices include:
  1. Alliteration
  2. Allusion
  3. Anaphora
  4. Antithesis
  5. Hyperbole
  6. Hypophora
  7. Litotes
  8. Metaphor
  9. Metonymy
  10. Onomatopoeia
  11. Parallelism
  12. Parenthesis
  13. Personification
  14. Points of view
  15. Repetition
  16. Rhetorical Question
  17. Simile
  18. Synecdoche
  19. Understatement

Stylistic devices can be grouped as:

  • Figures of speech, such as,
  • Synecdoche
  • Metonymy
  • Similes
  • Metaphors
  • Apostrophe, etc
  • Sound devices/techniques, such as:
  • Alliteration
  • Assonance
  • Consonance
  • Repetition

Figurative language

A figure of speech is any way of saying something other than the ordinary way. Figurative language is language using figures of speech. Irony- a subtle of meaning

Synecdoche

Synecdoche is a literary device that refers to a whole as one of its parts. For example, someone might refer to her car as her “wheels,” or a teacher might ask his class to put their eyes on him as he explains something. Or Workers can be referred to as ‘pairs of hands’, a vehicle as one′s ‘wheels’

Metonymy

Definition of Metonymy

  • Metonymy is a figure of speech in which something is called by a new name that is related in meaning to the original thing or concept. Common Examples of Metonymy

Examples In Regular Language

  • “He writes a fine hand.” – This means that the person writes neatly or has good handwriting.
  • The big house—Refers to prison
  • “The pen is mightier than the sword.” – This only means that words spoken well are mightier than military force.
  • “We have always remained loyal to the crown.” – The crown here stands to symbolize the king of a particular country. This metonymy is used commonly when talking or writing.
  • “The House was called to order.” – This means that the ‘members’ of a House, assuming that it’s a House of Parliament, were called to order.
  • ‘He is a man of cloth’ – This only means to say that the man who the sentence is talking about is a man who belongs to a religious sect; cloth signifies that ‘robes’ worn by religious men.
  • ‘She works with a newspaper’ – Newspaper  stands to represent a group of journalists and editors working together to churn out news items.

Consider the poem below:

 

We passed the School, where Children strove At Recess, in the Ring

We passed the fields of Gazing Grain

We passed the Setting Sun

Or rather, He passed Us

The Dews drew quivering and chill

For only Gossamer, my Gown

My Tippet, only Tulle

We paused before a House that seemed

A Swelling of the Ground

The Roof was scarcely visible

The Cornice in the Ground Since then ’tis

Centuries, and yet Feels shorter than the

Day I first surmised the Horses’ Heads

Were toward Eternity

Because I Could Not Stop For Death –

Emily Dickinson

 

At a quick look, it may be difficult to find out the use of metonymies in the poem above. However, when you take a deeper look at the poem and are better acquainted with a metonymy as a figure of speech, the whole process should be a breeze.

Look at the first line in the first poem: “We passed the School, where Children strove”. The word ‘school’ represents a building, but then school also stands to represent the children studying in that particular school.

 

Difference Between Metonymy and Synecdoche

  • Metonymy and synecdoche are very similar figures of speech, and some consider synecdoche to be a specific type of metonymy.
  • Synecdoche occurs when the name of a part is used to refer to the whole, such as in “There are hungry mouths to feed.” The mouths stand in for the hungry people. The definition of metonymy is more expansive, including concepts that are merely associated in meaning and not necessarily parts of the original thing or concept.

Function of Metonymy

Poets use metonymy:

  1. to address something in a more poetic and unique way.
  2. to make statements more concise.

Apostrophe

Similar to ‘personification’ but indirect. The speaker addresses someone absent or dead, or addresses an inanimate or abstract object as if it were human.

Sample Apostrophe Poem

DEATH BE NOT PROUD

JOHN DONNE (England)

Death be not proud, though some have called thee

Might and dreadful, for thou art not so;

For those whom thou think’st thou dost overthrow

Die not, poor Death, nor yet canst thou kill me.

From rest and sleep, which but thy pictures be,

Much pleasure- then from thee much more must flow;

And soonest our best men with thee do go,

Rest of their bones and soul’s delivery.

Thou art slave to fate, chance, kings and desperate men,

And dost with poison, war, and sickness dwell;

And poppy or charms can make us sleep as well,

And better than thy stroke. Why swell’st thou then?

One short sleep past, we wake eternally,

And death shall be no more; Death, thou shalt die.

Explanation

The poet addresses death but death does not respond. Eg he says ‘poor death’ which is an embarrassing way to talk to someone who considers himself tough.

Anaphora

  • This is where successive clauses or sentences start with the same word(s)
  • The same word or phrase is used to begin successive clauses or sentences.
  • This makes the reader’s / listener’s attention to be drawn directly to the message of the sentence.
  • The most common examples in prose are:
  • Every child must be taught these principles. Every citizen must uphold them. And every immigrant, by embracing these ideals, makes our country more, not less, American.
  • The beginning of wisdom is silence. The second step is listening. (unknown)
  • A man without ambition is dead. A man with ambition but no love is dead. A man with ambition and love for his blessings here on earth is ever so alive. (Pearl Bailey)

Sample Anaphora Poem

I have been one acquainted with the night.
I have walked out in rain – and back in rain.
I have outwalked the furthest city light.

I have looked down the saddest city lane.
I have passed by the watchman on his beat
And dropped my eyes, unwilling to explain.

I have stood still and stopped the sound of feet
When far away an interrupted cry
Came over houses from another street,

But not to call me back or say good-bye;
And further still at an unearthly height,
One luminary clock against the sky

Proclaimed the time was neither wrong nor right.
I have been one acquainted with the night.

Robert Frost

  • You realize there is realize that the phrase “I have” is the same in:

I have been one acquainted with the night.
I have walked out in rain—and back in rain.
I have outwalked the furthest city light.
I have looked down the saddest city lane.
I have passed by the watchman on his beat
And dropped my eyes, unwilling to explain.

 

Antithesis

Examples:

  • That’s one small step for a man, one giant leap for mankind. (Neil Armstrong)
  • To err is human; to forgive, divine. (Pope)
  • It is easier for a father to have children than for children to have a real father. (Pope)
  • Antithesis is a figure of speech which refers to the juxtaposition of contrasting ideas.
  • Antithesis emphasises the opposition between two ideas.
  • The structure of the phrases / clauses is usually similar in order to draw the reader’s / listener’s attention directly to the contrast.
  • It involves the bringing out of a contrast in the ideas by an apparent contrast in the words, phrases, clauses, or sentences, within a parallel grammatical structure.

Examples:

These are the common examples of antithesis:

  • “Man proposes, God disposes.”
  • “To err is human, to forgive, divine”
  • “Love is an ideal thing, marriage a real thing.”
  • “That’s one small step for man, one giant leap for mankind.”
  • It is easier for a father to have children than for children to have a real father.
  • “Many are called, but few are chosen.”

Sample Antithesis Poem

 

Read the part of John Donne’s poem “Community

“Good we must love, and must hate ill,
For ill is ill, and good good still;
But there are things indifferent,
Which we may neither hate, nor love,
But one, and then another prove,
As we shall find our fancy bent.”

  • Two contrasting words “love” and “hate” are combined in the above lines. It emphasizes that we love good because it is always good and we hate bad because it is always bad. It is a matter of choice to love or hate things which are neither good nor bad.

Function of Antithesis

  • When contrasting ideas are brought together, the idea is expressed more emphatically.
  • It helps bring forth a judgment on either the pros or the cons. This is after we examine the contrasting ideas.

 

Hyperbole

Ø It refers to a deliberate exaggeration

  • It draws the attention to a message that the poet wants to emphasise.

Example:I was so hungry, I could eat an elephant.

  • Do not confuse it with

 

Sample Hyperbole Poems

Appetite

In a house the size of a postage stamp
lived a man as big as a barge.
His mouth could drink the entire river
You could say it was rather large
For dinner he would eat a trillion beans
And a silo full of grain,
Washed it down with a tanker of milk
As if he were a drain.

What Am I?

I’m bigger than the entire earth
More powerful than the sea
Though a million, billion have tried
Not one could ever stop me.
I control each person with my hand
and hold up fleets of ships.
I can make them bend to my will
with one word from my lips.
I’m the greatest power in the world
in this entire nation.
No one should ever try to stop
a child’s imagination.

My Dog

His bark breaks the sound barrier
His nose is as cold as an ice box.
A wag of his tail causes hurricanes
His jumping causes falling rocks.
He eats a mountain of dog food
And drinks a water fall dry.
But though he breaks the bank
He’s the apple of my eye.

Poetry by Sharon Hendricks

Understatement

  • A statement is deliberately weakened to sound ironical or softened to sound more polite.

Examples are:

  • I know a little about running a company. (a successful businessman might modestly say.)
  • I think we have slightly different opinions on this topic. (instead of: I don’t agree with you at all.)
  • “It is just a little cool today” – when the temperature outside is 5° below zero.
  • “The food was tolerable” – on the food that was prepared by the best chef in the world.
  • “It was ok” – when a top ranker was asked about his exam results.
  • “I wouldn’t say he was thin” – describing a very obese person.
  • “He is a little on the old side” – describing a very old person.
  • “I wouldn’t say it tasted great” – on terrible food.

Sample Understatement Poem

Fire and Ice

By Robert Frost

Some say the world will end in fire,

Some say in ice.

From what I’ve tasted of desire

I hold with those who favor fire.

But if it had to perish twice,

I think I know enough of hate

To say that for destruction ice

Is also great

And would suffice.

Consider: I think I know enough of hate, to say that for destruction ice, is also great, and would suffice’

 

 

 

 

 

 

 

 

 

Rhetorical Questions

  • The author speaker raises a question that doesn’t need a direct answer.
  • The answer is seen as obvious.
  • Rhetorical questions are used to provoke, emphasise or argue.
  • The line following the rhetorical question is not answer to it.

Sample Rhetorical Questons Poems

 What happens to a dream deferred?

Does it dry up
like a raisin in the sun?
Or fester like a sore—
And then run?
Does it stink like rotten meat?
Or crust and sugar over—
like a syrupy sweet?

Maybe it just sags
like a heavy load.

Or does it explode?

How will i die

The more i think about it the more I wonderwill i die in a deep slumber?, or will the heavens send a angel for me.will a car come from around the corner and hit me will a robber stick me up at gun point……will he squeeze the trigger and will the bullet land in my skull and then i diewill i die of cancer or some unknown bacteria will i die of an heart attack will i get food poising from my favorite dishes will i die from a lack of sleep will i die at the hands of mercy or at the hand the hands of my greatest enemywill i be pushed from a tall mountain peekoh how frustrating will i die not knowing how i will die should i treat this question as a rhetorical questionoh how will i die urggggggggggg!!!!!!!!!!

Terrica Richards

 Hypophora

  • Hypophora is a figure of speech in which a writer raises a question and then immediately provides an answer to that question.
  • The speaker asks a question and answers it.
  • Hypophora is used to get the audience’s attention and make them curious.
  • The question is often raised at the beginning of a stanza and answered in the course of that stanza.
  • This kind of question can also be used to introduce a new topic of discussion.

Examples are:

  • Why is it better to love than be loved? It is surer.
  • How many countries have actually hit the targets set at Rio, or in Kyoto in 1998, for cutting greenhouse-gas emissions? Precious few.
  • Should not be confused with rhetorical questions.

Difference between Hypophora and Rhetorical Question

The basic difference between hypophora and a rhetorical question is that in a rhetorical question the answer is not provided by the writer since it does not require an answer.

However, in hypophora, the writer first poses a question and then answers that question immediately.

Functions of Hypophora

  • The major purpose of using hypophora is to create curiosity among the readers.
  • It helps to capture the attention of the audience.
  • However, hypophora can also be employed to introduce new discussions or topics of importance about which the readers might not have information.
  • It can also be used as a directional device to change the topic.

Litotes

Litotes is a form of understatement which uses the denied opposite of a word to weaken or soften a message.

Examples:

  • That’s not bad. (instead of: That’s good/great.)
  • Money isn’t easy to find . (instead of: Money is  hard/difficult to find.)
  • They aren’t the happiest couple around.
  • He’s not the ugliest fellow around!
  • She’s not the brightest girl in the class.
  • The food is not bad.
  • It is no ordinary city.
  • That sword was not useless to the warrior now.
  • He was not unfamiliar with the works of Dickens.
  • She is not as young as she was.
  • You are not wrong.
  • Einstein is not a bad mathematician.
  • Heat waves are not rare in the summer.
  • It won’t be easy to find crocodiles in the dark.
  • He is not unlike his dad.
  • That’s no small accomplishment.
  • He is not the kindest person I’ve met.
  • That is no ordinary boy.
  • He is not unaware of what you said behind his back.
  • This is no minor matter.
  • The weather is not unpleasant at all.
  • She’s no doll.
  • That was no small issue.
  • The city is not unclean.
  • Rap videos with dancers in them are not uncommon
  • Running a marathon in under two hours is no small accomplishment.
  • She’s no idiot.
  • That’s not a meager sum.
  • You’re not doing badly.
  • That’s no mean feat.

 

Simile

Two things are compared directly by using ‘like’ , ‘as’, ‘similar to’etc.

Sample Simile Poems

Berkley

Black as midnight,
Bad as the devil
With eyes like pieces of dark chocolate,
He thinks he’s king of the world,
My dog Berkley.
He’s very much like a pig
With his pudgy stomach and all.
Like a leech, he’s always attached
To his next meal.

Even though he’s as bad as the devil,
Berkley is my best fellow.

Stars

They are like flashlights in the night sky;
God’s little helpers guiding us on our journeys.
Stars are as bright as a lighthouse on an icy, ocean night;
they are like guardians committed to bringing you home.

Birds

Chirping non-stop, like a machine in the trees,
Building their nest like little worker bees.
They sing their songs, like chatter-boxes.
As regular as alarm clocks,
Waking people up each day.
They are silent at night,
Like snakes advancing on prey.

My Cat

Eyes like a green-yellow crayon,
Almost as bright as a ripe orange.
My cat rules my heart and my actions.
I am as a puppet on strings
When he purrs against me
Like I am a warm blanket heating him in the cold.

My Friend

She is as crazy as a rooster;
Still I love her like a sister.
Her hair is black and dark
Like the color of the midnight sky.
Her skin is as pale as flour
Placing her among the Twilight vampires.

Metaphor

Metaphor is a figure of speech where two things are compared in a figurative sense. Unlike in a simile (A is like B.), “like” is not used in metaphor (A is B.).

Example:

  • Truths are first clouds, then rain, then harvest and food. (Henry Ward Beecher)
  • Through much of the last century, America’s faith in freedom and democracy was a rock in a raging sea. Now it is a seed upon the wind, taking root in many nations.

Sample Metaphor Poem

A Broken Family Tree

Lori McBride

Published on February 2006

I am one of many
Small branches of a broken tree,
Always looking to the ones above
For guidance, strength and security.
One little branch trying
To keep the others from breaking away.
Who will fall?
And who will stay?
Now I stand alone,
Looking at the earth through the rain,
And I see the broken branches I knew
Scattered about me in pain.
There are those who have taken an ax
To the root of our very foundation
And who have passed this destruction
Down to every new generation.
If I could take that ax,
I would toss it deep into the sea,
Never to return again
To harm the generations that follow me.
I am one of many,
But alone I will go
And plant the new seeds
Where a beautiful tree will grow.

 Onomatopoeia

The pronounciation of the word imitates a sound. Onomatopoeia is used because it’s often difficult to describe sounds. Furthermore, a story becomes more lively and interesting by the use of onomatopoeia.

Examples:

  • The lion roared.
  • The steaks sizzled in the pan.
  • The bomb went off with a bang.
  • My son swooshed the basketball into the net.
  • Please do not whisper in the examination hall.
  • The owl hooted as it sat in the tree.
  • I scared you.
  • Meow, where’s my milk, cried the cat.
  • The jangle of her bracelets caught his attention.
  • Click, click. She made a sound with her tongue to show her displeasure.
  • The birds like to tweet outside my window.
  • Drip, drip, drip, went the faucet all day long.
  • Grandma loves to hear the pitter-patter of little feet around the house.

Sample Poems with Onomatopoeia

When The Lights Go Out

The door went creak
In the still of the night
The floor went bump
Oh what a fright
All of a sudden, we heard a chime
The grandfather clock was keeping good time
We turned down a hallway and heard a loud crash
It seems that someone had dropped all the trash
So many sounds when the lights go out
It’s enough to make you scream and shout!

Keys

I went for a ride with my Uncle Jay
He slammed the door shut and we were on our way
Then he revved the gas pedal, which was on the floor
What a lovely day, could I ask for more?
I wanted a hot dog so he slammed on the brake
It belched out a screech, for goodness sake
We finished our hot dogs in record time
We were out of that place by half past nine
He misplaced his keys
We were in a mess
I must admit, he started to stress
He picked up his soda started to sip
Eager to resume our wonderful trip
From inside the cup he felt something shake
A closer look he surely did take
His keys had fallen into that refreshing soft drink
And we figured it out when we heard the clink

The Precocious Teapot

The teapot whistled at the lovely young pot
She came to a boil at this unfortunate shot
When the pan saw what happened, he began to sizzle
The pot was his cousin and it made him grizzle
The faucet joined in with a whoosh of its water
It seems that the pot was his only daughter
The teapot was humbled and expressed his regret
It was an unfortunate gaffe he’d rather forget

The Construction Site

On my way home from school today
I stopped for sweets along the way
When I heard a thud that made me jump
It seems that a wrecking ball was in a dump
Bu this wasn’t a dump, I am not a fool
It was a construction site and it was so cool
I was startled by the rat-tat-tat of a loud jackhammer
It caused me to stumble and even to stammer
Then all of a sudden, I heard some loud taps
A hammer was banging
A new building perhaps?
One of the men had a brand new drill
It buzzed as it drilled; it was such a big thrill
Then I heard the beep beeping of a truck in reverse
What a great afternoon, I could sure have done worse!

City Street

Beep beep goes the taxi as the light turns green
He is in a hurry so he makes a huge scene
The police car let out a mighty wail
‘Honking your horn sir can land you in jail!’
Well that is not entirely true
But it can surely get you into a mighty big stew
A bus goes by with a mighty roar
With a sleeping passenger who started to snore
A traffic policeman whistled his whistle
The sound of that whistle made everyone bristle
It is true that the city is a busy, busy place
But said another way, It’s just a faster pace

Poetry by Alan Loren

 

Parallelism

  • Successive clauses or sentences are similarly structured.
  • This similarity makes it easier for the reader / listener to concentrate on the message.
  • The normal progression of a sentence is interrupted by extra information or explanations enclosed in commas, brackets or dashes.
  • The extra information can be a single word, a phrase or even a sentence.

 

Examples of Parallelism:

  • The mediocre teacher tells, The good teacher explains. The superior teacher demonstrates. The great teacher inspires.
  • The mistakes of the fool are known to the world, but not to himself. The mistakes of the wise man are known to himself, but not to the world.
  • Tell me and I forget. Teach me and I may remember. Involve me and I will learn.
  • We are bound by ideals that move us beyond our backgrounds, lift us above our interest, and teach us what it means to be citizens.
  • Like father, like son.
  • Parallelism is a useful device for instructions. Due to the parallel structure, the reader can concentrate on the message and will immediately know what to do.

Sample Parallelism Poem

The Tyger

By William Blake

Tyger Tyger, burning bright,

In the forests of the night;

What immortal hand or eye,

Could frame thy fearful symmetry?

 

In what distant deeps or skies.

Burnt the fire of thine eyes?

On what wings dare he aspire?

What the hand, dare seize the fire?

 

And what shoulder, & what art,

Could twist the sinews of thy heart?

And when thy heart began to beat,

What dread hand? & what dread feet?

 

What the hammer? what the chain,

In what furnace was thy brain?

What the anvil? what dread grasp,

Dare its deadly terrors clasp!

 

When the stars threw down their spears

And water’d heaven with their tears:

Did he smile his work to see?

Did he who made the Lamb make thee?

 

Tyger Tyger burning bright,

In the forests of the night:

What immortal hand or eye,

Dare frame thy fearful symmetry?

Personification

  • Animals, inanimate objects or abstractions are given attributes of human beings.
  • Such attributes could be behaviour, feelings, character etc.
  • Personification can make a poem more interesting and lively.

Examples:

  • Why these two countries would remain at each other’s throat for so long. (3)
  • I closed the door, and my stubborn car refused to open it again.
  • The flowers nodded their heads as if to greet us.
  • The frogs began their concert.
  • Every morning my alarm clock springs to life; I hate it when that happens.
  • The peaceful and fun loving town close to the coast was swallowed by an angry tsunami.
  • The picture in that magazine shouted for attention.
  • With a lot of anger, the lighting lashed out from the skies.
  • Art is a jealous mistress. Money is a bad master.
  • All of a sudden, the radio stopped singing and started to stare at me. It was hard to believe.

Sample Personification Poems

Dinnertime Chorus

The teapot sang as the water boiled
The ice cubes cackled in their glass
the teacups chattered to one another.
While the chairs were passing gas
The gravy gurgled merrily
As the oil danced in a pan.
Oh my dinnertime chorus
What a lovely, lovely clan!

My Town

The leaves on the ground danced in the wind
The brook sang merrily as it went on its way.
The fence posts gossiped and watched cars go by
which winked at each other just to say hi.
The traffic lights yelled, ”Stop, slow, go!”
The tires gripped the road as if clinging to life.
Stars in the sky blinked and winked out
While the hail was as sharp as a knife.

Nature’s Chorus

Willows bend to their partners
while the spruces curtsey in response.
Cherry trees form a circle
and the oaks dance just like debutantes
of woodpeckers tapping on their trunks
and squirrels chattering in the boughs.
Listen to the sounds of nature’s chorus
What fun it does arouse!

Games

Chipmunks chatter and scurry,
Blue jays scream and scold.
Robins talk and gossip
demanding their story to be told.
Squirrels skip and box one another
and rabbits play hop scotch.
The games they play, the sounds they make
Really are top notch.

The Gastronomic Gym

Pasta twirling and spinning,
peas do vertical jumps
mashed potatoes swimming.
meat doing bench press and pumps.
Food has begun to exercise
but it’s not in any gym.
My brother said its happening
right inside of him.

Repetition

Words or phrases are repeated throughout the text to emphasise certain facts or ideas.

Oxymoron

An oxymoron is a figure of speech that combines contradictory terms.

Examples:

An oxymoron can be made of an adjective and a noun:

  • Dark light
  • Deafening silence
  • Living dead
  • Open secret
  • Virtual reality

Oxymoron can also be a combination of a noun and a verb.

  • The silence whistles

Sample Oxymora Poem

Read the poem below by Christopher and identify the use of oxymora.

A blind man looks back

Into the future with the

Ear-splitting whispers of

Unconcealed ghosts

Thundering silently.

~~~~

A wealthy peasant marches

Weakly across a blazing glacier

As the stars in the cloudy sky

Glisten grimly.

~~~~ A hateful saint drowns afloat

Into the dismal heaven of peaceful war.

~~~~

Solid water surges down a

Minuscule mountain into

A celestial hell.

~~~~

A colossal dinghy raises

Its feather-light anchor

With vicious doves circling

In the bright winter sky.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

TONE IN POETRY

  • Tone expresses the poet’s or speaker’s attitude toward the subject, the reader, or herself or himself.
  • Tone can shift through a poem.

 

ADJECTIVES USED TO DESCRIBE TONE

  • The following are the common tone/attitude words:
  Word Meaning
1.

2.

3.

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

Accommodating

Accusatory

Humorous

Optimistic

Pessimistic

Sadistic

Bitter

Malicious

Respectful

Resigned

Loving

Understanding

Spiteful

Inferior

Nostalgic

Critical

Cynical

Ironical

Patronizing

condescending

Satirical

Slanderous

Supportive

Contemptuous

Judgmental

obsequious

callous

derisive

.     ribald

Willing to help. Also oblidging.

charging of wrong doing.

Making one laugh.

Having hope.

Having no hope.

Being cruel to others.

exhibiting strong animosity as a result of pain or grief.

Intending to harm.

Showing respect.

Reluctantly accepting something unpleasant.

Feeling or showing love.

Sympathetic to

Wanting to hurt /annoy/offend.

Feeling smaller before.

Have a look at the happy/good past.

Pointing out mistakes in.

Having little faith in.

Meaning the opposite of.

Belittle/consider inferior.

a feeling of superiority

Mocking/ ridiculing.

Making false statement about.

Giving help/encouragement

Despising/ looking down upon.

authoritative and often having critical opinions

polite and obedient in order to gain something

.        unfeeling, insensitive to feelings of others.

ridiculing, mocking

offensive in speech or gesture

 

 

Tone Poems

A FREEDOM SONG

BY Marjorie Oludhe Macgoye (Kenya)

 

Atieno washes dishes,

Atieno plucks the chicken,

Atieno gets up early,

Beds her sucks down in the kitchen,

Atieno eight years old

Atieno yo.

 

Since she’s my sister’s child

Atieno needs no pay

While she works my wife can sit

Sewing each sunny day,

With her earning I support

Atieno yo.

 

Atieno’s sly and jealous

Bad example to the kids

Since she minds them, like a school girl

Wants their dresses, shoes and beads.

Atieno ten years old,

Atieno yo.

 

Now my wife has gone to study

Atieno’s less free,

Don’t I feed her, school my own ones,

Pay the party, union fee

All for progress? Aren’t you grateful,

Atieno yo?

 

Visitors need much attention,

Specially when I work nights.

That girl stays too long at market

Who will teach her what is right?

Atieno rising fourteen,

Atieno yo.

 

Atieno’s had a baby

So we know that she is bad

Fifty-fifty it may live

To repeat the life she had,

Ending in post partum bleeding

Atieno yo.

 

Atieno’s soon replaced

Meat and sugar more than all

She ate in such a narrow life

Were lavished in her funeral

Atieno’s gone to glory

Atieno yo.

The tone is sympathetic to the child.

Ironic Tone

Read the poem “Building the Nation

The poet uses an ironic tone, and his choice of words clearly reflects his bitterness and anger about the pretence by leaders like the PS, who attempt to hide their greed and selfishness behind empty official meetings.

Nostalgia Poems

 

The two poems below have nostalgic tone.

  1. Nostalgia by Billy Collins

Remember the 1340s? We were doing a dance called the Catapult.

You always wore brown, the color craze of the decade,

and I was draped in one of those capes that were popular,

the ones with unicorns and pomegranates in needlework.

Everyone would pause for beer and onions in the afternoon,

and at night we would play a game called “Find the Cow.”

Everything was hand-lettered then, not like today.

 

Where has the summer of 1572 gone? Brocade and sonnet

marathons were the rage. We used to dress up in the flags

of rival baronies and conquer one another in cold rooms of stone.

Out on the dance floor we were all doing the Struggle

while your sister practiced the Daphne all alone in her room.

We borrowed the jargon of farriers for our slang.

These days language seems transparent, a badly broken code.

 

The 1790s will never come again. Childhood was big.

People would take walks to the very tops of hills

and write down what they saw in their journals without speaking.

Our collars were high and our hats were extremely soft.

We would surprise each other with alphabets made of twigs.

It was a wonderful time to be alive, or even dead.

 

I am very fond of the period between 1815 and 1821.

Europe trembled while we sat still for our portraits.

And I would love to return to 1901 if only for a moment,

time enough to wind up a music box and do a few dance steps,

or shoot me back to 1922 or 1941, or at least let me

recapture the serenity of last month when we picked

berries and glided through afternoons in a canoe.

 

Even this morning would be an improvement over the present.

I was in the garden then, surrounded by the hum of bees

and the Latin names of flowers, watching the early light

flash off the slanted windows of the greenhouse

and silver the limbs on the rows of dark hemlocks.

 

As usual, I was thinking about the moments of the past,

letting my memory rush over them like water

rushing over the stones on the bottom of a stream.

I was even thinking a little about the future, that place

where people are doing a dance we cannot imagine,

a dance whose name we can only guess.

2.     PATRIOT INTO TRAITOR BY ROBERT BROWNING

It was roses, roses, all the way,
With myrtle mixed in my path like mad:
The house-roofs seemed to heave and sway,
The church-spires flames, such flags they had,
A year ago on this very day.

The air broke into a mist with bells,
The old walls rocked with the crowd and cries.
Had I said, “Good fold, mere noise repels–
But give me your sun from yonder skies!”
They had answered, “And afterward, what else?”

Alack, it was I who leaped at the sun
To give it my loving friends to keep!
Nought man could do, have I left undone:
And you see my harvest, what I reap
This very day, now a year is run.

There’s nobody on the house-tops now–
Just a palsied few at the windows set;
For the best of the sight is, all allow,
At the Shambles’ Gate– or, better yet,
By the very scaffold’s foot. I trow.

I go in the rain, and more than needs,
A rope cuts both my writs behind;
And think, by the feel, my forehead bleeds,
For they fling, whoever has a mind,
Stones at me for my year’s misdeeds.

Thus I entered, and thus I go!
In triumphs, people have dropped down dead.
“Paid by the world, what dost thou owe
Me?”– God might question; now instead,
‘Tis God shall repay: I am safer so.

 

 

 

 

 

 

 

 

 

 

 

 

 

MOOD IN POETRY

  • Mood is referred to as the atmosphere , as it creates an emotional situation that surrounds the readers. Mood is developed in a literary piece through various methods. It can be developed through setting, theme, tone and diction.

Common Mood Words

Word Meaning
Agreeable

Angry

Nostalgic

Sad

Amusing

Defiant

Diffident

Festive

Indifferent

Sorrowful

Melancholic

Excited

Remorseful

Solemn

Serene

Violent

Pleasing.

Bitter with.

Looking at the happy past.

Not happy.

Making one laugh.

Opposing openly.

Lacking self-confidence.

Joyous

Not having interest in.

Sad because sth bad has happened.

Sad

Happy.

Showing regret/repentance.

Not happy/not smiling.

Calm/peaceful

Use force/fighting.

 

Poem with Sad Mood

Waiting For My Dad

By Amy

I sit alone in the darkness
Waiting…
Waiting for him to come back to me.
Can he hear my cries?
Can he feel my tears?
Can he sense my breaking heart?
God only knows such a fact.
How can this be that he can’t see me?
Is it because I’m sitting alone in the darkness?
I just walk past everyone as if I were invisible.
Can he see me now?
Can he see the pain he’s caused me?
Or does he look past it?
I think I should move on,
But something tells me to wait.
It’s my heart.
I’ll give him one more chance
He needs to prove his love to me.
As I return to sit alone in the darkness…
Waiting.

Melancholic Mood Poem

On longer evenings,
Light, chill and yellow,
Bathes the serene
Foreheads of houses.
A thrush sings,
Laurel-surrounded
In the deep bare garden,
Its fresh-peeled voice
Astonishing the brickwork.
It will be spring soon,
It will be spring soon —
And I, whose childhood
Is a forgotten boredom,
Feel like a child
Who comes on a scene
Of adult reconciling,
And can understand nothing
But the unusual laughter,
And starts to be happy.

“The Trees.”

The trees are coming into leaf
Like something almost being said;
The recent buds relax and spread,
Their greenness is a kind of grief.

Is it that they are born again
And we grow old? No, they die too,
Their yearly trick of looking new
Is written down in rings of grain.

Yet still the unresting castles thresh
In fullgrown thickness every May.
Last year is dead, they seem to say,
Begin afresh, afresh, afresh.

 

Happy Mood Poem

So fine a day it is today
To the world I send out my hope and care
Happy and joyful I am so to all I must say
That I will remember you all in my prayer.

I woke up with a smile upon my face
After knowing again that I was still alive
And I am not ill through GOD and his grace
Another year and night I did survive.

The morning has begun and I am awake
And my wallet is full of money
I owe no bills my finances I will not break
So I will go out while it is nice and sunny.

I will greet all comers with a wave or a hello
As I am walking down the street
I am living for the day and not for long ago
And no strangers today will I meet.

I will feed the birds and I will pet the dogs
If I decide to walk to the park
And I will not be a nuisance to any who jogs
My day is a day of happiness as I do embark.

Today is such a beautiful day
So I decided to start it off with a pleasant attitude
I wished today was a holiday
Since I woke up in a joyful and happy mood.

Randy L. McClave

 

ANALYSIS OF POETRY
to understand the analysis of poetry, study the diagram below.

 

 

Now try to analyse the two poems below following the guidelines.

“It Was Long Ago”
Eleanor Farjeon

I’ll tell you, shall I, something I remember?
Something that still means a great deal to me.
It was long ago.

A dusty road in summer I remember,
A mountain, and an old house, and a tree
That stood, you know.

Behind the house. An old woman I remember
In a red shawl with a grey cat on her knee
Humming under a tree.

She seemed the oldest thing I can remember,
But then perhaps I was not more than three.
It was long ago.

I dragged on the dusty road, and I remember
How the old woman looked over the fence at me
And seemed to know

How it felt to be three, and called out,
I remember ‘Do you like bilberries and cream for tea?’
I went under the tree

And while she hummed, and the cat purred, I remember
How she filled a saucer with berries and cream for me
So long ago,

Such berries and such cream as I remember
I never had seen before, and never see
To day, you know.

And that is almost all I can remember,
The house, the mountain, the grey cat on her knee,
Her red shawl, and the tree,

And the taste of the berries, the feel of the sun I remember,
And the smell of everything that used to be
So long ago,

Till the heat on the road outside again I remember,
And how the long dusty road seemed to have for me
No end, you know.

That is the farthest thing I can remember.
It won’t mean much to you. It does to me.
Then I grew up, you see.

Read “The African Beggar” below and try your hand at analysing the poem.

African Beggar
Raymond Tong

Sprawled in the dust outside the Syrian store,
a target for small children, dogs and flies,
a heap of verminous rags and matted hair,
he watches us with cunning, reptile eyes,
his noseless, smallpoxed face creased in a sneer.

Sometimes he shows his yellow stumps of teeth
and whines for alms, perceiving that we bear
the curse of pity; a grotesque mask of death,
with hands like claws about his begging-bowl.

But often he is lying all alone
within the shadow of a crumbling wall,
lost in the trackless jungle of his pain,
clutching the pitiless red earth in vain
and whimpering like a stricken animal.

 

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CHEMISTRY NOTES FORM 4 PDF

ORGANIC CHEMISTRY II: ALCOHOLS AND ALKANOIC ACIDS:

1. Alkanols

  • Nomenclature of alkanols
  • Primary secondary and tertiary alcohols
  • Preparation and properties of alcohols
  • Hydrolysis of halogenoalkanes
  • Hydration of alkenes
  • Fermentation of sugars and starches (Ethanol)
  • Physical properties of alcohols
  • Chemical properties of alcohols
  • Combustion
  • Reaction with metals (sodium)
  • Esterification
  • Oxidation
  • Reaction with acidified potassium dichromate
  • Reaction with acidified potassium permanganate
  • Reaction with copper metal
  • Dehydration reactions of alcohols
  • Uses of alkanols

DOWNLOAD THE FREE PDF NOTES HERE;

Free Chemistry notes, revision questions, KCSE past Papers, Exams, Marking Schemes, Topical revision materials, Syllabus and Many more

CHEMISTRY FORM FOUR NOTES: NEW

  1. Alkanoic acids
  • Nomenclature
  • Preparation and properties of alkanoic acids
  • Physical properties
  • Chemical properties
  • Reaction with sodium carbonate
  • Reaction with sodium hydroxide
  • Reaction with magnesium metal
  • Esterification
  • Uses of ethanoic acid

 

  1. Fats and oils

 

  1. Soaps and soapless detergents
  • Soaps
  • Preparation of soaps
  • Role of soap in cleaning
  • Effect of hard water on soap
  • Removal of hardness
  • Temporary hardness
  • Permanent hardness

 

  • Soapless detergents
  • Preparation
  • Advantages

 

  1. Polymers
  • Natural and synthetic polymers
  • Addition polymerization
  • Condensation polymerization

ALCOHOLS (ALKANOLS):

– They derivatives of alkanes in which a hydrogen has been replaced by a hydroxyl group (OH); which is the functional group.

– They form a homologous series of the general formula CnH2n+1OH, in which the OH can also be denoted as ROH, where R is an alkyl group.

Note: alkanol is the IUPAC name while alcohol is the common name;

 

Nomenclature of alcohols

  1. The ‘e’ of the corresponding alkane molecule is replaced with the suffix – ol;
  2. The parent molecule is the longest chain containing the – OH group;
  3. The numbering of carbon atoms is done such that the carbon atom with the hydroxyl group –OH- attains the lowest possible number
  4. The constituent branch is named accordingly;

 

Examples:                                         

CH3                                                      (ii). CH3CH2CH2CH2OH                    (iii). H     OH     H

|                                                 Butan-1- ol;

CH3-C-OH                                                                                          H     C       C       C        H

|

CH3                                                                                                                                  H       H      H

2,2 dimethylpropanol-1-ol;                                                                                                                                   Propan-2-ol;

 

(iv).                 CH3                 (v). H     H      OH     H     H             (vi).          H      H

 

CH3     CH2    CCH3                H    C     C        C      C      C    H                H    C       C      OH

 

OH                         H     CH2   H     H      H                              H       H

2 methyl butan-2-ol;                                                                                      Ethanol;

CH3

3,4 ethyl methylhexan-3-ol;

 

Isomerism in alkanols.

– Alcohols exhibit positional isomerism due to the fact that the position of attachment of the functional group varies within the carbon chain;

 

Examples of isomeric alcohols

(a). Isomers of propanol

(i).      H     OH     H                            (ii).      H     H            H

 

H     C     C         C     H                        H     C     C     O    C     H

 

H     H        H                                        H     H            H

Propan-2-ol                                                                              Ethyl methyl ether or methoxyethane;

 

(iii).      H     H     H                            (iv).

 

H     C     C     C     O     H

 

H     H    H

Propan-1-ol;

Note: Ethyl methyl ether is not actually an alcohol as it lacks the –OH group;

– All have the molecular formula: C3H7OH

 

(b). Isomers of butanol

(i). CH3CH2CH2OH                (ii). CH3CH2CHCH3               (iii).            CH3

                  Butan-1-ol;                                        

OH                            CH3     C      OH

                                                            Butan-2-ol;

CH3

                                                                                                2,2-dimethyl propan-2-ol

Note: All have the molecular formula: C4H9OH

 

(c). Draw the Isomers of pentanol, C5H11OH;

(i).       H    H    H          H   H                   (ii).                                   (iii)                               (iv)

 

H – C – C – C – O – C – C – H

 

H    H   H           H    H

                      Ethylpropylether

 

Primary, secondary and tertiary alcohols.

  • Primary alcohols:

– Are alcohols in which the OH group is attached to a carbon atom to which 2 hydrogen atoms are attached;

– Thus they contain a –CH2OH group;

Examples:

H    H    H    H

 

H – C – C – C – COH

                      H     H   H    H

    Butan-1-ol;

 

  • Secondary alcohols:

– Are alcohols in which the hydroxyl group is attached to the carbon atom to which only one other hydrogen atom is attached;

– The carbon atom with the OH group is thus bonded to two carbon atoms;

– They contain a CHOH group;

Example:

H    H    H    H

 

H – C – C – C – C – H

                      H     H   OH H

Butan-2-ol;

 

  • Tertiary alcohols:

– The hydroxyl group is attached to a carbon atom with no hydrogen atoms attached

– The carbon atom with the OH group is bonded to 3 other carbon atoms; hence sorrounded by the methyl groups;

– Tertiary alcohols thus a contain a –COH group;

Examples:

H   OH H

 

H – C – C – C – H

          H    CH3   H

         2-methyl propan-2-ol;

Preparation and properties of alkanols.

– Alkanols are prepared from three main methods.

  • Hydrolysis of halogenoalkanes;
  • Hydration of alkenes
  • Fermentation of starches and sugars (mainly for ethanol)

 

(a). Hydrolysis of halogenoalkanes;

– Halogenoalkanes are compounds in which one or more hydrogen atoms in an alkane are replaced by halogens;

– Addition of aqueous KOH or NaOH to a halogenoalkane and heating results to corresponding alcohol;

– Reaction involves replacement of the halogen atoms with the -OH from the alkali;

 

Examples:

(i). Preparation of methanol

CH3Cl     +       NaOH       heat              CH3OH    +    NaCl

Chloromethane          sodium hydroxide;                           methanol                sodium chloride

 

(ii). Preparation of propanol;

CH3CH2CH2Br    +      KOH               heat      CH3CH2CH2OH       +     KBr

1-bromopropane                          potassium hydroxide                       Propan-1-ol                                      Potassium bromide

 

Note: –the conversion of a halogenoalkane to an alcohol is known as hydrolysis;

– Reagent in this case is an alkali and condition for reaction is heat;

 

(b). Hydration of alkenes.

– Conversion of an alkene to an alcohol is known as hydration;

– Main reagent for the reaction is water;

– Conditions for the reaction are:

  • An acid catalyst, mainly conc. H2SO4 or phosphoric acid (H3PO4);
  • High temperatures of about 80oC;
  • High pressures of about 25-30 atmospheres;

 

Examples:

(i). Preparation of ethanol from ethene.

 

C2H4   +  H2O      Conc. H2SO4           C2H5OH;

80oC; 25-30 atm;

H     H                                                         H    H

 

 

H     C  = C     H     +     H     OH                H     C     C     H

Ethane                                   Water

H     OH

Ethanol

(ii). Preparation of butanol from butene.

 

CH3CHCHCH3   +    H2O       Conc H3PO4         CH3CH2CH2CH2OH

       Butene                                    80oC, 25-30 atm                              Butanol

 

(c). Preparation (of ethanol) by fermentation.

It is prepared from the fermentation of starches or sugars in the presence of yeast;

Fermentation: Is a chemical decomposition brought by bacteria or yeast (anaerobically) usually accompanied by evolution of carbon (IV) oxide and heat.

 

The chemical process

– Starch is broken into sugars by the action of the enzyme amylase or diastase;

Break up into

Starch molecule + water                                    sucrose molecules

Amylase

(C6H10O5) n  +  nH2O                         nC6H12O6

      Starch                      Water                                  many sucrose molecules

 

– When yeast is added to dilute sucrose solution (ordinary sugar); the enzyme sucrase in yeast catalytically breaks down sugar (sucrose) into the simplest sugars, glucose and fructose i.e.

 

Equation:

C12H22O11(aq) +  H2O(l)        Sucrase             C6H12O6(aq) +  C6H12O6

     Sugar                          Water                                                     glucose                      fructose

 

– Finally the enzyme zymase, also produced by yeast converts glucose and fructose into ethanol and carbon (IV) oxide.

 

Equation:

C6H12O6 (aq)        Zymase         2C2H5OH(aq) + CO2

Glucose/fructose                                    ethanol                         carbon (IV) oxide

 

Optimum conditions for fermentation:

  • Temperatures of 25-30oC;
  • Yeast catalyst;
  • Absence of oxygen (airtight);

 

Note:

– When the reaction mixture contains about 12% by volume of ethanol, the activity of yeast ceases.

– This is because higher ethanol concentrations kill the yeast cells;

– Fermentation provides about 10% alcohol by volume;

– The concentration of resultant ethanol can be increased by fractional distillation.

– During the process, ethanol distills over fast due to its lower boiling point (78°C)

– The distillate at below 95°C is first collected (leaving water behind).

– The resultant fraction will have 95% alcohol by volume; and is called rectified spirit;

– Absolute ethanol; which is 99.5% by volume can be obtained by re-distillation of rectified ethanol between 78-82oC to remove all the water in the mixture;

 

– This can be done in two main ways:

  • Addition of a small amount of benzene to the rectified spirit and then distilling; (benzene dissolves in the water in the alcohol)
  • Distillation of rectified spirit over a suitable drying agent like calcium oxide and then over calcium; (calcium reacts with steam, calcium oxide takes in condensed water)
Properties of alcohols (Ethanol)

(a) Physical properties

(i). It is a colourless, volatile liquid soluble in water in all proportions forming a neutral solution;

(ii). Has a characteristic smell and boils at78.5°C

 

Variation in physical properties of alkanols.

 

Name Molecular formula Molecular mass Boiling point (oC) Melting point (oC) Solubility in 100g of water
Methanol CH3OH 32 64.5 -94 Soluble
Ethanol C2H5OH 46 78.5 -117 Soluble
Propanol C3H7OH 60 97 -127 Soluble
Butanol C4H9OH 74 117 -90 Slightly soluble
Pentanol C5H11OH 88 138 -79 Slightly soluble
Hexanol C6H13OH 102 158 -52 Slightly soluble
Heptanol C7H14OH 116 175 -34.6 Very slightly soluble
Octanol C8H16OH 130 194 -16 Very slightly soluble;

 

Note:

– Solubility of alkanols decreases with increase in molecular mass;

– Both melting and boiling points increases with increase in the relative molecular mass; due to progressive increase in number of van der waals forces;

– Alkanols have higher melting and boiling points than their corresponding alkanes with the same molecular formula;

Reason:

– Alkanols have hydrogen bonding between their molecules, caused by the presence of the OH group; alkanes have van der waals between its molecules; Hydrogen bonds are stronger than weak van der waals;

 

Diagram: Hydrogen bonding between four ethanol molecules.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(b) Chemical properties /main reactions of ethanol.

Note; the main reactions of ethanol are those of its functional group –OH;

 

(iii). Combustion

Procedure;

– A few drops of ethanol are placed in a watch glass and lit.

– A dry gas jar is held over the flame and the gas collected tested with limewater.

 

Observation:

– It burns with a blue flame, which is almost colourless.

– The resultant gas turns limewater into a white precipitate, indicating it is carbon (IV) oxide.

 

Explanations:

– Ethanol (alcohols) burns in air (oxygen) producing carbon (IV) oxide, water and heat energy;

– The lower members of the homologous series burn with a blue or non-luminous flame leaving no residue;

– As the hydrocarbon chain increases the flame becomes more luminous and smoky and a black residue remains;

 

Equation

C2H5OH(l) +  3O2(g)                                         2CO2(g) + 3H2O(l)

 

Note:

– If an alkanol is burnt in a limited supply of oxygen, then the combustion is incomplete and the products include carbon (II) oxide or carbon and water

 

Equations

(i). C2H5OH(l) +  2O2(g)                                   2CO(g) + 3H2O(l)

 

(ii). 2C2H5OH(l) +  2O2(g)                                4C(s) + 6H2O(l)

 

(iv). Reaction with metals (sodium);

Procedure:

– Tiny pieces of sodium one at a time; are added 1cm³ of pure ethanol in a boiling tube.

 

Observation;

– Sodium metal darts on the surface of the ethanol and then dissolves /disappears;

– The beaker becomes warmer indicating an exothermic reaction.

Effervescence occurs and bubbles of a colourless gas are observed; gas burns with a pop sound.

 

Explanation:

– Sodium reacts with alcohol much as it does with water but the reaction is more gentle.

– Sodium reacts with ethanol to produce hydrogen gas, which on testing burns with a pop sound.

– The reaction is exothermic producing heat hence warmer beaker.

– A clear solution of sodium ethoxide is left formed in the boiling tube.

 

Equation;

2C2H5OH + Na (l)                                   2CH3CH2ONa(l)  + H2(g)

 

Note:

Alkanols react with electropositive metals such as sodium, potassium and aluminium to liberate hydrogen gas and form a solution of the metal salt, the metal alkoxide;

 

Examples

(i). 2CH3CH2CH2OH  +  Na                           2CH3CH2CH2ONa  +  H2

       Propanol                                       Sodium metal                           Sodium propoxide                         hydrogen

 

(ii). CH3CH2CH2CH2CHOH  +  K                              2CH3CH2CH2CH2CHOK  + H2

          Pentanol                                                     Potassium                                                 Potassium pentoxide                             Hydrogen

 

Note: The reactivity of alkanols with metals decreases as the hydrocarbon chain increases;

 

(v). Esterification;

– It is the production of esters (alkyl alkanoates) from the reaction between alcohols and carboxylic acids;

 

Procedure

– 2-3 drops of concentrated sulphuric acid are added to a mixture of equal proportions of ethanol and pure ethanoic acid in an evaporating dish.

– The mixture is warmed gently in a water bath for sometime.

– The mixture is poured into a beaker and smelt;

 

Observation

A fruity sweet smell (of ethyl ethanoate);

 

Explanation

– Ethanol reacts with ethanoic acid in the presence of a few drops of concentrated sulphuric acid to form ethyl ethanoate and water.

– The reaction is very slow and so catalyzed by the hydrogen ions from the sulphuric acid.

 

Equation                                      Conc. H2SO4; warm

C2H5OH(l) +  CH3COOH (aq)                                CH3COOC2H5(l)   +   H2O(l)

Ethanol                     ethanoic acid                                                                 ethyl ethanoate (ester)

 

Structurally;

H     H                          H    O                                 H     O            H     H

Conc. H2SO4

H        C      C     OH  +   H     C    C                          H     C     C     O     C     C    H   +   H2O

                                                                                                                                             Water

H     H                          H    OH                              H                    H      H

Ethanol                                                          Ethanoic acid                                                           Ethylethanoate                     

 

– The alkyl part of the ester is derived from the alkanol, while the alkanoate part is derived from the alkanoic acid;

– The alkanol attaches itself at the group in the carboxylic acid thereby displacing hydrogen atom;

– Under ordinary conditions the reaction takes place slowly; but in presence of concentrated sulphuric acid which act as a catalyst and warm (heat) conditions, the reaction is enhanced;

 

 

 

Further examples:

Write balanced equations for each of the following esterification reactions:

(i). Ethanol and propanoic acid;

 

(ii). Propanol and ethanoic acid;

 

(iii). Ethanol and methanoic acid;

 

Note:

– Generally a reaction between alcohol and a carboxylic acid (-COOH-) produces an ester and water in a process called Esterification;

 

General equation:

Conc. H2SO4, warm

Organic acid +  alcohol                               ester  + water….(Esterification), while;

 

 

Inorganic acid   + alkali                                        salt +  water….(neutralization)

 

Differences between neutralization and Esterification

  1. Esterification is slower than neutralization as the reaction is between molecules and not ions as in neutralization
  2. Esterification is reversible; the forward reaction is esterification and the backward reaction is hydrolysis.
  3. Esterification results to esters which are covalent compound; neutralization forms salts which are electrovalent.

 

Name and formulae of some common esters.

 

Alkanol Alkanoic acid Ester
Methanol

CH3OH

Propanoic acid

O

 

CH3CH2     C     OH

Methylpropanoate

O

 

CH3CH2     C     OCH3

 

Ethanol

C2H5OH

Methanoic acid

O

 

H     C     OH

Ethylmethanoate

O

 

H     C     OCH2CH3

Propan-1-ol Ethanoic acid

 

 

 

Butan-1-ol Ethanoic acid

 

 

 

 

 

 

(vi). Oxidation of primary alkanols;

– On heating in presence of oxidising agents, primary alkanols are oxidised to alkanoic acids;

Note:

– During oxidation of alcohols, they first lose hydrogen to form compounds called aldehydes (compounds ending in –al)

– The resultant aldehydes (alkanal) then gain oxygen to form alkanoic acids.

 

Equations:

(i). Alkanol – hydrogen                 alkanal + water

Then;

(ii). Alkanal +  oxygen                alkanoic acid;

 

General equation:

H                                                         O

 

R     C     OH  +   2 [O]                        R     C     OH   +   H2O

From oxidizing agent

H

 

 

Example: Oxidation of ethanol

– Ethanol like all other alcohols is oxidized by strong oxidizing agents such as potassium dichromate (VI) and potassium manganate (VII) to form ethanoic acid.

 

(a). Reaction with acidified potassium dichromate (VI):

Procedure

– A little solution of acidified potassium dichromate (VI) is added to a little solution of ethanol in a test tube and then warmed gently;

 

Observation:

– The solution (dichromate) changes from yellow to green;

 

Explanation

– The acidified potassium dichromate (VI) oxidizes the ethanol to ethanal then to ethanoic acid, while the dichromate undergoes reduction (chromate (VI) to Cr3+ changing colour from yellow to green;

 

Equations

O

H+ from Conc. H2SO4

C2H5OH(l)    +   [ O ]                                            CH3CH(aq)   +   H2O(l)

Ethanol                           From                        Slow reaction                            Ethanal                          water

K2Cr2O7

 

Structurally:

H     H                                                                 H    H

 

H     C      C     OH   +    [ O ]                              H     C     C    O   +   H2O

 

H     H                                                                  H

Then;

O                                                                     O

 

CH3CH(aq)  +   [ O ]                                         CH3C     OH

Ethanal                                       H+ reaction                     Ethanoic acid

General equation

H     H                                                                        O

 

H     C     C    OH(l)   +   2 [ O ]                                   CH3C   OH   +   H2O

 

H     H

(b). Reaction with acidified potassium manganate (VII).

Procedure

Acidified potassium permanganate solution is added to ethanol in a test tube end the mixture is warmed gently.

 

Observation

The  permanganate solution turns from purple to colourless.

The characteristic smell of ethanoic acid is felt.

 

Explanation

The ethanol is oxidized to ethanal then to ethanoic acid. The reduced permanganate decolorizes (turns from purple to colourless);

-The H+/KMnO4 is decolourised due to reduction of manganate (VII) ions to Mn2+

 

General equation:

 

H     H                                                                        O

 

 

H     C     C    OH(l)   +   2 [ O ]                                   CH3C   OH   +   H2O

H+/ KMnO4

H     H

 

 

(c). Catalytic oxidation of alkanols.

– Catalytic oxidation of alkanols results to a dehydrogenation (removal of hydrogen) resulting to the formation of an alkanal (aldehyde);

– These are compounds with the functional group COH;

 

Example: Catalytic oxidation of ethanol with hot copper metal

When ethanol is passed over heated copper at about 300 °C, it is dehydrogenated i.e. hydrogen is removed.

This results into formation of an ethanal; and hydrogen gas is liberated.

 

Equation                       Cu(s)

CH3CH2OH(g)                                    CH3CHO(l)   +   H2(g)

Ethanol                             250oC             Ethanal

(vii). Dehydration reactions of alcohols.

– Dehydration is the removal of water molecules from a compound;

– Excess concentrated sulphuric (VI) acid dehydrates alkanols and forms corresponding alkenes.

 

Conditions required: –

– High temperatures of 140-180°C.

– Catalysts such as conc. sulphuric acid, phosphoric acid and aluminium oxide;

 

Example: Dehydration of ethanol

(i). Apparatus

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure

– 15cm3 of absolute ethanol are put in around bottomed flask and 5cm3 of con H2SO4 added.

– The contents are mixed thoroughly by swirling the flask.

– The flask is then heated (warmed) gently with little shaking for about 1 minute.

– The gas collected is tested with acidified potassium manganate (VII) and bromine water.

 

(iii). Observations

– Evolution of a colourless gas that decolourises the purple acidified potassium manganate (VII)

– The resultant gas also decolourises the red brown bromine water.

 

(iv). Explanations

– On heating the alkanol (ethanol) undergoes an elimination reaction.

– It loses both a hydrogen (H) and a hydroxyl (OH) from two adjacent carbon atoms.

– The H and OH combine to form water; and the remnants form ethene.

– The sulphuric acid acts as a catalyst.

 

Note: Disadvantage of concentrated sulphuric acid over phosphoric acid.

– Being a strong oxidising agent the concentrated sulphuric acid oxidizes some of the alkanol formed to CO2 and it is itself reduced to SO2;

– This reduces the volume and purity of resultant alkanol; and is also a potential source of pollution;

 

 

Equation

H     H                                                           H    H

Conc. H2SO4

H     C     C     H                                            H     C = C     H   +   H2O

170oC

H     OH

 

Note:

With cold concentrated sulphuric (VI) acid, alkanols react to form alkyl hydrogen sulphates.

 

Example: Ethanol and cold conc. sulphuric acid.

 

C2H5OH   +  H2SO4(l)                          C2H5HSO4(l)  +   H2O(l)

Ethanol                                                                         Ethyl hydrogen sulphate

                                                                                                     (Ethoxyethane)

Uses of ethanol

  1. Used as solvents; in the preparation of drugs, perfumes, liquors, vanish and paints.
  2. Source of fuel e.g ethanol when blended with gasoline to form gasohol.

Note:

– Addition of ethanol to petrol improves the antiknock propertied of petrol, due to its low ignition point;

– The alkanol also absorbs any traces of moisture that may enter and damage the petrol system;

– Alkanols also ignite on their own to liberate heat;

Example:

CH3CH2OH(l)   +   3O2(g)                                 2CO2(g)   +   3H2O(l)  ∆H = -1368KjMol

 

  1. Starting material for the manufacture of polyvinyl chloride (P.V.C)
  2. Ethanol is used as a disinfectant (antiseptic) at special concentrations e.g for cleaning tissues and surgical equipments during operations and in dressing wounds;
  3. Manufacture of alkanoic acids e.g ethanoic acid
  4. Ethanol is used as thermometer liquid for measuring low temperatures;
  5. Manufacture of alcoholic drinks
  6. Large amounts of methanol are used in the manufacture of formaldehyde a chemical used in preservation of corpses;
  7. They are used as antifreeze mixtures in car radiators e.g a mixture of ethanol and water freezes at lower temperatures that pure water;
  8. Manufacture of esters to giver fruity flavourings for confectionery and drinks;

 

Tests for primary alkanols.

  1. Reaction with sodium metal

– Alkanols liberate hydrogen gas, a colourless gas that burns with a pop sound;

 

Equation:

2CH3CH2OH(l)  +   2Na(s)                    2CH3CH2ONa   +   H2(g)

 

  1. Reaction with phosphorus (V) chloride:

– Alkanols liberate misty fumes of hydrogen chloride gas;

 

Equation:

PCl5(s)   + CH3CH2OH                                   2CH3CH2OP   +   HCl2

Note:

– Both alkenes and alkanols decolourise the purple acidified potassium manganate (VII);

– However, alkenes decolourise the red bromine water; while alkanols do not.

 

Summary on preparation of alkanols (ethanol).

 

 

Ester                                                                                                 RCOOC2H5

 

 

 

                                                                                       Heat with NaOH(aq)

 

 

 

Ethyl iodide        Boil with KOH(aq)                     Ethanol                             Fermentation        Starch, sugars

(Iodoethane)                                                                                                                cellulose

CH3CH2I

 

 

                                                                                       H2O + H2SO4

 

 

 

                                                                           Ethene

                                                            CH2CH2

 

 

 

Summary on reactions of alkanols.

 

 

CO2(g)  +  H2O(l)

 

 

 

                                                                                          Combustion

 

 

 

Ester                      Esterification                           Ethanol                 Sodium metal                  Sodium ethoxide   + Hydrogen

RCOOC2H5          H+/ RCOOH                       CH3CH2OH                                        CH3CH2ONa  + H2

 

 

                              Conc. H2SO4 at 170oC                                    Warm with acidified K2Cr2O7 or KMnO4

 

 

 

 

Ethene                                                                       PCl5                                                          Ethanoic acid

CH2CH2                                                                                                                                     CHCOOH

 

 

 

                                                           

                                                            Chloroethane

                                                            CH3CH2Cl

 

 

 

 

 

 

ALKANOIC ACIDS (CARRBOXYLIC ACIDS)

– Also called organic acids and form a homologous series with a general formula of CnH2n-1OOH

– The formula can also be written as CnH2n + 1COOH; in which case n = (no. of carbon atoms – 1)

– Members differ from each other by an additional CH2 group.

– Their functional group is the carboxylic group (-COOH) which is attached to the alkyl group.

– Graphical representation of carboxyl group; O

 

C   OH

Note: All carboxylic acids have the –COOH as the functional group but alkanoic acids are strictly alkanoic acids derived from alkanes

 

Nomenclature of alkanoic acids

– The ending of the corresponding alkane is replaced by ¢oic acid’.

 

Examples;

– Methane to methanoic acid.

– Ethane to ethanoic acid.

 

– They are named as if they are derived from alkanes through replacement of one of the hydrogen atoms by the -COOH group.

 

Note: –Unlike alkanols the functional group (COOH) in alkanoic acids can only be at the end of the carbon chain.

– The C in the COOH is always given the first position, while the substituents are given locants (numbers in reference to the first position).

 

Examples:

IUPAC name Old (traditional ) name Structural formula
Methanoic acid Formic acid              OH

 

H     C = O

Ethanoic acid Acetic acid                OH

 

CH3 – C = O

Propanoic acid Propionic acid                     OH

 

CH3CH2 –  C = O

Ethanedioic acid Oxalic acid  

O                    O

C –   C

HO                  OH

 

Butanedioic acid Succinic acid                  O

CH2C

OH

O

CH2C

OH

Branched alkanoic acids

– The naming of branched alkanoic acids follow the same general rules like that of alkanes; as long as the carbon atom with the –COOH group is given the first position.

– The branch can either be an alkyl group or a halogen other than hydrogen.

 

Examples:

 

Compound IUPAC name
                            O

Cl  –  CH2 –  C

OH

 

2-chloroethanopic acid;

                             O

CH3  –  CH –  C

OH

CH3

 

3-methylpropanoic acid;

                            O

CH3  –  CH –  C

OH

OH

 

2-hydroxypropanoic acid;

 

OH

O

CH3  –  C –  C

OH

CH3

 

 

2-hydroxy, 2-methylpropanoic acid;

                                         O

CH2  – CH  –  CH2 –  C

OH

Br        Cl

 

4-bromo, 3-chlorobutanoic acid;

 

 

Isomerism in alkanoic acids.

– Due to the existence of branched alkanoic acids, it is possible to obtain various isomers for a given alkanoic acid;

 

Example:

Draw all the isomers of pentanoic acid.                                                                            (3 marks)

 

 

 

 

 

 

 

 

 

 

 

Preparation of alkanoic acids

(a). Industrial manufacture

– Is done by the oxidation of primary alkanols using air (oxygen) as the oxidising agent.

 

Conditions:

– Moderate temperatures

– 5 atm pressure

– Hot copper catalyst;

 

Laboratory preparation.

– Is done by the oxidation of primary alkanols using acidified potassium dichromate (VI).

 

Example: Laboratory preparation of ethanoic acid.

(i). Apparatus

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii). Procedure

– Acidified potassium dichromate (VI) is heated in a water bath and ethanol added slowly from a water bath.

-The mixture is heated further and then distilled.

– The distillate is collected at about 105oC.

 

(iii). Observations and explanations.

Colour of potassium dichromate (VI) changes from orange to green.

Reason:

The dichromate ions are reduced to chromium (III) ions.

 

Equation:

Cr2O72-(aq) +  14H+(aq)  +  6e-                      2Cr3+(aq)  +   7H2O(l)

 

– The ethanol is oxidised to ethanal (acetaldehyde); which is further oxidised to alkanoic acid.

Equations

O

H+ from Conc. H2SO4

C2H5OH(l)    +   [ O ]                                            CH3CH(aq)   +   H2O(l)

Ethanol                           From                        Slow reaction                            Ethanal                          water

K2Cr2O7

 

Structurally:

H     H                                                                 H    H

 

H     C      C     OH   +    [ O ]                              H     C     C    O   +   H2O

 

H     H                                                                  H

Then;

O                                                                     O

 

CH3CH(aq)  +   [ O ]                                         CH3C     OH

Ethanal                                       H+ reaction                     Ethanoic acid

General equation

H     H                                                                        O

 

H     C     C    OH(l)   +   2 [ O ]                                   CH3C   OH   +   H2O

 

H     H

 

Properties of alkanoic acids

 

Gradation in physic al properties of alkanoic acids

 

Name of acid Formula (structural) Molecular formula M.P °C B.P °C Solubility
Methanoic acid

Ethanoic acid

Propanoic acid

Butanoic acid

Pentanoic acid

Hexanoic acid

HC OOH

CH3COOH

CH3CH2COOH

CH3CH2CH2COOH

CH3CH2CH2CH2COOH

CH3(CH2)4COOH

CH2O2

C2H4O3

C3H6O2

C4H8O2

C5H10O2

C6H12O2

8.4

16.6

-20.8

-6.5

-34.5

-1.5

101

118

141

164

186

205

Most Soluble

 

increasing

     solubility

 

 

Least soluble

 

Physical properties of ethanoic acid

– A colourless liquid with a sharp pungent smell.

– B.P is 118oC and freezes at 17°c forming ice like crystals termed as glacial ethanoic acid.

– It is soluble in water and is weakly acidic with a P.H of approximately4.8.

 

Note

– Concentrated ethanoic acid is only slightly ionized and is a poor conductor of electricity

– On dilution its conductance steadily improves as the extent of ionization increases

 

 

 

 

Chemical properties (reactions) of alkanols.

(i). Reaction with carbonates.

– Alkanoic acids react with metal carbonates to form a salt (metal alkanoate), carbon (IV) oxide and water.

 

Examples:

Ethanoic acid and sodium carbonate

Ethanoic acid reacts with sodium carbonate to form sodium ethanoate and water with the liberation of carbon (IV) oxide gas.

 

Equation

Na2CO3  +   2CH3COOH                       2CH3COONa   +   CO2(g)   +   H2O(l)

Sodium carbonate                                                                       Sodium ethanoate

 

Zinc carbonate and ethanoic acid.

ZnCO3  +   2CH3COOH                        (CH3COO)2Zn   +   CO2(g)   +   H2O(l)

Zinc carbonate                                                                            Zinc ethanoate

 

(ii). Reaction with metal hydroxides (Neutralization)

Alkanols neutralize alkalis like sodium hydroxide forming a salt (metal alkanoate) and water only.

 

Examples:

Sodium hydroxide and ethanoic acid.

CH3COOH   +   NaOH                        CH3COONa   +   H2O(l)

Ethanoic acid                 Sodium hydroxide                      Sodium ethanoate

 

Potassium hydroxide and methanoic acid

HCOOH   +   KOH                               HCOOK   +   H2O(l)

Methanoic acid         Potassium  hydroxide                      Potassium methanoate

 

(iii). Reaction with metal oxides (neutralization).

– Alkanoic acids react with metal oxides to produce salt (metal alkanoate and water only.

 

Examples:

Ethanoic acid and copper (II) oxide.

2CH3COOH   +   CuO                         (CH3COO)2Cu   +   H2O(l)

Ethanoic acid                     Copper (II) oxide                      Copper ethanoate

 

(iv). Reaction with metals

Alkanoic acids react with reactive metals to form a salt (metal alkanoate) and hydrogen gas;

 

Examples:

Ethanoic acid and sodium metal

2CH3COOH  +  Na(s)                                       2CH3COONa   +   H2(g)

Ethanoic acid                                                                                             Sodium ethanoate

 

Propanoic acid and magnesium metal

2CH3CH2COOH  +  Mg                                  (CH3CH2COO)2Mg   +   H2(g)

Propanoic acid                                                                                            Magnesium propanoate

 

 

(v). Reaction with alkanols (Esterification)

– Alkanoic acids react with alkanols to form esters;

Conditions:

– Drops of concentrated sulphuric acid.

– Gentle warming.

 

Reaction of ethanoic

– Ethanoic acid reacts with ethanol in the presence of a few drops of concentrated sulphuric acid forming a sweet fruity smelling compound called ester. 

– The process is called esterification.

 

Procedure

– 2-3 drops of concentrated sulphuric acid are added to a mixture of equal proportions of ethanol and pure ethanoic acid in an evaporating dish.

– The mixture is warmed gently in a water bath for sometime.

– The mixture is poured into a beaker and smelt;

 

Observation

A fruity sweet smell (of ethyl ethanoate);

 

Explanation

– Ethanol reacts with ethanoic acid in the presence of a few drops of concentrated sulphuric acid to form ethyl ethanoate and water.

– The reaction is very slow and so catalyzed by the hydrogen ions from the sulphuric acid.

 

Equation                                      Conc. H2SO4; warm

C2H5OH(l) +  CH3COOH (aq)                                CH3COOC2H5(l)   +   H2O(l)

Ethanol                     ethanoic acid                                                                 ethyl ethanoate (ester)

 

Structurally;

H     H                          H    O                                 H     O            H     H

Conc. H2SO4

H        C      C     OH  +   H     C    C                          H     C     C     O     C     C    H   +   H2O

 

H     H                          H    OH                              H                    H      H

 

– The alkyl part of the ester is derived from the alkanol, while the alkanoate part is derived from the acid;

– The alkanol attaches itself at the group in the carboxylic acid thereby displacing hydrogen atom;

– Under ordinary conditions the reaction takes place slowly; but in presence of concentrated sulphuric catalyst and warm (heat) conditions, the reaction is enhanced;

 

If propanol were used in place of ethanol, the reaction would yield the ester propyl ethanoate, according to the following equation;

 

CH3COOH(aq) +  CH3CH2CH2OH(aq)                   CH3COOCH2CH3(aq) + H2O(l)

Ethanoic acid                    propanol                                                             Propylethanoate

 

 

 

Note: –

Esters react with water to form the respective alkanoic acid and alkanol.

This reaction is termed hydrolysis and occurs in presence of concentrated sulphuric acid and heat as conditions.

 

Example:                                Conc. H2SO4

CH3COOCH3   +   H2O                                           CH3COOH   +   CH3OH

Methyl ethanoate                                                      Heat                            Ethanoic acid                  Methanol

 

(vi). Reaction with ammonia.

Alkanoic acids react with ammonia to produce the ammonium salt of the acid.

General formula of the ammonium alkanoate salt is RCOONH4

 

Example:

Ethanoic acid and ammonia gas.

CH3COOH   +   NH3                           CH3COONH4

 

Uses of alkanoic acids

  1. In pharmaceuticals, for making medicines e.g ethanoic acid is used in the manufacture of aspirin.
  2. Manufacture of dyes and insecticides
  3. Seasoning food as vinegar
  4. Coagulation of rubber latex
  5. Preparation of polyethenyl ethanoate and cellulose ethanoate which are used are used to make artificial fibres such as rayon.
  6. Manufacture of soaps.
  7. Preparation of perfumes and artificial favours used in food manufacture.

 

Tests for alkanoic acids.

The following tests can be used to test for alkanoic acids.

Reaction with carbonates and hydrogen carbonates

Esterification

 

Summary: Draw a summary flow chart to show all the reactions of a named alkanoic acid.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

FATS AND OILS.

– Are esters of long chain carboxylic acids and glycerol

(a). Oils

– Oils occur naturally in plants and animals.

Examples: Whale oil, groundnut oil, corn oil and castor oil.

– Oils are liquids at room temperature.

Reason:

– They have high proportion of esters derived from the unsaturated oleic acid in them.

 

(b). Fats.

– These occur naturally in animals only.

Examples: Tallow, butter from milk, lard from pigs etc.

– Fats are solids at room temperature

– Oils can be converted to fats /hardened into fats by hydrogenation.

– This is the conversion of oils into fats by use of hydrogen; and forms the basis of margarine manufacture.

– During hydrogenation:

  • Hydrogen is bubbled into oils under high pressure and temperatures of about 400oC in the presence off a nickel catalyst

 

Note

– Fats and oils are important raw materials in the manufacture of soaps.

 

Soaps and soapless detergents

Soaps

Are a variety of compounds produced when oils or fats are reacted with sodium hydroxide.

– They are similar in that they contain a long hydrocarbon chain ending in a carboxylate anion to which is attracted a sodium cation.

– A typical soap is sodium stearate; C17H35COONa+

 

Structurally

                                                                                                                               O

CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2C

ONa+

 Preparation of soaps

– 2cm3 of castor oil and 10 cm3 of 4M sodium hydroxide are poured into a 100cm3 beaker.

– The mixture is then for about 10 minutes, stirring continuously and adding distilled water to make up for evaporation.

 

Explanation:

– On boiling an alkali with fat or oil, a hydrolysis reaction occurs.

 

 

 

 

 

 

Equation:

CH2 – COO – C17H35                                                                                                  CH2OH

 

 

CH –   COO – C17H35   +   3NaOH                             3C17H35COO Na+   +   CHOH

Sodium stearate (soap)

CH2 – COO – C17H35                                                                                                  CH2OH

Fat                                                                                                                                                                                                     Glycerol

 

– When hydrolysis reaction occurs in the presence of an alkali (sodium hydroxide), the process is known as saponification (the chemical reaction between a fat and an alkali)

– In the fat hydrolysis NaOH neutralizes the acid formed (i.e. stearic acid) to form the sodium salt of the acid removing it from the aqueous mixture.

– Thus in excess alkali, all the fat is utilized.

– The sodium salt (sodium stearate) of the acid is termed soap if the number of carbon atoms per molecule is more than eight.

 

Note:

– KOH may be used in place of NaOH as the alkali.

 

  1. To the boiled mixture 3 spatula measures of sodium chloride are added, stirred well and allowed to cool.

Reason:

– The NaCl helps in separating the soap from the glycerol.

– This step is called salting and it reduces the solubility of the soap in the aqueous layer.

– The lower layer consists of glycerol, salts and unused alkali solution.

 

General formula of ordinary soap

CnH2n+1COONa+ where n >8 (n is greater than eight)

 

  1. The solid is filtered off and washed with cold distilled water, to remove impurities like NaCl

 

  1. The solid sample is placed in a test tube with distilled water, then with tap water.

 

Note:

The resultant soap may not have lathered easily with tap water.

Reason:

Some tap water contains a high proportion of calcium or magnesium ions that make water hard.

Summary on soap preparation

 

Fat (oil)    Step I: saponification        Solution of soap and alcohol

 

 

Add NaOH and boil                                    Step II: salting

Add NaCl, stir well

And allow to cool

 

 

 

Ordinary soap

 

 

 

 

The role of soap in cleaning

Note: functions of soap in water

  1. It makes the water able to wet material more effectively by lowering the surface tension.
  2. Emulsification of oil and grease.

Soap molecules have two dissimilar ends:

  • A hydrocarbon chain which is non-polar and has no attraction for water- hence oil soluble
  • A Carboxylate end, which is polar and is attracted to water; hence is water-soluble.

Note: the Carboxylate end is in fact negatively charged in water because after dissolution, the sodium ion and the carboxylate ion exist as separate entities.

 

Illustration:

O

 

CH3(CH2)16 – C – ONa+

Non polar                Carboxylate (Polar)

 

Effects of soap on oil –water mixture (removal of oils and grease during washing)

Note: schematic representation of a soap molecule

 

 

 

 

 

  • On adding soap into oil water mixture the following build up occurs:

– A molecule of soap has a polar (hydrophilic) and non-polar (hydrophobic) parts;

– The non-polar end dissolves in oil and the polar end dissolves in water.

– When the mixture is agitated (thoroughly shaken) the hydrocarbon chain (tail) dissolves in grease while the carboxylate – sodium end of the soap molecule (the head) remains dissolved in water.

 

Diagrams: role of soap in cleaning.

 

 

 

 

 

 

 

 

– Each oil drop ends up with a large cloud of negative charge around it as the polar heads are negatively charged.

– Consequently the oil drops repel each other, hence preventing them from coalescing.

– The water –soluble sodium “heads” on the surface of the droplets keep the droplets emulsified (suspended) in the water.

– During rinsing the water carries away the oil droplets.

 

 

 

 

Effect of hard water on soap

– The Calcium and magnesium ions in hard water react with soap (sodium stearate) and remove it as an insoluble grey scum of magnesium or calcium stearate.

 

Equations:

(i). 2C17H35COONa+(aq)   +   Ca2+(aq)                          (C17H35COO)2Ca2+(s)   +   2Na+ (aq)

            Sodium stearate (soap)                                                                                                 Magnesium stearate

 

(ii). 2C17H35COONa+(aq)   +  Mg2+(aq)                        (C17H35COO)2Mg2+(s)  +  2Na+ (aq)

           Sodium stearate (soap)                                                                                                  Magnesium stearate

– Soap is wasted in this way until all the calcium (II) and magnesium (II) stearate has been removed.

– The resultant scum is deposited on fabrics, giving them an unsightly dull appearance.

– Thus in hard water districts it is obviously advantageous to remove hardness before washing.

 

Removal of hardness:

Depends on whether the hardness is temporary or permanent

 

(a). Temporary hardness

Cause:

– The presence of calcium hydrogen carbonate or magnesium hydrogen carbonate dissolved in water.

 

Removal

– By boiling the water

– During the process the soluble calcium or magnesium hydrogen carbonate is precipitated out as insoluble calcium or magnesium carbonate.

 

Equation:

Ca(HCO3)2(aq)        Heat            CaCO3(s) + CO2(g) + H2O(l)

 

(b). Permanent hardness

Cause:

– Presence of calcium or magnesium chlorides and sulphates

 

Removal

Can be removed using the following methods:

 

(i). Distillation

– The water is distilled and the dissolved substances are left behind as water is evaporated and condensed

 

(i). Addition of washing soda (Na2CO3)

The washing soda reacts with the Mg²+ and Ca²+(aq) ions precipitating them as the insoluble carbonates

Equations:

2Na+(aq)   +   CO32-(aq)   +   2Cl-   +  Ca2+ aq)                       CaCO3(s)   +   2Na+ (aq)  + 2Cl (aq)

 

CO32- (aq)    +   Mg2+                            MgCO3(s)

 

Disadvantage of washing soda as a water softener

– It is alkaline and can cause damage to wool and silk.

(iii). Ion exchange process (e.g. in the permutit water softener)

– It involves use of resins and compounds which will exchange their own Na+ for Ca²+or Mg²+ dissolved in hard water

– Thus as the Na²+ go into the water are left in the resin.

 

Equation:

2Na+ (resin)(aq)   +   Ca2+(aq)                           Ca2+ (Resin)2 (aq)   +   Na+(aq)

 

Advantages and disadvantages of hard water

Advantages of hard water

  1. It is good for drinking purposes as calcium contained in it helps to form strong bones and teeth.
  2. When soft water flows in lead pipes some lead is dissolved hence lead poisoning. However when lead dissolves in hard water insoluble PbCO3 are formed, coating the inside of the lead pipes preventing any further reaction
  3. It is good for brewing and the tanning industries;

 

Disadvantages of hard water

  1. Soap forms insoluble salts with magnesium and calcium ions; scum (calcium or magnesium stearate) thereby wasting soap.

Note: For these reason soapless detergents are preferred to ordinary soaps because they do not form scum; but rather form soluble salts with Mg2+ and Ca2+

– Examples of soapless detergents: brand names such as omo, perfix, persil, and fab e.t.c.

 

  1. Deposition of insoluble magnesium and calcium carbonates and sulphates formed from hard water result into blockage of water pips due to the formation of boiler scales

 

3.     Formation of kettle fur which make electrical appliances inefficient and increases running costs.

 

Soap and pollution effects.

  1. The wash water with soap and the dirt (grease) ends up in rivers and lakes thus affecting aquatic life; since plants do not grow well in soapy water.

 

Note: soaps are however biodegradable and so do not persist long in the environment.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Soapless detergents (synthetic detergents)

– Are cleansing agents lacking the carboxylate ions; but also act like soap in the cleaning process

– Instead they have sulphates (-OSO3Na+) or sulphonate groups (-SO3Na+) groups.

Are thus of two main types:

  • Sodium alkyl sulphates
  • Sodium alkylbenzene sulphonate.

 

(i). Sodium alkyl sulphates.

– Are detergents with the formula R – OSO3Na

 

Illustration.

– Consider the structure of sulphuric acid

                      O

 

HO     S      OH

 

O

– Replacing one of the hydrogen atoms with an alkyl group, R results into the compound

R – OSO3H;

– This is known as alkyl hydrogen sulphate;

– If the alkyl group (R) is a long chain such as dodecyl; CH3(CH2)10CH2 ; then the formula of the compound becomes CH3(CH2)10 – OSO3H (alkylhydrogen sulphate)

– Reacting the alkylhydrogen sulphate with an alkali (NaOH) results into a compound with the formula CH3(CH2)10 – OSO3Na.

– This is the soapless detergent, and is known as sodium dodecyl sulphate.

 

(ii). Sodium alkyl benzene sulphonates.

– Are formed when one of the OH groups in sulphuric acid is replaced with an alkyl benzene group.

 

 

– Have the general formula R –           – SO3Na

 

Illustration.

– If the alkyl benzene is of formula CH3(CH2)10CH2 ; then the resultant compound is of formula CH3(CH2)10CH2 –           – SO3H (hydrogen dodecyl benzene sulphonate)

 

– The sodium alkylbenzene sulphonate is neutralized using sodium hydroxide to obtain the detergent, CH3(CH2)10CH2 –           – SO3Na

 

The compound CH3(CH2)10CH2 –           – SO3Na is called sodium dodecyllbenzene sulphonate.

 

Conclusion:

The two main types of soapless detergents are:

  • Sodium alkyl sulphates; CH3(CH2)10 – OSO3Na
  • Sodium alkylbenzene sulphonate; CH3(CH2)10CH2 – – SO3Na

 

Note:

– Due to lack of carboxylate ions, soapless detergents do not form scum with hard water. Advantage over soap

– They are not affected by hard water, as they do not form scum in hard water as follows:

Preparation of soapless detergents

– Most are made from residues from crude oil distillation

– The hydrocarbons are treated with concentrated sulphuric acid  (instead of alkalis in cases of soap)

 

Procedure

– About 10cm³of olive oil in a small beaker, which is then stood in a larger one with ice-cold water.

– While stirring with a glass rod, concentrated sulphuric acid is carefully added to the olive oil using a dropping pipette.

– The acid is added until the yellow oil turns uniformly brown.

– 20cm³of 6M NaOH is then added; to neutralize the acid solution, resulting into a slightly basic product.

The soap is then tested with tap water and distilled water.

 

Explanation

– Some compounds like olive oil contain double bonds which can react concentrated sulphuric acid to form compounds of alkylhydrogen sulphate

 

Example:

 

R – CH = CH – CH – CH2 – Est   +   H2SO4 (l)                   R – CH – CH – CH – CH2 – Est

 

 

H   OSO3H

– In this case; R=alkyl group, either branched or straight; while Est = ester group.

– On adding NaOH, the alkyl hydrogen sulphate is neutralized where the hydrogen of the hydrogen sulphate is replaced by a sodium atom.

– The resultant compound is sodium alkylsulphate; R-CH  – CH-CH2-Est;

|          |

H     OSO3Na

Equation:

 

R – CH – CH – CH – CH2 – Est   +   NaOH(aq)          R – CH – CH – CH – CH2 – Est   +  H2O(l)

 

 

H   OSO3H                                                                     H     OSO3Na

 

– The R – CH  – CH – CH – CH2 – Est is the detergent and lathers easily with both tap and distilled water.

|          |

H     OSO3Na

Note:

– Alkylhydrogen sulphates can also be from alcohols.

 

Example:

 

R – CH – CH3   +   H2O(l)                             R – CH – CH3   +   H2O(l)

 

 

OH                                                                  OSO3H

– Most soapless detergents are sodium alkyl sulphates with a general formula ROSO3Na

 

– Sodium alkylbenzene soapless detergents are industrially manufactured from alkylbenzene;

– In this process alkylbenzene (a petroleum product) reacts with SO3 to form sulphuric acid

– Upon neutralization with NaOH, sulphuric acid forms sodium alkylbenzene sulphonate,

R –            – – SO3Na, a detergent.

 

Summary: manufacture of soapless detergents

 

Dodecene                                                   benzene

CH3(CH2)9CH = CH2   +

 

 

                              Chemical reaction

 

                                   

CH3(CH2)11                   Dodecylbenzene

 

 

Conc. sulphuric acid

 

 

 

CH3(CH2)11 –         SO3H

Hydrogen dodecylbenzene sulphonate

 

 

Sodium hydroxide

 

 

CH3(CH2)11            SO3 Na+

 

Sodium dodecylbenzene sulphonate

 

 

Mode of action of soapless detergent.

Soapless detergents have two ends; a long hydrocarbon part, the tail and a short ionic part the head.

 

Simple representation of a detergent molecule.

                                                                        SO3  Na+

 

 

 

Non-polar tail (water hating)                                                 Polar head (water loving)

 

Examples:

(i). Sodium lauryl sulphate

 

CH3(CH2)10CH2  – O – SO3Na+

 

 

       Tail                                                head

 

(ii). Sodium alkylbenzene sulphonate

 

CH3(CH2)10CH2  –              SO3Na+

 

 

            Tail                                                     head

– The tail is non-polar and dissolves in oil or grease (waterphobic) while the head is polar and dissolves in water (waterphilic).

– Each oil or grease gets sorrounded by the detergent molecules and hence a cloud of charged heads hence repel each other and do not coalesce.

– The dirt (grease loses its direct contact with the fabric being washed.

– Any agitation at this point then removes the dirt from the object.

 

Advantages of soapless detergents over soap;

  1. They lather easily with hard water since the corresponding calcium and magnesium salts are soluble in water due to lack of the carboxylate ions.
  2. Are mainly prepared from non-food raw materials.
  3. They do not react with acidic water.

 

Note:

– Soapless detergents with branched chain alkyl groups are not easily broken down by bacteria and are therefore the cause of frothing in sewerage plants, rivers etc.

– Consequently modern industry is overcoming this disadvantage by making the detergents from alkylbenzene with straight chain alkyl groups.

 

Pollution effects of soapless detergents.

  1. The active ingredients such as alkylbenzene sulphonates are non-biodegradable, hence accumulate in water sources and end up in human bodies.
  2. Some of the additives such as phosphates cause eutrophication hence excessive build up of algae (algal blooms) which change the water taste and odour, and also reduce oxygen supply hence poor growth of aquatic organisms.
  3. Because of their high lathering tendency, they cause excessive frothing and foaming in water sources especially after heavy rains.

 

Comparisons between soaps and soapless detergents.

 

Detergents Soaps
Have strong cleansing action Have weaker cleansing action
Are highly soluble in water; hence can be used in acidic or hard water. Are not very soluble in water, and tend to be wasted when used in hard water;

They cannot be used in acidic water.

Are made from byproducts of petroleum industry; which helps to conserve edible fats and oils Are made from edible fats and oils;
They cuse water pollution; Are biodegradable and have minimum pollution effects.
They are expensive. Are cheaper than detergents.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

POLYMERS

– A polymer is a macromolecule formed when two or more molecules link together to form a larger unit.

– Polymers have different properties different from those of the monomers.

– The process of polymer formation is called polymerization.

– Two types of polymers exist; natural and synthetic /artificial polymers.

 

(a). Natural polymers and fabrics

– Natural polymers occur naturally in living systems.

 

Examples:

– Rubber (latex), starch, cellulose, wool and proteins.

 

Natural rubber.

– Rubber trees give out a liquid called latex, which is collected from cuts in the trunks of rubber trees.

– Natural rubber is made out of latex from rubber trees.

– Latex is a hydrocarbon C5H8, called isoprene (2,methylbut 1,3 diene).

Formula of isoprene.

 

CH2 = C – CH = CH2

 

CH3

 

– Coagulation of latex leads to formation of a hydrocarbon polymer consisting of isoprene

(2, methylbut-1, 3-diene) units.

– This polymer is called polyisoprene with the formula; [ CH2 – C = CH – CH2 ]

 

CH3

Characteristics of natural rubber.

– Soft and sticky;

– Low resistance and low tensile strength; thus breaks easily upon stretching;

– Loses its rubber like properties at temperatures above 60oC;

 

Note:

– These are not good qualities and for industrial purposes this quality must be improved.

– This is done through vulcanisation of rubber.

 

Vulcanization of rubber.

– Is a chemical reaction in which raw rubber is heated with sulphur and is done purposely to improve the wear quality of rubber.

 

Process of vulcanization;

– Rubber is heated with sulphur

– The sulphur atoms form links between chains of rubber molecules.

– This reduces the number of double bonds in the polymer; making the material tougher, less flexible and less softer.

– During the process the sulphur atoms attach themselves to the rubber molecule in such a way that the molecules become locked in place and are prevented from slipping.

 

Note: –

– Soft rubber has about 2% sulphur, while toughened rubber about 10 % sulphur.

 

– Rubber can also be made artificially in industries and this gives a form of synthetic rubber.

– Other than undergoing vulcanisation it has chemicals that give it desired properties.

– It is made from byproducts of petroleum industry.

– Examples of synthetic rubber include neoprene and thiokol.

 

Note: Neoprene is made by polymerization of chloroprene (2, chlorobut 1, 3 diene), C4H5Cl

 

Equation:

CH2 = C – CH = CH2                                     CH2 – C = CH – CH2

 

Cl                                                                    Cl

Chloroprene                                                                                                                     Neoprene or polychloroprene

 

 

 

Properties of synthetic rubber.

  1. Are unreactive, hence they don’t react with industrial chemicals like oils, grease, petrol etc.
  2. Are non-inflammable, as they don’t catch fire easily.
  3. Capable of withstanding wide range of temperatures without changing shape.
  4. Have high mechanical strength.

 

Uses of synthetic rubber.

  1. Manufacture of insulating materials for electrical connections.
  2. Making conveyor and seat belts.
  3. Manufacture of car tyres and tubes.
  4. Making gaskets, flexible pipes.

 

Advantages and disadvantages of natural polymers.

– Are biodegradable hence not a likely cause of environmental pollution.

– Are made from renewable resources such as wool and trees; hence not easily exhaustible.

– Most are not easily flammable hence good materials for items like clothing.

 

  • Disadvantages

– Are often very expensive compared to synthetic polymers.

– Some do not last for very long.

– Are easily affected by acids, alkalis, air etc.

 

 

 

 

 

 

 

 

 

 

(b). Synthetic polymers

– They are man-made e.g. polythene, Perspex

– They are of two main types; thermoplastic and thermosetting polymers.

 

  • Thermoplastic;

– Softens on heating and becomes rigid on cooling.

Examples: nylon, polythene, polystyrene, etc.

 

  • Thermosetting:

– Are those that become hard on heating and cannot be softened by heating.

 

Advantages of synthetic polymers and fibres over natural polymers

  1. They can be made into different shapes easily.
  2. They are cheaper.
  3. Are often unaffected by ac ids, alkalis, water and air.
  4. Are usually less denser and yet stronger

 

Disadvantages

  1. Plastics are non-biodegradable hence causes a lot if problems in disposal
  2. Plastics burn more readily than natural material
  3. Some synthetic polymers give off poisonous gases when they burn. E. g. polyurethane gives off cyanide and carbon monoxide

 

Methods of polymerization:

  1. Addition polymerization;

– Occurs when unsaturated molecules (monomers) join to form a long chain molecule (polymer) without the formation of any other product.

– Usually the monomers must have at least a double or triple bond.

– One of the bonds in the double or triple bonds in the monomer opens up, and the unbonded electrons form bonds with neighbouring molecules.      

 

Conditions for polymerization.

– High pressures.

 

Examples.

Polymerization of ethene to Polythene

 

CH2 = CH2   +  CH2 = CH2                                          [-CH2 – CH2 – CH2 – CH2 -]

 

Calculations involving polymers.

  1. A polyisoprene molecule is represented as: -[ CH2 – C = CH – CH -]n

 

CH3

– Given that the relative formula mass of polyisoprene is 748 000, calculate the number of isoprene units in the polymer.

 

 

 

 

Summary on common addition polymers.

 

Monomer Polymer properties Uses
H                 H

C = C

H                 H

Ethene

   H     H

 

C –  C

 

H     H

Polythene

 

Light, tough, and durable – Polythene bags, bowls packaging, electrical insulation; plastic pipes etc
       H    H     H

 

H    C    C =  C – H

 

H

Propene

   H       H

 

C       C

 

CH3  H

Polypropylene

 

Light, tough and durable – Making crates, boxes; and plastic ropes
H                 H

C = C

H                 Cl

Chloroethene

   H     H

 

C –  C

 

H     Cl

Polychloroethene(polyvinyl chloride)

 

Strong and hard (not as flexible as polythene) –        Making plastic pipes,

-electrical insulators, floor tiles, credit cards,

                 CH = CH2

 

 

Phenyl ethene (styrene)

         H     H

 

C –   C

 

Cl

 

Polyphenylethene (polystyrene)

 

Light, poor heat conductor, brittle; -Insulation,

-packaging materials and food containers;

F                  F

C = C

F                 F

 

Tetrafluoroethene

      F       F

 

C –    C

 

F      F

Polytetrafluoroethene (Teflon)

 

Non-stick surface;

Withstands high temperatures

– Non-stick coatings on pans; Insulation;
                  O

 

CH2 = C – C – O – CH3

 

CH3

Methylmethacrylate

     H      CH3

 

C  –    C

 

H       C – O – CH3

 

O

Polymethylmethacrylate (Perspex)

  -Optical components

– transparent doors and windows;

– Display signs;

– dental fillings;

 

 

 

  1. Condensation polymerization

– Occurs when monomers (similar or different) combine to form a long chain molecule; with the loss of small molecules like ammonia or water.

– The monomers should have at least two functional groups.

Reason:

– For molecules to join at both ends permitting chain formation;

 

Illustration

– Consider two molecules A and B; each with 2 functional groups:

  • Molecule A; HO – A – OH, with two OH functional groups;
  • Molecule B; HOOC – B – COOH, with two -COOH functional groups

 

On condensation

HO – A – O H +  H O O C – B – COOH + HO – A – OH +…………..

                   

                       Lost to form water

Hence;

HO – A – OH  + HOOC – B – COOH                       [HO-A-OOC-B-CO]n + 2H2O(l)

 

Types of condensation polymers.

– Are of two types:

  • Polyesters

– Are polymerized formed by an ester linkage; usually with the liberation of a water molecule.

 

Examples

O                   O                                                                    O                     O

 

 

HO    C                   C – OH + H – OCH2CH2O – H                            –    C                   C – OH + H – OCH2CH2O –    +  H2O

A diol                                                                                                                                   n

          A dioic acid                                                                                                            Polyester

 

– Are condensation polymer involving monomers that have at least an amine group (in at least one of them); and thus usually result to the evolution of ammonia gas or water.

 

Example:

O                   O                              H                  H                                                  O                   O                             H                   H

 

H – C – (CH2)4 – C – OH   +  H – N – (CH2)6 – N – H                                                    – C – (CH2)4 – C – OH   +  H – N – (CH2)6 – N –     +   H2O

n

Examples of condensation polymers and their uses.

 

Polymer Monomer Uses
Polyvinyl chloride Vinyl chloride Making rain coats, plastic discs, plastic water pipes, and electric insulation;
Starch Glucose Laundry
Cellulose. Glucose Paper and clothing manufacture
Silk and wool Proteins (amino acids) Making clothes
Polyester (terylene) Terephthalic acid and Ethan 1,2  diol Clothes and seat belts
Bakelite Urea and methanol Electrical fittings
Perspex Methylmethacrylate Safety belts, windscreen, and plastic lenses;
Nylon Hexane-1,6-dioyl dichloride; and hexane- 1,6- diamine; Used in making safety glass; reflectors; contact lenses; false teeth

Further examples of polymerization reactions.

Formation of nylon.

                                                                                                              H                     H    O                    O

H                                                           H               O                                           O

NCH2CH2CH2CH2CH2CH2N           +            CCH2CH2CH2CH2C                                                            – N – (CH2)6 – N – C – (CH2)4 – C –

H                                                     H            Cl                                          Cl                                                                                                           n

 

Hexane – 1,6 – diamine                                   hexane – 1,6 – dioyl dichloride                                                                                 Nylon

 

 Formation of Terylene
UNIT 2: ACIDS, BASES AND SALTS

Unit Checklist:

  1. Acids;
  • Meaning
  • Strong and weak acids
  • Concentrated and weak acids
  • Comparing strength of acids
  • Using evolution of hydrogen and carbon (IV) oxides
  • Using electrical conductivity
  • Using PH
  • Role of solvents in acidic properties of solvents
  • Hydrogen chloride in water
  • Hydrogen chloride in methylbenzene

 

  1. Bases:
  • Meaning
  • Strong and weak bases (alkalis)
  • Measuring strength of alkalis
  • Using electrical conductivity
  • Using PH values
  • Effect of solvent type on properties of ammonia solution
  • Ammonia n water
  • Ammonia in methylbenzene

 

  1. Uses of acids and bases

 

  1. Oxides and hydroxides
  • Basic oxides
  • Acidic oxides
  • Neutral oxides
  • Amphoteric oxides
  • Meaning of amphoteric oxides
  • To verify amphoteric oxides
  • Reactions of amphoteric oxides

 

  1. Salts
  • Meaning
  • Preparation methods (summary)
  • Solubility of salts
  • Qualitative tests for cations using NaOH and NH4OH;
  • Effect of heat on metal oxides and hydroxides.
  • Effect of sodium carbonate on salt solutions
  • Properties of cations with sodium chloride, sodium sulphate and sodium sulphite

 

  1. Solubility and solubility curves
  • Definition of solubility
  • Factors affecting solubility
  • Solubility curves and related calculations.
  • Fractional crystallization

 

  1. Water
  • Hardness of water
  • Temporary hardness (Meaning; causes; and removal)
  • Permanent hardness (meaning; cause; and removal)

 

 

  1. Acids

– Are substances whose molecules yield hydrogen ions in water; or

– Are substances, which contain replaceable hydrogen, which can be wholly or partially replaced by a metal.

 

HCl (aq)                      H+(aq)  + Cl(aq)

 

OR: – Acids are proton donors i.e. a substance which provides protons or hydrogen ions.

 

Strength of Acids

– Acids can be categorized as either strong or weak acids;

 

  • Strong acids

– Are those which dissociate or ionize completely to a large extent in water, to yield many hydrogen ions.

– They yield to the solution as many protons as they possibly can.

 

  • Examples

Hydrochloric acid; HCl(aq)                             H+ (aq)  + Cl(aq)

Sulphuric acid; H2SO4(aq)                              2H+ (aq)  + SO4 2-(aq)

Nitric acid; HNO3(aq)                                    H+ (aq)    NO3 aq)

 

  • Weak acids.

– Are acids, which undergo partial dissociation to yield fewer hydrogen ions.

– They do not ionize in water completely or to a large extent i.e. some of their molecules remained unionized in solution.

 

Examples:

Carbonic acid:

H2CO3(aq)              water           H+(aq) +  HCO3(aq)

 

Ethanoic acid

CH3COOH (aq)            water                  H+(aq) + CH3COO(aq)

 

Note: – concentrated acids and dilute acids

Concentrated acids

– Is an acid with a high number of acid molecules per given volume.

 

Dilute acids:

Are acids with a low number of acid molecules per given volume.

 

– Thus there are concentrated strong acids or dilute strong acids; as well as concentrated weak acids and dilute weak acids.

 

 

 

 

Comparing the strength of acids

(i). Using rate of evolution of hydrogen

Apparatus:

– Boiling tubes; 1M HCl/ H2SO4/ HNO3; Methanoic acid/ tartaric acid; magnesium ribbon.

 

Procedure:

– One boiling tube is half filled with 1M HCl; while another is half filled with 1M Ethanoic acid.

– 2 pieces of magnesium ribbons are cleaned to remove a layer of oxide on the surface.

– One of the two pieces is put in each tube of the acid.

 

Observations:

– Hydrochloric acid evolves hydrogen much more quickly than Ethanoic acid yet they were of equal concentration.

 

Conclusion

– Hydrochloric acid is a strong acid;

– Ethanoic acid is a weak acid.

 

Note:

– The same experiment can be repeated with marble chips (CaCO3) in acids of same concentration.

-The marble chips dissolve more quickly in HCl, which is a strong acid.

 

(ii). Using electrical conductivity

Procedure:

– 50cm3 of 2M-hydrochloric acid solution is placed into a beaker and set up apparatus as shown below.

– The switch is closed and the brightness of the bulb noted.

 

Diagram: Electrolytic circuit

 

 

 

 

 

 

 

 

 

 

 

Observations

– Strong acids like HCl, HNO3 and sulphuric acid gave a brighter bulb light than weak acids like ethanoic, carbonic acids e.t.c

 

Explanations

– Strong acids are completely dissociated and have more H+ in solution and hence have got a higher electrical conductivity; than solutions of weak acids which are only partially ionized thus have fewer hydrogen ions in solution

 

 

(iii). Using PH

Procedure

– 2cm3 solutions of different acids of equal concentrations are paired into different test tubes.

– To each test tube 2 drops of universal indicator are added.

– Acids tested: HCl, H2SO4, HNO3; ethanoic acid, carbonic acid, and tartaric acid.

– All acids are of 2M solutions

– The indicator colour and hence the PH number of each is noted; by comparing against the indicator chart.

 

Observations

Substance (1M) Colour of universal indicator PH
Sulphuric acid

Hydrochloric acid

Nitric acid

Ethanoic

Carbonic acid

Tartaric acid

Red

Red

Red

Orange

Yellow

orange

3

3

3

5

6

5

 

Explanations

– Solutions of strong acids contain a higher concentration of hydrogen ions than those of weak acids

– Strong acids have low PH usually less than 3.

– Weak acids have higher PH values usually between 5 – 6.

 

Role of solvents on acidic properties of a solute.

Experiment: – to find out if solutions of HCl in different solvents display acidic properties

Apparatus and reagents

– Hydrogen chloride gas, water and methylbenzene

– Beakers and a funnel

– Blue and red litmus papers.

 

Procedure

– Solutions of hydrogen chloride gas are made by bubbling the dry gas from a generator into water and into methylbenzene contained in separate beakers.

– The hydrochloric gas is passed into the solution using an inverted funnel to prevent sucking back.

 

Apparatus

 

 

 

 

 

 

 

 

 

The resultant solutions are each separately subjected to various tests as shown below and observations recorded

 

Tests and observations

 

Test. Aqueous HCl solution Solution of HCl in methylbenzene
1.      A piece of dry blue litmus paper is dropped into solution

2.      Dry universal indicator paper

3.      Add magnesium ribbon

4.      Add small marble chips

5.      Electrical conductivity

Blue litmus turns red

 

– Turns red (strong acid)

– Evolution of hydrogen

– CO2 evolved

– Good conductor

No effect on litmus

 

– Turns green (neutral)

– No reaction

– No reaction

– Does not conduct

 

Explanation

– The results show that the aqueous solution of hydrogen chloride behaves as an acid; but the solution in methylbenzene lacks acidic properties

– When HCl gas dissolves in water it changes from molecules to ions;

 

Equation:

HCl (aq)        water                 H+(aq) + Cl(aq)

 

– It is the hydrogen ions which give the acidic properties and these can only be formed in the presence of water

– HCl in water conducts electric current due to presence of free ions in solution

– HCl gas in methylbenzene does not conduct electric current because the HCl exists as molecules hence lack free ions

 

Note: – hydrogen chloride gas dissolves in water because both HCl and water polar molecules;

– This causes mutual attraction of both ends of HCl molecule by different water molecules causing the dissociation of HCl molecules into ions.

 

Illustration:

 

 

 

 

 

 

Hence:

HCl (g) + water                       HCl (aq)

HCl (aq)                      H+(aq)  + Cl(aq)

– The presence of hydrogen ions in aqueous solution of hydrogen chloride explains the electrical conductivity and acidic properties of hydrogen chloride

  • Acidic properties: –
  • turns blue litmus paper red;
  • evolves hydrogen gas when reacted with magnesium;
  • evolves carbon dioxide on reaction with CaCO3;

 

2H+ (aq)  +  CaCO3 (s)                           Ca2+(aq)  +  CO2(g) + H2O(l)

 

2H+(aq)  +  Mg(s)                         Mg 2+(aq)   + H2  (g)

 

– Methylbenzene has a weak attraction for hydrogen chloride and hence hydrogen chloride remains as molecules in methylbenzene

 

  1. Bases

– Are substances which accept the protons donated by acids and are hence proton acceptors

NH3 (aq)   +   H+ (aq)                          NH4+(aq)

 

CuO(s) +  2H+ (aq)                         Cu 2+(aq)   +   H2O(l)

 

Alkalis

– An alkali is a soluble base i.e. a base that is soluble in water.

– They are compounds, which produce hydroxyl ions in aqueous solutions.

 

NaOH(aq)                           Na+(aq)   +  OH (aq)

 

Note: –

When an acid proton reacts with a base (hydroxyl ions) in aqueous solution, a neutralization reaction occurs.

 

Strength of an Alkali

– Alkalis can be grouped as either strong or weak alkalis.

 

(a). Strong alkalis

– Are alkalis that undergo complete dissociation in aqueous solution; yielding a large number of hydroxyl (OH-) ions

 

Examples:

– Sodium hydroxide.

– Potassium hydroxide.

 

(b). Weak alkalis

– Are alkalis that undergo only partial dissociation in aqueous solution (water) yielding fewer numbers of hydroxyl ions.

 

Examples

– Calcium hydroxide

– Ammonium hydroxide

 

Measuring the strength of alkalis

(i). Using electrical conductivity

Procedure

50 cm3 of 2 M sodium hydroxide solution is put into a beaker and the apparatus set as shown below

Apparatus

 

 

 

 

 

 

 

 

 

Procedure

– The same procedure is repeated using other alkalis like NH4OH; Ca (OH)2 e.t.c.

 

Observation

– The bulb lights brightly with KOH and NaOH as electrolyte than with NH4OH and Ca (OH)2

 

Explanation

– NaOH and KOH are strong alkalis and are completely dissociated and have more ions in solution and hence have got a higher electrical conductivity than the weak alkalis of NH4OH and Ca(OH)2(aq)

 

(ii). Using PH values

Procedure

– 2 cm3 of NaOH and 2 cm3 of NH4OH are each poured into 2 different test tubes separately

– Into each test tube 2 drops of universal indicator are added.

– The colour change is noted and the corresponding PH scale recorded

 

Observations

 

Alkali Colour of universal indicator PH
– Ammonium hydroxide (1M)

– Calcium hydroxide (1M)

– Sodium hydroxide (0.1M)

– Sodium hydroxide (1M)

– Potassium hydroxide

Blue.

Blue.

Purple.

Purple.

Purple.

11

10

13

14

14

 

 

 

 

 

 

 

Note: the PH scale

– Is a scale which gives a measure of the acidity of alkalinity of a substance.

 

Illustration: a PH scale.

 

1     2     3     4     5     6     7     8     9     10     11     12     13     14

 

 

Strong acid               Weak acids       Neutral   Weak alkali                                    Strong alkali

 

 

Increasing acidity                                                                                 Increasing alkalinity

(High H+ ion concentration)                                                               (Low H+ ion concentration)

 

 

Indicator colours:

 

PH 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Colour Red Orange/ red Yellow/ Green Green Green/ Blue Blue/ Purple Purple

 

Effects of type of solvent on the properties of ammonium solution

Procedure

– Ammonium solution is prepared by bubbling the gas from a generator into methylbenzene (toluene) and into water contained in separate beakers

– The solutions are each divided into 3 portions and tested with litmus paper; universal indicator and for electrical conductivity

 

Apparatus

 

 

 

 

 

 

 

 

 

Observations

Test Solution of NH3 in water Solution of NH3 in methylbenzene (toluene)
Dry litmus paper Red litmus paper turns blue; No effect
Dry universal indicator paper Colour turns purple (alkaline PH) Turns green (Neutral PH)
Electrical conductivity Poor conductor Non-conductor

 

Explanations

– When NH3(g) dissolves in water it changes from molecules to ions.

Equation:

NH3(g) + H2O(l)                 NH4+(aq) + OH(aq)

 

– It is the hydroxide ions that cause alkaline properties.

– Since ammonium hydroxide is a weak alkali, it dissociates partially releasing fewer hydroxide ions hence the poor electric conductivity.

– Ammonium gas in methylbenzene or trichloromethane exists as molecules without free ions hence no alkaline properties and the electrical conductivity.

Uses of acids and bases

  1. Acids

– Refer to the various acids for uses of sulphuric, nitric and hydrochloric acids.

 

  1. Bases/ alkalis

– Some weak bases e.g milk of magnesia, are used to relieve stomach disorders.

 

Amphoteric oxides and hydroxides.

  • Oxides

– An oxide is a binary compound of oxygen and another element.

– Are of four categories:

  • Basic oxides
  • Acidic oxides
  • Neutral oxides
  • Amphoteric oxides

 

(i). Basic oxides

– Are usually oxides of metals (electronegative elements)

– They react with acids to form salt and water only.

Examples

CaO, MgO, CuO etc.

 

(ii). Acidic oxides

– Are usually oxides of non metals (electronegative elements).

– Many of them react with water to form (give) acids and are known as acid anhydrides

Examples

CO2; SO2; SO3; P2O5 and NO2

 

(iii). Amphoteric oxides

– Are oxides, which behave as both bases and acids.

– Are mainly oxides of certain metals in the middle group of the periodic table.

Examples

Oxides of Zn, Al, Pb

 

Experiment: – To verify amphoteric oxides

Procedure

– A small sample of aluminium oxide is placed in a test tube and 5 cm3 of 2M nitric acid added to it and the mixture shaken.

– The procedure is repeated in different test tubes with ZnO, PbO, CuO and CaO.

– The experiments are repeated using excess 2M sodium hydroxide in place of nitric acid

 

Observations

 

Name of solid Observations when
acid is added hydroxide is added
Aluminium oxide

Zinc oxide

Lead II oxide

Zinc hydroxide

Lead hydroxide

Aluminium hydroxide

Oxide dissolves

Oxide dissolves

‘’

Hydroxide dissolves

‘’

Oxide dissolves

‘’

‘’

Hydroxide dissolves

‘’

Explanations

– These oxides are soluble in acids as well as in the alkalis (NaOH)

  • Reaction with acids

– Oxides react with acids to form a salt and water only in a reaction called neutralization reaction.

 

Equations

  • Oxides

(i).PbO(s) + 2H+(aq)                   Pb2+(aq) + H2O(l)

 

(ii). Al2O3(s) + 6H+(aq)                 2Al3+(aq) + 3H2O(l)

 

(iii). ZnO(s) + 2HCl(aq)               ZnCl2(aq) + H2O(l)

 

  • Hydroxides:

(i). Pb(OH)2(s) + 2HCl(aq)                          PbCl2(aq) + 2H2O(l)

 

(ii). Zn(OH)2(s) + 2HCl(aq)                   ZnCl2(aq) + 2H2O(l)

 

(iii). Al(OH)3(s) + 3HCl(aq)                      AlCl3(aq) + H2O(l)

 

Note: in these reactions the metal oxides are reacting as bases

 

  • Reaction with alkalis

– These oxides and hydroxides also react with alkalis e.g sodium hydroxide in which case they are reacting as acids.

– Their reactions with alkalis involve the formation of complex ions; M(OH)2-4

 

Equations

  • Oxides

(i). PbO(s) + 2NaOH(aq) + H2O(l)                   Na2Pb(OH)4(aq)

 

Ionically: PbO(s) + 2OH(aq) + H2O(l)                 [Pb(OH)4]2-(aq)

 

(ii). Al2O3(s) + 2OH(aq) +  3H2O (l)                   2[Al(OH)4](aq) + 3H2O(l)

 

Ionically: Al2O3(s) + 2OH(aq) + 3H2O(l)                       2[Al(OH)4](aq)

 

(iii). ZnO(s) + 2NaOH(aq) + H2O(l)                     Na2Zn(OH)4(aq)

 

Ionically: ZnO(s) + 2OH(aq) + H2O(l)                 [Zn(OH)4]2-(aq)

 

  • Hydroxides:

(i). Al(OH)3(s) + NaOH(aq)                              NaAl(OH)4(aq)

 

Ionically: Al(OH)3(s) + OH(aq)                     [Al(OH)4](aq)

 

(ii). Zn(OH)2(s) + 2NaOH(aq)                           Na2Zn(OH)4(aq)

 

Ionically: Zn(OH)2(s) + 2OH(aq)                      [Zn(OH)4]2-(aq)

 

(iii). Pb(OH)2(s) + 2NaOH(aq)                            Na2Pb(OH)4(aq)

 

Ionically: Pb(OH)2(s) + 2OH(aq)                      [Pb(OH)4]2-(aq)

 

 

Salts

– Is a compound formed when cations derived from a base combine with anions derived from an acid.

– Salts are usually formed when an acid reacts with a base i.e. when the hydrogen ions in an acid re wholly or partially by a metal ion or ammonium (NH4+) radical.

 

Laboratory preparations of salts

– Salts are prepared in the laboratory using various depending on property of the salt especially solubility

Examples

(a). Preparations by direct synthesis

Equation:

Fe(s) + Cl2(g)                          2FeCl3(s)

 

(b). Reactions of acids with metals, metal oxides, metal hydroxides and metal carbonate

Equations:

Zn(s) + H2SO4(aq)                         ZnSO4(aq) + H2(g)

 

CuO(s) + H2SO4(aq)                      CuSO4(aq) + H2O(l)

 

NaOH(aq) + HCl(aq)                     NaCl(aq) + H2O(l)

 

PbCO3(s) + 2NHO3(aq)                 Pb(NO3)2(aq) + CO2(g) + H2O(l)

 

Note:

– Acid + metal method will not be suitable if:

  • The metal is too reactive e.g. sodium or potassium.
  • The salt formed is insoluble; as it will form an insoluble layer on the metal surface preventing further reaction.
  • The metal is below hydrogen in the reactivity series.

 

(c). Double decomposition/ precipitation

– Mainly for preparations of insoluble salts

– Involves formation (precipitation) of insoluble salts by the reaction between two solutions of soluble salts.

 

Equations:

Pb(NO3)2(aq) + 2NaCl(aq)                    PbCl2(s) + 2NaNO3(aq)

 

AgNO3(aq) + HCl(aq)                  AgCl(s) + HNO3(aq)

 

Types of salts:

– Are categorized into three main categories:

  • Normal salts
  • Acid salts
  • Double salts

 

 

 

 

Solubility of salts: – a summary

– All common salts of sodium, potassium and ammonium are soluble.

– All common nitrates are soluble.

– All chlorides are soluble except silver, mercury and lead chlorides.

– All sulphates are soluble except calcium, barium, lead and stomium sulphates.

– All carbonates are insoluble except sodium, potassium and ammonium carbonates.

– All hydroxides are insoluble except sodium, potassium ammonium and calcium hydroxides is        sparingly soluble.

 

Note:

– Lead(II) chloride is soluble in hot water.

– Calcium hydroxide is sparingly soluble in water.

 

Reactions of some cations with NaOH(aq)  and NH4OH(aq) and solubilities of some salts in water

 

Cation Soluble compounds in water Insoluble compounds in water Reaction with NaOH(aq) Reaction with NH4OH solution
K+ all None No reaction No reaction
Na+ all None No reaction No reaction
Ca2+ Cl; NO3 CO32-; O2-; SO42-; OH; White precipitate insoluble in excess No precipitate
Al3+ Cl; NO3; SO42-; CO32-; O2-; OH; White precipitate

soluble in excess

White precipitate insoluble in excess
Pb2+ NO3; ethanoate; All others White precipitate soluble in excess White precipitate insoluble in excess
Zn2+ Cl; SO42-; NO3; CO32-; OH; White precipitate soluble in excess White precipitate soluble in excess
Mg2+ Cl; SO42-; NO3; CO32-; OH; White precipitate insoluble in excess No precipitate
Fe2+ Cl; SO42-; NO3; CO32-; O2-; OH; (dark) green precipitate insoluble in excess Green precipitate insoluble in water
Fe3+ Cl; SO42-; NO3; CO32-; O2-; OH; (red) brown precipitate insoluble in excess Brown precipitate insoluble in excess
Cu2+ Cl; SO42-; NO3; CO32-; O2-; OH; Pale blue precipitate insoluble in excess Pale blue precipitate soluble in excess forming a deep blue solution
NH4+ all none Ammonium gas on warming Not applicable

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Explanations

– In these experiments NaOH forms insoluble hydroxides with ions of Zn2+, Al3+, Cu2+, Fe2+, Ca2+, Mg2+, Fe3+, and Pb2+.

– These hydroxides have a characteristic appearance, which form the basis of their identification

Examples

Equations:

Zn2+(aq) + 2OH(aq)                        Zn(OH)2(s)

(White ppt).

 

Cu2+(aq) + 2OH(aq)                        Cu(OH)2(s)

(Pale blue ppt).

 

Fe2+(aq) + 2OH(aq)                         Fe(OH)2(s)

(Dirty green ppt).

 

Fe3+(aq) + 2OH(aq)                         Fe(OH)3(s)

(Red-brown ppt).

 

Pb2+(aq) + 2OH(aq)                         Pb(OH)2(s)

(White ppt).

 

– The hydroxides of aluminum, zinc and lead dissolves in excess sodium hydroxide solution because of complexes are formed

Equations:

Al(OH)3(s) + OH(aq)                      [Al(OH)4](aq)

(Tetra-hydroxyl-aluminium (III) ion)

 

Pb(OH)2(s) + 2OH(aq)                   [Pb(OH)4]2-(aq)

(Tetra-hydroxyl-lead (II) ion)

 

Zn(OH)2(s) + 2OH(aq)                       [Zn(OH)4]2-(aq)

(Tetra-hydroxyl-zinc (II) ion)

 

Note: – in these reactions KOH(aq) may be used instead of sodium hydroxide

 

With ammonia solution

– Insoluble metals hydroxides are similarly formed.

Zn2+(aq) + 2OH(aq)                        Zn(OH)2(s)

(White ppt).

 

Cu2+(aq) + 2OH(aq)                        Cu(OH)2(s)

(Pale blue ppt).

 

Fe2+(aq) + 2OH(aq)                         Fe(OH)2(s)

(Dirty green ppt).

 

Fe3+(aq) + 2OH(aq)                         Fe(OH)3(s)

(Red-brown ppt).

 

Pb2+(aq) + 2OH(aq)                         Pb(OH)2(s)

(White ppt).

 

– However hydroxides of copper and zinc dissolve in excess ammonia solution due to formation of complex ions/ salts

Equations:

Zn(OH)2 (s) + 4NH3(aq)                      [Zn(NH3)4]2+(aq)  + 2OH(aq)

(White ppt)                                             (Tetra-amine zinc (II) ion; colourless solution)

 

Cu(OH)2(s) + 4NH3(aq)                      [Cu(NH3)4]2+(aq)  + 2OH(aq)

(Pale blue ppt)                                              (Tetra-amine copper (II) ion; deep blue solution)

Effects of heat on metal hydroxides

Procedure

– Hydroxides of Zn, Ca, Pb, Cu e.t.c are strongly heated in a test tube each separately

 

Observation

– Most metal hydroxides are decomposed by heat to form metal oxides and water

– Sodium and potassium hydroxides only decompose at very high temperatures.

– Hydroxides of metals lower in the reactivity series are readily decomposed by heat than those metals higher in the series.

 

Examples

Cu(OH)2(s)            Heat               CuO(s) + H2O(l)

(Blue)                                                                     (Black)

 

Pb(OH)2(s)            Heat               PbO(s) + H2O(l)

(White)                                                                  (Red brown when hot; yellow when cold)

 

Zn(OH)2(s)            Heat               ZnO(s) + H2O(l)

(White)                                                                  (Yellow when hot; white when cold)

 

Ca(OH)2(s)            Heat               CuO(s) + H2O(l)

(White)                                                                  (White)

 

Note:

– Both iron (II) and iron (III) hydroxides give iron (III) oxide when heated.

Equations:

2Fe(OH)2(s)  + ½ O2(g)          Heat                Fe2O3(s) + 2H2O(l)

(Green)                                                                                           (Red-brown)

 

Fe(OH)3(s)             Heat               Fe2O3(s) + 3H2O(l)

(Brown)                                                                 (Red-brown)

 

These oxides do not decompose on further heating

 

Effects of sodium carbonate on various salt solutions

Procedure

– 3 drops of NaOH(aq) are added to 2cm3 of 1M solution containing magnesium ions in a test tube

the procedure is repeated with salt solutions containing

 

Solution containing Observations after adding sodium carbonate
Mg2+ A white precipitate is formed
Ca2+ A white precipitate
Zn2+ A white precipitate
Cu2+ A green precipitate
Pb2+ A white precipitate
Fe2+ A green precipitate
Fe3+ A brown precipitate and a colourless gas that forms a white ppt. in lime water;
Al3+ A white precipitate and a colourless gas that forms a white ppt. in lime water;

 

Explanations:

– Sodium carbonate, potassium and ammonium carbonate are soluble in water; all other metal carbonates are insoluble

– Hence their solutions may be used to precipitate the insoluble metal carbonates.

Ionic equations:

Ca2+(aq) + CO32-(aq)                         CaCO3(s)

 

Note:

Iron (III) and Aluminium salts hydrolyse in water giving acidic solutions which react with carbonates to liberate carbon dioxide gas; hence effervescence.

 

Reaction of metal ions in salt solutions with sodium chloride, sodium sulphate and sodium sulphate

(i). Procedure

– 2cm3 of a 0.1M solution containing lead ions is placed in a test tube.

– 2-3 drops of 2M sodium chloride solution are added and the mixture warmed;

– The procedure is repeated using salt solutions containing Ba2+; Mg2+; Ca2+; Zn2+; Cu2+; Fe2+ and Fe3+

– Each experiment (for each salt) is repeated using Na2SO4 and Na2SO3 respectively, in place of sodium chloride.

 

(ii). Observations

 

Solution containing Sodium sulphate Sodium chloride Sodium sulphate
Zn2+ – Colourless solution – Colourless solution Colourless solution
Mg2+ – Colourless solution – Colourless solution Colourless solution
Cu2+ – Blue solution – Blue solution Blue solution
Fe2+ – Greenish solution – Green solution
Fe3+ – Yellow solution – Yellow/ dark brown solution
Pb2+ – White precipitate – White precipitate which dissolve on warming White precipitate
Ba2+ – White precipitate – White precipitate White precipitate

 

Explanations

– All the listed cations soluble salts except Ba2+ and Pb2+

– Lead sulphate and barium sulphate are insoluble in water;

– Lead chloride and barium sulphite are insoluble; however PbCl2(s) dissolves on warming

 

Equations:

Pb2+(aq) + 2Cl(aq)                     PbCl2(s)

 

Pb2+(aq) + SO42-(aq)                   PbSO4(s)

 

Ba2+(aq) + SO42-(aq)                   BaSO4(s)

 

Ba2+(aq) + SO32-(aq)                   BaSO3(s)

 

Note:

– To distinguish the precipitate of barium sulphate from barium sulphite; dilute HNO3 (aq) or HCl(aq) is added to both;

– BaSO3(s) will dissolve in the dilute acid but barium sulphate will not.

 

Uses (importance) of precipitation reactions.

– Precipitation of metal carbonate from aqueous solutions is useful in softening hard water; usually by removing calcium and magnesium ions from water as insoluble carbonate

 

Useful information on salts (qualitative analysis)

Colours of substances in solids and solutions in water.

 

COLOUR  
SOLID AQUESOUS SOLUTION

(IF SOLUBLE)

1. White Colourless Compound of K+; Na+, Ca2+; Mg2+; Al3+; Zn2+; Pb2+; NH4+
2. Yellow Insoluble Zinc oxide, ZnO (turns white on cooling); Lead oxide, PbO (remains yellow on cooling, red when hot)
Yellow Potassium or sodium chromate;
3. Blue Blue Copper (II) compound, Cu2+
4. Pale green

 

Green

Pale green (almost colourless)

Green

Iron (II) compounds,Fe2+

 

Nickel (II) compound, Ni2+; Chromium (II) compounds, Cr3+; (Sometimes copper (II) compound, Cu2+)

5. Brown Brown (sometimes yellow)

 

Insoluble

Iron (III) compounds, Fe3+;

 

Lead (IV) oxide, PbO2

6. Pink Pink (almost colourless)

Insoluble

Manganese (II) compounds, Mn2+;

Copper metal as element (sometimes brown but will turn black on heating in air)

7. Orange Insoluble Red lead, Pb3O4 (could also be mercury (II) oxide, HgO)
8. Black Purple

Brown

Insoluble

Manganate (VII) ions (MnO) as in KMnO4;

Iodine (element)-purple vapour

Manganese (IV) oxide, MnO2

Copper (II) oxide, CuO

Carbon powder (element)

Various metal powders (elements)

 

Reactions of cations with common laboratory reagents and solubilities of some salts in water

 

CATION SOLUBLE COMPOUNDS (IN WATER) INSUOLUBLE COMPOUNDS (IN WATER) REACTION WITH AQUEOUS SODIUM HYDROXIDE REACTION WITH AQUEOUS AMMONIA SOLUTION
Na+ All None No reaction No reaction
K+ All None No reaction No reaction
Ca2+ Cl; NO3; CO32-; O2-; SO42-; OH; White precipitate insoluble in excess White precipitate insoluble in excess, on standing;
Al3+ Cl; NO3; SO42- O2-; OH; White precipitate soluble in excess White precipitate insoluble in excess
Pb2+ NO3; ethanoate; All others; White precipitate soluble in excess White precipitate insoluble in excess
Zn2+ Cl; NO3; SO42- CO32-; O2-; SO42-; OH; White precipitate soluble in excess White precipitate soluble in excess
Fe2+ Cl; NO3; SO42- CO32-; O2-; OH; (Dark) pale green precipitate insoluble in excess (Dark) pale green precipitate insoluble in excess
Fe3+ Cl; NO3; SO42- CO32-; O2-; OH; (Red) brown precipitate insoluble in excess (Red) brown precipitate insoluble in excess
Cu2+ Cl; NO3; SO42- CO32-; O2-; OH; Pale blue precipitate insoluble in excess Pale blue precipitate soluble in excess forming a deep blue solution
NH4+ All None; Ammonias gas on warming Not applicable.

 

 

 

Qualitative analysis for common anions.

 

  SO42-(aq) Cl(aq) NO3(aq) CO32-(aq)
TEST Add Ba2+(aq) ions from Ba(NO3)2(aq); acidify with dilute HNO3(aq) Add Ag+(aq) from AgNO3(aq).

Acidify with dilute HNO3

Alternatively;

Add Pb2+ from Pb(NO3)2 and warm

Add FeSO4(aq);

Tilt the tube and carefully add 1-2 cm3 of concentrated H2SO4(aq)

Add dilute HNO3(aq); bubble gas through lime water;
OBSERVATION The formation of a white precipitate shows presence of SO42- ion; The formation of a white precipitate shows presence of Cl ion;

Formation of a white precipitate that dissolves on warming shown presence of Cl(aq) ions

The formation of a brown ring shows the presence of NO3 ions Evolution of a colourless gas that forma a white precipitate with lime water, turns moist blue litmus paper red; and extinguishes a glowing splint shows presence of CO32- ions
EXPLANATION Only BaSO4 and BaCO3 can be formed as white precipitates.

BaCO3 is soluble in dilute acids and so BaSO4 will remain on adding dilute nitric acid

Only AgCl and AgCO3 can be formed as white precipitates.

AgCO3 is soluble in dilute acids but AgCl is not;

– PbCl2 is the only white precipitate that dissolves on warming

Concentrated H2SO4 forms nitrogen (II) oxide with NO3(aq) and this forms brown ring complex (FeSO4.NO) with FeSO4; All CO32- or HCO3 will liberate carbon (IV) oxide with dilute acids

 

Checklist:

  1. Why is it not possible to use dilute sulphuric acid in the test for SO42- ions;
  2. Why is it not possible to use dilute hydrochloric acid in the test for chloride ions?
  3. Why is it best to use dilute nitric acid instead of the other two mineral acids in the test for CO32- ions?
  4. How would you distinguish two white solids, Na2CO3 and NaHCO3?

 

What to look for when a substance is heated.

 

1. Sublimation White solids on cool, parts of a test tube indicates NH4+ compounds;

Purple vapour condensing to black solid indicates iodine crystals;

2. Water vapour (condensed) Colourless droplets on cool parts of the test tube indicate water of crystallization or HCO3 (see below)
3. Carbon (IV) oxide CO32- of Zn2+; Pb2+; Fe2+; Fe3+; Cu2+;
4. Carbon (IV) oxide and water vapour (condensed) HCO3
5. Nitrogen (IV) oxide NO3of Cu2+; Al3+; Zn2+; Pb2+; Fe2+; Fe3+
6. Oxygen NO3 or BaO2; MnO2; PbO2;

 

 

 

 

 

 

Reduction-oxidation (Redox reactions)

(a). Displacement reactions.

(i) More reactive halogens metals will displace less reactive metals from solutions of their salts in the series:

Zn                               Fe  Pb Cu

More reactive                                Less reactive

 

Example:

– Zinc powder placed in a solution of copper (II) sulphate, which contains Cu2+(aq) ions, will become Zn2+(aq) ions and brown copper solid (metal) will be deposited.

– The Cu2+(aq) is reduced to copper by addition of electrons and the zinc is oxidized to Zn2+(aq) by removal of electrons.

 

(ii). More reactive halogens will displace less reactive halogens from solutions of their salts in series:

Cl2                             Br2 I2

More reactive                              Less reactive.

 

Example:

– Chlorine bubbled into a solution of potassium iodide (colourless), which contains I(aq) ions will turn grey (black) as iodine is liberated.

– The chlorine is reduced to Cl(aq) ions by addition of electrons and the I(aq) ions are oxidized to iodine by removal of electrons.

 

(b). Decolourisation of purple potassium manganate (VII) ions.

When a few drops of purple KMnO4 solution are added to a compound and the purple colour disappears, then this shows that the MnO4(aq) ions have been reduced to almost colourlessMn2+ (aq) ions

The substance in the solution has been oxidized.

 

Example:

– KMnO4 will oxidize Fe2+(aq) ions to Fe3+(aq) ions; pale green solution turns red-brown.

– KMnO4 will oxidize Cl(aq) ions to Cl2(g); colourless solution results to a green gas with a bleaching action;

 

(c). Orange potassium chromate (VI) turning to a green solution.

– Orange solution of dichromate ions, Cr2O72-(aq), changes to green Cr3+(aq) ions when the dichromate is reduced.

– The substance causing this change is oxidized.

 

Example:

– K2Cr2O7 will oxidize Fe2+(aq) to Fe3+(aq)

– K2Cr2O7 will oxidize SO2(g) to SO42-(aq)

– Formation of sulphate ions in solution from sulphur (IV) oxide gas is often used in the test for sulphur (IV) oxide gas.

 

(d). Oxidation of Fe2+(aq) to Fe3+(aq) ions by concentrated nitric acid.

 

 

 

 

Solubility and solubility curves

  • Solubility

– Is the maximum number of grams of a solid which will dissolve in 100g of solvent (usually water) at a particular temperature

– A solution is made up of two parts: – a solute and a solvent.

 

Solute

– The solid part of a solution usually dispersed in the solvent e.g. a salt.

 

Solvent

– The liquid part of the solution into which the solute is dissolved.

 

Experiment: to determine the solubility of potassium nitrate at 20oC.

(i). Materials

– Beakers, evaporating dish, measuring cylinder, burner, scales, thermometer, distilled water and potassium nitrate

 

(ii). Apparatus

 

 

 

 

 

 

 

 

 

 

 

(iii). Procedure

– About 50cm3 of distilled water is placed in a beaker

– Potassium nitrate is added to it a little at a time stirring continuously.

– The nitrate is added until no more will dissolve and there is an excess undissolved salt present. This is the saturated solution of KNO3 at the temperature.

Note:

  • Saturated solution: solution that cannot dissolve any more of the solid/ solute at a particular temperature

– The solution is allowed to settle and it is temperature recorded.

– About 25 cm3 of clear solution is poured in a previously weight evaporating dish.

– The mass of the dish and solution is recorded.

– The dish is then heated in a water bath (to avoid spurting) till the solution is concentrated.

– The concentrated solution is allowed to cool and the dish weighted with its contents.

 

Results and calculations

 

Temperature 20.0oC
Mass of evaporating dish + solution 100.7g
Mass of evaporating dish 65.3g
Mass of solution 35.4g
Mass of evaporating dish + dry salt 73.8g

Calculating:

Mass of salt dissolved = (73.8 – 65.3)g = 8.5g;

Mass of water (solvent) = (35.4 – 8.5)g = 26.9g

Thus:

If 26.9g of water dissolves 8.5g of KNO3 at 20oC;

Then 100g of water will have ? = 100 x 8.5 = 31.6g of salt;

26.9

Therefore the solubility of KNO3 at 20oC = 31.6g per 100g of water

 

Factors affecting solubility.

(i). Temperature

– For most salts solubility increases with rise/ increase in temperature.

Reason

– Increased temperature increases the kinetic energy, and hence the momentum and velocity of the solvent molecules so that they can disintergrate the solute molecules more effectively.

– However solubilities of certain salts remain almost constant with temperature change

– Solubility of gases however decreases with increase in temperature;

Reason:

Increase in temperature causes the gas molecules to expand and hence escape from the solvent.

 

Experiment: To investigate the effect the effect of temperature on solubility.

Requirements: potassium nitrate, distilled water, test tube, thermometer, stirrer, bunsen burner, 250 cm3 glass beaker, 4.5g of potassium nitrate.

 

(ii). Procedure.

Using a 10ml measuring cylinder, measure 5 cm3 of distilled water and add it to the boiling tube containing solid potassium nitrate.  Insert a thermometer into the boiling tube and heat the mixture gently in a water bath or while shaking to avoid spillage. Continue heating until all the solid has dissolved.  Stop heating and allow the solution to cool while gently stirring with a thermometer. Record the temperature at which the crystals of potassium nitrate first appear. Note this in the table below.

Retain the boiling tube and its contents for further experiments.

Measure 2 cm3 of distilled water and add to the mixture in the boiling tube. Heat until the crystals dissolve, then cool while stirring with a thermometer. Record the temperature at which the crystals again first stat to reappear. Repeat this procedure, each time adding more 2 cm3 of distilled water, heating, cooling and recording the crystallization temperature until the table is completely filled.

 

Table 2:

 

Experiment number I II III IV V
Volume of water added 5 7 9 11 13
Temperature at which crystals appear (oC)
Solubility of K in g/100g of water

 

 

 

 

Questions:

(a). Complete the table and calculate the solubility of solid X in g/100g of water at different temperatures.                                                                                                                                (2 marks)

 

(b). Using the table above, plot a graph of solubility of solid X in g/100g of water against temperature.                                                                                                                                             (5 marks)

(c). From the graph:

(i). calculate the mass of K that would be obtained if the saturated solution is cooled from 60oC to 40oC.                                                                                                                                            (2 marks)

 

(ii). determine the solubility at 70oC.                                                                                                        (1mark)

 

(iii). at what temperature would solubility of K be 100g/100g of water?                                    (1mark)

 

(ii). Stirring

– Stirring increases the solubility of a solid

Reason

– Stirring causes the molecules of solvent and solute to move faster causing the solute particles to disintergrate more effectively

 

Solubility curves

– Are curves showing the variation of solubility with temperature.

 

Uses / importance of solubility curves

– Can be used to determine the mass of crystals that would be obtained by cooling a volume of hot saturated solution from one known temperature to another.

– Solubility differences can be used to separate substances i.e. recrystallization or fractional crystallization (refer to separation of mixtures)

– Separation of salts from a mixture of salts with differing solubilities e.g. extraction of sodium carbonate from Trona (refer to carbon and its compounds)

– Manufacture of certain salts e.g. sodium carbonate by the Solvay process (refer to carbon and its compounds)

 

Worked examples

  1. An experiment was carried out to determine the solubility of potassium nitrate and the following results were obtained.

 

Temperature 10 15 30 40 50 60
Mass of KNO3 per 100g of water 20 25 45 63 85 106

 

(a). What is meant by solubility?                                                                                                 (1 mark)

 

(b). Plot a graph of mass of potassium nitrate against temperature.                                                     (3 marks)

 

(c). From the graph work out the mass of KNO3 that would crystallize if a solution containing 70g of KNO3 per 100g of water was cooled from 45oC to 25oC.                                                                  (2 marks)

 

(d). Explain what would happen if 100g of KNO3 was put in cold water and heated to 50oC.                                                                                                                                                                  (2 marks)

  1. The table below shows the solubility of sulphur (IV) oxide at various temperatures.

 

Temperature (oC) 0 5 10 15 20 25 35 40 45 50 55 60
Mass of SO2 per 100g of water 22 18.4 15.4 13.0 10.8 9.05 7.80 6.80 5.57 4.80 4.20 3.60

 

(a). On the grid provided plot a graph of solubility against temperature.                                  (3 marks)

 

(b). From the graph determine:

(i). The lowest temperature at which 100cm3 of water would contain 11.6g of sulphur dioxide.(1 mark)

 

(ii). The maximum mass of sulphur (IV) oxide that would dissolve in 2 litres of solution at 10oC. (Assume that the density of the solution is 1gcm-3)                                                                      (3 marks)

 

(c) (i). Sulphur (IV) oxide reacts with sodium hydroxide solution to form sodium sulphite and water.                                                                                                                                     (1 mark)

(ii). Write the equation for this reaction.                                                                                     (1 mark)

 

(iii). Using the information from the graph, determine the volume of the saturated sulphur (IV) oxide solution that can neutralize 153 cm3 of 2M sodium hydroxide solution at 25oC.                                                                                                 (3 marks)

Water

– Can be pure or impure

Pure water

– Is a pure substance which is a compound of hydrogen and oxygen; that boils at 100oC; melts at 0oC and has a density of 1gcm-3 at sea level.

 

Impure water

– Are the natural waters constituted of dissolved solutes in pure water.

 

Hardness of water

– Water without dissolved substances (salts) hence lathers easily with soap is referred to as soft water while water with dissolved substances that does not lather easily with soap is termed as hard water

 

Experiment: effect of water containing dissolved salts on soap solution

Procedure

– 2 cm3 of distilled water is put in a conical flask.

– Soap solution from a burette is added into the water and shaken until formation of lather is noted.

– If the soap fails to lather more soap solution is added from the burette till it lathers and the volume of the soap required for lathering recorded.

– The procedure is repeated with each of the following: tap water, rain water, dilute solutions of MgCl2, NaCl, CaCl2, a(NO3)2, CaSO4, MgSO4, Mg(HCO3)2, Ca(HCO3)2, ZnSO4 , NaHCO3, and KNO3.

– The procedure is repeated with each of the solutions when boiled.

 

 

 

 

 

 

Observations

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Explanations

– Distilled water requires very little soap to produce lather because it lacks dissolved salts and hence termed soft water.

– Solutions containing NaCl, ZnSO4, KNO3 and NaHCO3 do not require a lot of soap to form lather

Water containing Ca2+ and Mg2+ ions do not lather easily (readily) with soap

Reason:

– These ions react with soap (sodium stearate) to form an insoluble salt (metal stearate) called (Mg and Ca stearate respectively); which is generally termed scum.

 

Equations:

With Ca2+

2C17H35COONa+(aq) + Ca2+(aq)                                 (C17H35COO)2Ca(s) + 2Na+(aq)

Sodium stearate                                                                                                            calcium stearate

 

With Mg2+

2C17H35COONa+(aq) + Mg2+(aq)                                (C17H35COO)2Mg(s) + 2Na+(aq)

Sodium stearate                                                                                                            Magnesium stearate

 

– Thus water with Mg and Ca is termed hard water and can only be made soft by removing these ions upon which the water will lather easily with water

– When Ca(HCO3)2(aq) and Mg(HCO3)2(aq) are boiled the amount of soap required for lathering decreases than before boiling

 

Reason

– Boiled decomposes the 2 salts into their respective carbonate s which precipitates from the solution leaving soft water which leathers easily with water

 

– The amount of soap solution used with  solutions containing sulphates and chlorides of calcium and magnesium did not change significantly even after boiling

Reason

– The soluble sulphates and chlorides of Mg and Ca do not decompose upon boiling hence can not be precipitated out.

 

Types of water hard ness

Temporary hardness

– Is hardness due to the presence of CaHCO3 or Mg(HCO3)2 in water; and can usually be removed by boiling.

 

Removal of temporary hardness in water:

(i). Boiling:

– Boiling decomposes and an insoluble chalk of CaCO3 and MgCO3 respectively is deposited in the sides of the vessel.

– This forms an encrustation commonly known as furr the process being furring.

 

Equations:

Ca(HCO3)2(s)           Heat        CaCO3(s) + 2CO2(g) + H2O(l)

 

Mg(HCO3)2(s)           Heat        MgCO3(s) + 2CO2(g) + H2O(l)

 

(ii). Distillation:

– Water containing dissolved salts is heated in a distillation apparatus;

– Pure water distils over first leaving dissolved salts in the distillation flask (refer to separation of mixtures)

– Is of less economic value as it is too expensive hence disadvantageous.

 

(iii). Addition of calcium hydroxide:

– Involves adding correct amount of lime water where CaCO3 is precipitated out.

– This method is cheap and can be used on large scale at water treatment plants.

– However if excess lime (Ca2+) ions is added this will make water hard again.

 

Equation:

Ca(HCO3)2(aq) + Ca(OH)2(aq)                           2CaCO3(s) + 2H2O(l).

 

(iv). Addition of ammonia solution:

– Addition of aqueous ammonia to water containing calcium and magnesium hydrogen carbonates (temporary hard) precipitates calcium and magnesium ions as corresponding carbonates.

 

Equations:

Ca(HCO3)2(aq) + 2NH4OH(aq)                           CaCO3(s) +2H2O(l) + (NH4)2CO3(aq)

 

 

(ii). By permutit softener (ion exchange).

– Uses a complex sodium salt (NaX), such as sodium aluminium silicate commonly known as sodium permutit.

– Permutit is a manufactured ion exchange resin.

 

  • Iron exchange resin: materials that will take ions of one element out of it’s compounds and replace it with ions another element

 

 

 

 

Working principle

– The permutit is contained in a metal cylinder

– The hard water is passed through the column of permutit in the cylinder and it emerges softened at the other end

– As hard water passes through the column ion exchange takes place.

– The Ca2+ and Mg2+ remain in the column while sodium ions from the permutit pass into water thus softening it.

 

Diagram: permutit water softener.

 

 

 

 

 

 

 

 

 

 

Equations:

NaX(aq) + Ca2+(aq)                       CaX(s) + 2Na+(aq)

 

NaX(aq) + Mg2+(aq)                      MgX(s) + 2Na+(aq)

 

– When all the Na+ ions in the permutit have been replaced by Ca2+ and Mg2+ ions the permutit can not go on softening water.

– It is then regenerated by washing the column with brine (a strong NaCl solution); during which calcium and magnesium chlorides are washed away.

 

Equation:

CaX(s) + 2NaCl(aq)                       CaCl2(aq) + Na2X(s)

 

MgX(s) + 2NaCl(aq)                      MgCl2(aq) + Na2X(s)

 

Permanent hardness

– Is that due to soluble sulphates and or chlorides of calcium and or magnesium and cannot be removed by boiling

 

Removal of permanent hardness

(i). By the addition of washing soda (sodium carbonate)

– Washing soda softens hard water by causing the formation of insoluble CaCO3 or MgCO3

– The soluble sodium salts left in water do not react with soap.

 

Equations:

Na2CO3(aq) + CaCl2(aq)                            2NaCl(s) + CaCO3(S)

 

Ionically:

Ca2+(aq) + CO32-(aq)                   CaCO3(s)

 

Na2CO3(aq) + MgSO4(aq)                         Na2SO4(s) + CaCO3(S)

 

Ionically:

Mg2+(aq) + CO32-(aq)                  MgCO3(s)

– This method is very convenient and economical on large scale. It softens both temporary and permanent hardness

 

(ii). By permutit softener (ion exchange); explanations as before

(iii). Distillation.

 

Advantages of hard water

(i). It is good for drinking purposes as calcium ions contained in it helps to form strong bones and teeth.

 

(ii). When soft water flows in lead pipes some lead is dissolved hence lead poisoning. However when lead dissolves in hard water insoluble PbCO3 are formed, coating the inside of the lead pipes preventing any further reaction; this reduces any chances of lad poisoning.

 

(iii). It is good for brewing and the tanning industries; it improves wine or beer flavour in brewing industries.

 

Disadvantages of hard water

(i). Soap forms insoluble salts with magnesium and calcium ions; scum (calcium or magnesium stearate) thereby wasting soap.

– For these reason soapless detergents are preferred to ordinary soaps because they do not form scum; but rather form soluble salts with Mg2+ and Ca2+

  • Examples of soapless detergents: omo, perfix, persil, fab e.t.c.

 

(ii). Deposition of insoluble magnesium and calcium carbonates and sulphates formed from hard water result into blockage of water pips due to the formation of boiler scales

(iii). Formation of kettle fur which makes electrical appliances inefficient hence increasing running costs.

 

(iv). Formation of scum on clothing reduces their durability and aesthetic appearance

 

 

UNIT 3: ENERGY CHNAGES IN CHEMICAL AND PHYSICAL PROCESSES (THERMOCHEMISTRY)

 

Unit Checklist.

1. Introduction

2. Specific heat capacity and enthalpy of a system

3. Endothermic and exothermic reactions.

4. Activation energy

5. Determination of heat changes

Ø  Heat of combustion

  • Meaning
  • Determination
  • Calculations

Ø  Fuels

  • Meaning
  • Heating value of a fuel
  • Fuel pollution
  • Choice of a fuel

Ø  Heat of neutralization

  • Meaning
  • Determination
  • Calculations

Ø  Enthalpy of solution

  • Meaning
  • Determination
  • Calculations
  • Enthalpy of displacement
    • Meaning
    • Determination
    • Calculations
  • Enthalpy of precipitation.
    • Meaning
    • Determination
    • Calculations
  1. Energy level diagrams
  2. Thermochemical cycles
  3. Hess’s law
  4. Enthalpy of formation.
  5. Heat of formation and bond energies.
  6. Latent heat
  • Latent heat of fusion
  • Latent heat of vapourisation
  • Relationship between vapourisation, fusion and structure.

 

Introduction:

– Most chemical and physical process are accompanied by energy changes which occur in the form of heat measured in joules (J) and kilojoules (KJ).

– The heat results from the motion of atoms and molecules.

 

Specific heat capacity

– Specific heat capacity of a substance is the number of joules required to raise the temperature of one gram of the substance by one degree Kelvin e.g. – SHC for water is 4.18Jg-1K-1

 

The heat content (Enthalpy) of a system.

– Heat content is denoted by H; while heat change is denoted as ∆H1;

– And; ∆H = HProducts – HReactants

= H2-H1

– Heat changes in chemical reactions can either be exothermic and endothermic.

 

Endothermic and exothermic reactions.

(a). Endothermic reactions

– Is a reaction accompanied by a fall in temperature and energy is absorbed from the surroundings.

– The enthalpy change is normally positive since heat of products (HProducts) is higher than heat of reactants (Hreactants).

– Thus H2-H1 is a positive value; since HI is less than H2.

 

Graphically this can be denoted as follows;-

 

 

 

 

∆H = +Ve; reaction is endothermic

 

 

Reactants; H1

 

 

 

 

Reaction path

 

Examples:

(i). When NH4NO3 dissolves in water, the temperature of the solution drops.

 

Procedure

– A spatula end-full of ammonium nitrate is dropped into a test tube of water.

– The bottom of the test tube is felt with the hand.

 

Observations:

– The hand feels cold.

Reason:

– Energy is absorbed by the products, cooling the test tube.

 

Thus H1 is less than H2 giving a positive ∆ H1 showing an endothermic reaction.

 

 

(ii) N2(g) +  O2 (g)                2NO(g); ∆H=+91KJMol-1

 

(b). Exothermic reactions.

– Are reactions accompanied by a rise in temperature and energy is liberated to the surroundings.

– The enthalpy change is normally negative, since heat of reactants (H1) is normally higher than heat of products (H2).

– Thus, H2-H1 gives a negative value since H1 is higher than H2.

 

Exothermic reactions can be represented graphically as follows;-

 

 

 

 

 

∆H = Ve; reaction is exothermic

 

 

Products; H2

 

 

 

 

Reaction path

 

– In exothermic reactions H2 is lower than H1.

 

Examples:

(i). Manufacture of ammonia in the Haber process;

i.e. N2(g) +  3H2 (g)                     2NH3(g); ∆H=-46Kjmol

 

(ii) Dissolving sodium hydroxide pellets in water,

-The temperature of the resulting NaOH(aq) is higher than the temperature of water at room temperature.

-This implies the internal temperature of products is lower than the reactants’ original temperature.

-Thus (H products – H reactants) = -ve value, an exothermic reaction.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Activation energy.

– Is the energy required to activate the reactants before a reaction can take place.

– Thus activation energy is the energy required to initiate a reaction.

– The size of activation energy will differ from one reaction to another and so will be the gap between the energy of reactants and the energy is products.

 

Examples:

(i). Exothermic reactions

 

 

E1

A               B      Activation energy

Reactants

C(s) + O2(g)

 

∆H = -ve.

E2

 

 

Products

CO2(g)

 

 

Reaction progress (pathway)

 

  • At A; bonds are being broken and the energy is absorbed.
  • At B; bonds are now being formed, and so energy is evolved.

 

Endothermic reactions

 

 

A          B

 

E2

Products

CaO(s) + CO2(g)

 

∆H = +ve

 

 E1

Reactants

CaCO3(s)

 

Reaction progress (pathway)

 

  • At A; bonds are broken and energy is absorbed.
  • At B; bonds are formed and energy is evolved.

 

 

 

 

 

 

 

 

 

 

 

Determination\ measurement of Enthalpy (heat) changes.

Calorimeters are sued and have to be insulated to reduce heat loss to the surrounding.

 

Source of errors:

– Heat loss to the surrounding.

– Absorption of heat by the calorimeter (vessels).

 

Main heat changes under consideration:

  • Enthalpy of combustion (∆Hc)
  • Enthalpy of neutralization (∆Hneut)
  • Enthalpy of solution (∆Hsoln)
  • Enthalpy of precipitation (∆Hprecip)
  • Enthalpy of displacement (∆HDisp)
  • Enthalpy of formation (∆Hf)

 

  1. Heat of combustion ∆Hc

– Is the heat changes when the mole of a substance is completely burned in oxygen, at one atmospheric pressure.

– Since heat is usually evolved and hence ∆H is usually negative.

 

Example;

C(s) +  O2(g)                             CO2(g) ; ∆ H= -394 Kjmol-1

 

 

Apparatus for finding ∆Hc of a fuel.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Heat evolved = Specific heat capacity of water x Mass of water x Temperature rise

 

= CM∆T; joules, where; C = specific heat capacity.

M = mass of water.

∆T= temperature change.

 

 

 

 

Examples

  1. Assume

Volume of water in calorimeter   = 100cm3

Initial temperature of water         = 225oC

Final temperature of water           = 50.5oC

Change in temperature of water   = 28.0oC

Mass of water                              = 100g

Mass of lamp before burning       = 30.46g

Mass of lamp after burning          = 30.06g

Mass of ethanol burnt                  = 30.46 -30.06

                                                     = 0.40g

(a). Determine the heat evolved;

= ∆ H= CM∆ T

=4.2 x 100 x 28

=11760 Joules

=11.760 KJ

 

(b). Hence calculate the molar heat of combustion of ethanol;

 

Moles of ethanol burnt  = Mass of ethanol burnt; 0.4 = 0.008695 moles.

RFM of ethanol            46

Thus if 0.008695 moles = 11.76 kilojoules,

Then 1mole = ?

 

1 mole = 11.76 x 1 = -1352.50 KJMol

0.008695

 

= -1352.5 KJMol-1 (negative value as the reaction is exothermic since heat is evolved).

 

  1. When ethanol was burnt in the apparatus shown (in example 1), the results were: mass of fuel burnt, M1=1.50g; mass of water, M2=500g; ∆T= 13.0oC. (C=12; H=1; O=16; SHC of water=4.18KJKgK). Find the molar enthalpy of combustion of ethanol.

Compare the experimental value with the listed value of –1368 KJMol and explain any difference.

 

Solution:

Heat evolved = CM∆T

= 4.18 x 500 x 13

1000

= 27.17KJ

Molar mass of ethanol, C2H5OH = [(12 x 2) + (1 x 6) + (1 x 16)] = 46

Thus if 1.5g = 27.17KJ

Then 46 g = ?

= 46 x 27.17

1.5

= 833.213KJMol

 

 

  1. In an experiment to determine the heat of combustion of methanol, CH3OH a student used a set up like the one shown in the diagram below. Study the set up and the data and answer the questions that follow.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Data:

Volume of water       =500 cm3

Final temperature of water =27.0oC

Initial temperature of water =20.0oC

Final mass of lamp + menthol = 22.11g

Initial mass of lamp + methanol =22.98g

Density of water- 1gcm-3

(Heat change = mass x temperature change x 4.2Jg-1oC1

 

Questions:

(a). Write an equation for the combustion of methanol.                                                                {1mark}

 

(b). calculate:

(i). The number of moles of methanol used in the experiment                                                    {2marks}

 

(ii). The heat change for this experiment.                                                                                      {1mark}

 

(iii). The heat of combustion per mole of methanol.                                                                   {2marks}

 

(c) Explain why the molar heat of combustion for methanol obtained in this experiment is different from the theoretical value.                                                                                                           {2marks}

 

  1. When 0.6g of element J were completely burned in oxygen, all the heat evolved was used to heat 500 cm3 of water, the temperature of the water rose from 23.0oC to 32.0oC. Calculate the relative atomic mass of element J given that the specific heat capacity of water = 4.2jg-1k-1, density of water=1.0gcm-3 and molar heat of combustion of J is 380JMol                           {3marks}

 

  1. Hydrogen peroxide contained in 100cm3 of solution with water was completely decomposed. The heat evolved was 1380J. Determine the rise in temperature due to the reaction.

(SHC of water = 4.2Jg-1k-1; density of water = 1gcm-3; O =16; H = 1)                                    {3 marks}

 

Fuels:

– Are compounds which produce a high heat of combustion.

– Fuels can be:-

  • Solids; such as charcoal, wood, coal.
  • Liquids; such as ethanol, gasoline
  • Gaseous; such as methane, water, gas etc.

 

Basic concepts:

  • Calorific value

– Is the energy content of a fuel;

– Is the heat evolved when a given mass of fuel is completely burnt in oxygen;

 

Note:

– Sometimes fuels may undergo incomplete combustion.

– Incomplete combustion of fuel is disadvantageous in that:-

  • It reduces the energy content.
  • It leads to pollution.

 

  • Heating value:

– Is the amount of heat energy given out when a unit mass or unit volume of a fuel is completely burnt in oxygen.

 

Fuel pollution.

– Is commonly caused by internal combustion engine.

– Fuel in engine (e.g. petrol) burns completely to water and carbon (IV) oxide, only under ideal conditions.

 

Causes of fuel pollution:

(i). Incomplete combustion which causes production of (produces) CO and unburnt carbon (soot).

 

(ii). Some fuels contain sulphur and nitrogen and on combustion release SO2 and NO2. These gases are acidic, resulting to acidic rain which corrodes buildings and affects trees and animals in various ways.

 

(iii). Fuel additives; e.g. tetraethyl lead, Pb (C2H5)4 added to petrol to enhance burning efficiency produces volatile lead compounds in the exhaust fumes.

– Lead is very poisonous and affects the nervous system and the brain in children.

 

Factors to consider when choosing a fuel.

(i) Heating value,

(ii) Ease and rate of combustion.

(iii) Availability

(iv) Ease of transportation

(v) Ease of storage

(vi) Environment effects.

 

 

 

 

 

  1. Heat of neutralization;

Is the heat evolved when acid and a base react to form and mole of water.

– Alternatively;

– It is the heat evolved when one mole of hydrogen ions from an acid reacts with one mole of hydroxide ions from an alkali to form\ give one mole of water.

 

Equation:

OH(aq) +  H+(aq)                     H2O(l)

 

– Neutralization reactions are exothermic.

– Are determined by:

– measuring the temperature rise produced when a known volume of acid is neutralized by a known volume of alkali.

Examples.

(a). Strong acids reacting with strong alkalis.

The ∆H is always about –57KJMoland higher than that for weak acids – weak alkalis.

Reason:-

– Strong acids and alkalis are already fully ionized in aqueous solution and no heat is lost in ionizing the acid or the alkali.

 

Consider:

– Reaction between sodium hydroxide and hydrochloric acid

 

NaOH(aq)                     Na+(aq) + OH(aq)

HCl(aq)                  H+(aq) +  Cl(aq)

 

On reacting; OH(aq) + H+(aq)                H2O(l)

 

Diagrammatically (energy level diagram)

 

 

 

 

 

 

∆H = -57KJMol-1

 

H2O(l)

 

 

 

 

 

 

Reaction progress (pathway)

 

(b). Neutralization reactions involving weak acids or weak alkalis.

– The ∆H is lower than expected, e.g. only –52Kjmol-; and hence lower than that for strong acids-strong alkalis.

 

Reason:

– Weak acids and weak alkalis are NOT fully ionized in aqueous solutions and some heat is used in ionizing them.

Consider; reaction between ethanoic acid and ammonia solution.

 

 

NH4+(aq) + OH(aq) + CH3COO(aq) + H+(aq)

 

 

 

∆H = +5KJ

 

∆H = -57KJ

NH3(aq) + H2O(l) + CH3COOH(aq)

 

 

 

 

CH3COO(aq) + NH4+(aq) + H2O(l)

 

 

 

Reaction pathway

 

Therefore;

∆H is given by;

∆H = ∆HI + ∆HII

= 5 + (-57)

= -52KJMol-1

 

Note:

– Dibasic acids e.g. H2SO4 contains two replaceable hydrogen atoms hence on incomplete neutralization, they form two moles of water.

 

 

– Therefore, H neutralization =  ½ x ∆H reaction.

 

Experiment: – To determine the heat of neutralization of hydrochloric acid by sodium hydroxide.

(i) Procedure:-

– A clean dry 250 cm3 glass or plastic beaker is wrapped with a newspaper leaf.

– Exactly 50 cm3 of 2.0M hydrochloric acid solution is transferred into the beaker.

– The temperature T1 of the acid solution is noted.

– Using another clean dry measuring cylinder, exactly 50 cm3 of 2.0M NaOH solution is measured and its steady temperature T2 is noted.

– The contents of the beaker (acid), are carefully stirred with a thermometer while adding NaOH.

– The highest temperature T4 attained by the resulting mixture is noted.

 

(ii) Apparatus

 

 

 

 

 

 

 

 

 

 

(iii). Results:

– Temperature of the acid, T1 = oC;

– Temperature of the hydroxide, T2 = oC;

– Average temperature of the two solutions; T1 + T2 = T3 = oC

2

-The highest temperature of the mixture; T4 = oC;

-The temperature change, ∆T= (T4 –T3) = oC;

 

Sample calculations:

Given: Temperature of hydrochloric acid solution T1= 22.75oC;

Temperature of sodium hydroxide solution T2 = 22.80oC;

Average temperature of acid and alkali T1 + T2 = T3 = 22.78oC;

2

Highest temperature of alkali and acid mixture, T4 = 36.40oC;

Temperature change, ∆T= T4 – T3 = (36.40 – 22.78)oC;

=13.62oC;

 

  • Assumption: the specific heat capacity of the solution= 4.2KJKg-1K-1.

 

– In the experiment, 50 cm3 of 2M HCl are neutralized by 50 cm3 of NaOH, thus; volume of the mixture= (50 + 50) cm3 = 100 cm3.

– Taking density of the resultant solution to be 1gcm-3, then;-

Mass of solution M;

=density x volume

=1g/cm3 x 100cm3= 100g.

=0.1 kg i.e. (100/1000)

Thus heat evolved,

Mass of solution x specific heat capacity x temperature change

= MC∆T,

= 0.1Kg x 4.2kj/kg.k x 13.62oC

=- 5.7 KJ.

 

– But 50 cm3 of 2M HCl contains 2 x 50 moles, = 0.1 moles of H+ ions.

1000

– Similarity 50 cm3 of 2M NaOH contains 2 x 50 = 0.1 moles of OH ions.

1000

– This implies that the two solutions neutralize each other completely.

Therefore;

Molar heat of neutralization (heat liberated when one mole of each reagent is used;-

  • Mole = 5.72KJ

1mole = 1 x 5.72 = -57.2 KJMol-1

0.1

– Since heat is evolved, the reaction is exothermic, the molar heat of neutralization= -57.2 KJMol-1.

Thus, enthalpy change;

H+(aq) + OH(aq)                    H2O(l); ∆H(neut) = -57.2 KJMol. (Thermochemical equation)

1mole            1mole                                   1mole

Note:

Thermochemical equation: refers to a chemical equation which shows the enthalpy change.

 

Worked examples:

  1. Given, T1= 21.0oC; T2=22.0oC; T4 = 34.5oC; volume of hydrochloric acid= 100 cm3, volume of NaOH(aq) =100 cm3, molarity of the solutions are each 2M. Calculate the heat of neutralization for the two reagents. (Assumptions, density of the mixture=1gcm-3 and SHC=4.2KJKg-1K-1)

 

2 (a). When 100g of water at 94.0oC were added to a calorimeter at 17.5oC, the temperature rose to 80.5oC. Determine the heat capacity of the calorimeter. What assumption did you make in your calculations?

 

Solution:

Heat given out by water = heat received by calorimeter of heat capacity C.

Heat = MC∆T,

0.100 x 4180 x (94.0 – 80.5) = C (80.5 – 17.5)

C = 0.100 x 4180 x 13.5

62.5

= 90.288Jg-1k-1

Assumption; specific heat capacity of water is 4180jg-1k-1

 

(b). 250 cm3 of sodium hydroxide were added to 250 cm3 of hydrochloric acid in the calorimeter. The temperature of the two solutions was 17.5oC initially and rose to 20.1oC. Calculate the standard enthalpy of neutralization.

 

Solution:

– Assuming the specific heat capacities of the solutions is the same as that of water, 4180jg-1k-1.

Heat from neutralization = heat by calorimeter + solutions.

= (C∆T)   + MC∆T,

Mass of solutions = 250+250) =500 cm3

= density x volume; = 1gcm-3 x 500 cm3=500g

Hat from neutralization = 90(20.1-17.5) + (0.500 x 4180 x (20.1-17.5)

= 5670J.

 

  1. The following results were obtained in an experiment to determine the heat of neutralization of 50 cm3 2M HCl and 50 cm3 2M sodium hydroxide.

Mass of plastic cup = 45.1g

Initial temperature of acid = 27.0oC

Initial temperature of alkali = 23.0oC

Mass of plastic cup + HCl + NaOH = 145.1g

Temperature of the mixture of acid and alkali = 38.5oC.

 

(a) Define heat of neutralization;                                                                                                  {1mark}

Solution:-

Enthalpy change when one mole of water is formed from a reaction between an acid and a base.

 

(b) Write an ionic equation for the neutralization of hydrochloric acid and sodium hydroxide.  {1mark} Solution:

H+(aq) + OH(aq)                      H2O(l)

 

 

 

(c) Calculate;

(i) The amount of heat produced during the experiment.                                                            {3marks}

(Specific heat capacity of solution=4.2kjkg-1k-1, density of solution= 1gcm-3).

 

Solution:

Amount of heat = MC∆T

Mass of solution = (145.1 – 45.1) = 100g

Temperature change; ∆T = (38.5)-(27.0 +23.0) = 38.5 – 25 =13.5oC;

2

Heat produced ∆H = 100g x 4.2kjg-1k-1 x 13.5oC

=5670 joules

=5.67KJ

 

(ii). Molar heat of neutralization for the reaction.                                                                    {3marks}

 

Solution:

Number of moles involved; taking only NaOH or HCl’

1000 cm3 = 2moles

50 cm3= 2 x 50 = 0.1moles

1000

If 0.1 mole produced 5.67 kg; 1 mol = ?

0.1=5.67kg

1mol= (5.67 x 1) = 56.7 KJMol-1

0.1

 

(d). Explain why the molar heat of neutralization of NaOH and ethanoic acid of equal volume and molarity would be less than the value obtained in c (ii) above.                                                     (1mark)

 

Solution:

– Some of the heat produced during neutralization is used up by the weak acid to dissociate fully hence the lower value.

 

(e). Write down the Thermochemical equation for the reaction between NaOH and dilute hydrochloric acid above.

 

Solution

NaOH(aq) + HCl(aq)                     NaCl(aq) + H2O(l); ∆H= -56.7KJMol-1

 

(f). Draw an energy level diagram for the neutralization reaction in 4 (c ) above.

 

Na+(aq) + Cl(aq)

 

 

∆ H=-56.7KJMOL-1

H2O(l)

 

 

 

 

 

Reaction pathway

 

  1. Heat of solution/ enthalpy of solution.

– Is the heat change when a given mass (moles) of a substance is dissolved in a stated amount of solvent (water).

 

Molar enthalpy of solution:

– Is the heat change when one mole of a substance is dissolved in a stated amount of solvent (water).

– Alternatively;

– It is the heat change when one mole of a substance dissolves in water to give an infinitely dilute solution i.e. (a solution which shows no change in its properties when more water is added).

 

– Is determined as;

 

 

∆Hsolution = – ∆Hlattice + ∆Hhydration    ; when ∆Hlattice is given as a negative value;

 

Alternatively;

 

 

 ∆Hsolution = ∆Hlattice + ∆Hhydration   ; where ∆Hlattice is given a s appositive value;

 

Where,

  • Lattice energy

– Is the leaf evolved when one mole of a compound is formed from its separate gaseous ions.

Alternatively;

– It is the energy required to break the ionic bonds within a crystal (solid) lattice.

 

  • Hydration (solvation) energy,

– Is the heat evolved when one mole of ions are hydrated by water molecules.

 

Note:

– Water is a good solvent because it has a high negative(-ve) enthalpy of solvation resulting from powerful interaction between polar molecules and solute ions due to large dipoles on water molecules.

 

– Therefore,

– Heat of solution can either be exothermic or endothermic depending on the magnitude of hydration and lattice energies.

 

– Solution process occurs into two stages.

 

Example:

– Consider the dissolution of sodium chloride solid.

(i). Energy is taken in to break the crystalline lattice.

 

NaCl(s)                       Na+(g) +  Cl(g)

∆H = +∆Hlattice

=+776KJ.

 

(ii). Heat is evolved when one mole of icons are hydrated by water molecules.

Na+(g) + Cl(g)   H2O(l)          Na+(aq) + Cl(aq)

 

∆H hydration=-771KJ

 

Illustrations:

(a) Energy level diagram for an endothermic dissolving process for a solid MX(s)

 

 

M+(g) + X(g) + H2O(l)

 

Step II

Step I                                                             ∆Hhydration

 

Final state

∆Hlattice                                 M+(aq) + X(aq)

 

 

∆Hsolution

 

MX(s) + H2O(l)

 

 

 

 

 

-The lattice energy is larger than the hydration energy hence;-

∆Hsolution = +ve

 

(b) Energy level for the diagram for an exothermic dissolving process for solid MX(s)

 

 

M+(g) + X(g) + H2O(l)

 

Step I                                                            Step II

∆Hlattice                                                         ∆Hhydration

 

Initial state

MX(s) + H2O(l)

 

 

∆Hsolution

 

 

Final state

M+(aq) + X(aq)

 

 

 

 

The lattice energy is smaller than hydration energy hence;

∆H solution = -ve.

 

Note: All gases dissolve with evolution of heat;

Reason:

– There are no intermolecular forces or bonds to break before hydration occurs.

 

 

 

 

 

 

 

 

Worked examples:

  1. The equation below represents changes in physical states for iron metal.

 

Fe(s)                         Fe(l); ∆H = 15.4KJMol

Fe(l)                         Fe(g); ∆H = 354KJMol

 

Calculate the amount of heat required to change 11.2g of solid into iron to gaseous iron. (Fe=56.0)

 

Solution:

Total heat needed to convert 1 mole of iron from solid to gas = (15.4Kj + 354Kj); = 369.4Kj;

1 mole = 56g;

Thus if 56g requires 369.4Kj;

Then 11.2g =?

= 11.2 x 369.4; = 73.88Kj;

56

 

  1. The lattice and hydration enthalpies for lithium chloride and potassium chloride are given below.

 

Salt ∆Hlattice ∆Hhydration.
LiCl -861 -884
KCl -719 -695

 

(a). Calculate the enthalpy of solution, ∆Hsolution; Kj mol for;

  1. Lithium chloride. (2 marks)

 

Solution:

 

Hsolution = -∆Hlattice + ∆Hhydration

= – (-861) + (-884)

= 861- 884;

= -23Kjmol-

 

  1. Potassium chloride: (2 marks)

 

Solution:

 

Hsolution = -∆Hlattice + ∆Hhydration

= – (-719) + (-695)

= 719- 695;

= -24Kjmol-

 

(b) Which of these two salts above has a higher solubility in water? Give reasons for your answer.

 

Solution:

Potassium chloride; the difference in ∆Hsolution is greater;

 

  1. Given that the lattice energy of NaCl(s) is +77kjmol-1 and hydration energies of Na+(g) and Cl(g) are

-406kjmol-1 and -364kjmol-1 respectively. Calculate the heat of solution, ∆Hsoln for one mole of NaCl(s)

 

 

 

 

Experiment: To measure the molar enthalpy change for the dissolution (enthalpy of solution) for various compounds.

(i) Procedure;

– A clean 250 cm3 glass or plastic beaker is wrapped with a newspaper leaf.

– About 50 cm3 of tap water is measured into the beaker and the steady temperature noted.

– The beaker is held in a tilted position and 2 cm3 of and sulphuric acid added into the water.

 

(ii) Apparatus.

 

 

 

 

 

 

 

 

 

 

 

 

Caution:

– Concentrated sulphuric acid should always be added to water and never vise versa.

Reason:

 

 

– The mixture is then carefully but vigorously stirred using a thermometer and the highest temperature of the solution recorded.

– The experiment is repeated with other compounds:

  • Ammonium nitrate solid
  • Potassium nitrate solid
  • Sodium hydroxide pellets.
  • Sodium nitrate solid

 

Results and observations.

 

Concentrated sulphuric acid Ammonium nitrate solid Potassium nitrate solid Sodium hydroxide pellets
Temperature of water(˚c)
Highest temperature of solution (˚c)
Change in temperature(˚c)

 

Calculate the enthalpy change (∆Hsoln) in each experiment done.

 

Additional information;

– Density of concentrated Sulphuric acid =1.84gcm-3

– Density of water =1gcm-3

– No change in volume when any of the solids dissolved in water

– The specific heat capacities of the solutions=4.2kjkg-1k-1

 

 

(a). Sulphuric acid:

– Total volume of solution = (50 +2) =52 cm3

– Temperature change =A rise of 1ºC

– Enthalpy change; ∆Hsoln; =MC∆T

M = 52 cm3 x 1gcm-3 = 52g = 0.52kg.

C = 4.2kjkg-1k-1

∆T = 5K.

Thus enthalpy change = 0.52 x 4.2 x 1

= – 2.184kj

 

Calculating molar enthalpy of solution of concentrated sulphuric acid:

Mass of acid used = density x volume

= 1.84 x 2 = 3.68g;

Formula mass of Sulphuric acid = 98g

3.6g of acid liberated 2.184kj

Thus 98g of acid liberated; 2.198 x 98 =56Kj;

3.68

Because heat was liberated (+ve ∆T), the ∆Hsoln = -56Kjmol-1

 

(b). Sodium nitrate:

Data:

– Volume of water used = 100 cm3

– Mass of sodium nitrate = 2.5g

– Initial temperature of water = 21.0ºC

– Final temperature of water =19.5ºC

– Temperature change; = A fall of 1.5ºC i.e. (19.5 – 21.0) ºC;

– Mass of 100 cm3 of water = 100g = 0.1kg;

 

Enthalpy change i.e. heat absorbed = MC∆T

= 0.1 x 4.2 x 1.5

= 0.63Kj.

Molar heat of dissolution;

– Molar mass of NaNO3 =23 + 14 + 48 = 85g;

– 2.5g of NaNO3 absorbed 0.63Kj;

– 85g of NaNO3 will absorb =?

= 0.63 x 85

2.5

= +21.42Kjmol-1

Since heat is absorbed (-ve heat change), the reaction is endothermic thus ∆Hsoln = +21.42Kjmol-1

 

 

 

 

 

 

 

 

 

 

Note: energy level diagram for the dissolution of sodium nitrate:

 

NaNO3(aq)

 

 

 

 

∆Hsoln

 

 

 

 

 

Reaction path

 

 

  1. Potassium hydroxide;

– Calculate the molar enthalpy of dissolution of potassium hydroxide (based on values obtained).

 

  1. Heat of displacement.

– This refers to the enthalpy change that occurs when one mole of a substance is displaced from a solution of its ions.

 

Experiment: To determine the molar enthalpy change in the reaction between Cu2+ ions and zinc or iron.

 

(i) Procedure:

– A plastic cup or glass beaker is wrapped with a newspaper leaf.

– 25 cm3 of 0.2M copper (II) sulphate solution is transferred into the beaker.

– The steady temperature of the solution is noted.

– 0.5g of zinc powder are carefully transferred into the plastic cup and stirred with a thermometer.

– The highest temperature attained by the solution is recorded.

 

 

Spatula

(ii). Procedure:

 

Thermometer

 

 

 

 

 

 

 

 

 

Observations:

– The blue colors of copper (II) sulphate fades.

– Brown deposits of copper metal are formed in the plastic cup.

 

Explanation:

– Zinc is higher in the electrochemical series than copper;

– Zinc therefore displaces copper ions from its solution.

 

Equations:

Zn(s) + CuSO4(aq)                      ZnSO4(aq) + Cu(s)

 

Ionically:

Zn(s) + Cu2+(aq)                   Zn2+ + Cu(s)

 

– During the reaction the blue Cu2+ in the solution are replaced by the colorless Zn2+.

– Consequently the blue colour of the solution fades, as brown deposits of copper metal are formed in the plastic cup.

– Excess solid (zinc powder) was used to ensure complete displacement of Cu2+.

 

Results:

– Initial temperature of copper sulphate solution, T1 = 23˚C

– Highest temperature of the mixture T2 = 33ºC

– Temperature change, ∆T; = T2-T1 =10ºC

– Volume of copper sulphate solution used = 25.0 cm3.

– Mass of zinc powder taken = 0.6g

– Density of the solution = 1gcm-3.

– Specific heat capacity of the solution = 4.2Kjkg-1k-1.

 

Assumptions:

– The volume of the solution remains unchanged after the reaction.

(Cu = 63.5, S = 32, O =16, Zn = 65, Fe =56)

 

Question:

Using the above data, calculate, calculate the;

(i). Heat change for the above reaction.

(ii). The molar heat of displacement of copper (II) ions.

 

Solutions:

(i) Heat change = MC∆T

Mass = density x volume

1gcm-3 x 25cm3 =25g.

∆H = 25kg x 4.2 kjkg-1k-1x 10k

1000

= 1.050Kj;

 

(ii). Molar heat of displacement of Cu2+ displacement.

1000 cm3 = 0.2 moles.

25 cm3 = 0.2 x 25 = 0.005moles

1000

Thus,

0.005moles =1.050Kj

1mole =1 x 1.050 = 1050 = 210Kj;

0.005          5

– Since the final temperature of the mixture is higher than initial temperature; it means the Hproducts is lower than Hreactants.

– The reaction is thus exothermic.

– Thus molar heat of displacement of Cu2+ = -210Kjmol

Thermochemical equation:

 

Zn(s) + Cu2+(aq)                    Cu(s) + Zn2+(aq); ∆H=-210kJmol-1

 

Note:

– The experimental value for the heat liberated in this reaction is lower than the theoretical value of 216Kjmol-1.

Reasons:

– The heat lost to the surroundings and the heat absorbed by the apparatus is not accounted for in the calculations.

– This reaction is typically a Redox reaction.

 

i.e.                               Oxidation

 

 

Cu2+(aq) + Zn(s)                        Cu(s) +  Zn2+(aq)

 

 

Reduction.

 

  1. Heat of precipitation:

– Is the heat change which occurs when one mole of a substance is precipitated from the solution.

 

Experiment: To determine the heat of precipitation of silver chloride.

Procedure:

– 50 cm3 each of 2M AgNO3 and 2M NaCl are left in separate beakers and their constant temperatures noted and recorded.

– The NaCl solution is added to the silver nitrate solutions.

– The beaker is covered with a card board and shaken gently to allow mixing of the solution.

– The highest temperature of the mixture is noted and recorded.

 

 

(ii). Apparatus:                                                                                                                           50cm3 2M NaCl(aq)

 

Thermometer

 

 

 

 

 

 

 

Beakers

 

 

 

 

 

   50cm3 2M AgNO3(aq                    50cm3 2M HCl(aq)                                                      50cm3 2M AgNO3(aq)                                                                 Mixture (1ooml)

 

(ii) Data (results)

– Temperature of AgNO3 solution = 21.0ºC

– Temperature of NaCl solution =19.0ºC

– Mean temperature of the solutions = {19+21} = 20.0ºC

2

– Final temperature after mixing = 34.0ºC;

– Temperature change = a rise of 14oC;

– Volume of solution mixed =100 cm3

– Mass of solution mixed =100g (assuming density=1gcm-3)

Questions:

– From the above data, calculate;-

(i) Energy change/ heat change for the reaction

(ii) The molar ∆Hprecipitation for AgCl.

 

Solution:

(i) Heat change, = MC∆T

= 100 x 4.2 x 14

1000

= – 5.88kj

(ii) Equation,

AgNO3(aq) + NaCl(aq)                      AgCl(s) + NaNO3(aq)

Mole ratio                1   :               1       :                       1       :       1

 

Moles of AgNO3;

1000 cm3 = 2 moles

50 cm3 = 2 x 50 = 0.1mole;

 

Since reaction ratio AgNO3: AgCl = 1:1, then moles of AgCl(s) precipitated = 0.1mole

If 0.1mole = 5.88kj

1 mole = (1 x 5.88) = 58.8kJmol

0.1

Reaction being exothermic, ∆Hprecip = -58.8kJmol

 

Thermochemical equation:

 

Ag+(aq) + Cl(aq)             AgCl(s); ∆H= -58.8kjmol-

 

Suitable energy level diagram,

 

 

                        Ag+(aq) + Cl(aq)

 

 

 

∆Hprep = -58.8kJmol

 

                                                                              

AgCl(s)

 

 

 

                        Reaction pathway

 

 

  1. Enthalpy of formation.

– Is the heat change when one mole of a substance is formed from its constituent elements under standard conditions.

Note: this will be dealt with under thermochemical cycles and Hess’s law.

 

 

 

Standard conditions for measuring enthalpy changes.

– Experimental evidence shows that enthalpy changes are affected by physical stats of the reactants, temperature, concentration and pressure.

– Consequently, for comparison purposes certain conditions have been chosen as standard for measurement and determination of enthalpy changes.

– These conditions are:

  • A temperature of 25oC (298K)
  • Pressure of 1 atmosphere (101.325K)

 

Standard enthalpy changes:

– Are values of enthalpy changes that are measured under standard conditions.

– They are denoted by he symbol ∆H0

– A subscript is also added to the symbol to indicate the type of enthalpy change involved.

 

Common standard enthalpy changes.

 

(i). ∆H0f; refers to the standard molar enthalpy change of formation.

 

Example:

H2(g) + ½ O2(g)                       H2O(l); ∆H0f (H2O) = -286KJmol

 

(ii). ∆H0c; refers to the standard molar enthalpy change of combustion;

 

Example:

CH4(g) + 2O2(g)                            CO2(aq) + 2H2O(l); ∆H0c(CH4)  = -890KJmol

 

(iii). ∆H0neut; refers to the standard molar enthalpy change of neutralization.

 

Example:

HCl(aq) + NaOH(aq)                        NaCl(aq) + H2O(l); ∆H0neut = -58KJmol

 

(iv). ∆H0soln; refers to the standard molar enthalpy change of solution (dissolution).

 

Example:

NaNO3(s) + H2O(l)                             NaNO3(aq); ∆H0soln (NaNO3) = +21KJmol

 

(v). ∆H0hyd; refers to the standard molar enthalpy change of hydration.

 

Example:

Na+(s) + Cl(s)    + H2O(l)                         Na+(aq) + Cl(aq); ∆H0hyd (NaCl) = -774KJmol

 

(vi). ∆H0latt; refers to the molar enthalpy change of lattice formation.

 

Example:

Na+(g) + Cl(g)                        NaCl(aq); ∆H0latt (NaCl) = -774KJmol

 

(vi). ∆H0at; refers to the standard molar enthalpy change of atomization.

 

Example:

Na(s)                       Na(g); ∆H0at (Na) = 108.4 KJmol

 

 

 

Energy level diagrams, thermochemical cycles and Hess’s law.

Hess’s law (the law of constant heat summation);

States that:

– If a reaction can occur by more than one route the overall change in enthalpy is the same,           whichever route is followed.

 

Route 1; ∆H1

Illustration:

 

 

 

Route 2

Route 2       ∆H2                              ∆H3

 

B

 

 

 

 

Therefore,∆H1= ∆H2 + ∆H3

 

Worked examples:

  1. Use the information below to determine the enthalpy of combustion of carbon (formation of carbon (IV) oxide).

C(s) + ½O2(g)                    CO(g); ∆H = -110.45KJ

CO(g) + ½O(g)                 CO2(g); ∆H = -282.0KJ

 

Solution:

 

 

 

Route 2

Route 2       ∆H2                              ∆H3

 

CO(g)

 

 

 

 

 

Therefore; ∆H1 = ∆H2 + ∆H3

= -110-45 + (-282.0) = – 392.45kJ

∆H1 = ∆Hcombustion

= -392.45KJ;

 

Note: – Enthalpy of formation, ∆Hf;

– Is the heat absorbed or evolved when one mole of a substance is formed from its elements under standard conditions.

– This can not be determined experimentally because reactions cannot take place under standard conditions.

– In such cases the enthalpy change is determined theoretically using other measurable enthalpies.

– In the case of enthalpy of formation, the enthalpies of combustion of the compound and that of its constituent elements are linked using an energy level or energy cycle diagram.

– The enthalpy of formation is then calculated using the law of constant heat summation (Hess’s law).

 

 

 

 

 

 

Worked examples:


  1. One mole of butane(C4H10) burns completely in oxygen and liberates 2877kj.

(a) Write an equation for the combustion of butane.

2C4H10(g) + 13O2(g)                          8CO2(g) + 10H2O(l)

 

(b) Draw an energy cycle, hence an energy level diagram for the reactions concerned.

 

Solution:

Equations for combustion:

 

C(s) + O2(g)                       CO2(g); then  8C(s) + 😯2(g)                     8CO2(g) (based on balancing for butane)

H2(g) + ½ O2(g)                H2O(g); then  10H2(g) + 5O2(g)                10H2O(g); (based on balancing for butane)

2C4H10(g) + 13O2(g)                          8CO2(g) + 10H2O(l)

 

Then, energy cycle for the changes,

 

 

 

 

 

      ∆Hc;       ∆H1 = ∆H2 + ∆H3                         ∆H3 = Hc        

 

8CO2(g) + 10H2O(l)

 

 

 

 

 

 

Thus energy level diagram for the reactions.

8C(s) + 10H2O(l)

 

 

 

∆H2 = ∆Hf

 

2C4H10(l)

+½13O2(g)

∆H1 = ∆H2 + ∆H3                          ∆H3 = ∆Hc

+13O2(g)

 

8CO2(g) + 10H2O(l)

 

 

Reaction pathway (progress)

 

(c) If the following heats of combustion are given

∆Hºc (graphite) = -393kjmol

∆Hºc (H2(g)) = -286KJmol

∆Hºc (C4H10) = -2877kjmol

 

Calculate the heat of formation of butane, C4H10 from its elements in their normal states at S.T.P.

 

 

 

Solution:

 

∆H1=∆H2 + ∆H3 where:

∆H2 = ∆Hformation of C4H10;

∆H1 =∆Hºc(graphite) + ∆Hºc(hydrogen);

While ∆H3= ∆Hºc(butane)

 

From the equation,

∆H1 = ∆H2 + ∆H3,

∆H2 = ∆H1 – ∆H3

= {8(-393) +10(-286)]-[2(-2877)]

Dividing all through by 2;

= {4(-393) +5(-286)]-2877]; i.e. in order to work with 1 mole of butane;

 

= (-1572) + (-1430) – (-2877)

= -3002 + 2877

= -125KJmol

 

2 (a). Define standard heat of combustion of a substance.

 

Solution:

– The heat change when one mole of a substance is completely burnt in oxygen under standard conditions;

 

(b). Ethanol, CH3CH2OH; cannot be prepared directly from its elements and so its standard heat of formation must be obtained indirectly.

(i) Write an equation for the formation of ethanol from its elements in their normal physical states at standard conditions of temperature and pressure.

 

Solution:

 

2C(s) + 3H2(g) +  ½O2(g)                       C2H5OH(l)

 

(ii). Draw an energy level diagram linking the heat of formation with its heat of combustion and the heats of combustion of its constituent elements.

 

Solution:

Equations

(i). C2H5OH(l) + 3O2(g)                         2CO2(g) + 3H2O(l)

Balancing based on equation for combustion of ethanol

(ii). 2C(s) + 2O2(g)                                2CO2(g)

(iii). 3H2O(g) + 1½O2(g)                          3H2O(l)

 

Then, energy cycle for the changes,

 

 

 

 

+3O2(g)

      ∆Hc;       ∆H1 = ∆H2 + ∆H3                         ∆H3 = Hc        

+3O2(g)

2CO2(g) + 3H2O(l)

 

 

 

Thus energy level diagram for the reactions.

2C(s) + 3H2(l) + ½ O2(g)

 

 

 

∆H2 = ∆Hf (ethanol)

 

C2H5OH(l)

+3O2(g)

∆H1 = ∆H2 + ∆H3                          ∆H3 = ∆Hc (ethanol)

+3O2(g)

 

2CO2(g) + 3H2O(l)

 

 

Reaction pathway (progress)

 

(iii). If the following heats of combustion are given:

∆Hc(graphite) =-393kJmol

∆Hc(hydrogen) =-286kJmol

∆Hc(ethanol) = -1368kJmol

 

Calculate the heat of formation of ethanol, ∆H0f (ethanol)

 

Solution:

∆H1 = ∆H2 + ∆H3

∆H2 = ∆H1 – ∆H3

= [2(-393 + 3(-286)]-[-1368]

= [(-786) + (-856)] – [-1368]

= -164 + 1368

= – 276kJmol

 

  1. Given the standard enthalpies of combustion as

C(s) + O2(g)                CO2(g); ∆H1 = -394kJmol;

H2(g) + ½O2(g)                    H2O(l); ∆H2 = -286kJmol

C2H2(g) +  1½O2(g)                    2CO2(g) + H2O(l); ∆H = -1300kJmol.

 

Find the standard enthalpy of formation of ethyne, C2H2(g).

 

Solution:

– Considering the Hc(ethyne), 2 moles of CO2 and 1 mole of H2O(l) are produced;

– Rebalancing the equations for Hc(carbon) and Hc(hydrogen) likewise:

2C(s) + O2(g)                2CO2(g);

H2(g) + ½O2(g)                    H2O(l);

 

 

 

 

 

 

 

 

 

Then, energy cycle for the changes,

 

 

 

 

+2½O2(g)

      ∆Hc;       ∆H1 = ∆H2 + ∆H3                         ∆H3 = Hc (ethyne)        

+2½O2(g)

     2CO2(g) +  H2O(l)

 

 

 

 

Thus energy level diagram for the reactions.

 

2C(s) + H2(l)

 

 

 

∆H2 = ∆Hf (ethanol)

 

C2H2(l)

+2½O2(g)

∆H1 = ∆H2 + ∆H3                          ∆H3 = ∆Hc (ethanol)

+2½O2(g)

 

2CO2(g) + H2O(l)

 

 

Reaction pathway (progress)

 

Thus ∆H1 = ∆H2 + ∆H3

∆H2 = ∆H1 – ∆H3

= [2(-394) + (-286)]-[-1300]

= (-788-286) + 1300

= -1074 + 1300

= +226kJmol

 

 

Since ∆Hf is +ve, ethyne is described as an endothermic compound.

 

Note:

– In the determination of ∆Hf of ethyne, Hess law is used because attempts to make ethyne from carbon and hydrogen {2C(s) + H2(g)                C2H2(g) }will result to the formation of a mixture of hydrocarbons.

 

 

 

 

 

 

 

 

 

 

 

 

Heat of formation and bond energies

– Chemical reactions involve:

  • Bond breaking; this requires energy.
  • Bond formation; releases energy.

 

Example:

– Formation of methane from carbon and hydrogen.

C(s) + 2H2(g)                CH4(g); ∆H = -74.9kJmol

 

Energy level diagonal for the formation of CH4.

 

 

C(s) + 4H(g)

 

 

II   ∆Hat;

 

C(s) + 2H2(g)

 

 

 

I   ∆Hat = +715KJmol–                    III   ∆H = -4(C-H); bond energy

C(s) + 2H2(g)

 

 

∆HIV = I + II + III

Final enthalpy               = -74.9Kj

CH4(g)

 

 

Reaction pathway (progress)

 

Energy changes involved are:

  • ∆HI; Enthalpy of atomization – this is the energy absorbed when a substance decomposes to form one mole of gaseous atoms.
  • ∆HII; enthalpy of atomization of H2; i.e. dissociation of hydrogen molecules to free hydrogen atoms.
  • ∆HIII; enthalpy of formation of four C – H bonds, during which heat is liberated;

 

Therefore, enthalpy of reaction ∆HR = ∆HIV

∆HR = ∆H1+ ∆HII + ∆HIII

= +715 + 4(218) + -4(C-H)

-74.9 =+715 + 872 +-4 (bond energy)

4 (C-H) =1587 + 74.9

=1661.9

C –H =1661.9 = 415.4kjmol-

4

– This is the amount of energy released when one C-H bond is formed;

 

 

 

 

 

 

 

 

Worked examples.

 

  1. Use the bond energies given below to calculate the heat of reaction for;-

H2(g) + Cl2(g)                     2HCl(g)

 

Bond Bond energy
H-H 435kjmol
Cl-Cl 243kjmol
H-Cl 431kjmol

 

Solution:

 

Bond breaking,

Total energy absorbed = +678 KJ

H2(g) = H-H                  2H(g); ∆H = +435Kj;

Cl2(g) = Cl-Cl                2Cl(g); ∆H= +243Kj

 

Total energy released = -862 KJ

Bond formation,

2H(g) + 2Cl(g)                  2(H-Cl)(g); ∆H=2(-431) = -862Kj;

 

Heat of reaction:

Energy absorbed in bond breakage + energy released in bond formation.

= [+435 + 243] + [-862]

= [+678-862] kJ

= -184kj

Thus, H – H + Cl – Cl                 2[H – Cl]; ∆H=-184Kj

 

Note: energy level diagram.

 

 

2H(g) + 2Cl(g)

 

 

∆HII  ∆Hat (chlorine)

+243KJ

2H(g) + Cl2(g)

 

 

 

∆HI   ∆Hat = +435KJmol–                          ∆HIII = -2 (H-Cl); bond energy

H2(g) + Cl2(g)

 

 

∆HIV = I + II + III

Final enthalpy               = -74.9Kj

2HCl(g)

 

 

Reaction pathway (progress)

 

 

 

 

 

 

 

 

  1. Study the information given in the table below and answer the questions that follow;-

        

Bond Bond energy(kJmol)
C – H 414
Cl – Cl 244
C – Cl 326
H – Cl 431

(a). Calculate the enthalpy change for the reaction

    CH4(g) + Cl2(g)                      CH3Cl(g) + HCl(g)

 

Solution:

 

Bond breaking

(i). 4(C-H)                  4CH(g); ∆H =4(+414) = +1656kj;

(ii). Cl – Cl                2Cl(g); ∆H1 = +244kj

 

Bond formation

(i). 3(CH)           3(C-H); ∆H= 3(-414) =-1242kj

(ii). CCl              (C-Cl); ∆H= -326kj

(iii). HCl             (H-Cl); ∆H= -431kj

 

Heat change = ∆Hbond breakage + ∆Hbond formation

= [(+1656) + (+244)] + [(-1242) + (-326) + (-431)]

= -1900 + (-1999)

∆H = -99kJ

 

(b). Draw an energy level diagram for the reaction

 

 

 

C4H(g) + 2Cl(g)

 

 

∆HII  ∆Hat (chlorine)

+244KJ

C4H(g) + Cl2(g)

 

 

 

∆HI   ∆Hat = +1656Kj                                 ∆HIII = -2 (H-Cl); bond energy

CH4(g) + Cl2(g)                                                = -99Kj;

 

 

∆HIV = I + II + III

Final enthalpy             

CH3Cl(g) + HCl(g)

 

 

Reaction pathway (progress)

 

 

 

 

 

 

 

  1. Study the information given in the table below and answer the questions that follow.

 

Bond Bond energy
C – C 346
C = C 610
  C – Br 280
  Br – Br 193
C – H 413

 

(i) Calculate the enthalpy change for the reaction.

 CH2=CH2 + Br2                       CH2-CH2

                                                     

                                                  Br     Br

Solution:

 

Bond breaking,

4(C-H)                4CH(g); ∆H = 4 (+413) = +1652kj

Br2                      2Br       ∆H = +193kj

C = C                       C = C; ∆H= + 610kj;

 

Bond formation:

4(C-H),                 4CH; = 4(-413) = -1662kj

2(C-Br),                2CBr; = 2(-280) = -560kj

(C-C )                   2C; = -346kj

 

Heat change:

∆H bond breakage + ∆H bond formation

= [+1652 + 193 + 610] + [-1652 + -560 + -346kJ]

= [+2455] + (-2558)

∆H change = -103kJ;

 

Molar heat of fusion, ∆Hf;

– Is the amount of heat required to change one mole of a solid into a liquid at its melting point.

– It is a measure of intermolecular forces between solid particles since it is the heat energy used to separate the solid particles.

 

Molar heat of vaporization, ∆Hvap;

– Is the amount of heat required to change one mole of liquid into vapor at its boiling point.

– It is a measure of the intermolecular forces between the liquid particles since it is the heat energy used to separate liquid particles.

 

 

 

 

 

 

 

 

Relationship between heats of fusion and vaporization to structure.

They are;-

(a). Higher for substances with giant structures e.g.

  • Ionic solids
  • Metals
  • Giant covalent structures, e.g. diamond, graphite and silicon (IV) oxide

(b). Lower in simple molecular structures whose particles are held together by weak intermolecular forces.

Examples:

  • Iodine crystals
  • Sulphur

 

Examples:

 

NaCl C(graphite) Cu H2O I(s) S
Structure type Giant ionic Giant atomic Giant Metallic Molecular Molecular Molecular
∆Hfusion kjmol 28 13 6 8
∆Hvap kjmol 171 717 305 41 21 10

 

 

UNIT 4: RATES OF REACTION AND REVERSIBLE REACTIONS

UNIT OUTLINE:

  1. Reactions rates:

Ø  Measurements of reaction rates

  • Rate of product yield pre unit time
  • Rate of disappearance of reactants

Ø  Apparatus for measuring reaction rates

  • Collision theory and activation energy
  • Factors affecting reaction rates

§  Temperature

  • Particles size (surface area)
  • Catalysts
  • Pressure

 

  1. Reversible reactants
  • Meaning
  • Chemical equilibrium

Ø  Factors affecting position of equilibrium

§  Temperature

  • Pressure
  • Concentration (of reactants or products)
  • Catalysts
  • Le’Chartaliers principle.

 

Reaction rate:

– The rate of a chemical reaction is taken as the rate at which products are formed or the rate at which reactants disappear.

 

Measurements of reaction rates

Reaction rates are measured in terms of

  • How much product appears in a given time.
  • How much reactant disappears in a given time

– Rates curves are then plotted

– These are plots of concentration or properties, which are functions of concentration against time

 

Graph: concentration of products against time.

 

 

 

 

 

 

 

 

 

 

 

 

Time

 

Graph: concentration of reactants against time.

 

 

 

 

 

 

 

 

 

 

 

 

 

Time

 

 

 

 

 

 

 

Apparatus used to determine rates of reaction

Examples:

– Reaction between dilute HCl and marble chips.

– This can be by:

(a). Measuring the decrease in mass with time

Apparatus:

 

 

 

 

 

 

 

Graph:

 

 

 

 

 

 

 

 

 

 

 

 

Time

 

(b). Measuring the volume of the gas produced with time and then plotting a graph.

Apparatus:

Graph:

 

 

 

 

 

 

 

 

 

 

 

Time

Similar experiments maybe done using zinc granules and zinc powder with an acid

 

Worked example:

Marble chips and dilute hydrochloric acid were mixed. The mass of the reaction mixture was measured and recorded with time. The results are shown in the table below.

 

Time (sec) 0 30 60 90 12 150 180 21 240 270 300
Mass of mixture (g) 42.0 41.5 41.0 40.7 40.4 4.02 40.1 40.0 40.0 40.0 40.0
Loss in mass (g)

 

(a). Complete the table.                                                                                                                (5 marks)

 

(b). Draw a graph of loss of mass against time.                                                                          (3 marks)

 

(c). On the same axes sketch the curves that would be obtained if:

(i). the acid was more concentrated.                                                                                              (1mark)

 

(ii). larger marble chips were used.                                                                                               (1mark)

 

Collision theory and activation energy.

The collision theory of reacting particles postulates that:

Ø  Reacting particles must collide before a chemical reaction occurs

  • Not all collisions are effective/ result in chemical reaction
  • Only those particles with sufficient energy result in effective collisions i.e. energy equal to or greater than the activation energy.
  • Any factor, which increases the rate of a chemical reaction, does so by increasing the number of effective collisions;

 

Activation energy (EA):

– Is the minimum amount of energy required by reacting particles to cause a successful collision to form products;

– It refers to the energy an energy barrier that must be overcome by the reactants to be converted to products;

– This energy barrier determines the magnitude of the activation energy of the reactants;

 

Factors determining the value of activation energy:

– Strength of the bonds in the particles of the reactants;

– Whether the reaction is exothermic or exothermic;

– Presence of catalysts;

 

 

 

 

 

 

 

 

 

Graph: activation energy:

 

 

 

 

 

 

 

 

 

 

 

 

 

Factors affecting the rate of reaction

– Factors that increase the rate of a reaction does so by:

  • Increasing number of effective collisions;
  • Lowering the activation energy; since a reaction with high activation energy is slow at ordinary conditions;

 

These factors are:

  • Concentration of reactants
  • Temperature
  • Surface area (size of the particles
  • Catalyst Light
  • Pressure

 

(a).  Concentration of reactants.

– Increase in concentration of one of the reactants increases the rate of the reaction;

Reason:

– Higher concentration results in a greater probability of collision hence a higher rate of reaction

 

Example: reaction between Mg s) and 2M and 4M hydrochloric acid.

 

Graphical representation

 

 

 

 

 

 

 

 

 

 

 

 

 

Time (seconds)

– The maximum volume of a gas collected in both experiments is the same because the number of moles of HCl is the same

– Curve for 4M HCl is steeper indicating a faster rate because the acid is more concentration

– The 4M HCl contains more H+ ions per unit volume than 2M HCl. This reaction takes a shorter time to go to completion

– The start of the flat part of the curve indicates the end of the reaction;

 

Diagram:

Magnesium and hydrochloric acid mixture

Note: Rate of reaction can also be verified by measuring the rate of disappearance of reactants per unit time.

– Reactants with 4M HCl will disappear faster;

– Curve for a faster reaction rate is always to the left of the reference curve;

 

Worked example:

In an experiment to determine the effect of concentration on the rate of a reaction, various concentrations of sodium thiosulphate were reacted with equal volume of hydrochloric acid and the time taken for the precipitate to obscure a cross on paper put under the reaction beaker was determined.

The results are shown in the table below.

 

Volume of S2O32- (cm3) Volume of H2O, cm3 Volume of HCl, cm3 Time for cross to disappear, (s)
50 0 10 20
40 10 10 25
30 20 10 34
20 30 10 52
10 40 10 70

 

(a). Plot a graph of time (vertical axis) against volume of aqueous sodium thiosulphate.         (3marks)

 

(b). From the graph estimate the time for the cross to disappear when 10 cm3 of hydrochloric acid is added to a mixture of 35cm3 of aqueous thiosulphate and 15cm3 of water.                                (2 marks)

 

(c). What is the effect on the rate of the reaction of adding more water to the aqueous sodium thiosulphate? Explain your answer.                                                                                            (2 marks)

 

(d). On the same graph, plot the curves you would obtain if the experiment were repeated at:

(i). at 45oC;                                                                                                                                     (1mark)

 

(ii). using 10 cm3 of less concentrated hydrochloric acid.                                                            (1mark)

 

 

(b). Effect of temperature.

– An increase in temperature increases the rate of a chemical reaction.

Reasons:

– Increase in temperature increases the kinetic energy of the reacting particles;

– The particles therefore move faster leading to more frequent and more effective collisions.

– The rate of reaction therefore increases;

 

Example: reaction between sodium thiosulphate and dilute HCl(aq)

Stoichiometric equation:

Na2S2O3(aq)  +  2HCl(aq)                      2NaCl(aq)  +  H2O(l)  +  S(s)  +  SO2(g)

 

Ionic equation:

S2O32- (aq)  +   2H+(aq)                    S(s)   +   SO2(g)   +  H2O(l);

 

– The time taken to precipitate the same amount of sulphur by the same volume of acid at different temperatures is measured.

– Concentration of thiosulphate solution and the acid are kept constant.

 

Graph: relationship between time and temperature in a reaction.

 

 

 

 

 

 

 

 

 

 

 

 

 

Temperature (oC)

 

– The rate is proportionate to 1/ t;

  • Hence;

– The higher the temperature the shorter the time for completion of the reaction and the higher the rate of the reaction;

  • Alternatively:

You can plot a graph of 1/ t against temperature;

 

Graphically:

Temperature (oC)

 

 

 

 

 

 

 

 

 

(c). Effect of particle size (surface area).

– A decrease in particle size (surface area) increases the rate of a chemical reaction; and an increase in particle size decreases the rate of a chemical reaction.

Reason:

– A smaller particle size means a larger total surface area and this offers a large surface on which the reacting particles can collide;

– This means more collisions, hence more chances of effective collisions leading to a higher rate of reaction;

 

Example: Reaction between dilute HCl and marble chips

The larger CaCO3 granules undergo a slower rate of reaction than the finely powdered CaCO3

 

Equation:

CaCO3  (S)     +  2H+ aq)                  Ca2+(aq)  +   CO2(g)   +   H2O(l)

 

Graph 1: volume of gas produced against time (for granules and powder)

 

 

 

 

 

 

 

 

 

 

 

 

 

Time (seconds)

 

Graph 2: loss in mass of calcium carbonate against time

 

 

 

 

 

 

 

 

 

 

 

 

 

Time (seconds)

 

 

 

 

 

Worked example:
1. exactly 3.0g of powdered carbonate of metal M of formula MCO3 were mixed with excess dilute hydrochloric acid. The mass of the reaction vessel and its contents were recorded at various times. From these readings, the total loss in mass of the reaction vessel and its contents was calculated and recorded as shown in the table below. The experiment was carried out at room temperature.

 

Time (secs) 0 30 60 90 120 150 180 210
Total loss in mass (g) 0 0.08 0.37 0.90 1.19 1.28 1.32 1.32

 

(a) (i). Plot a graph of total loss in mass against time.                                                                 (1mark)

 

(ii). Determine the total loss in mass after 100seconds.                                                             (2 marks)

 

(b) (i). Write an equation for the reaction that occurs.                                                                 (1mark)

 

(ii). calculate the number of moles of:

  • Carbon (IV) oxide gas produced.       (2 marks)

 

  • MCO3       (2 marks)

 

(iii). Calculate the relative atomic mass of metal M.                                                                             (3 marks)

 

(c). On the same axis sketch a curve that you are likely to obtain if the experiment was repeated:

(i)9. at 50oC. Explain.                                                                                                                  (2 marks)

 

(ii). Using MCO3 granules instead of the powder. Explain.                                                       (2 marks)

 

(d). Effect of catalysts

– Catalysts are substances which alter the rate of a chemical reaction without being consumed, e.g. finely divided iron in the harber process;

– They alter the rate of a chemical reaction but remain unchanged at the end of the reaction;

– They can therefore be reused at the end of the reaction;

 

Modes of action of catalysts:

  • Provision of a surface over which particles can react.

Note: Due to this solid catalysts are more effective when used in powder form;

  • Formation of short-lived intermediate compounds with products which then break up to give the products and the catalyst;
  • By adsorption;

 

Note: – Generally, a catalyst increases the rate of a reaction by providing a different pathway of lower activation energy.

– This means more collisions can overcome this energy barrier and result in a reaction.

– It lowers the activation energy of the reaction;

– Catalysts are also reaction-specific;

 

 

 

Some common catalysts:

 

Catalyst Reaction catalyzed.
Manganese (IV) oxide Dissociation of hydrogen peroxide to give oxygen gas and water;
Vanadium (V) oxide or platinum Contact process for the conversion of sulphur (IV) oxide to sulphur (VI) oxide;
Platinum Manufacture of nitric acid in the Oswald’s process;
Finely divided iron or vanadium Haber process in the manufacture of ammonia;
Nickel Hydrogenation of oils to make fats;

Hydrogenation of unsaturated hydrocarbons

Copper (II) oxide Oxidation  of ethanol to produce ethanal;
Titanium (IV) chloride Ziegler method of polymerizing alkenes

 

Graph: reaction rates with and without a catalyst.

 

 

 

 

 

 

 

 

 

 

 

 

 

Time (seconds)

 

Worked example:

Hydrogen peroxide decomposes slowly to water and oxygen under normal conditions.

When a little manganese (IV) oxide is added to the solution the rate of decomposition is enhanced. The results of an experiment on the decomposition of hydrogen peroxide in presence of manganese (IV) oxide are shown below. Study the results and answer the questions that follow.

 

Volume of oxygen (cm3) 0 19 27 33 36 38 39 40 40
Time (seconds) 0 30 60 90 120 150 180 210 240

 

(a). Plot a graph of volume of oxygen produced (vertical axis) against time.                           (3 marks)

 

(b). What is the rate of reaction:

(i). during the first 30 seconds.                                                                                                    (2 marks)

 

(ii). between 30 seconds and 60 seconds.                                                                                    (2 marks)

 

(c). Explain why: (i). the slope of the curve is steeper at the start of the reaction.                    (2 marks)

 

(ii). the curve goes flat at the end of the reaction.                                                     (2 marks)

(d) (i). Suggest the role of manganese (IV) oxide I the decomposition of hydrogen peroxide.     (1mark)

 

(ii). Comment on change in mass of manganese (IV) oxide at the end of the reaction. Explain.(2 marks)

 

(e). Effect of light on the rate of reaction

-The effect of heating and illuminating substances is the same;

– In both cases the constituent particles absorb radiant energy leading to an increase in the number of particles with activation energy resulting in increased rate of reaction;

– Light energizes the particles involved in a reaction;

– This increases the chances of effective collisions per unit time thus increasing the rate of reaction.

– Light of higher frequencies give higher reaction rates e.g. UV Light;

– Examples of reactions affected by light are those involving halogens:

 

Examples of reactions affected by light.

  • Reaction between Cl2(g) and H2(g) does not take place in the dark but is explosive in bright light;

Cl2(g)  +  H2(g)    bright light            2HCl (g)

 

  • For the reaction between methane and bromine; decolourisation of bromine only occurs in presence of light;

CH4(g)  +   Br2(l)           Light        CH3Br(g)  +  HBr(g)

 

Experiment: Effect of light on the decomposition of silver bromide.

  • Procedure:

– About 20cm3 of 0.1M potassium bromide is put in a glass beaker.

– 5cm3 of 0.05M silver nitrate solution is added.

– The resulting pale yellow precipitate is divided into three portions in 3 separate test tubes.

– One of the test tubes in immediately placed in a dark cupboard; the second on a bench and the third is placed in a direct source of light e.g. sunlight.

 

– Formation of a pale yellow precipitate of silver bromide when silver nitrate reacts with potassium bromide.

  • Equation:

KBr (aq) +  AgNO3(aq)                          AgBr(s)   +    KNO3(aq)

Pale yellow

  • Test tube in light: precipitate changes colour from pale yellow to grey.
  • Test tube on the bench: slight change in colour from pale yellow to slight grey.
  • Test tube in dark cupboard: no observable (noticeable) colour change in precipitate.

 

Explanation:

– Light decomposes silver bromide to metallic silver (hence the grey colour) and bromine.

 

Equation:

Light

2AgBr(aq)                       2Ag(s)  +   Br2(g)

 grey

– No observable change in test tube placed in darkness due to lack of light;

– The degree of decomposition and hence change depends on the light intensity falling on the test tubes;

– The rate of decomposition of silver bromide increases with increase in light intensity.

 

 

 

Conclusion:

– Light affects the rate of some chemical reactions by energizing the particles involved in a reaction hence increasing the chances of effective collisions per unit time thus increasing the rate of reaction.

 

Application of effect of light on reaction rate.

– Processing of black and white photographic films is done in dark to prevent the decomposition of the silver bromide that is usually used to coat photographic plates;

 

Further examples of reactions affected by light:

(i). Cl2(g) + H2(g)           UV Light             2HCl(g)

 

CH4(g) + Br2(g)       UV Light     CH3Br(g) + HBr(g)

 

(iii). 6CO2(g) + 6H2O(l)             Light       C6H12O6(aq) + 6O2(g)

 

(f). Pressure

– Increase in pressure increases the rate of reaction involving the formation of small volume of product because of increase in concentration and slight increase in temperature

– In a given volume the higher the number of molecules of a given gas in a container, the greater the pressure;

– Increase in pressure causes the same effect as increase in concentration;

– Thus the rate of reactions involving gases can be increased by increasing the pressure of the gases

 

Note:

From gas laws, decreases in volume results in increase in pressure. Thus, reactions accompanied by decrease in volume move faster to completion.

 

 

 

 

 

 

 

 

 

 

 

Reversible reactions

– Are reactions which can be made to go to either direction (forward or backward) by changing conditions such as temperatures, pressure etc.

– In such cases none of the reactants is completely used up and so the reaction does not go into completion.

 

Note:

Reversible reactions are of two types:

  • Reversible physical changes e.g. heating ice, iodine etc.
  • Reversible chemical changes e.g. heating hydrated copper (II) sulphate, Haber process, decomposition of limestone, heating blue cobalt chloride, ammonium chloride etc.

 

Examples:

– Reaction of nitrogen and hydrogen to give ammonia in the Haber process.

N2(g)  +  3H2(g)                     2NH3 (g)

 

– The reversible sign means the reaction can reach a state of equilibrium if left undisturbed.

 

Explanations of reversible reactions:

Consider the curves below:

 

 

 

 

 

 

 

 

 

 

Time

 

Point (a): forward reaction.

– At the start of the reaction, the rate of the forward reaction is faster.

– The concentration of the reactants is greatest at the beginning.

– The rate of forward reaction and the concentration of reactants decreases with time as the reaction proceeds.

 

Point (b): backward reaction.

– At zero time the rate of the backward reaction is zero.

– The concentration of the products is lowest at this time.

– The rate of the backward reaction and concentration of products increases as the forward reaction proceeds.

 

Point (c): equilibrium.

– It comes to a time (t) when the rate of forward reaction is equal to the rate of backward reaction.

– This balance is called equilibrium.

 

 

 

Equilibrium:

– Is the state at which the rate of forward reaction is equal to the rate of backward reaction.

 

Types of equilibrium:

Static equilibrium:

– refers to a situation when two opposing forces balance each other and whatever was happening before comes to a standstill.

 

Dynamic equilibrium:

– Refers to a situation when two opposing processes, the forward and reverse, continue taking place but at the same rate.

– The equilibrium is said to be dynamic and this state of balance can be reached from either direction.

 

Examples:

3Fe (s)  + 4H2O (g)                    Fe3O4 (s)  + 4H2 (g)

– The Equilibrium state here is established when:

  • Either steam is passed over heated iron in a closed container; or
  • When hydrogen is passed over heated iron oxide;

 

Characteristics of equilibrium systems.

(i) The concentrations of reactants and products do not change after the equilibrium has been reached unless the system is disturbed.

 

Note:

– Observable properties of reactants or products such as colour, mass, volume, PH, temperature etc can be used to detect whether a system is in equilibrium or not.

Examples:

 

Property How it can be used to detect attainment of equilibrium
Colour The colour intensity remains constant at equilibrium
Volume. Total volume of solution remains constant at equilibrium;
Precipitate The height of the precipitate remains constant;

 

(ii). The equilibrium can be reached from either direction.

(iii). All the reactants and products are present in the system.

 

 

 

Factors affecting the position of equilibrium

– Such factors act as a strain on the state of equilibrium and the system reacts in a way to oppose the change.

– The factors are:

  • Temperature
  • Pressure
  • Concentration

 

(a). Temperature

  • Endothermic reactions

– Increase in temperature favours the endothermic reactions;

 

Example:

N2O4 (g)                        2NO2 (g) ; ∆H = + ve

Pale yellow                                dark brown

 

– Increase in temperature shifts the equilibrium to the right; since the reaction is endothermic and hence more yield of nitrogen (IV) oxide; hence dark brown fumes will be observed;

– Decrease in temperature shifts the equilibrium to the left;  since the reaction is endothermic favoured by low temperatures; hence formation of N2O4; and the pale yellow colour is observed;

  • Exothermic reactions:

– Decrease in temperature favours exothermic reactions.

 

Example:

2 SO2(g)   +    O2(g)                    2SO3(g); ∆H = -ve

 

Thus:

– Decrease in temperature causes more yield of sulphur (VI) oxide; because the equilibrium shifts to the right; since the reaction is exothermic favoured by low temperatures;

– Increase in temperature causes SO3 to decompose decreasing its yields. Increase in temperature shifts the equilibrium to the left; because the reaction is exothermic which is favoured by low temperatures;

 

(b). Pressure

-Increase in pressure favours the side with fewer numbers of gaseous molecules since this is the side where pressure is reduced (low).

 

Example:

  • N2 (g) + 3H2 (g) 2NH3 (g)

 

– Increase in pressure favours the production of ammonia; by shifting the equilibrium position to the right; because the volume of gaseous reactants is higher than the volume of gaseous products.

– Decrease in pressure leads to the production of less ammonia (ammonia decomposes); by shifting the equilibrium position to the left; because the volume of gaseous reactants is higher than the volume of gaseous products.

 

 

 

  • N2O4 (g) 2NO2 (g)

Pale yellow                                                        dark brown

 

Increase in pressure:

– Shifts the equilibrium to the left; leading to formation of more N2O4 hence mixture turns pale yellow; because the volume of gaseous products (NO2) is higher than the volume of gaseous reactants (N2O4).

Decrease in pressure:

– Shifts the equilibrium position to the right; leading to formation of more NO2 hence colour turns dark brown; because the volume of gaseous reactants  (N2O4) is lower than the volume of gaseous products (NO2);

 

Note: –

  • Change in pressure has no effect on the equilibrium mixture where gaseous molecules on the two sides are equal

Example:

N2 (g)  +  O2 (g)                            2NO(g)

 

– Increase or decrease in pressure has no effect on position of equilibrium; because the volume of gaseous reactants is equal to volume of gaseous products;

 

Note:

– Generally reactions involving only solids and, or liquids are not affected by pressure change because they are not compressible;

 

Summary: effects of pressure on equilibrium:

 

Reaction Effects of pressure change on position of equilibrium
Increase Decrease
N2O4(g)                    2NO2(g) Equilibrium shifts to the left; more N2O4 is formed; Equilibrium shifts to the right; more NO2 is formed;
N2(g) + 3H2(g)                     2NH3(g) – More NH3 is formed; equilibrium shifts to the right; since volume of gaseous reactants is higher than volume of gaseous products – More N2 and H2 are formed; equilibrium shifts to the left; since volume of gaseous reactants is higher than volume of gaseous products
2SO2(g) + O2(g)                   2SO3(g) – More SO3 is formed; equilibrium shifts to the right (forward reaction is favoured); since volume of gaseous reactants is higher than volume of gaseous products – Less SO3 is formed (more SO2 and O2 are formed); equilibrium shifts to the left (backward reaction is favoured); since volume of gaseous reactants is higher than volume of gaseous products
4NH3(g) + 5O2(g)               4NO(g) + 6H2O(g) – More NH3 and O2 is formed; equilibrium shifts to the left; since volume of gaseous reactants is lower than volume of gaseous products – Less NH3 and O2 is formed (more NO and H2O are formed); equilibrium shifts to the right; since volume of gaseous reactants is lower than volume of gaseous products
H2(g) + Cl2(g)                     2HCl(g) – No effect on the equilibrium; because volume of gaseous reactants and volume of gaseous products is the same; – No effect on the equilibrium; because volume of gaseous reactants and volume of gaseous products is the same;

 

 

 

(c). Concentration

– Consider the equilibrium in bromine water system:

 

Br2(aq)  +  H2O(l)                              OBr(aq)+ Br(aq) + 2H+(aq)

Yellow orange                                                                       Colourless

 

  • Addition of sodium hydroxide to the equilibrium:

– When NaOH(aq) is added to the equilibrium, the concentration of H+  decrease and the rate of forward reaction are favoured (equilibrium shifts to the right); the reaction of bromine with water increases and there is loss of colour of bromine water.

 

Explanation:

– Addition of sodium hydroxide provides hydroxyl ions into the equilibrium.

– The hydroxyl ions react with the hydrogen ions (on the right of equilibrium) to form water.

 

Equation:

OH(aq) + H+(aq)                        H2O(l)

 

– This process removes hydrogen ions from the equilibrium mixture.

– This shifts the equilibrium to the right hence formation of more products;

– This leads to a change in colour from yellow orange to colourless;

 

  • Addition of hydrochloric acid:

– Addition of HCl(aq) is added concentration of H+ increases; more bromine is formed and the orange-yellow colour of bromine water becomes more intense;

 

Explanation:

– Addition of hydrochloric acid introduces more hydrogen ions into the equilibrium;

– The hydrogen ions react with the colourless bromide and hypobromite ions to form yellow-orange aqueous bromine;

– This shifts the equilibrium to the left hence the increase in the intensity of the yellow-orange colour of bromine water;

 

Further examples:

 

(i). Given the equilibrium:

 

SO2(g) +  O2(g)                  2SO3(g)

 

  • Removal of sulphur (VI) oxide:

– Reducing the concentration of SO3 by removing it causes more sulphur (IV) oxide to be converted to sulphur (IV) oxide.

 

  • Addition of either oxygen or sulphur (IV) oxide.

– Addition of either sulphur (IV) oxide or oxygen to the equilibrium shifts the equilibrium to the right; due to increase of concentration of products hence more yield of sulphur (VI oxide;

 

  • Addition of pyrogallic acid:

– Addition of pyrogallic acid into the equilibrium shifts the equilibrium to the left; since pyrogallic acid dissolves oxygen gas reducing concentration of reactants on the left hence less yield of sulphur (VI) oxide;

(ii). Given the equilibrium:

 

2CrO42-(aq) + 2H+(aq)                          2Cr2O72-(aq) + H2O(l)

Yellow                                                                                 Orange

 

  • Addition of sodium hydroxide to the equilibrium:

– When NaOH(aq) is added to the equilibrium, the concentration of H+  decrease and the rate of backward reaction are favoured (equilibrium shifts to the left); the reaction of dichromate ions with water to form chromate ions and hydrogen ions increases and there is change in colour to yellow.

 

Explanation:

– Addition of sodium hydroxide provides hydroxyl ions into the equilibrium.

– The hydroxyl ions react with the hydrogen ions (on the left of equilibrium) to form water.

 

Equation:

OH(aq) + H+(aq)                        H2O(l)

 

– This process removes hydrogen ions from the equilibrium mixture.

– This shifts the equilibrium to the left hence formation of more reactants (chromate and hydrogen ions);

– This leads to a change in colour from orange to yellow;

  • Addition of hydrochloric acid:

– Addition of HCl(aq) is added concentration of H+ increases; more dichromate solution is formed and the orange colour of dichromate ions become more intense;

 

Explanation:

– Addition of hydrochloric acid introduces more hydrogen ions into the equilibrium;

– The hydrogen ions react with the yellow chromate solution to form orange dichromate solution (and water);

– This shifts the equilibrium to the right hence the increase in the intensity of the orange colour of dichromate solution;

 

Generally:

– A change in concentration disturbs the already established equilibrium by making the reaction rate in one direction faster

– The reaction then proceeds predominantly in that direction until equilibrium is re established

 

(d). Effect of catalyst

– Presence of a catalyst has no effect on the position of the equilibrium but alters the rate at which the equilibrium is attained.

– Catalysts usually allow the equilibrium to be reached in a shorter period of time by increasing the rates of the reactions

 

 

 

 

 

 

 

 

Le Chatelier’s Principle

– States that:

 

 

When stress is applied to a system in equilibrium, the system reacts so as to oppose the stress.

 

 

– This implies that when a change in condition is applied to a system in equilibrium, the system moves so as to oppose the change.

Note:

– The effect of different factors on equilibrium was first investigated in 1888 by a French Chemist; Henri Louis Le Chatelier.

– All explanations so far described are based on Le Chatelier’s principle.

 

Industrial applications of chemical equilibrium.

– The ability to change the position of an equilibrium by varying the conditions has been important in industrial processes as industrialists aim at obtaining maximum products at minimum cost and shortest time possible.

– Conditions required to obtain greatest yield of products at minimum costs and shortest time possible are called optimum conditions.

– Such optimum conditions are obtained by continuous removal of products hence reduction in its concentration or varying external factors like temperature and pressure.

 

Summary:

Optimum conditions for common industrial processes.

 

Condition Optimum condition for:
Ø  Haber process

N2(g) + 3H2(g)          2NH3(g); ∆H=- 92KJMol-

Ø  The contact process

2SO3(g) + O2(g)           2SO3(g); ∆H = -197Kjmol-

Temperature – The reaction is exothermic hence favoured by low temperatures; which shift the equilibrium to the right hence more yield of ammonia gas;

– However rate at which the NH3(g) will be produced would be too slow and thus uneconomical; an optimum temperature of about 450oC is thus normally used;

– Forward reaction is exothermic thus increase in temperature favours backward reaction; shifts the equilibrium to the left hence less yield of SO2(g);

– Low temperature will favour forward reaction thus shifts the equilibrium to the right; hence more production of SO2(g); However rate of yield will be too slow and hence optimum temperature for maximum yield are set at about 450oC;

Pressure Increase in pressure favours the forward reaction; shifting the equilibrium to the right hence more yield of ammonia; because the volume of gaseous reactants is higher than the volume of gaseous products;

– However the cost of producing and maintaining high pressures in a system is very high;

– Thus for maximum yield the optimum pressure is 200 atmospheres;

– An increase in pressure will shift the equilibrium to the left leading to  low yield of sulphur (IV) oxide; since the volume of gaseous reactants is lower than the volume of gaseous products;

– Thus the optimum pressure used is atmospheric pressure, which gives a percentage conversion of SO2(g) to SO3(g) of about 96%.

Concentration – Removal of ammonia gas shifts the equilibrium to the right hence more yield of ammonia; since this removal lowers concentration of NH3(g)

– thus for maximum yield ammonia is removed as soon as it is produced; so that more H2(g) and N2(g) can continue reacting;

– Removal of sulphur (VI) oxide gas shifts the equilibrium to the right hence more yield of SO3(g); since this removal lowers its concentration;

– thus for maximum yield SO3(g) is removed as soon as it is produced; so that more SO2(g) and O2(g) can continue reacting;

Catalyst – Main catalyst used is finely divided iron; platinum is a better catalyst but is very expensive and easily poisoned by impurities, hence may increase cost of production; – A catalyst of vanadium (V) oxide is used to increase the rate of reaction;

 

UNIT 4: ELECTROCHEMISTRY

Checklist:

  1. Meaning of electrochemistry
  2. Displacement and Redox reactions.
  3. Oxidation and Redox reactions in terms of electron gain and electron loss
  4. Oxidation numbers (states)
  • Rules of assigning oxidation numbers
  • Calculating oxidation numbers
  1. Redox reactions involving halide ions (halogens)
  • Further examples of Redox reactions.
  1. Tendency of metals to from ions.
  2. Measurements of tendency of metals to ionize
  • Electrochemical cells
  • Salt bridge
  • Cell diagrams
  • Cell equations
  1. Standard electrode potential
  • Definition
  • Standard hydrogen electrode
  • The hydrogen half cell
  • Electrode potentials (negative and positive values)
  • Calculating E0 values from Redox reactions.
  1. Voltaic cells
  • Primary cells
    • Structure
    • Reactions
    • Zinc (negative) terminal
    • Brass (positive) terminal
    • Functions of various components
  • Secondary cells (lead acid accumulators)
    • Structure
    • Reactions
    • During discharge
    • During recharging;
  1. Electrolysis
  • Definition
  • Terminologies used in electrolysis (basic concepts)
  • Preferential discharge of ions
  • Electrolysis of various substances
    • Dilute sulphuric acid
    • Dilute sodium chloride
    • Concentrated sodium chloride (brine)
    • The mercury cathode cell
    • Copper (II) sulphate
      • Using inert electrodes
      • Using copper electrodes
    • Factors affecting electrolysis and electrolysis products
    • Applications of electrolysis
      • Extraction of reactive metals
      • Purification of metals
      • Electroplating
      • Anodizing aluminium utensils
      • Manufacture of sodium hydroxide, chlorine and hydrogen;
        • The mercury cathode cell
        • Diaphragm cell;
        • Membrane cell;
      • Quantitative aspects of electrolysis
        • Basic terminologies
          • The ampere
          • The coulomb
          • The faraday
        • Faradays laws of electrolysis
          • Statements

Definition:     

– Electrochemistry is the chemistry of electrochemical reactions; which deal with the relationship between electrical energy and chemical reactions.

– Electrochemical reactions involve transfer of electrons and are essentially REDOX reactions.

 

Displacement and REDOX reactions

Experiment 1:- Displacement reactions among metals

(i). Procedure

(a). 5 cm3 of 1M CuSO4 (aq) is put in a test-tube and its temperature recorded.

– In the solution, a spatula end-full of iron fillings is added.

– Any observations and temperature change are determined and recorded.

– The procedure is repeated with fresh samples of CuSO4 with Zn, Mg, and Cu powders.

 

(b). The procedure is repeated with 1M magnesium sulphate solution instead of CuSO4 (aq).

 

(ii). Observations:

 

 

 

Metal solid

Copper (II) Sulphate Magnesium Sulphate
Iron fillings – A red brown solid (Cu) is formed

– The blue colour of the solution (Cu2+) fades then changes to green (Fe2+)

– No reaction

 

Zinc powder – A red brown solid, copper metal is deposited.

– The blue colour of the solution (Cu2+) fades then turns colourless;

– No observable reaction (change)

 

Copper powder – No reaction

 

– No reaction

 

 

(iii). Explanations

– Reactions between metals and ions of another metal involve transfer of electrons from the metal to the other metal ion in solution.

 

Examples:

  • Fe(s) and CuSO4 (aq)

– Copper being lower in the electrochemical series accepts electrons easier (than Fe) to form copper atoms (brown solid);

 

Half equations

Fe (s)                         Fe2+(aq) + 2e

 

Cu2+(aq) + 2e                       Cu (s)

 

Overall reaction       

Cu2+(aq)  +   2e-  + Fe(s)                           Fe2+(aq)  + 2e +  Cu(s)

Then;

 

Cu2+(aq)  +  Fe(s)                                 Fe 2+ (aq)  +  Cu  (s)

(Blue)                                                                                  (Green)                     (Brown solid)

 

 

Oxidation and reduction in terms of electron loss and gain

– The loss of electrons is oxidation and the species that gains electrons (causes electron loss); Cu2+ in this case is called oxidizing agent; and is itself reduced.

– Reduction: refers to gain of electrons, and the species that donates electrons (iron solid in this case) is called a reducing agent and is itself oxidized.

– Displacement reactions generally involve reduction and oxidation simultaneously and are thus termed Redox Reactions.

 

Further examples

(i). Zinc solid and CuSO4(aq)

 

(i). Zn(s)                       Zn2+(aq)  +  2e(Oxidation)

(ii). Cu2+(aq)   +  2e             Cu(s) (Reduction)

 

Oxidation

 

 

Then:   Cu2+(aq) +  Zn(s)                        Zn2+(aq) + Cu(s) (Redox)

                    Blue                                                                     Colourless;       Red brown

 

 

Reduction

 

(ii). Magnesium solid and copper (II) sulphate

 

(i). Mg(s)                      Mg2+(aq)  +  2e(Oxidation)

(ii). Cu2+(aq)   +  2e             Cu(s) (Reduction)

 

                           Oxidation

 

 

Then:   Mg(s) + Cu2+(aq)                        Mg2+(aq) + Cu(s) (Redox)

                                          Blue                                               Colourless;       Red brown

 

 

                                                   Reduction

 

(iii). Silver nitrate and copper solid.

 

(i). Cu(s)                       Cu2+(aq)  +  2e(Oxidation)

(ii). 2Ag+(aq)   +  2e            2Ag(s) (Reduction)

 

                                   Oxidation

 

 

Then:   Cu(s) + 2Ag+(aq)                        Cu2+(aq) + 2Ag(s) (Redox)

                    Brown       Colourless                                     Blue                Grey

 

 

                                                   Reduction

 

Note:

– Amount of heat evolved in these redox reactions depends on the position of the metal in the activity series relative to the metal ion in solution.

– The closer the metals are in the activity series; the less readily displacement occurs and the lower the heat evolution during the displacement.

E.g.:    Heat evolved Mg//Cu2+ is higher than that evolved between Fe//Cu2+.

 

Conclusion:

– Metals displace from solutions, those metals lower than themselves in the activity series.

Note:

– The more the reactive a metal is; the stronger a reducing agent it is and the weaker an oxidizing agent it is.

 

Example:

Potassium is stronger reducing agent; but weaker oxidizing agent than silver, gold etc.

 

Summary:

Strength of reducing/ oxidizing agent

 

 

                                         Weakest oxidizing agent        Potassium, K        Strongest reducing agent

Sodium, Na

Magnesium, Mg

Aluminium, Al

Zinc; Zn

Iron, Fe

Lead, Pb

Copper, Cu

Silver, Ag

                                       Strongest oxidizing agent        Mercury, Hg         Weakest reducing agent

 

Summary on displacement reactions

 

 

 

Metal ion (in solution)

Mg Al Zn Fe Pb Cu
K+ C C C C C C
Na+ C C C C C C
Ca2+ C C C C C C
Mg2+ C C C C C C
Al3+ C C C C C
Zn2+ C C C C
Fe2+ C C C
Pb2+ C C
Cu2+ C
Ag2+

 

Key:

A cross (x) indicates no reaction hence no redox reaction occurs.

A tick indicates redox reaction occurs.

 

 

 

 

 

 

 

 

Oxidation numbers/ oxidation state.

– Is the apparent charge that atoms have in molecules or ions.

– For monoatomic ions, the oxidation number (state) is the magnitude and sign of charge;

 

Example:

Oxidation no of Aluminium in Al3+ is +3;

 

Importance of oxidation numbers:

– Helps in keeping track of electron movement in redox reactions; hence determination of the reduced and oxidized species.

 

Oxidation and reduction in terms of oxidation numbers

  • Oxidation            

– Is an increase in oxidation number.

 

  • Reduction

Refers to a decrease in oxidation number

 

Rules in assigning oxidation numbers

  1. Oxidation number of an uncombined element is zero (0)
  2. The charge on a monoatomic ion is equivalent to the oxidation number of that element;
  3. The oxidation number of hydrogen in all compounds is +1 except in metal hydrides where its –1;
  4. The oxidation number of oxygen in all compounds is –2 except in peroxides where it is –1 and 0F2 where it is +2.
  5. In complex ions the overall charge is equal to the sum of the oxidation states of the constituent elements.
  6. In compounds, the sum of oxidation numbers of all constituent atoms is equal to zero.

 

Worked examples

  1. Calculate the oxidation number of nitrogen in:

(i). NO3

 

Solution:

N + (-2 x 3) = -1

N = -1 + 6

= +5

Note: thus nitric acid with a nitrate ion (NO3) is called nitric (V) acid since the oxidation number of nitrogen in it is +5;

 

(ii). NO2;

 

Solution:

N + (-2 x 2) = 0

N = 0 + 4

= +4

Note: Thus the gas NO2 is referred to as nitrogen (IV) oxide because the oxidation umber of nitrogen in it is +4

 

(iii). NO2;

 

Solution:

N + (-2 x 2) = -1

N = -1 + 4

= +3

Note: thus nitrous acid containing nitrite ion is called nitrous (III) acid since the oxidation number of nitrogen in it is +3.

 

(iii). AgNO3;

 

Solution:

1 + N + (-2 x 3) = 0

1 + N + (-6) = 0

N = 0 – 1 + 6;

N = +5

 

  1. Determine the oxidation number of manganese in each of the following, and hence give the systematic names of the compounds.

(i). MnSO4

 

Solution:

Mn + 6 + (-2 x 4) = 0

Mn = 0 – 6 + 8;

Mn = +2

Systematic name: Manganese (II) sulphate;

 

(ii). Mn2O3;

 

Solution:

2Mn + (3 x -2) = 0;

2Mn = 0 + 6

Mn = ½ x 6;

Mn = +3;

Systematic name: Manganese (III) oxide;

 

(iii). KMnO4

 

Solution:

1 + Mn + (-2 x 4) = 0

Mn = 0 – 1 + 8;

Mn = +7;

Systematic name: Potassium manganate (VII) oxide

 

(iv). MnO3;

 

Solution:

Mn + (-2 x 3) = -1

Mn = -1 + 6;

Mn = +5

Systematic name: Manganese (V) ion;

 

 

 

 

 

 

 

 

 

 

Determination of redox reactions using oxidation numbers.

Worked examples:

 

 

 

 

 

Redox reactions involving Halide ions and halogens

Experiment

(i). Procedure:

– 2 cm3 of chlorine gas are bubbled into each of the following solutions: – KI, KCl, KBr, and KF.

– The observations are made and recorded.

– The procedure is repeated using fluorine, bromine and iodine in place of chlorine.

 

Precaution:   

– Chlorine and bromine are poisonous.

 

(ii). Observations

 

 

 

Halogen

Potassium fluoride Potassium Chloride Potassium Bromide Potassium  Iodide
Fluorine (F2) No visible, change Green-yellow gas is evolved Colourless solution  changes to red brown Colourless solution turns black;
Chlorine (Cl2) No visible colour change; No visible colour change; Colourless solution turns red brown Colourless solution turns  black
Bromine (Br2) No visible colour change; No visible colour change; No visible colour change; Colourless solution turns black;
Iodine (I2) No visible colour change; No visible colour change; No visible colour change; Colourless solution turns black;

 

Note:  Colours of halogens in tetra – chloromethane:-

 

Halogen Colour in tetrachloromethane
Fluorine
Chlorine Yellow
Bromine Red-brown
Iodine Purple

 

(iii). Explanations

– Fluorine displaces all the other halogens; Cl2, Br2 and I2 because it has a greater tendency to accept electrons than all the rest.

– Chlorine displaces both Bromine and Iodine from their halide solutions

– Cl2 takes electrons from the bromide and iodide ions i.e. oxidizes them, to form bromine and iodine respectively.

 

Equations:

(i). Chlorine and potassium bromide:

Cl2 (g) + 2KBr (aq)                          2KCl (aq) + Br2 (l)

 

Ionically:

Cl2 (g) + 2Br(aq)                               2Cl(aq) + Br2 (l)

Green-yellow                                                                                           Red brown

 

Redox equation:                    Reduction

 

 

Cl2 (g) + 2Br(aq)                                    2Cl(aq) + Br2 (l)

 

 

Oxidation

 

(ii). Chlorine and Potasium iodide:

Cl2 (g) + 2KI (aq)                        2KCl (aq) + I2 (l)

 

Ionically:

Cl2 (g) + 2I(aq)                                  2Cl(aq) + I2 (l)

Green-yellow                                                                                           Black

 

Half cell reactions:

  • Oxidation: 2I(aq)        I2(aq) + 2e-
  • Reduction: Cl2 (g) + 2e- 2Cl(aq)

 

Redox equation:                    Reduction

 

 

Cl2 (aq) + 2I(aq)                         2Cl(aq) + I2 (l)

 

 

Oxidation

 

– Bromine takes electrons form iodide ions but not from fluorine and chlorine.

– Iodine is formed i.e. due to oxidation of iodide ions by the Bromine.

 

Equations

(ii). Bromine and Potasium iodide:

Br2 (g) + 2KI (aq)                        2KBr (aq) + I2 (l)

 

Ionically:

Br2 (g) + 2I(aq)                                  2Cl(aq) + I2 (l)

Red brown                                                                                               Black

 

Half cell reactions:

  • Oxidation: 2I(aq)        I2(aq) + 2e-
  • Reduction: Br2 (g) + 2e- 2Br(aq)

Redox equation:                    Reduction

 

 

Cl2 (aq) + 2I(aq)                         2Cl(aq) + I2 (l)

 

 

Oxidation

Note: oxidation number of chlorine decreases from 0 to -1 hence reduction; while oxidation number of iodine increases from -1 to 0; hence oxidation;

 

(iv). Conclusion:        

– The stronger the tendency of an element to accept electrons, the stronger is its oxidizing power.

– Fluorine is the strongest oxidizing agent of the 4 halogens considered.

 

 

 

 

 

Order of oxidizing power for halogens.

Fluorine; F2;

Chlorine; Cl2     Increasing oxidizing power.

Bromine; Br2

Iodine; I2

 

Further examples of redox reactions

(a). Action of acid on metals

Mg (s) + 2HCl (aq)                        2MgCl (aq) + H2 (g)

 

Ionically:

Mg (s) + 2H+(aq)                                Mg2+(aq) + H2 (g)

 

Half cell reactions:

  • Oxidation: Mg(s)       Mg2+(aq) + 2e-
  • Reduction: 2H+(g) + 2e- H2(g)

Redox equation:                    Reduction

 

 

2H+ (aq) + Mg(s)                          H2 (aq) + Mg2+ (aq)

 

 

Oxidation

Note: oxidation number of hydrogen decreases from 1 to 0 hence reduction; while oxidation number of magnesium increases from 0 to 2; hence oxidation;

 

(b). Reaction of active metals with water

Example

2Na (s) + 2H2O (l)                           2NaOH (aq) + H2 (g)

 

Ionically:

2Na(s) + 2H2O (l)                                 2Na+(aq) + 2OH(aq) + H2 (g)

 

Half cell reactions:

  • Oxidation: 2Na(s)        2Na+(aq) + 2e-
  • Reduction: 2H2O (g) + 2e- 2OH(aq) + H2(g)

Redox equation:                    Oxidation

 

 

2Na(s) + 2H2O (aq)                     2Na+ (aq) + 2Cl(aq) + I2 (l)

 

 

Reduction

Note: oxidation number of water decreases from 0 to -1(total) hence reduction; while oxidation number of sodium increases from 0 to 1; hence oxidation;

 

 

 

 

 

 

(c). Reaction of heated Iron with dry chlorine

2Fe (s) + 3Cl 2(g)                           2FeCl 3(s)

 

Ionically (assumed):

2Fe (s) + 3Cl2(aq)                            2Fe3+ (aq) + 6Cl (g)

 

Half cell reactions:

  • Oxidation: 2Fe(s)       2Fe3+(aq) + 6e
  • Reduction: 3Cl2(g) + 6e H2(g)

Redox equation:                    Oxidation

 

 

2Fe (s) + 3Cl2(g)                         2Fe3+ (aq) + 6Cl (aq)

 

 

Reduction

Note: oxidation number of chlorine decreases from 0 to -1 hence reduction; while oxidation number of iron increases from 0 to 3; hence oxidation;

 

 

(d). Reaction between Bromine and Iron (II) ions

Ionically:

Br2(l) + 2Fe2+(aq)                              2Fe3+(aq) + 2Br (aq)

 

Half cell reactions:

  • Oxidation: Br2(l)       2Br(aq) + 2e-
  • Reduction: 2Fe2+(g) + 2e- 2Fe3+(aq)

Redox equation:                    Reduction

 

 

Br2 (aq) + 2Fe2+(s)                        2Br (aq) + 2Fe3+ (aq)

 

 

Oxidation

 

Note: oxidation number of bromine decreases from 0 to -1 hence reduction; while oxidation number of Fe2+ increases from 2 to 3 (in Fe3+); hence oxidation;

 

(e). Oxidation by potassium Manganate (VII) (KMnO4)

Procedure:    

– Purple Potassium manganate (VII) is added into a solution containing iron (II) ions in a test tube.

– A few drops of concentrated sulphuric (VI) acid are added.

 

Observations:           

– The purple solution (containing Manganate (VII) ions) turns to colourless (manganate (II) ions) i.e. the purple solution is decolourised;

 

Explanation:

– The Manganate (VII) ions which give the solution a purple colour are reduced to Manganese (II) ions which appear colourless.  This is a redox reaction.

 

 

Equations

Ionically:

MnO4 (aq) + 8H+(aq) + 5Fe2+(aq)                                         Mn2+(aq) + 5Fe3+(aq) + 4H2O (l)

Purple                                                                                                                     Colourless

 

Half cell reactions:

  • Oxidation: 5Fe2+(s)       5Fe3+(aq) + 5e; (since number of electrons is 5)
  • Reduction: MnO4(aq) + 8H+(aq) + 5e                     Mn2+(aq) + 4H2O(l)

Redox equation:                    Reduction

 

 

MnO4 (aq) + 5Fe2+(aq)                   Mn2+(aq) + 5Fe3+ (aq)

 

 

Oxidation

Note:

  • oxidation number of Manganate ions in KMnO4 decreases from 7 to 2 (in Mn2+) hence reduction; while oxidation number of iron increases from 2 (in Fe2+) to 3 (in Fe3+); hence oxidation;

 

  • The presence of Fe3+ at the end of the reaction can be detected by adding sodium hydroxide solution to form a red brown precipitate of Fe(OH)3;

 

(f). Action of potassium dichromate (VI) on iron (II) ions (Fe2+):

(i). Procedure:

– A solution containing iron (II) ions is added into a solution of oxidized potassium dichromate (VI).

 

(ii). Observations:

– The orange solution of potassium dichromate turns green.

 

(iii). Explanations:    

– The iron (II) ions are oxidized to iron (III) ions

– The chromium (VI) ions (orange) are reduced to chromium (III) ions

– This is thus a REDOX reaction.

 

Equations

Ionically:

Cr2O72- (aq) + 14H+(aq) + 6Fe2+(aq)                             2Cr3+(aq) + 6Fe3+(aq) + 7H2O (l)

Orange                                                                                                                    Green

 

Half cell reactions:

  • Oxidation: 6Fe2+(s)       6Fe3+(aq) + 6e; (since number of electrons is 6)
  • Reduction: Cr2O72-(aq) + 14H+(aq) + 5e            2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)

Redox equation:                    Reduction

 

 

Cr2O72- (aq) + 6Fe2+(aq)                 2Cr3+(aq) + 6Fe3+ (aq)

 

 

Oxidation

 

Note:

  • The oxidation number of dichromate ions in K2Cr2O7 decreases from 6 to 3 (in Cr3+) hence reduction; while oxidation number of iron increases from 2 (in Fe2+) to 3 (in Fe3+); hence oxidation;

 

  • The presence of Fe3+ at the end of the reaction can be detected by adding sodium hydroxide solution to form a red brown precipitate of Fe(OH)3;

 

(g). Action of acidified potassium permanganate on Hydrogen Peroxide.

– The overall reaction is a Redox reaction

Redox equation

2MnO4 (aq) + 5H2O (l) + 6H+ (aq)                              2Mn2+ (aq) + 5H2O (l) + 5O2 (g)

Purple                                                                                                                     Colourless

 

Half cell reactions:

  • Oxidation: 5H2O2(aq)            10H+(aq) + 5O2(g) + 10e;
  • Reduction: 2MnO4(aq) + 16H+(aq) + 10e              2Mn2+(aq) + 8H2O(l)

Redox equation:                    Reduction

 

 

2MnO4 (aq) + 5H2O2 (aq)                2Mn2+(aq) + 10H+ (aq) + 5O2(g)

 

 

Oxidation

 

Note:

  • The oxidation number of manganate ions in KMnO4 decreases from 7 to 2 (in Mn2+) hence reduction; while oxidation number of hydrogen increases from -1 (in H2O2) to 1 (in H+); hence oxidation;
  • Thus the acidified potassium manganate (VII) oxidizes hydrogen peroxide to water and hydrogen;

 

(h). H2O2 oxidizes Iron (II) salts to Iron (III) salts in acidic medium

Oxidation:

2Fe2+ (aq)                           2Fe3+ (aq) + 2e

 

Reduction:

2H+ (aq) + H2O2(aq) + 2e                    2H2O(l)

 

Overall redox:

2Fe2+ (aq) + H2O2 (aq) + 2H+ (aq)                               2Fe3+ (aq) + 2H2O (l)

 

 

 

 

 

 

 

 

 

Terms used in describing oxidation – reduction

 

Term Electron change Oxidation number change
Oxidation

Reduction

Oxidizing agent

Reducing agent

Substances oxidized

Substance reduced

loss of electrons

gain of electrons

receives electrons

loses electrons

loses electrons

gains electrons

increases

decreases

decreases

increases

increases

decreases

 

The tendency of metals to form ions

– When a  metal  is placed in an aqueous solution of its ions, some of the metal dissolves;

Equation:

M(s)                          Mn+ (aq) + ne

 

– Dissolution of the metal causes electron build up on its surface; making it negatively charged, while the surrounding solution becomes positively charged.

 

Diagram: dissolution of metal and electron build up.

 

 

                                                            Metal rod

 

 

 

 

                                                                 Solution containing metal ions

 

 

 

– The positive charge of the solution increases and some of the cations start recombining with the electrons on the metal surface to form atoms.

 

Equation:

Mn+ (aq) + ne                         M(s)

 

– Consequently, an electric potential difference is created between the metal rod and the positively charged ions in solution.

– This arrangement of a metal rod (electrodes) dipped in a solution of its ions constitutes a half – cell.

 

Note: 

– The tendency of a metal to ionize when in contact with the ions differs form one metal to another.

– This difference can be measured by connecting two different Half – cells to make a full cell.

– The electrodes of the 2 half – cells are connected by a metallic conductor; while the electrolytes (solutions) of the half cells are connected through a salt bridge.

 

 

 

 

 

 

 

 

 

Diagram: Connection of two half cells to a full cell.

Experiment: – to measure the relative tendency of metals to ionize

(i). Procedure:

– A Zinc rod is placed into 50 cm3 of 1M zinc sulphate in a beaker;

– Into another beaker containing 50 cm3 of 1M CuSO4 (aq), a copper rod is dipped;

– The two solutions are connected using a salt bridge.

– The two metal rods are connected through a connecting wire connected to a voltmeter

– The experiment is repeated using the following half-cells instead of the Zinc-half cells:-

  • Mg rod dipped in 1M MgSO4 (aq)
  • Lead dipped in 1M lead Nitrate
  • Copper dipped in 1M CuSO4 (aq)

 

(ii). Apparatus

Diagram:

(iii). Observation

– The zinc rod in the zinc-zinc ions half – cell dissolves;

– The blue colour of the copper (II) Sulphate solution fades/ decrease;

– Red-brown deposits of copper appear on the copper rod in the copper-copper ions half-cell.

– A voltage of 1.10 V is registered in the voltmeter.

 

 

 

(iv). Equations/ reactions, at each half cell

  • In zinc-zinc ions half cell

Zn(s)                             Zn2+ (aq) + 2e

 

  • In the copper-copper ion half cell

Cu2+ (aq) + 2e-                          Cu(s);

 

(v). Explanations:     

– Zinc rod has a higher tendency to ionize than the copper rod, when the metal rods are placed in solutions of their ions.

– Thus the zinc rod has a higher accumulation of electrons than the copper rod.

– This makes it more negative compared to the relatively more positive copper rod, which has a lower accumulation of electrons.

– On connecting the 2 half cells; electrons will flow form the zinc rod to the copper rod through the external wire.

– The copper rod gains the electrons lost by the Zinc rod.

 

Roles of the slat bridge:       

Note: It forms a link between the 2 half cells, thereby completing the circuit

– It compete the circuit by:

  1. Allowing its ions to carry charge from one half – cell to the other.
  2. Maintaining the balance of charge in the two half-cells; by providing the ions which replace those used up at the electrodes.

 

– The overall reaction in the cells is a Redox  reaction

Half cell reactions:

Zn(s)                             Zn2+ (aq) + 2e (oxidation)

 

Cu2+ (aq) + 2e-                          Cu(s); (reduction)

 

Overall redox equation

                                                  Oxidation

 

 

Zn (s) + Cu2+ (aq)                          Zn2+ (aq) + Cu (s)

 

 

Reduction

 

– The voltage of 1.10 V registered in the voltmeter is a measure of the difference between the electrode potential (Eθ) of Zinc and Copper electrodes, i.e. the potential difference/ the Electromotive force.

– Thus: An electrochemical/ voltaic cell;

  • Is the combination of two half –cells to give a full cell capable of generating an electric current from a redox reaction.

 

 

 

 

 

 

Cell diagram for a voltaic cell

Rules/conventions for cell representation

  1. A vertical continuous line (/); represents the metal-metal ion or metal ion-metal interphase.
  2. Vertical broken line ( ­); between the 2 half-cells; represents the salt bridge.

 

Note: The salt bridge may also be represented by two unbroken parallel lines (//).

 

Example: – cell diagram for the above cell;

Zn(s) / Zn2+ (aq)     Cu2+ (aq) / Cu(s);

 

Alternatively:

Zn(s) / Zn2+ (aq) // Cu2+ (aq) / Cu(s);

 

Electrode potential E0, values of other metal – metal ions relative tot he Cu/Cu 2+ half cell

Metal / metal ion half cell Electrode potential Eθ relative to Cu2+ / Cu half cell
Mg(s) / Mg2+ (aq) +2.04
Zn(s) / Zn2+ (aq) +1.10
Pb(s) / Pb2+ (aq) +0.78
Cu(s) / Cu2+ (aq) +0.00
Ag(s) / Ag+ (aq) -0.46

 

Positive and negative E values

 

(i). Positive E values –          

– If the Eθ value for a metal / metal ion is positive then the metal undergoes oxidation (loses electrons) while the reference electrode undergoes reduction (accepts electrons)

 

Example:

– The Eθ value for Zn(s) / Zn2+ (aq) relative Cu2+ (aq) / Cu(s) is positive because the zinc metal is oxidized to zinc ions while the copper ions are reduced to copper metal.

 

(ii). Negative E values:

– Implies that the reference half cell undergoes oxidation (donates electrons) while the other metal ions in the other half cell undergoes reduction (accepts electrons)

 

Example:

– The E value for Ag (s) / Ag+ (aq) is negative because Cu is more reactive than silver and gives out electrons (oxidation); while the less reactive Ag has its ions accepting electrons (reduction) to form Ag solid.

 

(iii). The 0 (Zero) E value:

Always indicate the reference electrode / half-cell; in which case there would be no potential difference (with itself)

 

Example:

– The 2 half  cells of Cu(s) / Cu2+(aq) or Cu2+(aq) / Cu(s)  have no potential difference in between them hence a zero (0) E value.

 

Note:

  1. – Any other element could be chosen as the reference electrode in place of copper Cu and difference electrode potentials values would be obtained for the same elements.

 

  1. – The electrode potential of a single element is usually determined by measuring the difference between the electrode potential of the element and a chosen standard electrode.

 

  1. – This gives the standard electrode potential (E0) of the element.

 

The standard electrode potential (E0)

Definition

– Is the potential difference for a cell comprising of a particular element in contact with 1 molar solution of its ions and the standard hydrogen electrode.

– It is denoted with the symbol E0.

 

Importance

– It is useful in comparing the oxidizing and reducing powers of various substances.

 

The standard Hydrogen Electrode

– Is the hydrogen half-cell, which has been conventionally chosen as the standard reference electrode.

– It has an electrode potential of zero at:-

  • A temperature of 25oC
  • A hydrogen pressure of 1 atmosphere
  • A concentration of 1M hydrogen ions

 

Note:  The ions in the other half-cell must also be at a concentration of 1 molar.

 

Components of the Hydrogen half – cell;

– Consist of an inert platinum electrode immersed in a 1M solution of Hydrogen ions

– Hydrogen gas at 1 atmosphere is bubbled into the platinum electrode.

– The hydrogen is adsorbed into the platinum surface and an equilibrium (state of balance) is established between the adsorbed layer of molecular hydrogen and hydrogen ions in the solution.

Equation:

½ H2 (g)                                   H+ (aq)   + e

 

Platinised platinum

– Is platinum loosely coated with finely-divided platinum.

– This enables it to retain comparatively large quantity of hydrogen due to its porous state.

– Platinised platinum also serves as a route by which electrons leave or enter the electrode.

– The hydrogen electrode is represented as: H2 (g)  / H+(aq); 1M

 

 

 

 

 

 

 

 

 

Diagram: The standard hydrogen electrode:

 

 

 

 

 

 

 

 

 

 

 

– The electrode potential of any metal is taken as the difference in potential between the metal electrode and the standard hydrogen electrode.

 

Negative and positive electrode potentials

(a). Negative electrode potential

– If the metal electrode has a higher/ greater tendency to loose electrons than the hydrogen electrode; then the electrode is negative with respect to hydrogen electrode; and its electrode potential is negative.

 

Examples: Zinc, Magnesium etc.

 

(b). Positive electrode potential

If the tendency of an electrode to loose electrode is lower than the hydrogen electrode, then the electrode is positive with respect to the hydrogen electrode; and its potential is positive.

 

Examples: copper, silver etc

 

Reduction potentials

– Is a standard electrode potential measured when the electrode in question is gaining electrons.

– The lower the tendency of an electrode to accept/ gain electrons; the lower (more negative) the reduction potential and vise versa.

 

Examples:

K+ (aq)   + e (aq)                                     K(s);         E0 = -2.92V

F2 (g)      + e (aq)                                      2F(aq);     E0 = +12.87V

Mg (s)    + 2e                                       Mg2+ (aq); Eθ = -2.71 V

 

– Thus potassium ions with E0 = -2.92V have a lesser tendency to gain electrons than magnesium ions.

– Thus Potassium is the weakest oxidizing agent; but the strongest reducing agent, since it has the greatest tendency to donate electrons.

 

Note:

– Oxidation potentials will be the potentials of electrodes measured when they are losing electrons hence undergoing oxidation.

 

 

 

Standard electrode potentials  reduction potentials) of some elements

 

 

  Reduction equation Eθvolts
Least readily reduced

 

Most readily reduced

 

 

 

 

  

                                – Increasing strength of oxidizing agent

                                – Decreasing strength of oxidizing agent

 

F2 (g)     +  2e                           2F (aq)

+2.87
Cl2 (g)   + 2e                            2Cl (aq) + 2.87
Br2(g)      +  2e                            2Br (aq) + 1.36
Ag+(aq)+ 2e                              Ag (s) + 0.80
I2 (g)      +  2e                           2I(aq) + 0.54
Cu2+(aq)+ 2e                            Cu (s) + 0.34
2H+ (aq) +  2e                           H2 (g) + 0.00
Pb2+(aq) +2e                             Pb (s) – 0.13
Fe2+(aq) +  2e                           Fe (s) – 0.44
Zn2+ (g) +  2e                           Zn (s) – 0.76
Al 3+ (aq)+   3e–                          Al (s) – 1.66
Mg2+ (aq) + 2e                         Mg (s) – 2.71
K+ (aq)    + e                             K (s) – 2.92

 

Note:

– The standard electrode potentials in the above table are reduction potentials.

– The greater the tendency to undergo reduction, the higher (more positive) the Eθ value.

– The reverse reaction (oxidation) would have a potential value equal in magnitude but opposite in sign to the reduction potential.

 

Example:

Zinc

Reduction potential

Zn2+ (aq)    +   2e-                                Zn (s); E0 = -0.76V;

 

Oxidation potential

Zn (s)                                    Zn2+ (aq)   +   2e; Eθ = + 0.76V;

 

Uses of E0 values

  1. Comparing the reducing powers and oxidizing powers of various substances;
  2. Predicting whether or NOT a stated REDOX reaction will take place.

 

 

 

 

 

 

The E0 value for a REDOX reaction

– Is usually calculated as the sum of the E0 value for the half cells involved.

 

Note:

– If the sum is positive then the reaction can occur simultaneously;

– If the value of the sum is negative the reaction cannot occur;

 

Sample calculations

  1. Cu 2+ can oxidize Zinc but Zn 2+ cannot oxidize Cu;

(a). Cu 2+/ Cu // Zn / Zn2+

Cu 2+ (aq) + 2e                               Cu (s); E0 = + 0.34 V

Zn (s)                                2e + Zn 2+ (aq); E0 = + 0.76 V

Cu 2+ (aq) + Zn (s)                      Cu (s) + Zn (aq); E0 = – 1.10 V

 

– The overall reaction is positive, hence zinc can be oxidized by copper (II) ions, and hence reaction occurs;

 

(b). Zn 2+ (aq) / Zn (s) // Cu (s) / Cu 2+ (aq)

Zn2+ (aq) + 2e-                                        Zn(s);  E0 = – 0.76 V

Cu(s)                             2e + Cu2+ (aq);  E0 = – 0.34 V

 

Zn 2+ (aq) + Cu(s)                   Zn (s) + Cu 2+ (aq) ; E0 = – 1.10 V

 

 

– The overall E0 is negative; thus Zn 2+ cannot oxidize Cu to Cu2+ (Cu cannot reduce Zn 2+ to Zn); and hence the reaction cannot occur.

 

  1. Can chlorine displace bromine form bromide solution?

Cell diagram: Cl2 (g) / Cl(aq) // 2Br(aq) / Br2 (g)

Cl2 (g) + 2e                                2Cl (aq); E0 = + 1.36 V

2Br (aq)                                2e + Br2 (g); E0 = – 1.09 V

Cl2 (g) + 2Br (aq)                    2Cl(aq) + Br2 (g); E0 = +0.27 V

 

– The overall E0 for the reaction is positive, so chlorine can displace bromine from a bromide solution.

 

 

 

 

 

 

 

 

 

 

 

Worked examples

  1. The diagram below represents part of the apparatus to be used for the determination of the standard electrode potential of Aluminum, Eθ Al 3+(aq) /Al (s)

 

 

 

 

 

 

 

 

 

 

 

 

(a). Name the solutions which  could be placed in beakers A and B; specifying their concentrations.

Answer:

– In beaker A; 1 M HCl (aq) i.e. any solution with 1 M hydrogen ions

– In beaker B; 1 M Al (NO3) (aq); i.e. any aqueous solution with 1M Al 3+.

 

(b). One essential part of the cell has been omitted.  Name the missing part and give its functions.

Answer:-       

Missing part: salt bridge

Function: completes the circuit by;

  1. Allowing its ions to carry charge form one half cell to another.
  2. Providing ions which repose those used up of the electrodes, hence maintaining a balance of charge in the 2 half – cells.

 

(c). The voltmeter reading was found to be –1.66 V.

(i). Give the standard electrode potential for the aluminum electrode.

Solution: it is -1.66 since the Hydrogen-hydrogen ions half-cell = 0.00V.

 

(ii). Show the direction of flow of electrons in the circuit;

Solution: From Al3+ / Al half cell to the H2 / 2H+ (aq) half cell

 

(d). Give the half-cell equations and the overall cell equation

Solution:

3H+ (aq) + 3e                               1½ H2 (g); E0 = + 0.00V;

Al (s)                                    3e + Al3+ (aq); E0 = + 1.66V;

Overall:

Al (s) + 3H+ (aq)                      Al3+ (aq) + 1½H2 (g); E0 = +1.66 V

 

 

 

 

 

  1. You are given the following half-equations;

Mg2+ (aq) + 2e-                   Mg (s);  Eθ = – 2.71 V

Zn2+(g)  + 2e-                     Zn (s);   Eθ = – 0.76 V

 

(i) (a). Obtain an equation for the cell reaction.

 

Mg (s)                              2e + Mg2+ (aq); E0 = +2.71V

Zn2+(aq) + 2e                                 Zn (s); E0 = – 0.76V

Mg (s) + Zn2+ (aq)                   Mg2+ (aq) + Zn(s); E0 = +1.95V

 

Thus equation:

Mg (s) + Zn2+ (aq)                   Mg2+ (aq) + Zn(s); E0 = +1.95V

 

(b). Calculate the E0 value for the cell.

Mg (s)                              2e + Mg2+ (aq); E0 = +2.71V

Zn2+(aq) + 2e                                 Zn (s); E0 = – 0.76V

Mg (s) + Zn2+ (aq)                   Mg2+ (aq) + Zn(s); E0 = +1.95V

 

(c). Give the oxidizing species.

– Reducing species

Magnesium i.e. The species that undergoes oxidation since its oxidation number increases (from 0 to 2); as it reduces the other;

 

– Oxidizing species

Zinc/Zinc ions; – the species that undergoes reduction; since its oxidation number decreases (from 2 to 0) as it oxidizes the other species (Mg).

 

  1. Given the following half-equations

I2 (g) + 2e-                          2I (aq); Eθ = + 0.54V

Br2 (g) + 2e-                       2Br (s); Eθ = +1.09 V

 

(a). Obtain an equation for the all reaction

2I(aq)                               2e + I2 (g); E0 = – 0.54V

Br2 (g) + 2e                          2Br (s); E0 = +1.09V

Br2 (g) + 2I (aq)                      2Br (aq) + I2 (g); E0 = +0.55V

 

(b). Calculate the E0 value for the cell

2I(aq)                               2e + I2 (g); E0 = – 0.54V

Br2 (g) + 2e                          2Br (s); E0 = +1.09V

Br2 (g) + 2I (aq)                      2Br (aq) + I2 (g); E0 = +0.55V

 

 

(c). Give the oxidizing species;

Oxidation

 

Br2 (g) + 2I (aq)                      2Br (aq) + I2 (g); E0 = +0.55V

Reduction

0             -1                                   -1            0

 

Oxidizing species: Bromine; Br2 (aq)

 

  1. Consider the following list of electrodes and electrode potential values.

 

Electrode reaction Eθ volts
A2+/ A

B2+/ B

C2+/ C

D+/ D

E2+/ E

F2+/ F

+0.34

-0.71

-0.76

+0.80

-2.87

-2.92

 

(a). Which of the ions is the strongest oxidizer?

D+; because it is most readily reduced/ have the highest tendency to accept electrons as evidenced by its highest positive Eθ value when the ions change to element (D+/ D)

 

(b). Which of the ions is the strongest reducer?

– Is least readily reduced hence lowest E0 value (-2.92 V); accepts electrons least readily i.e. shows the lowest E0 when its ions gain electrons/ are reduced (F+/ F, = -2.92 V)

 

  1. The following is a list of reduction standard electrode potentials.

 

Metal Eθ volts(standard electrode potential)
Magnesium

Zinc

Iron

Hydrogen

Copper

Silver

-2.36

-0.76

-0.44

0.00

+0.34

+0.79

 

(a). Which two metals, if used together in a cell would produce the largest e.m.f?

Magnesium-silver cell;

Mg2+ (aq) + 2e                    Mg (s);  Eθ = – 2.36 V

2Ag+(g)  + 2e                    2Ag (s); Eθ = – 0.79 V

 

Thus;

Mg (s)                              2e + Mg2+ (aq); E0 = +2.36V

2Ag+(aq) + 2e                               2Ag (s); E0 = – 0.79V

Mg (s) + 2Ag+ (aq)                  Mg2+ (aq) + 2Ag(s); E0 = +3.15V

(b). What would be the voltage produced by:-

(i). Zinc-copper cell

Cu2+ (aq) + 2e-                    Cu (s);  Eθ = + 0.34 V

Zn2+(g)  + 2e-                      Zn (s);  Eθ = – 0.76 V

 

Thus; Zn(s) / Zn2+ (aq) // Cu2+ (aq) / Cu(s);

 

Zn (s)                               2e + Zn2+ (aq); E0 = + 0.76V

Cu2+(aq) + 2e                               Cu (s); E0 = + 0.34V

Zn (s) + Cu2+ (aq)                    Zn2+ (aq) + Cu (s); E0 = +1.10V

 

(ii). Copper-silver cell

Cu2+ (aq) + 2e-                    Cu (s); Eθ = + 0.34 V

Ag+ (g) + 2e-                       Ag (s); Eθ = + 0.79 V

 

Thus; Cu(s) / Cu2+ (aq) // 2Ag+ (aq) / 2Ag(s);

 

Cu (s)                               2e + Cu2+ (aq); E0 = – 0.34V;

2Ag+(aq) + 2e                               2Ag (s); E0 = + 0.79V;

Cu (s) + 2Ag+ (aq)                   Cu2+ (aq) + 2Ag(s); E0 = + 0.45V

 

(c). Explain the meaning of the positive and negative signs;

 

Positive signs

– The metal in question has a lower tendency to loose electrons than hydrogen hence more relatively positive to hydrogen;

– They are stronger oxidizing agents but weaker reducing agents but weaker reducing agents than hydrogen;

 

Negative signs

– The particular metal has a higher tendency to loose electrons than hydrogen; hence relatively more negative than hydrogen.

– They are weaker oxidizing agents but stronger reducing agents than Hydrogen.

 

 

 

 

 

 

 

 

 

 

 

 

  1. The following are some half-cell electrode potentials of some elements.

 

 

 

 

Reaction

 

(a). Select two half cells which when oxidized give the  burst E value; and fill the cell representation.

Solution:         The silver – copper cell; i.e.

Cell representation

 

(b). Calculate the E0 value

 

(c). Give the strongest reducing  agent and strongest oxidizing agent.

 

Strongest reducing agent

Strongest oxidizer      –           silver   –           has highest reduction potential

 

  1. Study the table below and answer the questions that follow.

(a). Which two metals would form a metallic couple with the highest EMS.

 

(b). Calculate the e.m.f. of the cell that would be produced by (i) above.

 

(c). Write down the cell representation for the cell above.

 

(d). Which metal is the strongest reducing agent in the above list

Metal A           –           have the lowest reduction potential.

 

8 (a). The table below gives reduction potentials obtained when the half-cells for each of the metals represented by letters J, K, L, M and N where connected to a copper half- cell as the reference electrode.

 

 

(i). What is metal L likely to be?  Give a reason

 

Copper            It has an E0 value/ reduction potential of 0.00, with copper as the                                       reference electrode

 

(ii). Which of the metals cannot be displaced from the solution of its salt by any other metal in the table?  Give a reason.

 

Metal J:          Has the lowest reduction potential; meaning it least readily accepts                                                 electrons (most readily donates electrons) than any other metal.

(iii). Metal K and M were connected to form a cell as shown in the diagram below;

 

  1. Indicate on the diagram, the direction of flow of electrons. Explain.

from K to M.

K is a stronger reducing agent than M, as evidenced by its lower reduction potential.

It thus loses electrons faster becoming more Negative than M; hence electrons move from K through  external wires to M.

 

  1. Write the equations for the half-cell reactions that occur at:-

Metal K electrode:-

            Metal M electrode

 

III. If the slat bridge is filled with saturated sodium Nitrate solution, explain how it helps to complete the circuit.

 

Answer

It allows its ions (Na+ (aq) , and No– 3 (aq) ) to carry charge from one half cell to another. Providing ions which replace those used up at the electrodes.

 

VOLTAIC CELLS

– Are also called electrochemical cells.

Are cells in which electrical energy is generated from chemical reactions.

 

Types of electrochemical cells

(i). Primary cells        –          

Electrochemical cells which are not rechargeable

 

(ii). secondary cells- 

– Voltaic/electrochemical cells which are rechargeable.

 

  1. Primary cells/dry cells:

– Are of various types and an example is the Le’ clanche dry cell.

 

The Le’clanche dry cell

 

  1. Structure

 

  • Consist of a Zinc can with carbon rod at the centre.
  • The central graphite/carbon rod is surrounded by powdered Manganese (IV) oxide and carbon; which are inturn surrounded by a paste of NH4Cl (s) and Zinc Chloride.
  • The protruding portion of the carbon rod is covered with a brass cap; and the zinc can covered with a sealing material.

 

Chemical reaction

  • The Zinc can is the negative terminal; while the carbon/graphite rod is the positive terminal;
  1. a the Zinc can/negative terminal
  2. positive terminal /brass cap

 

Hydrogen gas

(NH3 (g) + 2e                         2NH3 (g) + H2 (g)

 

Note:  These gases (NH3 (aq)  and H2 (g) ) are NOT used immediately but are used in more             complex reactions.

  • The ammonia gas forms a complex with the zinc chloride in the paste.
  • The hydrogen gas is oxidised to water by the Manganese (IV) oxide.

 

Functions of the various components.

 

  1. a) Brass cap –           Functions as the positive terminal where the reduction reaction

 

  1. b) Zinc can –           Is the Negative terminal; where the oxidation reaction occurs.

 

  1. c) Carbon rod –           It serves as the positive electrode.
  • Acts as the connecting wire between the positive and negative terminal through it electron flow form the Zinc can to the brass cap.
  1. d) Manganese (IV) Oxide –           To oxidise the Hydrogen gas produced at the                                                             anode/positive terminal/brass cap to water.

 

  • A single dry cell can produce a potential of 1.5 V

 

Note:  Dry ammonium chloride does not conduct electric current.  This explains why a paste,   which is a conductor, is used.

 

The dry cells cannot provide a continuous supply of electricity for an undulate period of time.

 

Reason            –           The reactants (electrolytes) are used up and cannot be replaced.

 

  1. Secondary cells

 

Are voltaic/electrochemical cells that are rechargeable.

A common example is the lead – acid accumulator.

 

The lead – acid accumulator.

 

Structure:        –           The positive plate is a lead grill filled with lead (IV)  oxide; while                                                 the negative plate  consists of a similar lead grill filled with                                                             spongy lead.

 

–           The grills are immersed in sulphuric acid;  which serves as the             electrolyte.

 

Reactions

 

  1. i) During discharge/when in use

–           At the  negative terminal/lead           –           The lead dissolves forming lead (II) ions

 

equation

 

  • At the positive terminal (lead (IV)

Lead (IV) oxide reacts with the                                                                                 Hydrogen ions (H+) in sulphuric acid; also forming lead (II) ions.

Then; the lead (II) ions formed at both electrodes react instantly with the Sulphate ions     to form lead (II) Sulphate.

 

–           The  insoluble lead Sulphate adheres to the electrodes.

 

Overall reaction

 

Note:   The Lead Sulphate should NOT be left for too long to accumulate on the electrodes

 

Reason:          The fine PbSO4 (s)  will charge to coarse non – reversible and inactive form and                  the accumulator will become less efficient.

  • During use/discharge; the lead and the lead (IV) oxide are depleted, and the concentration of sulphric acid declines.

 

  1. ii) During recharging –           Is usually done by applying a suitable voltage to                                                      the terminals of the

–           At the negative  terminal       –           the lead ions                                                               accept electrons to form lead solid.

Overall reaction

 

This process restores the original reactants.

 

 

ELECTROLYSIS

 

Defination

 

Is the decomposition of molten or aqueous solutions by passage of electric current through it.

 

Terminologies used in electrolysis

  1. i) Electrolyte

–           Is a solution which allows electric current to pass through while it gets decomposed.

  • Electric current transfer in electrolyte occur through ions.
  • The electrolyte can be aqueous solutions or molten solutions
  • Electrolyte with may ions are called strong electrolyte; while those with few ions are called weak electrolytes.

 

Examples

 

  1. ii) Electrodes –           Are the solid conductors, usually roots, which usually complete                                             the circuit between electrolytes and cell/battery

–           Are of two types

 

  1. a) Anode – The electrode connected to the positive terminal f a battery/cell
  2. b) Cathode– Electrode connected to the negative terminal of the battery/cell.

 

Note:  Graphite rods are commonly preferred as electrodes in most access.

Reasons:         They are inert/unreactive

Are cheap

  • Platinum is also relatively inert; but not less preferred  to Graphite because its expensive.

 

 

Preferential discharge  of ions

 

  • The products of electrolysis of any given electrolyte depend on the ions present in an electrolyte.
  • Commonly most molten electrolytes have only two ions; a cation and an anion and are termed Binary electrolytes.
  • As the electrolyte decomposes, ions collect/move to the opposite poles.
  • Negatively charged ions move to the Anode; the positive electrode, while the positively charged  ions move to the cathode, the negative  electrode;
  • Regardless of how many ions move to an electrode; only one can be discharged ot give a product.
  • Both cations and anions have a preferential discharge series.

 

  1. Discharge for cations

 

Cations  are discharged by reduction (accepting electrons) to form their respective products.

 

The ease of reduction of cations depends on their position of electrochemical series.

 

Thus Ag+ is most readily discharged as it’s the weakest reducing agent.

 

  1. Discharge for anions

 

Anions are discharged by oxidation (electron loss) to form their respective products.

 

Discharge of anions is viz.

 

 

 

ELECTROLYSIS OF VARIOUS SUBSTANCES

 

  1. Electrolysis of various substances
  2. i)

 

 

 

  1. ii) Procedure –           An electric current is passed through the dilute sulphuric acid.

 

iii)       Observation

 

At the Anode

–           A colourless gas; collects

  • The gas collected relights a glowing splint; and its volume is half the volume of the gas at cathode.  The gas is oxygen.

 

At the cathode

  • A colourless gas collects
  • The collected gas burns with a pop-sound; and its volume is double the volume of gas at the anode.
  • The gas is Hydrogen gas.

 

  1. Explanations –           Ions present in the electrolyte
  2. i) Hydrogen ions and Sulphate ions form sulphuric acid.
  3. ii) Hydrogen ions and hydroxide ions from water.

 

At the Anode (Positive electrode)

 

–           The negatively charged Sulphate ions and hydroxide ions migrate to the anode.

 

Reason:          OH (aq) ions have a greater tendency to loose electrons than the SO2 – 4 (aq) ions

 

Anode equation

 

 

At the cathode (positive electrodes)  –           The positively charged hydrogen ions migrate to the cathode.

 

Equation

 

 

Note: 

 

  1. The volume of oxygen produced at the anode is half the volume of hydrogen produced at the cathode.

 

Reason:          The 4 electrons lost by the hydroxide ions to form 1 mole (1 volume) of                     oxygen molecules are gained by the four hydrogen ions which form 2                         molecules (2 volumes) of hydrogen  molecules.

 

  1. During the electrolysis, the concentration of the electrolyte (H2SO4 (aq), increases

 

Reasons:         The volumes of hydrogen and oxygen gas liberated are in the same ratio                as they are combined in water.

 

Thus the amount of water in the electrolyte progressively decrease; hence the increased electrolyte concentration.

 

Conclusion

 

Electrolysis of dilute sulphuric acid is thus the electrolysis of water.

 

Note:   The Hoffmans voltmeter can be used instead of the circuit above.  Viz.

 

 

 

 

Electrolysis of dilute sodium chloride

 

  1. i) Apparatus

 

 

 

  1. ii) Procedure
  • An electric current is passed through dilute sodium chloride solution; with carbon rods     as the electrodes.
  • Gases evolved of each  electrode  are collected and tested.

 

iii)       Observations

  • At the anode:
  • A colourless gas is collected
  • The gas relights a glowing  splint, and its volume is half the volume  of the gas collected at the cathode.
  • The gas is Oxygen, O2

 

  • At the Anode
  • A colourless gas collects
  • The gas burns with a pop sound; and its volume is twice the volume of the gas collected at the anode.
  • The gas is hydrogen gas;

 

  1. iv) Explanations
  • The Ions present in the electrolyte

 

 

 

  1. At the Anode
  • Cl and OH migrate to the anode
  • OH- are preferentially discharged – coz they have greater tendency to lose electrons than the chloride ions;  – the OH (aq)  lose electrons to form water and O2 (g) at anode.

 

Anode equations

 

 

  1. At the Cathode

–           The positively charged Na+ (aq) , and H+ (g)   migrate to the cathode

  • the H+(aq) are preferentially discharged.

 

Reason:          They  H+ (aq) have a greater tendency to gain electrons than Na+ (aq) ions

The H+ (aq) gain elect5rons to form Hydrogen atoms (H) which ten form molecules of hydrogen which bubble off at the electrode.

 

Cathode equations

 

 

  1. Conclusion

Ratio of the volumes of H2 (g) and O2 (g) evolved at cathode and anode is 2:1 respectively.

 

Electrolysis f dilute NaCl is thus the electrolyze of water since only water is decomposed.

 

 

 

  1. Electrolysis of Brine/concentrate sodium chloride.

 

  1. i) Apparatus

 

 

  1. ii) Procedure

An electric current is passed though concentrated sodium chloride/brinc

 

iii)       Observation

  1. a) At the Anode
  • A greenish – yellow gas is evolved.
  • The gas has a pungent irritating smell; and its volume is equal to the volume of the gas evolved at he cathode.
  • The gas is chlorine Cl2 (g)

 

  1. b) At the Cathode
  • A colourless gas is liberated
  • The gas burns with a pop sound; and its volume is equal to volume of gas evolved at the anode.
  • The gas is Hydrogen gas; H2 (g)

 

  1. iv) Explanations
  • The ions present in the electrolyte are:-
  • Na+ (aq) and Cl from sodium chloride
  • H+ (aq) and OH- (aq) from water.

 

  1. At the Anode
  • Cl (aq), and OH (aq), migrate to the anode
  • The chloride ions are preferentially discharged.

 

Reason;           –           OH (aq) have higher tendency to lose electrons  than Cl ions.

  • However coz of the higher concentration Cl (aq) , relative to OH (aq), the Cl (aq), are preferentially discharged hence the evolution of Chlorine gas.

 

  1. A the Cathode
  • Na+ (aq) and H+ (aq) migrate to the cathode.
  • H+ with a higher tendency to gain electrons are preferentially discharged; hence the evolution of hydrogen gas at the cathode.

 

NB:     1.         The pH of the electrolyte becomes alkaline/increases with time.

 

Reason:          The removal of H+ (aq) which come form water leaves excess                                              hydroxide ions  (OH (aq), hence the alkalinity.

 

  1. Evolution of chlorine gas at anode soon stops after sometime and is replaced by O2 (g)

 

Reason:          Evolution of Cl2 (g) decrease/lowers the concentration of Cl (aq) in                         the electrolyte.

                       

As soon as the Cl (aq) concentration  becomes equal to that of OH (aq)

 

The mercury cathode cell

 

Is an electrolytic arrangement commonly used for the large scale manufacture of chlorine and sodium hydroxide.

 

  1. i) Apparatus
  • Electrolyte in the mercury cell is Brine (concentrated NaCl)
  • Anode is carbon or titanium
  • Cathode is a moving mercury film.

 

 

  1. ii) Reactions
  2. At the Anode
  • Both Chloride and Hydroxide ions are attracted
  • Due to their high concentrations the chloride ions are preferentially discharged.
  • The Cl (aq) lose electrons to form Chlorine gas. (greenish yellow)

 

  1. Cathode (moving mercury)
  • The Na+ (aq) and H+ (aq) are attracted
  • The discharge of H+ (aq) is more difficult than expected
  • Hydrogen has a high over voltage at the moving mercury electrode and so sodium is discharged.

 

Equation

 

  • The discharged sodium atoms combine with mercury to form sodium amalgam

 

Equation

 

  • The sodium amalgam reacts with water to form sodium hydroxide, hydrogen and mercury.

 

Equation

 

  • Hydrogen is pumped out while the Mercury is recycled.
  • The resultant NaOH is of very high party.

 

 

Limitations/disadvantages of the Mercury Cathode cell.

 

  1. Its expensive due to the high cost of mercury.
  2. Pollution form Mercury; i.e. Mercury is poisonous and must be removed from the effluent.

 

 

  1. Electrolysis of Copper (II) Sulphate solution.

 

NOTE:            The products of electrolysis of copper (II) Sulphate solution depends on the nature of the elctrodes used.

 

  1. i) Apparatus

 

 

 

 

  1. ii) Ions present in the elctrolyte:
  • From copper (II)  Sulphate

 

  • From water

 

  • During electrolysis

 

  1. Using carbon/platinum electrodes

 

iii)       Observations

 

  1. At the Anode:
  • A colourless gas is liberated
  • The gas relights a glowing splint; hence its oxygen.

 

  1. At the cathode
  • A reddish – brown coating (of Cu solid) is deposited.

 

  1. In the electrolyte
  • The blue colour of the solution (CuSO4) (aq) / becomes pale and finally colourless after a long time.

 

Reason:          The blue colour is due to Cu2+.  As the Cu2+ (aq) are continuously being                            discharged at the cathode; the concentration of CU 2+ decreases i.e. decrease in                     the concentration of Cu 2+ (aq) in the solution

 

The electrolyte become acidic/pH decreases (declines)

 

Reason:          Accumulation of H+ (aq) in the solution since only OH (from water) are being                         discharged (at the anode).

 

 

  1. iv) Explanation

 

At the anode

 

  • The SO 2- 4 (aq) and OH (aq) migrate to the anode.
  • The hydroxide ions have a higher tendency to lose electrons than the SO 2- 4 (aq)
  • They (OH) easily loose electrons to form the neutral and unstable hydroxide radical (OH)
  • The hydroxide radical (OH) decomposes to form water and Oxygen.

 

 

At the cathode

  • Copper ions and H+ (aq) migrate to the cathode
  • Cu2+ (aq) have a greater tendency to accept electrons than H+ (aq)
  • The Cu 2+ (aq) are thus reduced to form copper metal which is deposited as a red-brown coating on the cathode.

 

Cathode equation.

 

 

  1. Using copper rods electrodes

 

  1. observations

 

At the anode   –           Mass of the anode (Copper anode) decreases

 

At the cathode –           reddish – brown deposit

  • cathode increases in mass

 

Electrolyte – no apparent change

 

Note:   The gain in mass of the cathode is equal to the loss in mass of the anode.

 

Explanations

 

At the anode

–           The SO 2- 4(aq) and OH (aq) are attracted to the anode.

  • However, none of them is discharged;
  • Instead; the copper anode itself gradually dissolves; hence the loss in mass of the anode;

 

Reason:          it’s easier to remove electrons form the copper anode itself than format the                         hydroxide ions

 

At the cathode

 

  • H+ (aq) migrate to the cathode
  • The Cu 2+ (aq) are preferentially discharged; because they have a greater tendency to accept electrons
  • The copper cathode is thus coated with a reddish brown deposit of copper metal hence increase in mass.

 

 

Cathode equation

 

Factors affecting electrolysis  and electrolytic products.

 

  1. Electrochemical series
  • Electrolytic products at the anode and cathode during electrolysis depends on its position in the Electrochemical series.

 

Cations:          The higher  the cation in the electrochemical series; the lower the tendency of                     discharge at the cathode.

 

Reason:          Most electropositive cations require more energy in order to be reduced and                         therefore are more difficult to reduce.

 

Reduction order

 

 

Anions:           Discharge is through oxidation ad is as follows.

 

 

  1. Concentration of electrolytes

A cation or anion whose concentration is higher is preferentially discharge if the ions are close in the electrochemical series.

 

Example:  dilute and concentrated NaCl

 

Product at the anode.

 

 

  1. The electrodes used: Products obtained at electrodes depend on the types of electrodes used

 

Examples:      in the electrolysis of CuSO4 (aq) using carbon and copper rods separately.

 

 

 

APPLICATIONS OF ELECTROLYSIS

 

  1. Extraction of reactive metals

 

Reactive metals/elements like sodium, magnesium, aluminum are extracted form their compounds by electrolysis.

 

Example:        Sodium is extracted from molten sodium chloride using carbon                             electrodes.

 

  1. Purification of metals

 

It can be used in refining impure metals

 

Examples:      Refining copper

 

  • The impure copper is made of the anode.
  • Their strips of copper are used as the cathode
  • Copper (II) Sulphate are used as the electrolyte.

 

  • During the electrolysis the anode dissolves and pure copper is deposited on the cathode.

 

  • The impurities (including valuable amounts of silver and gold) from the crude copper collect as a sludge become the anode.

 

  1. Electroplating

 

Is the process of coating one metal with another, using electrolysis so as to reduce corrosion or to improve its appearance.

 

During electrolysis:

  • the item to be electroplated is made the cathode
  • the metal to be used in electroplating is used as the anode
  • the electrolyte is made from a solution containing the ions of the metal to be sued in electroplating.

 

Examples

  • Gold-plated watches; silver – plated utensils
  • Steel utensils marked EPNS. I.e. Electroplated Nickel Silver.

 

  1. Anodizing Aluminum

 

Is the reinforcement of the oxide coating on Aluminum utensils/articles

Is done by electrolysis of dilute sulphuric acid using Aluminum articles as anode.

 

Importance:   Prevention arrosion of Aluminum articles

 

  1. Manufacture of sodium hydroxide, chlorine and hydrogen

 

  • Sodium hydroxide is prepared by the electrolysis of brine, for which 3 methods are available
  • The method depends o the type of electrolytic cell.
  • These cells are
  • The mercury cell
  • The diaphragm cell
  • The membrane cell

 

  1. The mercury cell

 

  • Components
  • The electrolyte is concentrated sodium chloride
  • The anodes are made of graphite or titanium, which are placed above the cathode.
  • The cathode consists of mercury, which flows along the bottom of the cell.

 

 

 

Chemical reactions

 

Anode

 

  • Both chloride and hydroxide ions are attracted.
  • Chloride ions are preferentially discharged due to their high concentration
  • The chloride ions undergo oxidation to form green – yellow chlorine gas.

 

Equation

 

 

 

At the cathode (flowing mercury)

  • Na+ (aq) migrate to the cathode
  • Sodium ions are preferentially discharged.
  • They undergo reduction to form sodium solid.

 

Equation

 

  • the discharged sodium atoms combine  with mercury to form sodium diagram

 

Equation

 

 

  • The sodium amalgam is then passed into another reactor containing water.
  • The amalgam reacts with water forming hydrogen and sodium hydroxide.

 

The mercury is regenerated and it is recycled into the main cell.

–           Main product:                        Sodium hydroxide

–           By products:                           Sodium and chlorine.

 

Advantages of mercury cathode cell

  • The resultant sodium hydroxide is very pure; as it has no contamination from sodium chloride.
  • It is highly concentrated; e. about 50%.

 

Disadvantages

  • Some of the mercury said its way into the environment leading to mercury pollution; a common case of brain damage in humans.
  • At the operating temperatures (700C – 800C), mercury vapours escape into the atmosphere and cause irritation and destruction of lungs tissues.
  • Its operation requires highly skilled man power.

 

  1. Diaphragm cell

 

Components

  • An asbestos diaphragm; to separate the electrolytic cell into two compartments; thus preventing mixing of H2 and Cl2 molecules
  • The anode compartment contains a graphite rod.
  • The cathode compartment contains a stainless steel cathode.

 

 

Diagram

 

 

Chemical reactions

  • The asbestos diaphragm is permeable only to ions, but not to the hydrogen or chlorine molecules.
  • It thus prevents H2 (g) and Cl (g) form mixing and reacting to yield HCl (g)
  • It also separates NaOH and Cl2 which would otherwise react.

 

Chemical reactions

 

At the anode

  • Chloride ions undergo oxidation to form chlorine gas.

 

Equation

 

–           At the cathode

  • H= and Na+ (a) migrate to the cathode compartment.
  • The H+ are preferentially discharged.
  • They (H+ (aq) undergo reduction to form hydrogen gas.

 

 

Equation

 

  • the discharge of H+ causes more water molecules to dissociate, thus increasing the concentration of OH- in the solution.
  • therefore, the Na+ and OH- ions also react in the cathode compartment to form sodium hydroxide.

 

Equation

 

Advantage

  • Does not result into pollution

 

Disadvantage

  • The resultant NaOh is dilute (12% NaOH)
  • It is also not pure due to contamination with NaCl – (12% NaOH + 15% NaCl by mass.

 

Note:  –           The concentration of the NaOH can be increased by evaporating excess water,              during which NaCl with a lower solubility crystallizes out first, leaving NaOH at                      a higher concentration.

–           The solid NaCl (Crystals) are then filtered off.

  • This is a case of fractional crystallization.

 

 

 

 

  1. The membrane cell
  • Is divided into 2 compartments by a membrane
  • Most commonly used type of Membrane is the cation – exchange membrane.
  • This membrane type allows only cations to pass through it.

 

Components

  • A cation exchange that divides the cell into 2 compartments; an anode and a cathode compartments.
  • Both electrodes are made of graphite
  • The electrolyte in the anode compartment is purified brine
  • The electrolyte in the cathode compartment is pure water.

 

Diagram

 

Chemical reactions

 

The anode

 

Chloride ions undergo oxidation to form green – yellow chlorine gas.

 

The cathode

 

As current passes through the cell H+ and Na+ pass across the membrane to the cathode

  • H+ are preferentially discharged.
  • They undergo reduction to liberate hydrogen gas.

 

  • Continuos discharge of H+ leaves the OH- at a higher concentration

 

  • The OH- react with Na+ to form sodium hydroxide

 

Advantages

 

  1. Resultant sodium hydroxide is very pure, since it has no contamination from NaCl.
  2. The sodium hydroxide has a relatively high concentration; at about 30 – 35% NaOH by mass.

 

 

Uses of sodium hydroxide, chlorine, and hydrogen

 

  1. Sodium Hydroxide

 

  • React with chlorine to form sodium chlorate sodium hypochlorite or I, NaOCl. This is a powerful oxidising agent which is used for sterilization and bleaching in textiles, paper and textile industries.

 

i.e. 2NaOH (aq)  +   Cl2 (g)                                NaOCl (aq)  +  NaCl (g)  + H2O (l)

 

  • Manufacture of sodas, detergents and cosmetics.
  • Neutralization of acidic solutions in the laboratories.

 

  1. Hydrogen

 

  • For hydrogenation in the manufacture of margarine
  • Manufacture of ammonia
  • Production of hydrochloric acid

 

  1. Chlorine
  • Formation of sodium chlorate I; for bleaching in pulp, textile and paper industries.
  • Sewage and water treatment
  • Manufacture of polymers such as polyvinyl chloride.
  • Manufacture of pesticides.

 

QUANTITATIVE ASPECTS OF ELECTROLYSIS

 

Basic terminologies and concepts

 

  • Ampere

Is the standard unit used to measure an electric  current; the flow of electrons

Is usually abbreviated as amps.

 

  1. Coulomb

Is the quantity of electricity, when a current of 1 ampere flows for one second. I.e.  1 Coulomb = 1 Ampere x 1 second

Generally:

Quantity of     Electricity       =          current x Time in seconds

A         =          It; Where

Q         =          Quantity of electricity in coulombs

I           =          Current in Amperes

T          =          time in seconds

 

  1. Faraday

Is the quantity of electricity produced by one mole of electrons; and is usually a constant equivalent to 96487 (approx. 96500) coulombs

 

Faradays laws of Electrolysis.

First law;

The mass of substance liberated during electrolysis is directly proportional to the quantity of electricity passed.

 

Worked examples

  1. A current of 2.0 Amperes was passed through dilute potassium sulphate solution for  two minutes    using  inert electrodes.
  2. Write the equation for the reaction at anode.

 

 

  1. Work out the mass of the product formed at the cathode.  (H = 1.0, Faraday = 96,000 C)

Solution;         quantity of electricity =          current x time

=          2 x 2x 60

=          240 coulombs

cathode reaction

 

1 mole of electrons     =          96000 C

4 moles of e-   =          4 x 96000 C

=          384,000 C

 

 

  1. What mass of copper would be coated on the cathode from a solution of copper (II) Sulphate  by a current of 1 amp flowing for 30 minutes.

(Cu = 63.5;      Faraday constant = 96487 Cuo/-)

 

solution

 

Cathode  reaction

 

Cu 2+ (aq)       +          2e-                               Cu (s)

 

–           1 mole of Cu requires 2 moles of electrons

  • Quantity of electricity passed; = 1 x 30 x 60 coulombs; = 1800C
  • 1 mole of electrons carriers a charge of 96487 coulombs = 192974 coulombs
    • coulombs deposit 63.5 g of Cu.

 

Thus 1800 C will deposit   63.5 x 1800         =          0.592 grams

192974

 

  1. An element x has relative atomic mass of 88g. when a current of 0.5 amperes was passed through a solution of x chloride for 32 minutes, 10 seconds;  44 g  of x was deposited at the cathode.  (1 faraday = 96500 c)

Calculate  the charge on the ion of x.

 

 

  1. In the electrolysis of dil CuSO4 solution, a steady current of 0.20 Amperes was passed for 20 minutes. (1 Faraday = 96, 500 C Mol-, Cu = 64)

Calculate

  1. The number of Coulombs of electricity used

 

  1. The mass of the substance formed at the cathode.

2 moles of electrons liberate 1 mole of Cu.  i.e.  Cu 2+ + 2e                                     Cu (s)

 

  1. An element p has a relative atomic mass of 44. When a current of 0.5 Amperes was possed through a fused chloride of p for 32 minutes and 10 seconds; 0.22g of p were deposit

 

 

 

 

 

 

UNIT 5: METALS: EXTRACTION PROPERTIES AND USES.

Introduction:

  1. Introduction
  2. Extraction methods
  3. Concentration of the ores
  4. Metal extraction
  • Sodium metal
  • Main ores
  • Extraction process
  • Properties of sodium
  • Uses of sodium

 

  • Aluminium metal
  • Main ores
  • Qualitative analysis
  • Extraction from bauxite.
  • Electrolysis of purified bauxite
  • Properties of aluminium
  • Uses of aluminium

 

  • Zinc metal
  • Main ores
  • Qualitative analysis
  • Extraction process
  • By oxidation
  • By electrolysis
  • Properties of zinc
  • Uses of zinc

 

  • Iron metal
  • Main ores
  • Qualitative analysis
  • Extraction from haematite
  • Properties of iron
  • Uses of iron

 

  • Copper metal
  • Main ores
  • Qualitative analysis
  • Extraction process from copper pyrites
  • Properties of copper
  • Uses of copper

 

  • Lead metal
  • Main ores
  • Extraction process
  • Properties of lead
  • Uses of lead

Introduction:

Only most unreactive metals occur naturally in their elementary form.

Examples: – Gold, Silver, Platinum.

– Other elements occur as ores i.e. metal – bearing rocks.

Examples:

– Oxides

– Sulphides

– Carbonates

– Chlorides.

 

Note:
An ore is a mineral deposit with reasonable composition of a desired metal.

 

Methods of Extraction

Depend on position of the metal in the reactivity series.

– Main methods are:

  1. Electrolytic Method:

– Used for metals high up in the reactivity series

E.g. – Sodium and Potassium

– Calcium and Magnesium

– Aluminium.

– These metals occur in very stable ores

 

  1. Reduction method:

– For less reactive metals.

E.g. Iron, Zinc, and Copper.

 

– Is achieved using;

(i) Carbon in form of coke.

(ii) Carbon (II) oxide

(iii) Hydrogen

– Oxidation is also used followed by reduction.

 

Preliminary steps before extraction.

– Minerals (mineral) are usually mined with several impurities which lower the concentration of the metal per given mass or volume.

– Thus the ore is first concentrated before the actual extraction.

– Concentration is possible due to difference in properties between the mineral compound and the earthy materials.

 

Methods of ore concentration.

  1. Physical methods.

(a). Optical sorting.

– Used to separate ore particles that have sufficiently different colours to be detected by the naked eye.

– It involves physical handpicking of the desired particles.

– Mainly used for minerals containing transition elements such as chromium.

 

 

 

(b). Hydraulic washing.

– Also called sink and float separation.

– Utilizes the difference in density between the minerals and the unwanted materials

– The ore is washed with streams of water.

– The denser ore particles will sink to the bottom of the washing container and can then be collected.

– Examples in ores of tin and lead.

 

(c). Magnetic separation.

– Is used when either the ore particles or the earthy materials (unwanted materials) are magnetic.

– A strong magnet is used to attract the magnetic components and leaving the non-magnetic materials behind.

– Examples: in ores like magnetite (Fe3O4) and chromite which are magnetic.

 

(d). Electrostatic separation.

– Used to separate particles which have different electric charges.

– The particles are subjected into an electric field.

– The oppositely charged particles follow different paths and can then be separated.

 

(e). Froth floatation

– Is mainly use for sulphide ores.

– Takes advantage of two facts.

  • Oil can wet the surfaces of ores.
  • Oil floats on water

The process:

– The ore is ground into a fine powder; to increase the surface area for upcoming reactions.

– It is then mixed with water and a suitable oil detergent e.g pine or eucalyptus;

– The mixture is then agitated by blowing compressed air through it;

– Small air bubbles attach to the oiled ore particles; which are thenn buoyed up and carried to the surface where they float.

– A froth rich in mineral is formed at the top while impurities sink at the bottom.

– The froth is skimmed off and dried.

– Froth floatation process is used for copper, lead and zinc metals;

 

Diagram: froth floatation apparatus.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Chemical concentration.

– Involves the use of chemical reactions to concentrate the ores.

Examples:

– Bauxite, the main aluminium ore is chemically concentrated by a process known as Bayer’s process.

– This takes advantage of the amphoteric nature of aluminium oxide, which can thus react with both acids and bases.

– Chemical concentration can also be done by leaching.

– This involves reacting the ore with a compound such as sodium cyanide;

– The cyanide ions form complex ions with the metal.

– The complex ions formed are water soluble, and can be separated by filtration, leaving the unwanted materials in the residue.

 

 

The metals

  1. Sodium

Main ores;

  1. Rock salt / sodium chloride; NaCl
  2. Chile saltpetre / sodium Nitrate; NaNO3

iii. Soda ash/sodium carbonate; Na2CO3.

 

Other ores include;

(i). Borax; Na2B4O7.10H2O

(ii). Sodium Sulphate, Na2SO4;

 

Extraction;

– Sodium is obtained by the electrolysis of fused sodium chloride in the electrolytic cell.

– Calcium chloride and calcium fluoride are added to the electrolyte.

Reasons;

– To lower the melting point of sodium chloride from 800oC to 600oC;

 

– Once molten, the electrical resistance within the cell is sufficient to maintain the temperature without external heating.

Steel or iron is used as the cathode, while carbon/graphite is used as the anode.

– Thus steel is not used as the anode.

Reason;

– At high temperatures, steel would react with chloride formed at the anode, but graphite is inert even at high temperatures.

 

– Steel wire gauze separates the electrodes.

Reason;

– To prevent products of electrolysis (sodium and chlorine) from mixing and reacting to form sodium chloride.

 

– The electrolytic apparatus used in sodium extraction is called the Downs cell.

 

Diagram: The Downs cell.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

– During electrolysis, fused sodium chloride dissociates according to the equation;

 

NaCl(l)                        Na+(l) +Cl(l)

 

At the cathode:

  • Observation;

– Soft silvery metal

 

Explanation

– Na+ ions are attracted and undergo reduction (accept electrons) to form/ produce molten sodium metal.

 

Equation;

Na+(l) + e                      Na(l)

 

– Molten sodium is lighter than fused sodium chloride and floats on the surface where it overflows into a separate container / sodium reservoir.

 

Note;

– The resultant sodium is usually collected in liquid / molten state, floating on top of the electrolyte.

Reasons;

– Less dense than molten sodium chloride

– Has a low melting point.

 

At the anode;

Observations;

– Evolution of a green-yellow gas.

 

Explanation:

– Chlorine gas is evolved as a by – product and collected separately.

– Negatively charged Cl ions migrate to the positive anode and undergo oxidation to form chlorine gas;

 

Equation:

2Cl(l)                    Cl2(g) + 2e-

 

 

Properties of Sodium;

– Is a soft silvery metal with low density; 0.979gcm-3

– Has a low melting point, 97oC, and a low boiling point of 883oC.

 

Chemical reactions

(a) With air

– Na is very reactive and tarnishes in moist air to form an oxide layer.

4Na(s)+ O2(g)                2Na2O(s);

 

– The oxide layer reacts with more air moisture to form hydroxide

Na2O(s) + CO2 (g)                  Na2CO3(s) + H2O(l)

 

Note

– Due to those series of reactions sodium is stored under a liquid hydrocarbon e.g. petroleum, kerosene.

 

– Sodium burns in oxygen with a golden yellow flame to form sodium peroxide

Equation:

2Na(s) + O2(g)                         Na2O2(s)

(White)

 

(b). With water

– Na reacts vigorously with water to form NaOH and Hydrogen.

Equation:

2Na(s) + 2H2O(l)                       2NaOH(aq) + H2(g)

 

– The resulting solution is highly alkaline with a PH of 14.

– Sodium is stored under oil to prevent contact with moisture from the atmosphere.

Note:

– The reaction between Na and dilute acids would be explosive and not safe to investigate

 

(c). With chlorine

– Sodium burns in chlorine gas;

2Na(s) + Cl2(g)                    2NaCl(s)

 

(d) With ammonia gas;

– Sodium forms hydrogen and a solid;

2Na(s) + NH3(g)                    2NaNH2(s) + H2(g)

            Sodamide

And;

NaNH2(s) + H2O(l)                    NaOH(aq) + NH3(g)

 

Uses of sodium;

  1. Is alloyed with lead in the preparation tetraethyl (IV) lead, which is added to petrol as an anti-knock.
  2. Provides the glow in sodium vapours lamps, for street lighting (orange-yellow street lights).
  3. Is an excellent conductor of heat and electricity with low melting point hence used;
  • In nuclear reactors to conduct away heat.
  • Modern aeroplane engines.
  1. Manufacture of sodium peroxide, and sodium cyanide used in the extraction of silver and gold.

 

Question;

– Although electrolysis is an expensive way of obtaining metals, it must be used for some metals. Explain.

Solution;

– Group 1 and 2 metals together with Al are themselves such powerful reducing agents that their oxides cannot be reduced by chemical reducing agents.

 

 

 

 

 

 

Worked example

  1. Below is a simplified diagram of the Downs cell in which sodium metal is manufactured.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(a) (i) Identify; –

Electrolyte X: – Molten sodium chloride

Gas Y: -Chlorine gas

 

(ii) Write an equation for the reaction at the cathode.

Na+(l) + e-                      Na(l).

 

(iii). In what state is sodium collected?

– Molten state/liquid state.

 

(iv). Give two properties of Na that makes it possible to be collected as in (b) (iii) above.

– Its less dense than molten sodium chloride.

– Has a low melting point.

 

(v). The cathode is made of steel but the anode is made of graphite.

Why is this yet steel is a better conductor?

– At high temperature steel would react with chlorine formed but graphite is inert even at high temperatures.

 

(vi). In this process, the naturally occurring, raw material is usually mixed with another compound. Identify the compound and state its use.

Compound; – Calcium chloride

Use;-To lower melting point of Nacl2 from 800oc to 600oc

 

(vii). What is the function of the steel gauze cylinder?

– Prevents sodium reacting with chlorine forming NaCl

 

(viii). Give one industrial use of sodium

– A coolant in nuclear reactors;

– Alloy with lead in tetraethyl (IV) lead;

 

(ix). Explain why sodium metal is stored under paraffin;

– Keep it out of air; reacts very fast with air forming a dull surface.

– Can react with water.

 

(b). State an industry that can be built next to a sodium extracting plant.

 

(c). A current of 100 Amperes flows through an electrolyte of molten sodium chloride for 15 hours. Calculate the mass of sodium produced in kg (Na = 23; 1F = 96500C)

Solution:

Q = It

=100 x 15 x 60 x 60

=5400000C.

Cathode equation

Na+(l) + e                     Na(l)

 

96500C = 23g of Na

5400000 C =23 x 5400000 =1287.04g

96500

=1.287kg

 

(d) .For the same quantity of electricity as in (c) above ; calculate the volume of the gaseous product produced in the cell at 150c and 800mmhg.(Molar gas volume at s.t.p = 22.4dm3)

 

 

  1. Aluminium

– Forms 7% of the earth’s crust and is the most common metal.

 

Main ores;

– Bauxite; Al2O3.H2O

– Mica; K2Al2S6016.

– China clay;Al2S2O72H2O

– Corundum;Al2O3.

 

Chemical test;

– Crush the ore into a fine powder;

– Add dilute nitric (V) acid to the powder

– Filter to obtain a solution of the ore;

– To a solution of the ore add NaOH(aq) dropwise till in excess, and then repeat the same procedure using Ammonia solution, NH4OH.

 

Observations:

  • With NaOH(aq):

– White precipitate in soluble in excess;

 

  • With NH4OH(aq):

– White precipitate insoluble in excess;

 

Extraction from Bauxite;

-Involves two main processes;-

  • Purification of Bauxite.
  • Electrolysis of purified bauxite (alumina)

 

  1. Purification of bauxite

– Chief impurities are small quantities of silica and iron (III) oxide.

– The oxide ore is ground and treated under pressure/ dissolved in hot aqueous sodium hydroxide.

 

During the process;

– The amphoteric bauxite dissolves in NaOH forming sodium aluminate;

Equation:

2NaOH(aq) + Al2O3.3H2O(s)                      2NaAl(OH)4(aq).

 

Ionically:

Al2O3(s) + 2OH(aq) + 3H2O(l)                           2[Al(OH)4](aq);

 

– Silica impurities also dissolve forming sodium silicate

Equation:

SIO2(s) + 2NaOH(aq)                       Na2SIO3(aq) + H2O(l)

 

– The iron impurities (mainly iron (III) oxide) DO NOT dissolve.

– This mixture is then filtered, during which iron (III) oxide remain as residue of red mud while a filterate of sodium aluminate and sodium silicate is collected.

– Carbon (IV) oxide is bubbled through the filterate, followed by dilution then addition of a little aluminium hydroxide to cause precipitation (seeding) of Aluminium hydroxide.

Ionically:

2[Al(OH)4](aq) + CO2(g)                         2Al(OH)3(s) + H2O(l);

 

Seeding

Al(OH)3

Alternatively:

Al(OH)4(aq)                               2Al(OH)3(s) + OH(aq);

 

General equation:

NaAlO2(aq) + 2H2O(l)    hydrolysis  NaOH(aq) + Al(OH)3(s).

 

– The precipitated Aluminium hydroxide is then filtered off, washed and ignited to give pure aluminium oxide (Alumina);

Equation:

2Al(OH)3(s)                        Al2O3(s) + 3H2O(l)

Alumina

 

  1. Electrolysis of purified bauxite (alumina)

The Alumina (Al2O3), has a high melting point, 2015oC and a lot of heat would be required to melt it.

– Additionally the molten compound is a very poor conductor of electricity.

– Consequently, cryolite (Na3AlF6) is mixed with the oxide.

Reason;

– To lower the melting temperature of Al2O3 from 2015oC to around 800oC;

– At this lower temperature the molten oxide also conducts well.

 

– The molten alumina mixed with bauxite is then electrolysed in a steel cell lined with carbon graphite as the cathode.

Note;

– Other than being an electrolyte the graphite cathode lining also prevents alloy formation, as it ensures no contact between the resultant aluminium and the steel tank;

-The anodes also made of Graphite dip into the steel tank at intervals.

Diagram: electrolytic steel cell for the extraction of Aluminium.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Electrolytic reactions;

– The Aluminium oxide dissociates to give constituent ions;

Equation:

Al2O3(l)                          2Al3+(l)  +3O2-(l)

At the cathode;

Observation;

– A silvery white metal which quickly becomes dulled.

 

Explanation:

– Aluminium ions move to the cathode and are reduced to form aluminium metal.

 

Equation;

2Al3+(l) + 6e-                2Al(l)

 

At the anode;

Observation;

– Effervescence of a colourless gas.

 

Explanations:

– Oxygen ions migrate to the anode and get oxidized to form oxygen gas.

– The resultant oxygen gas reacts further with the graphite anode to form carbon (IV) oxide.

– This is due to the high temperatures involved during the process.

Note;

– Consequently the carbon anode should be replaced from time to time.

 

Equations;

3O2-(l)                 3O2(g) + 6e-

Then;

C(s) (anode) + O2(g)                        CO2(g)

 

Note:

– Cryolite usually adds Na+; and F ions into the electrolyte.

– Thus the anions are O2- and F ions into the electrolyte.

– However oxygen is discharged in preference to Fluorine.

Reason;

– Fluorine is a stronger oxidizing agent than oxygen. Thus oxygen easily gives electrons than fluorine, hence discharge.

– Aluminium is discharged in preference to sodium.

 

 

 

 

 

 

 

 

 

 

Summary: – Flow chart on the Extraction of Aluminium from bauxite.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Properties of Aluminium

Physical properties

– Is a silvery white metal which quickly becomes dulled with a thin oxide layer.

– Has very low density (2.7gcm-3), with ability to be rolled into wires / foil.

– Is a good conductor of heat and electricity.

 

Chemical properties.

  1. Reaction with air;

– In air it acquires a continuous very thin coating of oxide, which resists further reaction.

– Removal of this protective cover renders the metal reactive.

– Consequently steel wool or wood ash should NOT be used in aluminium utensils.

– Usually, salty water attacks the oxide film allowing the aluminium to corrode and for this reason, ordinary aluminium is not used for marine purposes.

– Aluminium will burn in air at 800oC to form is oxide and nitrate.

Equations:

4Al(s) + 3O2(g)        2Al2O3(s);

2Al(s)­ + N2(g)                    2AlN(s)

 

  1. Reaction with Acids.

Note:

– The protective Aluminium oxide (being covalent and insoluble) layer makes its reactivity with acids less than expected.

 

  • With nitric (V) acid;

– Has hardly any effect on the metal, at any concentration.

Reason:

– Being a powerful oxidizing agent, it simply thickens the oxide layer thereby preventing further reaction.

 

  • With sulphuric (VI) acid;

– Only hot concentrated sulphuric (VI) acid breaks down the oxide layer and reacts with the metal.

Equation:

2Al(s) + 6H2SO4(l)                  Al2(SO4)3(aq) + 6H2O(l)  + 3SO2(g)

 

  • With Hydrochloric acid;

– Dilute HCl dissolves aluminium slowly; liberating hydrogen.

Equation:

2Al(s) + 6HCl(l)            2AlCl3(aq) + 3H2(g)

With concentrated HCl the rate of reaction is increased.

 

  1. Reaction with chlorine;

– Hot aluminium burns in chlorine gas with a white light, forming dense white fumes of Aluminium (III) Chloride.

-The white fumes cool and collect on the cooler parts of the apparatus as a white solid.

Equation:

2Al(s) + 3Cl2(g)                     2AlCl3(g)

 

Note:

– The apparatus for the preparation of AlCl3 is kept very dry.

Reason:

– Aluminium chloride is readily/easily hydrolysed by water/moisture, and so it fumes in damp air with the evolution of hydrogen chloride gas.

Equation:

AlCl3(s) + 3H2O(l)                     Al(OH)3(s)+3HCl(g)

 

  1. Reaction with water.

– Aluminium does not react with cold water, due to the formation of an insoluble coating of Aluminium oxide.

Note;

– If the oxide film is removed, the metal reacts slowly with cold water.

 

  1. Reaction with caustic soda.

– The metal, especially in powder form, reacts with caustic soda solution, liberating hydrogen and leaving sodium aluminate in solution.

– The reaction is exothermic and once started, it is very vigorous.

Equation:

2NaOH(aq) + 2Al(s) + 2H2O(s)                       2NaAlO2(aq) + 3H2(g)

 

Ionically:

2Al(s) + 2OH(aq) + 2H2O(l)                         2AlO-2(g)  + 3H2(g)

Note:

– Thus aluminium has an amphoteric nature as it reacts with both acids and alkalis.

Uses of aluminium

  1. Making parts of airplanes, railway, trucks, buses, tankers, furniture, and car e.t.c.

Reason;

– It is Very light due to a very low density.

 

  1. Making cooking vessels/ utensils such as sufurias.

Reason:

– It is a good conductor of heat and electricity.

– It is not easily corroded by cooking liquids due to the unreactive coating of aluminium oxide.

 

  1. Making overhead cables

Reason:

– It is a good conductor of electricity.

– It is light hence can easily be supported by poles and ductile to be rolled into wires (cables).

 

  1. Aluminium powder mixed with oil is used as a protective paint.

 

  1. Making Aluminium foils due to its high malleability. The foil is used in cooking; packaging and for milk bottle tops.

 

  1. Making alloys, which have high tensile strength and yet light.

Examples:

 

Alloy Component
Duralumin Aluminium, copper, manganese and magnesium
Magnalium Aluminium (70%) and magnesium (30%)

 

  1. As a reducing agent in the Thermite process in the production of some elements such as chromium, cobalt manganese and titanium.

Example:

Cr2O3(s) + 2Al(s)                       2Cr(s) + Al2O3(s)

 

Note:

  • The thermite process

Is a process of reducing oxides of metals which are ordinarily difficult to reduce using Aluminium powder.

 

Examples:

  • Iron (III) Oxide (Fe2O3)
  • Chromium (III) Oxide (Cr2O3)
  • Compound oxide of manganese (Mn3O4)

 

– The oxide and the Al powder are well mixed together, forming Thermite.

– The thermite is ignited using magnesium ribbon fuse, since the reaction will not start at low temperatures.

– The high heat of formation of Aluminium oxide , results into a vigorous exothermic reaction that leads to a molten metal.

 

Example: -In the reduction of chromium (III) Oxide.

Cr2O3(s) + 2Al(s)                         2Cr(l) + Al2O3(s) + Heat.

 

Sample question:

 

  1. Zinc

Main ores;

(i). Zinc blende; ZnS

(ii). Calamine; ZnCO3.

 

Extraction:

– Is done by electrolysis or reduction of its oxide using carbon.

 

Preliminary steps:

– The ore is first concentrated by froth floatation.

– The ore is roasted in air to convert it to the oxide.

Equations:

  • From Zinc blende:

2ZnS(s) + 3O2(g)                 2ZnO(s) + 2SO2(g)

 

  • From Calamine:

ZnCO3(s)                          ZnO(s) + CO2(g)

 

– After obtaining the oxide the metal is extracted by either reduction or reduction:

 

(a). The reduction method.

– The oxide is mixed with coke and limestone and heated in a furnace.

 

Diagram: Furnace for zinc extraction by reduction:

 

 

– The limestone (CaCO3) decomposes to liberate CO2 which is then reduced by coke to form carbon (II) oxide.

 

Equations:

Heat

 

 

CaCO3(s)                      CaO(s) + CO2 (g)

Then:

CO2 (g) + C(s)                 2CO (g)

 

– The resultant carbon (II) oxide and coke are the reducing agents in the furnace, at about 1400oC.

– They reduce the oxide to the metal; which is liberated in vapour form.

 

Equations:

  • ZnO(s) + C(s)                      Zn (g)  + CO (g)
  • ZnO(s) + CO(s)                       Zn (g)  + CO2 (g)

– At the furnace temperatures zinc exists in vapour form, and leaves at the top of the furnace.

– Liquid zinc being lighter settles above molten lead and is run off;

– The vapour is condensed in a spray of molten lead to prevent re-oxidation of zinc.

– The resultant zinc is 98-99% pure and can be further purified by distillation.

– SO2 is a by-product and is the main source of pollution in the extraction of zinc.

– Usually it is channeled to a contact process plant for the manufacture of sulphuric acid.

– Alternatively it can be scrubbed off to prevent pollution of the environment.

– Less volatile impurities remain in the furnace.

– The silica impurities combine with the quicklime/ calcium oxide (CaO) from limestone to form calcium silicate.

 

Equation:

CaO(s)  + SiO(s)                             CaSiO3(s)

 

– The silicates together with other less volatile impurities form slag, at the bottom of the furnace from where it is run off.

 

Summary: Flow chart and the extraction of zinc

 

                            Sulphur (IV) oxide                  Coke and limestone                       CO2 and excess CO(g)

ZnO(s)

 

 

 

 

ZnS (ore)

 

 

 

                                                                                                                                                             Zinc + impurities

Separation chamber
      Slag

 

 

                                     Air

 

 

 

                                                                                                                                                             Zinc liquid (pure)

 

 

 

 

 

(b). Electrolytic extraction of zinc.

– Zinc metal is obtained from the oxide via a series of steps:

 

Step I: Preparation of electrolyte:

– The ZnO obtained from roasting the ore is converted to zinc sulphate by reacting it with dilute sulphuric (VI) acid.

 

Equation:

ZnO(s) + H2SO4(aq)                           ZnSO4(aq) + H2O(l)

 

– Any lead (II) oxide impurity present in the zinc oxide reacts with the acid to form lead (II) sulphate.

Equation:

PbO(s) + H2SO4(aq)                          PbSO4(s) + H2O(l)

 

– The insoluble lead (II) sulphate is then precipitated and separated by filtration;

– The zinc sulphate is then dissolved in water and the solution electrolysed.

 

Step II: The electrolytic process:

Electrolyte:

– Zinc (II) sulphate solution;

 

Ions present:

– Zn2+ and H+ as cations; and SO42- and OH as anions;

 

Cathode:

Lead containing 1% silver.

 

Anode:

– Aluminium sheets;

 

Chemical reactions:

Cathode:

Observations:

– Deposits of a grey solid.

 

Explanations:

– Zn2+ and H+ migrate to the cathode.

– The Zn2+ are discharged in preference to H+;

Reason:

– The cathode is relatively reactive. Thus since zinc is more reactive thn hydrogen, its ions undergo reduction faster;

 

Equation:

Zn2+(aq) + 2e-                     Zn(s);

 

Note:

If graphite electrodes were used, hydrogen gas would have been evolved instead;

 

 

Anode:

Observations:

– Evolution of a colourless gas that relights a glowing splint;

 

Explanations:

– OH and SO42- migrate to the cathode.

– The OH are discharged in preference to SO42-; giving off oxygen gas

Reason:

The OH ions have a higher oxidation potential than SO42- and therefore easily giving electrons for reduction at the cathode

 

Equation:

4OH(aq)                             2H2O(l) + O2(g) + 4e-

 

Note:

– Over 80% of zinc is extracted by the electrolytic method.

– Zinc extracted by the electrolytic method is much more pure.

 

Note: – Industrial plants that can be set up near the zinc extraction plant.

– Contact process plant, to make use of the SO2 by-product.

– Lead accumulators factories, to utilize the zinc produced.

– Paper factory using, SO3 and hence SO2 in bleaching.

– Brass factory for alloying zinc and copper.

– Steel factory to use zinc in galvanization.

 

Properties of zinc

Physical properties

– Is a blue-grey lustrous metal with:

  • Density of 7.1gcm-3
  • Melting point of 420oC
  • Boiling point of 907oC

 

Chemical reactions

(a). with air;-

– Zinc tarnishes slowly forming a protective layer which prevents further reaction i.e. oxide layer or basic carbonate.

 

Equations:

2Zn(s) + O2(g)                           2ZnO(s)

Then;

ZnO(s) + CO2(g)                    ZnCO3(s)

 

– It burns with a blue-green flame when strongly heated in air to form an oxide which is yellow when hot and white when cold.

Equation:

2Zn(s) + O2(g)                          2ZnO(s)

 

 

 

(b). With water

– Zinc does not react with water

– Steam reacts with red-hot zinc, forming zinc oxide and liberating hydrogen gas.

Equation:

Zn(s) + H2O(g)                          ZnO(s) + H2(g)

 

(c). with dilute acids

– Zinc is above hydrogen in the reactivity series hence displaces hydrogen from steam (water) and dilute acids like H2SO4 and HCl.

Equation:

Zn(s) + 2H+(aq)                            Zn2+(aq) + H2(g)

 

– Pure zinc reacts slowly while impure zinc reacts faster/ more quickly.

– Copper (II) sulphate is used as a catalyst to speed up the reaction.

 

(d). Concentrated acids.

(i). Concentrated sulphuric (VI) acid.

Equation:

Zn(s) + 2H2SO4(l)                        ZnSO4(aq) + 2H2O(l) + SO2(g)

 

(ii). 50% concentrated nitric (v) acid.

– It reacts with 50% concentrated nitric (V) acid to liberate nitrogen (II) oxide.

Equation:

3Zn(s) + 8HNO3(l)                        Zn(NO3)2(aq) + 2H2O(l) + 2NO(g)

 

(iii). Concentrated nitric (V) acid.

– It reduces concentrated nitric (V) acid to nitrogen (IV) oxide.

Equation:

Zn(s) + 4HNO3(l)                        Zn(NO3)2(aq) + 4H2O(l) + 2NO2(g)

 

(e). with alkalis.

– Zinc is amphoteric and dissolves in hot alkalis to give the zincate ion and hydrogen gas.

Equation:

Zn(s) + 2OH(aq) + 2H2O(l)                          H2(g) + [Zn(OH)4]2-(aq)

 (Zincate ion)

(f). Other reactions.

(i). Zinc burns in chlorine to give zinc chloride

Zn(s) + Cl2(g)                       ZnCl2(s)

(ii). Zinc combines with sulphur

Zn(s) + S(s)                      2ZnS(s)

 

 

 

 

 

 

Uses of zinc:

  1. Galvanization of iron sheets to prevent corrosion and rusting.

Note:

Rusting does not occur even when galvanized iron sheets are scratched and exposed.

Reason:

– The rest of the zinc protects the iron from rusting. This is because zinc being more reactive gets oxidized in preference to iron, and is hence “sacrificed” in the protection of iron.

– This method is referred to as cathodic or sacrificial protection.

  1. Making brass; an alloy of copper and zinc.
  2. Making outer casings of dry batteries;
  3. Die-castings contain zinc and aluminium, and are used for making radio and car parts;
  4. Zinc cyanide is used for refining silver and gold;

 

Sample question:

 

  1. Iron

– Is the second most abundant metal after aluminium, forming about 7% of the earth’s crust.

 

Main ores

– Haematite, Fe2O3;

– Magnetite, Fe3O4;

– Siderite, FeCO3;

 

Qualitative analysis for presence copper in an ore sample.

– Crush the ore into fine powder;

– Add dilute nitric (V) acid to the ore, to dissolve the oxide filter to obtain the filtrate.

– To the filtrate add aqueous sodium hydroxide / ammonium hydroxide dropwise till in excess.

– Formation of a red brown / brown precipitate in both cases indicates presence of Fe3+

 

Extraction from haematite (Fe2O3).

Summary of the process

The ore-haematite is crushed and mixed with coke and limestone.

– The mixture is called charge.

– The charge is loaded into the top of a tall furnace called blast furnace.

– Hot air-the blast is pumped into the lower part of the furnace.

– The ore is reduced to iron as the charge falls through the furnace.

– A waste material called slag is formed at the same time.

– The slag floats on the surface of the liquid iron produced.

– Each layer can be tapped off separately.

 

Details of the extraction process.

(i) The blast furnace

Is a tall, somewhat conical furnace usually made of silica and lined on the inside with firebrick.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Details of the extraction process

Raw materials.

– Iron ore; i.e. haematite

– Coke; C

– Limestone; CaCO3;

– Hot air;

 

Conditions

– Temperature at the bottom of furnace, 1400-1600oC

– Temperature at the top of the furnace, 400oC

 

Reactions and processes

Step-1 – crushing and loading

– The ore is crushed into powder form, to increase the surface area for the upcoming reduction/ redox reactions.

– It is then mixed with coke and limestone and then fed at the top of the furnace using the double bell (double-cone devise) changing system

Note:

– The double bell charging system ensures that the furnace can be fed continuously from the top with very little heat loss, by preventing any escape of hot gases.

– This in turn reduces production costs.

 

Step 2: -Pre heating of the blast furnace.

– Air that has been preheated to about 700oC is blown/ fed into the base of the blast furnace through small pipes called tuyers.

– This provides the required temperatures for the reactions in the blast furnace.

– This results into highest temperatures, about 1600oC at the hearth (bottom of the furnace) which then decreases upwards the furnace.

 

Step 3: -Generation of reducing agents.

– Two reducing agents are used in this process: Coke and carbon (II) oxide; with carbon (II) oxide being the main reducing agent.

 

(i). Oxidation of coke;

– Coke burns in the blast at the bottom of the furnace.

– The reaction temperatures is about 1600oC and the product is Carbon (IV) oxide gas

– This reaction is exothermic, producing a lot of heat in the blast furnace.

 

Equation:

C(s) + O2(g)                CO2(g)

 

(ii). Decomposition of limestone;

– The limestone in the charge decomposes in the blast furnace to calcium oxide (Quicklime) and carbon dioxide.

 

Heat

Equation:

CaCO3(s)                           CaO(s) + CO2(g)

 

 

– The calcium oxide will be used in the removal of the main ore impurity/ silicates/ silica in the form of silicon (IV) oxide.

– The CO2 then moves up the blast furnace to regenerate carbon (II) oxide, the chief reducing agent.

 

(iii). Production of carbon monoxide

– The CO2 from oxidation of coke and decomposition of limestone (calcium carbonate) react with (excess) coke, to form carbon (II) oxide

– The reaction occurs higher up in the blast furnace at about 700oC;

 

Equation:

CO2(g) + C(s)               2CO(g)

 

Step 4: The actual reduction process

– Reduction of the ore is by either CO or coke, depending on temperatures.

 

(i). Reduction by coke

– This occurs much lower down the furnace at higher temperatures of about 800oC and above.

– This reaction is ordinarily slow and thus serves to only reduce the part of the ore reduced by CO at lower temperatures in the upper parts of the furnace.

 

Equation:

2Fe2O3(s) + 3C(s)                   4Fe(s) + CO2(g)

 

Note:

– The resultant CO2 is quickly reduced to CO by the white-hot coke to more carbon (II) oxide as per step 3(iii) above.

 

(ii). Reduction by carbon (II) oxide

– This is the main reducing agent.

– The reaction between CO and Fe2O3 is relatively faster and occurs at lower temperatures of 500oC-700oC, higher up the furnace.

 

Equation:

Fe2O3(s) + 3CO (g)                           2Fe(s) + CO2 (g)

 

– The resultant carbon (IV) oxide is also quickly recycled by being reduced to CO by coke to from more reducing agent

 

(iii). Melting

– The iron produced in both of the reduction processes is in solid state.

– As the iron drops / falls down the furnace, it melts as it passes through the melting zone/ molten zone (1500oC-1800oC)

– The molten iron runs to the bottom of the furnace.

– Temperatures at the hearth (bottom of the furnace) is maintained at approx. 1400oC and yet pure iron melts at about 1525oC.

– Consequently the molten iron would easily solidify at the base (Temp =1400oC)

– However this is not usually the case;

Reason:

-Impurities absorbed by iron during melting (mainly carbon) reducing the melting point to below 1400oC.

– The molten iron is then easily tapped off.

 

Step 5: -Removal of earthy impurities.

– The earthy impurities in the ore (mainly silica) react with calcium oxide from decomposition of limestone to form calcium silicate.

 

Equation:

CaO(s) + SiO2(s)                     CaSiO3(s)

 

– These earthy impurities form molten slag whose main component is calcium silicate.

– The slag does not mix with iron but rather floats on top of it, at the base of the furnace.

 

Importance of the slag

– As it floats on top of molten iron it protects it from being re-oxidized by the incoming hot air.

 

Uses/application of the slag

  1. Light-weight building material.
  2. Manufacture of cement.
  3. Road building material.

 

Step 6:- Removal of furnace (waste) gases.

– Hot unreacted/waste gases leave at the top of the furnace.

– Main components include Nitrogen, unreacted CO2, unreacted CO, oxygen and Argon (Noble gases)

– Additionally they contain dust particles.

Note:

– Upon removal of dust particles, the furnace gases, being hot can be used to pre-heat the air blown in at the base.

 

Summary: flow chart for the extraction of iron.

 

Properties of iron:

Physical properties:

– It has a melting point of 420oC and a boiling point of 907oC;

– Have a good thermal and electrical conductivity;

– It is ductile and malleable;

 

Chemical properties.

(i). Reaction with air.

– It readily rusts in presence of moist air hydrated brown iron (III) oxide; Fe2O3.H2O(s)

Equation:

4Fe(s) + 2H2O (l) + 3O2 (g)                          2Fe2O3.H2O (l)

 

– When heated it reacts with oxygen to form tri-iron tetroxide; Fe3O4;

Equation:

 

3Fe(s) + 2O2 (g)                             Fe3O4(s)

 

(ii). Reaction with water.

– It does not readily react with cold water.

– It however reacts with steam liberating hydrogen gas and forming tri-iron tetroxide.

Equation:

3Fe(s) + 4H2O(g)                          Fe3O4(s) + 4H2(g)

 

(iii). Reaction with chlorine.

– Hot iron glows in chlorine without further heating, forming black crystals of iron (III) chloride;

– Iron (III) chloride sublimes on heating and will thus collect on the cooler parts of the apparatus;

Equation:

2Fe(s) + 3Cl2(g)                           2FeCl3(s)

 

Note:
– Iron (III) chloride fumes when it is exposed to damp (moist) air;

Reason:

– It is readily hydrolysed by water with evolution of hydrogen chloride gas;

 

Equation:
FeCl3(s) + 3H2O(l)                       Fe(OH)3(s) + 3HCl(g)

 

(iv). Reaction with acids:

  • Hydrochloric acid:

– Iron reacts with hydrochloric acid to liberate hydrogen gas.

Equation:

2Fe(s) + HCl(aq)                         FeCl2(aq) + H2(g)

 

  • Sulphuric (VI) acid:

Fe(s) + H2SO4 (aq)                        FeSO4 (aq) + H2 (g)

 

Note: With hot concentrated H2SO4;

– The iron reduces hot concentrated H2SO4 to sulphur (IV) oxide and it is itself oxidized to iron (III) sulphate.

Equation:

2Fe(s) + 6H2SO4 (l)                              Fe2 (SO4)3(aq) + 6H2O (l) + 3SO2 (g)

 

  • Nitric (V) acid.

– Iron reacts with dilute nitric (V) acid to form nitrogen (IV) oxide and ammonia which then forms ammonium nitrate.

Equation:

10HNO3 (aq) + 4Fe(s)                          4Fe(NO3)2(aq) + NH4NO3(aq) + 3H2O(l)

 

– Warm dilute nitric (V) acid gives iron (II) nitrate.

– Concentrated nitric (V) cid renders the iron unreactive.

Reason:

– Formation of iron oxide as a protective layer on the metal surface.

 

(vi). Reaction with sulphur.

– Iron when heated in sulphur forms iron (II) sulphide.

Equation:

Fe(s) + S(s)                      FeS(s)

 

Uses of iron.

– Iron exists in different types and alloys, depending on percentage composition of iron, and other elements.

– Each type of alloy of iron has different uses depending on properties.

 

Iron alloy or type Properties. Uses
Cast iron – Refers to iron just after it has been produced in the blast furnace;

– Contains 3-5% carbon, 1% silicon, and 2% phosphorus;

Disadvantage: very brittle hence easily breaks;

Advantage: It is extremely very hard;

– Making:

Ø  Furnaces;

Ø  Grates;

Ø  Railings;

Ø  Drainage pipes;

Ø  Engine blocks;

Ø  Iron boxes;

Note: This is due to its very hard nature;

– Manufacture of wrought iron and steel;

Wrought iron – Refers to cast iron with 0.1% carbon.

– It is malleable hence can easily be moulded or welded;

– making iron nails; horse shoes; agricultural implements like pangas;

Note:

Its use is declining due to increased use of mild steel

Steel – Are alloys whose main component is iron;

– Other components may be carbon; vanadium; manganese; tungsten; nickel and chromium;

Examples:

Mild steel: has about 0.3% carbon, 99.75% iron;

Special steel: has a small percentage of carbon together with other small substances;

 

Stainless steel:

-Contains 74% iron, 18% chromium, and 8% nickel;

 

 

 

Cobalt steel:

– Contains about 97.5% iron and 2.5% cobalt;

– Very tough and hard;

– Slightly magnetic;

– Mild steel is used for making:

Ø  Nails; Car bodies;

Ø  Railway lines; Ship bodies;

Ø  Rods for reinforced concrete, pipes;

Note: Advantage of mild steel:

– It is easy to work on;

 

 

– That with 10-12% chromium and some nickel is used to make: cutlery; sinks; vats;

– Steel containing 5-18% tungsten is used for: making high speed cutting and drilling tools;

 

– For making electromagnets;

  1. Copper

Description: – A red brown metal.

Distribution: – Canada, USA, Zambia, and Tanzania.

 

Main ores;

– Copper pyrites, CuFeS2;

– Cuprite, CuO

– Chalcocite, Cu2S

– Malachite, CuCO3.Cu (OH)2

 

Qualitative analysis / test for presence in an ore sample.

– Crush the ore and then add dilute nitric or hydrochloric or sulphuric acid to dissolve the ore.

– Filter to obtain Cu2+ filtrate.

– Divide filtrate into 2 different test tubes.

– To one sample add aqueous Ammonium hydroxide dropwise till in excess formation of a pale blue precipitate soluble in excess NaOH to form a deep blue solution.

 

Equations

With little NaOH:-

Cu2+(aq) + 2OH(aq)                        Cu(OH)2(s)

Pale blue ppt.

In excess:

Cu(OH)2(aq) + 4NH3(aq)                 [Cu(NH3)4]2+(aq)  + 2OH(aq)

Deep blue solution

 

-To the second portion add sodium hydroxide solution dropwise till in excess, formation of a pale blue precipitate insoluble in excess confirms presence of Cu2+.

 

Extraction- from copper pyrites.

  1. Crushing the ore

– The ore is crushed to increase the surface area for the succeeding chemical reactions.

– The ore is then concentrated.

 

  1. Concentration of the ore.

– The ore is concentrated by froth floatation.

– The fine ore powder is mixed with water and oil, after which air is blown into the mixture, usually from below.

– Bubbles of the air forms froth, resulting to concentration of the ore.

– The lighter oil floats on top of the water, with the ore floating on top of the oil.

– The denser water sediments the earthy impurities like soil particles.

– The concentrated ore is then tapped off.

– This process involves formation of an oil froth onto which the ore floats hence the name froth formation.

 

  1. First Roasting

-The concentrated copper pyrite, CuFeS2 is then roasted in air to remove some of the sulphur impurities as sulphur (IV) oxide.

Equation:

2CuFeS2(s) + 4O2 (g)                 3SO2 (g) + 2FeO(S) + Cu2S(s);

Note:

– During 1st roasting limestone and silica (SiO2) are added to the roasted ore and the mixture heated in the absence of air.

 

Importance

– Removal of iron impurities.

– The iron (II) so formed during roasting is converted to iron (II) silicate, FeSiO3.

– The iron (II) silicate constitutes the major portion/component of the slag.

 

Equation:

FeO(s) + SiO2(g)                 FeSiO3(s)

 

– The slag separates itself from the copper (I) sulphide.

– The sulphur (IV) oxide escapes into the atmosphere and is the major pollutant in this process.

 

Pollution control mechanisms.

– Scrubbing the gas using calcium hydroxide;

 

Equation:

SO2(g) + Ca (OH)2(s)                CaSO3(s) + H2O(l)

 

– Construction of a contact process nearby.

 

  1. Second Roasting.

– The CuS is heated in a regulated supply of air where some of it is converted to Cu2O

 

Equation:

2Cu2S(s) + 3O2(g)                      Cu2O(s) + 2SO2(g)

 

  1. Reduction of copper (II) oxide

Note: – Not all the Cu2S was oxidized to copper (I) oxide Cu2O.

– The unreacted (unoxidized) Cu2S serves as the reducing agent in this step.

i.e. The copper (I) oxide formed in step 4 is reduced to copper metal by the (unreacted) copper (I) sulphide.

– This is called blister copper.

 

Equation:

Cu2S(s) + 2Cu2O(s)                6Cu(s) + SO2 (g);

 

  1. Electrolysis

-The copper metal from reduction in step 6 is impure and is thus purified by electrolysis.

 

Main impurities

– Traces of gold

– Traces of silver

– Iron

– Sulphur

 

 

 

Electrolytic apparatus

Anode: Impure copper

Cathode: Pure copper plates/ sheets;

Electrolyte: Dilute copper (II) sulphate solution (containing Cu2+; H+; SO42- and OH)

 

Diagram of electrolytic apparatus.

 

 

 

 

 

 

 

 

 

 

 

 

Electrolytic reactions;

At the cathode.

Observations:

– Deposition of a brown solid.

 

Explanations:

– The copper (II) ions, Cu2+ move to the cathode, where they accept electrons and undergo reduction.

– Cations in the electrolyte are Cu2+ and H+ but Cu2+ are preferentially discharged due to their easy tendency to undergo reduction.

 

Equation:

Cu2+(aq) + 2e-                     Cu(s);

 

At the anode

Observations:

– Dissolution of the anode, hence the impure copper rod decreases in size.

 

Explanation

– Since the metal rod is dipped into a solution of its ions, the copper solid undergoes oxidation, losing electrons to form copper ions, Cu2+

– Consequently as more copper ions, Cu2+ get reduced at the cathode; more are released by the dissolving anode.

 

Equation:

Cu(s)                          Cu2+(aq) + 2e-

 

Overall reaction

Cu(s) + Cu2+(aq)                     Cu2+(aq) + Cu(s)

 

– The electrolytic product is 99.98% copper.

– Traces of silver and gold collect as sludge at the bottom of the cell.

Note:-To improve purity of the product of electrolysis, the following steps are advisable;

(i). Increase the dilution of the electrolyte/ use a very dilute electrolyte.

(ii). Reduce the amount of current / use a low current.

 

Summary of extraction of copper from copper pyrites.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Uses of copper

  1. Making copper wires and contacts in switches, plugs and sockets

Reason:-Copper is a good conductor of electricity

Note:-For this purpose, pure copper is necessary, since impurities increase electrical resistance.

 

  1. Making soldering instruments.

Reason:-Copper has a high thermal conductivity.

 

  1. Manufacture of alloys.

Examples

Alloy Components
Brass Copper and zinc
Bronze Copper and tin
German silver Copper, zinc and nickel;

 

  1. Making coins and ornaments.

Reason:-it is durable and aesthetic.

 

 

Properties of copper.

  1. Physical properties

– Soft red-brown metal.

– Melting point of 1083oC and a boiling point of 2595oC

– Density is 8.92gcm-3 and electrical conductivity is about 5.93 x 10-9 Ώ-1m-1

 

  1. Chemical properties

(i). It does not react with cold water or steam.

 

(ii). Heating in air

– When heated in air it forms a black layer of copper (II) oxide on its surface.

– Finely divided copper burns with a blue flame.

 

Equation:

2Cu(s) + O2(g)                     2CuO(s)

(Red brown)                                           (Black)

 

(iii). Reaction with chlorine

– Cu reacts with chlorine in presence of heat to form green copper (II) chloride.

 

Equation:

Cu(s) + Cl2(g)             CuCl2(s)

 

(iv). Reaction with Acids.

– Copper does not react with dilute hydrochloric, nitric and sulphuric acids.

– However it reacts with 50% Nitric acid, concentrated Nitric acid and concentrated sulphuric acid.

 

  • With 50% Nitric (V) acid

– Copper reduces the nitric acid to nitrogen monoxide.

 

Equation;

3Cu(s) + 8HNO3(l)                    3Cu(NO3)2 (aq) + 4H2O(l) + 2NO(g);

 

  • With concentrated Nitric (V) acid

– Copper reduces the acid to brown nitrogen (IV) oxide/Nitrogen dioxide gas.

 

Equation:

Cu(s) + 4HNO3(l)               Cu (NO3)2(aq) + 2H2O(l) + 2NO2(g)

 

  • With concentrated sulphuric acid.

-The sulphuric acid is reduced to sulphur (IV) oxide.

 

Equation:

Cu(s) + 2H2SO4(aq)                 CuSO4 (aq) + 2H2O(l) + SO2 (g)

 

 

Worked example.

The flow chart below outlines some of the processes involved in the extraction of copper from copper pyrites. Study it and answer the questions that follow.

 

Hot air                                       Air

(a) (i). Write an equation for the reaction in the first roasting chamber.                                  (1mark)

 

(ii). Name gas K.                                                                                                                          (1mark)

 

 

(iii). Give the name and formula of slag M.                                                                                  (1mark)

 

(iv). Give the name of the reaction in chamber N.                                                                        (1mark)

 

(v). Name the impure copper X.                                                                                                     (1mark)

 

(b). Pure copper is obtained from impure copper by electrolysis.

(i). Name the anode the cathode and the electrolyte.                                                                              (3 marks)

 

(ii). Write equations for the reactions at the anode and cathode.                                        (2 marks)

 

(iii). calculate the time taken for a current of 10 amperes to deposit 32kg of pure copper. (Cu = 64, 1F = 96000C)                                                                                                                                (3 marks)

(c). Draw a diagram to show how you would plate an aluminium spoon with copper

 

6. Lead.

– Is a transition element that combines with other elements to form compounds with 2 oxidation states.

– It is among the group 4 elements;

 

Main ores:

– Galena, PbS (lead sulphide); the main ore;

– Cerrusite, PbCO3 (lead carbonate)

– Anglesite, PbSO4 (lead II) sulphate);

Qualitative analysis / test for presence of Zn2+ in an ore sample.

– The ore is crushed and then dilute nitric or hydrochloric or sulphuric acid added to dissolve the ore.

– It is then filtered to obtain Zn2+ filtrate.

– The filtrate is divided into 2 different test tubes.

– To the first portion sodium hydroxide solution is added dropwise till in excess, formation of a white precipitate soluble in excess confirms presence of either Zn2+; Al3+ or Pb2+.

– To the second sample aqueous ammonium hydroxide is added dropwise till in excess; formation of a white precipitate soluble in excess NH4OH(aq) confirms presence of Zn2+ only;

 

Equations

With little NH4OH:-

Zn2+(aq) + 2OH(aq)                        Zn(OH)2(s)

White ppt.

In excess:

Zn(OH)2(aq) + 4NH3(aq)                 [Zn(NH3)4]2+(aq)  + 2OH(aq)

Colourless solution

Extraction of lead:

– Occur in three main steps:

Ø  Ore concentration;

Ø  Extraction by reduction;

Ø  Purification (refining) by electrolysis;

1. Ore concentration:

– Is done by selective froth floatation;

– The ore is ground into a fine powder, then water and a suitable oil added;

– Air is then blown into the mixture; facilitating formation of a low density froth that floats on top;

– Additionally, chemicals such as sodium cyanide and zinc sulphate are added to facilitate separation of zinc sulphide present in the ore.

– The separated PbS is then dried and broken into smaller pieces, then subjected to reduction;

2. Reduction:

Step I: Roasting the ore:

– The crushed and concentrated ore is roasted in a furnace to convert it to lead (II) oxide;

Equation:

2PbS(s) + 3O2 (g)                       2PbO(s) + 2SO2 (g)

– During roasting some of the lead (II) sulphide is converted to lead (II) sulphate;

Equation:

PbS(s) + 2O2(g)                          PbSO4(s);

– Any lead sulphate formed is converted to lead silicate by silicon (IV) oxide;

– The fate of lead (II) silicate;

Note:

– Additionally the lead (II) sulphate may further react with lead sulphide to form lead metal;

Step II: Ore reduction:

– The lead oxide obtained is mixed with coke, limestone and silica and some scrap iron;

– The mixture is fed into the top of the Imperial smelting furnace (ISF); where it is melted using hot air blasts introduced near the bottom of the furnace;

 

Diagram: The Imperial Smelting furnace for Lead extraction.

Main reactions:

(i). The lead (II) oxide is reduced to lead by the coke.

Equation:
PbO(s) + C(s)                                Pb(s) + CO(s);

 

(ii). The resultant carbon (IV) oxide produced in reaction (i) above further reduces any remaining lead (II) oxide;

 

Equation:
PbO(s) + CO(s)                                Pb(s) + CO2(g);

 

(iii). The scrap iron is added so as to react with any lead sulphide that may be present.

Equation:

Fe(s) + PbS(s)                       Pb(s) + FeS(s)

 

 

 

(iv). The limestone undergoes decomposition to give calcium oxide and liberate carbon (IV) oxide;

Equation:

CaCO3(s)                       CaO(s) + CO2 (g);

 

– The carbon (IV) oxide gets reduced by coke to form more carbon (IV) oxide for reduction as in reaction (ii);

 

Equation:
CO2 (g) + C(s)                    2CO (g);

 

(v). The calcium oxide reacts with silica in form of SiO2 to form calcium silicate;

 

Equation:

CaO(s) + SiO2 (g)                       CaSiO3 (l);

 

Waste gases and residues.

– The iron sulphide and calcium silicate form a molten slag which is less dense and floats on top of molten lead at the bottom of the furnace;

– From here the slag is separately tapped off;

– Excess gases and air that did not react in the blast furnace escape through outlets at the top of the furnace;

– These waste gases can be trapped and recycled;

– These gases include: excess CO; excess CO2; oxygen; nitrogen; some SO2; and argon;

 

Pollution effects:
– Main pollutant is sulphur (IV) oxide from roasting of the ore.

 

Pollution control:
– It is directly fed into a contact process plant or scrubbed using calcium hydroxide forming calcium sulphite;

 

  1. Purification (refining) of lead:
    – The molten lead obtained in this process contains impurities such as gold, silver, copper, arsenic, tin and sulphur;

– The impure lead is refined by electrolysis.

 

Electrolysis of molten lead.

(i). Electrolyte:

Any aqueous solution containing lead ions;

 

(ii). The anode:

– Impure lead;

 

(iii). The cathode:

– Pure lead;

 

 

 

 

Electrolytic reactions;

At the cathode.

Observations:

– Deposition of a grey solid.

 

Explanations:

– The lead (II) ions, Pb2+ move to the cathode, where they accept electrons and undergo reduction.

– Cations in the electrolyte are Pb2+ and H+ but Pb2+ are preferentially discharged due to their easy tendency to undergo reduction.

 

Equation:

Pb2+(aq) + 2e-                     Pb(s);

 

At the anode

Observations:

– Dissolution of the anode, hence the impure lead rod decreases in size.

 

Explanation

– Since the metal rod is dipped into a solution of its ions, the impure lead solid undergoes oxidation, losing electrons to form lead (II) ions, Pb2+

– Consequently as more lead (II) ions, Pb2+ get reduced at the cathode; more are released by the dissolving anode.

 

Equation:

Pb(s)                          Pb2+(aq) + 2e-

 

Overall reaction

                       Oxidation at anode;

 

 

Pb(s) + Pb2+(aq)                     Pb2+(aq) + Pb(s)

 

 

Reduction at cathode;

 

Summary: flow chart on extraction of lead.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Properties of lead:

Physical properties;

– Has a low melting point but a high density;

– The unusually low melting point of lead is difficult to explain using simple metallic bonding theory;

– It is rather soft and pliable;

– Relatively malleable;

 

Chemical properties.

Note: – Lead is fairly unreactive to most other metals;

 

  1. Uses of lead:
  1. It is used in several alloys e.g. solder and also added to bronze alloys to make them stronger;
  2. Lead ingots are used in the manufacture of accumulators;
  3. Being so malleable and so chemically inert lead sheeting was used for roofing (look at roofs of old churches and cathedral)-cost and pollution effects have however brought this to a stop;
  4. Making lead pipes for water supply; this is also discouraged particularly in soft water areas due to threat of lead poisoning;
  5. Making tetraethyl lead (IV) which for many years was used as a fuel additive to increase octane rating of fuels;
  6. Used in weights, clock pendulums, plumb bobs etc; due to its high density;
  7. It absorbs X-rays and hence lead aprons and lead glass are used to shield hospital radiographers;
  8. Used for safe disposal or storage of radioactive substances since no radioactive emission has been known to pass through thick lead blocks;

 

Sample question:

 

 

 

 

UNIT 6: RADIOACTIVITY

Checklist

  1. Meaning of radioactivity
  2. Natural and artificial radioactivity
  3. Nuclear equations and chemical equations.
  4. Types of radiations
  • Alpha particles
  • Beta particles
  • Gamma rays
  1. Half life of a radioisotope
  2. Radioactive decay curves and half life
  3. Radioactive disintegration and nuclear equations.
  4. Radioactive decay series, nuclear fission and nuclear fusion;
  5. Uses of radioactivity;
  6. Dangers and pollution effects of radioactivity;

 

 

Definitions:

  • Radioactivity is an automatic spontaneous disintegration of nuclei of some heavy elements emitting some kinds of radiant energy
  • Elements which exhibit this are said to be radioactive and emit different types of radiation

Types of radiation from radioactive elements

They can be identified by

  1. Measuring their penetration power
  • They were directed to thin leaves of paper, aluminium, lead plates of varying thickness.
  • Observations:
  1. One penetrated only thin foils of paper but not aluminium of 0.05mm thick and lead
  2. One penetrated both paper and Al but not lead
  3. One penetrated paper Al and lead
  4. Directing them to air
  • Observations
  • Penetrated less through air but caused a lot of ionization to the air molecule
  • Penetrated more through air but caused less ionization
  • Penetrated most through air but caused less ionization

These penetrated just as far as X-rays

Directing through magnetic /electric fields

 

 

 

 

 

 

                                                                                                                    S                                                                                           N

–           +

 

 

 

 

 

 

The radioactive element radiation emits all the 3types of radiation when they were directed to poss through electric magnetic fields to the fields

Were slightly deflected towards the negative
Were strongly deflected towards the positive
  1. Were not deflected

Therefore

A and B –are charged particles

-Have opposite charges

A is positive, B is negative

And

Deflections obeyed through L.H Rule

  • A were deflected less because they were moving with higher momentum (MV) because of high mass

Experimental evidence

  • Experiments carried out by Rattlerford revealed
  1. Were positively charged and had a mass of 4 units and thus a charge of +2 these are Alpha- particles therefore double charged helium ions,
  2. Were negatively charged and had similar properties to cathode rays. Measurement of charge/ mass confirmed they were electrons and carries a unit of –1

They are beta particles

  1. Were carrying no charge. They are electromagnetic waves similar to light rays and X- rays with a wave length of only 10 metres

Their emission enables a nucleus to lose surplus energy

They are gamma rays

Evidence for the nature of gamma rays

  • Gamma rays
  • Are unaffected by an electric field
  • Are an affected by a magnetic field
  • Can penetrate several centimeters of lead
  • Can be diffracted by the lattice of a crystal
  • Have no change in atomic number mass of the atoms emitting them

 

NATURE AND PROPERTY OF α , β ,AND γ

 

Nature α- particles helium nuclei

4He (He )

Β-particles electrons Rays electromagnetic waves radiation
Relative penetrating power Least 5cm in air stopped by paper and Al foils Several metres of air thin Al foil (100) Penetrate air, Al and many mm of Pb (10000)
Range in air

 

 

 

A few cm A few m A few km
Effect of electric and magnetic fields Small deflection Large deflection No deflection
Ionization of gases Cause much ionization

Discharges electroscopes rapidly

Cause less ionization

Discharges electroscopes slowly

Negligible ionization

 

Radioactive decay curve

Is always a symptotic to the x-axis

 

 

 

 

 

Number of atoms of the

Radioactive element

 

 

 

 

– The number of atoms disintegrating per unit of time /second is always proportional to the number of atoms, N at that time

This number N decreases slowly / exponentially with time

  • At any time to the number of atoms of a radioactive element is N/ No and at time T½, only ½ the total number of atoms of the original radioactive element will be present i.e.

T½ = ½N

This time is referred to as the half-life period of a radioactive element

Therefore

Half-life period T½ of a radioactive element is defined as the time taken for ½ the atoms to disintergrate

Thus

In T½ the radioactivity of the element diminishes to half its value

Example

Radioactive decay curve of radon

With half life of 4 days

 

No = 6 X 103 atoms

T½ = 4days

 

No. Of atoms

103 x 6

 

 

103 x 5

 

 

103 x 4

 

 

103 x 3

 

 

 

 

 

 

 

Radon emits α- particles

Initial no of radon atoms = 103 X 6

After 4 days the number of atoms present=103 X 3

“ “ “ “8 days “ “ “ “ “ “ “ “ “ “ “ “ “ “ “ “ =103 X  1.5

“ “ “ “12 days “” “ “ “ “ “ “ “ “ “ “ “  “ “ = 103 X 0.75

It is impossible to predict the atoms that will disintergrate next

 

The half lives of some radioactive isotopes

Radioactive isotopes Half life
Uranium-238 4.5 X 109  years
Radium-226 1.6 X 103 years
Carbon-14 5.7 X 103 years
Strontium-90 28 years
Iodine-131 8.1days
Radon-222 4days
Bismuth-214 19.7minutes
Polonium-218 3minutes
Polonium-214 1.5 X 10 -4sec

-The half-life of a radioactive isotope provides aquantitave measure of its stability

Thus

-The shorter the half-life the faster the isotope decays and the more unstable it is

-The longer the half-life the slower the decay process and the more stable the isotope

RADIOACTVE DISINTERGRATION EQUATIONS

-Different radioactive elements disintergrate to emit different radiations

  1. Emission of α particles

-Occurs in isotopes with an atomic number greater than 83 (Z> 83) because they are unstable since their nuclei are so heavy and their atomic mass is too large

-They attain stability by ejecting an alpha particle containing two protons and two neutrons

Examples

  • 238U-α partially-

 

(b) Emission of β particles

  • Occur in isotopes with more neutrons than stable isotopes of the same element, therefore heaviest isotopes of an element  are likely to emit β particles to attain stability
  • During the β-decay process, a neutron splits up forming a proton &an electron

 

 

The proton remains in the nucleus while the electron /β particles is ejected.

Result

Number of neutrons in the isotope decreases by one while number of protons increase by one.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

© Emissions of α-& β alternatively

 

 

 

 

 

 

 

 

(2) Emissions of beta particles in two –stage

 

Application of Radioactivity /Using radioactive isotopes.

  1. Carbon dating
  • Radio active Carbon –14 nuclide produced when Nitrogen –14 is bombarded with radiocum from the sum is used.
  • When they die carbon-14 resent starts to reduce as decay takes place by emission of β particles.
  • Using the decay curve of carbon –14 it is possible to estimate age of the animals/plants since the carbon –14 is present in their tissues.

 

2.Medication/treatment of cancer

  • Gamma rays are used to treat /kill cancer cells when the tumour is subjected to the radiations

E.G Gamma rays /penetrating from CO are used in treating inaccessible growths.

Superficial /skin cancers can be treated by less penetrating radiation from or in plastic sheets strapped on the affected

 

3.Studying metabolic pathways

-Radioactive Isotopes can be used to trace the uptake of metabolism of various elements by animals /plants.

E.g Uptake of phosphate & metabolism of phosphorus by plants can be studied using a fertilizer containing

Radioactive tracer studies using have helped in the exudation of photosynthesis and protein synthesis

I have been used in the diagnosis & treatment of thyroid diseases &in research into thyroid gland functioning.

 

4.Thickness gauge and empty packet detectors

-Radiation passing through a material decreases as the material gets thicker

Hence:

Amount of penetrating Beta –or-gamma- radiation can be used to estimate the thickness of various materials like paper, metal or plastic

-Radiation thickness gauges can be used to control the thickness of sheet steel emerging from a high-speed rolling mill.

-β –Rays measure thickness upto ~ 0.2cm of steel γ-rays can be used with steel upto 10cm thick.

-Level gauges are used to measure amount of liquid in fire extinguisher &gas cylinders

-Empty –packet detectors can be set to reject empty/insufficiently filled packets filled packets of biscuits/cigarettes.  \

5.decting pipe bursts

  • Can be underground pipes carrying water /oil
  • If the water / oil is mixed with radioactive substances form the mixture will leak at the point where there is a burst and the radiations can be detected if a detector is passed
  1. Effect on static electricity
  • In textile industry the presence of static charges can attract dust and cause fires
  • When a radioactive element is placed in such industries the radiations emitted will ionize air and ions formed will attract the static charges : this minimises problems due to static charges

Hazards of radioactive

  • Arise from
  1. Exposure of the body to external radiation
  2. Ingestion/ inhalation of the radioactive matter
  • They damage body cells/ tissues
  • Cause mutation/deformities

Precautions

  • Protect the body with lad/ concrete shielding
  • Never pick/ hold radioactive elements with bare hands ; use forceps and well protected tongs
  • Use radiation absorbers

SAMPLE QUESTIONS

  • Particles from a radioactive source move through 7cm in air at ordinary pressure
  • The radioactive emission of radium are α-, β- and γ. Draw labelled diagram to show how the rays can be separated
  • the table below shows nuclides which are radioactive products of   Their ½ lives and K.E during decay are shown
nucleide Half life energy
Th 1.39 X 1010 3.98
Th 1.9yrs 5.42
Ra 3.64days 5.66
Rn 54.5 sec 6.28
Po 0.16 sec 6.77
At 3 X 10-4 sec 7.64
Bi 60.5 min X
Po 2.9 X 10-7 sec 8.78

 

 

  1. Identify pairs of isotopes of same element
  2. Identify 2 nuclides by in this table, which have been produced directly by α- decay of other nuclides in the table. Use equations
  • Identify 2 nuclides in this table, which have been produced by β- decay of other nuclides in the table. Use equations
  1. Suggest a series of decays for formation of
  2. Deduce how ½ life varies with energy of emitted rays

Suggest the value of X for

Grade 2 Schemes For Term 2 (11 Weeks)

In need of Grade 2 Term 2 CBC Schemes Of Work for the revised school calendar? Well. Get all the schemes here for term 2 having eleven weeks. Remember, these resources are available at no cost.

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GRADE TWO CBC SCHEMES FOR TERM 2 (11 WEEKS).

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FUNCTIONAL COMPOSITION REVISION TOOL KIT FOR FORM FOUR

FUNCTIONAL COMPOSITION REVISION TOOL KIT FOR FORM FOUR 

NAME……………… ADM. NO: …………………..CLASS…………….

Instructions

1. Using this handout together with class notes and available text books, make distilled notes on each of the compositions. Learning points in your distilled notes should be the format, basic content and l anguage. Sketch of each composition can be drawn as part of your mind maps.

2. In a different book, write each of the composition during the session you will be at home for any school break before end of July.

3. Ensure each composition is discussed and/or marked before closing day in August.

4. Ensure the right use of language, be neat, fluent and use simple language with apt tone.

Composition (year tested) Sample Question for guide in interpreting content Major learning point expectations used in awarding of scores during marking
1 2 Shopping list Imagine that your sibling is to join form one next month. You have to help in the planning for her shopping. The shopping to be made is to have: toilet paper, utensils, soap, bedding, stationery and sanitary material. Write the shopping list. Format and content  to have:
·    Title to have the owner’s name, the phrase ‘shopping list’, the event for which the shopping is being done and the date for shopping
DAU’S SHOPPING LIST FOR JOINING FORM ONE ON 17TH JULY 2020
·    Body- in table form with items under categories as hinted at in the question.
·    The content to have a table that is fully filled having the following column headers:
a) serial number- number of categories, b) Category- list categories with the most important category numbered first, c) items- column for list of items starting with the most important item in each category, d) quantity- column to quantify each item using apt units of measurement, e)description- column for brief description of each item (of opinion, size, shape, age, colour, originality, material, purpose), f) price- column showing the costing of each item in Kenyan Shillings or dollars and
g) Shop- column for the shop you expect to get the packed
No Category Item Quantity Description Price Shop
1 Clothing Shirt 2 pairs White @Ksh. 450 Wahu’s shop
Trouser 3 pairs Navy blue @Ksh. 700 Kilombelo’s shop
·    Language is brief and in note form.
2 1 Packing list. Imagine you are one of the members of prefect’s council. You are to go for a students’ retreat at Soya Palace. Your teacher has asked you to pack some items for your use during the trip. Write the packing list that you will take to your teacher for approval before packing. Format and content to have:
·    Title to have the owner’s name, the phrase ‘packing list’, the event for which the packing is being done and the date for writing
·    Body- in table form with categories as per the question
·    The content to have a table with columns with the following headers: a) serial number- number of categories, b) Category- list categories with the most important category numbered first, c) items- column for list of items starting with the most important item in each category, d) quantity- column to quantify each item using apt units of measurement, e) description- column for brief description of each item being packed. Table is same to shopping list but without price and shop.
·    The language is in note firm and in table format
3 1 Inventory Imagine you came back to Tumtum Academy from home last week. You brought stationary, clothing, bedding and sanitary material. Write the inventory you had kept by the end of week the first week. Your items should not be more than fifteen. Format and content to have:
·    Title to have owner’s name, phrase ‘inventory’ and the date of last modification.
TUMTUM ACADEMY.
LANDO’S SCHOOL INVENTORY AS ON 17TH JUNE 2020.
·    Body- to have basic rubrics in the question included under categories
·    Serial number- column with number for each category
·    Category- column with list of the categories
·    Date in- column for the date the item was entered
·    Item- column for names of items
·    Quantity- column for the quantity of each item
·    Description- column for the description (opinion/ condition, size, shape, age colour, originality, material, purpose)
·    Price/Value- the price in Ksh or the sourcing method such as donation or gift
·    Date out- Column for the date the item was taken out by self or others
·    Quantity out- column for the quantity taken
·    balance- Column for what remained after quantity out
·    Sign/name- column for the sign or name of the person (self or others) who took
No Date in Category Item Quantity in Price Date out Quantity out Balance Sign
The language is that of note form in table format.

CBC Free Grade Four, Five and Six Topically Arranged Revision Resources

Free Grade 4, 5 and 6 Free CBC Topical Revision Questions and Answers

CBC AGRIC & NUT {G4-6) TOPICAL REVISION (2).pdf
CBC CAS {G4-6} TOPICAL TOPICAL REVISION (2).pdf
CBC CRE {G4-6} TOPICAL TOPICAL REVISION (2).pdf
CBC ENG {G4-6} TOPICAL TOPICAL REVISION (2).pdf
CBC IRE {G4-6} TOPICAL TOPICAL REVISION (2).pdf
CBC KIS {G4-6} TOPICAL TOPICAL REVISION (2).pdf
CBC MAT {G4-6} TOPICAL TOPICAL REVISION (2).pdf
CBC SCI & TECH {G4-6} TOPICAL TOPICAL REVISION (2).pdf
Copy of CBC AGRIC & NUT {G4-6) TOPICAL REVISION (2).pdf
Copy of CBC CAS {G4-6} TOPICAL TOPICAL REVISION (2).pdf
Copy of CBC CRE {G4-6} TOPICAL TOPICAL REVISION (2).pdf
Copy of CBC ENG {G4-6} TOPICAL TOPICAL REVISION (2).pdf
Copy of CBC IRE {G4-6} TOPICAL TOPICAL REVISION (2).pdf
Copy of CBC KIS {G4-6} TOPICAL TOPICAL REVISION (2).pdf

STANDARD 8 KISWAHILI SCHEMES OF WORK TERM 1-3

AZIMIO LA KAZI

DARASA LA NANE 

MUHULA WA I

               ASILIA

  1. KISWAHILI MFUTI
  2. Mwongozo wa Mwalimu
  3. Oxford
  4. Kamusi
JUMA KIPINDI FUNZO MADA MALENGO SHUGHULI ZA MWALIMU SHUGHULI ZA MWANA FUNZI NYENZO ASILIA MAONI
1 MATAYARISHONA KUFUGUA SHULE  
2 1 Kusikiliza na kuzungumza/kongea Maamkizi Kufikia mwisho wa kipindi mwanafunzi aweze, kutumia baadhi ya msamiati wa maakizi, k.m. hujambo :sijambo. Habari nzuri. -kusoma

-kusikiliza

-kutambua

-kuzungumza

-kusikiliza

-kutambua

-kuzungumza

-Picha kiabuni

-Kamusi

KS MFUTI

UK 2

MWM UK1

 
2 kusoma

Ufahamu

Chada chema Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  ufahamu chada chema -kueleza

-kutaja

-kuandika

-kuiga

-kueleza

-kutaja

-kuandika

-kuiga

-kadi

-picha

michoro

KS MFUTI

UK 3

MWM UK3

 
3 Maombo ya lugha Shairi ‘saiti kwenda wema’ Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kutamka na kukariri shairi  ‘saiti kwenda wema,’ kusoma

-kusikiliza

-kutambua

-kuzungumza

-kusoma

-kuandika

-kutamka

-kutumia

-kadi

-picha

michoro

KS MFUTI

UK 4

MWM UK4

 
4 Sarufi Viambishi ngeli Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kulikariri shairi ‘umuhimu wa kutenda wema’ -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi KS MFUTI

UK5

MWM UK4

 
5 Kuandika (insha) Heshima Kufikia mwisho wa kipindi mwanafunzi aweze kuandika na kutumia  maneno ya heshima ifaayo katika insha -kueleza

-kutaja

-kuandika

-kuiga

-kueleza

-kutaja

-kuandika

-kuiga

-Picha kiabuni

-Kamusi

KS MFUTI

UK5-7

MWM UK5

 
                   
3 1 Kuongea/kusikiliza Akisami Kufikia mwisho wa kipindi mwanafunzi aweze kubainisha akisami pammoja na maelozo yake, kuandika akisami kwa usahihi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

-kadi

-picha

michoro

KS MFUTI

UK7

MWM UK7

 
2 Ufahamu Majaaliwa Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kujibu na kuandika maswali yaufahamu kusoma

-kusikiliza

-kutambua

-kuzungumza

kusoma

-kusikiliza

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK8

MWM UK8

 
3 Mapambo ya lugha Misemo na methali Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kutambua na kutumia misemo na methali katika sentensi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi KS MFUTI

UK9

MWM UK9

 
4 Sarufi Vivumishi visivyochukua viambishi ngeli Kufikia mwisho wa kipindi mwanafunzi aweze kusoma, kubainisha na kutumia vivumishi katika sentensi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-Picha kiabuni

-Kamusi

KS MFUTI

UK11

MWM UK10

 
5 Kuandika (insha) Siku ambayo sitaishahau Kufikia mwisho wa kipindi mwanafunzi aweze kumbuni, na kuandika insha akitumia maneno aliyopewa kwa usahihi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kutumia

-kueleza

-kuuliza

-kadi

-picha

michoro

KS MFUTI

UK12

MWM UK10

 
4 1 Kusikiliza na kuzungumza tarakimu Kufikia mwisho wa kipindi mwanafunzi aweze kubainisha tarakimu  milion kumi hadi milioni mia moja na kuandika tarakimu kwa usahihi kusoma

-kusikiliza

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK 12

MWM UK11

 
2 Ufahamu Sayari Kufikia mwisho wa kipindi mwanafunzi aweze, kusoma kutaja na kutumia msamiati wa sayari kwa usahihi   kusoma

-kusikiliza

-kutambua

-kuzungumza

-Picha kiabuni

-Kamusi

KS MFUTI

UK20

MWM UK12

 
3 Mapabo ya lugha Vitate Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,  kutambua na kutumia vitate katika sentensi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

  -kadi

-picha

michoro

KS MFUTI

UK 14

MWM UK14

 
4 Sarufi Vihisishi Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kueleza, kubainisha na kutumia vihisishi katika sentensi kwa usahihi   -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK22

MWM UK26

 
5 Kuandika  (insha) Sayari Kufikia mwisho wa kipindi mwanafunzi aweze kumbuni, na kuandika isha akitumia maneno aliyopewa kwa usahihi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kutumia

-kueleza

-kuuliza

-kadi

-picha

michoro

KS MFUTI

UK 23

MWM UK18

 
5 1 Kusikiliza na kuzungumza Dira Kufikia mwisho wa kipindi mwanafunzi aweze  kutambua , kuchora na kutumia msamiati wa dira kwa usahihi -kusoma

-kutumia

-kueleza

-kuuliza

  -Picha kiabuni

-Kamusi

KS MFUTI

UK 25

MWM UK20

 
2 Ufahamu Nyota njema Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kubainisha na kujibu maswali ya ufahamu kwa usahihi   -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK 26

MWM UK22

 
3 Mapambo ya lugha Silabi- cha Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,  kutambua na kutumia methali katika sentensi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi KS MFUTI

UK28

MWM UK23

 
4 Sarufi Vihisishi Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kueleza, kubainisha na kutumia vihisishi katika sentensi kwa usahihi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

  -Picha kiabuni

-Kamusi

KS MFUTI

UK29

MWM UK24

 
5 Kuandika (insha) Barua ya kirafiki Kufikia mwisho wa kipindi mwanafunzi aweze kuandika insha kwa kujaza mapengo ukitumia maneno uliyopewa -kusikiliza

-kutamka

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK30

MWM UK

 
6 1 Kusikiliza na

kuzungumza

 

Mali ya sili Kufikia mwisho wa kipindi mwanafunzi aweze kueleza, kuorodhesha na kufafanua  faida za maliasilina wajibu wake kuzihusu -kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK 12

MWM UK25

 
2 Ufahamu Maliasili Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,  kutambua  na kujibu maswali yaufahamu -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusikiliza

-kutamka

-kutambua

-kuzungumza

-Picha kiabuni

-Kamusi

KS MFUTI

UK38

 
3 Mapambo ya lugha Methali zinazopingana kimaana Kufikia mwisho wa kipindi mwanafunzi aweze kusoma, kutumia na kuandika methali zinazo pingana kimaana -kusoma

-kutumia

-kueleza

-kuuliza

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK39

MWM UK

 
4 Sarufi Viunganishi

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kueleza, kubainisha na kutumia viunganishi katika sentensi kwa usahihi -kusoma

-kusikiliza

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK 40

MWM UK28

 
5 Kuandika (insha) Maliasili Kufikia mwisho wa kipindi mwanafunzi aweze  kubunikuandika insha kwa kujaza mapengo ukitumia maneno uliyopewa -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

-Picha kiabuni

-Kamusi

KS MFUTI

UK 13

MWM UK

 
7 MTIHANI WA KATI WA MUHULA  
8 1 Kusikiliza na kuzungumza Majina ya wizara mbalimbali

 

Kufikia mwisho wa kipindi mwanafunzi aweze kutaja na na kueleza shughuli za wizara mbalimbali -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK 40

MWM UK29

 
2 Ufahamu Lau ningekuwa waziri ya elimu

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,  kutambua na kujibu  maswali ya ufahamu kwa usahihi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi KS MFUTI

UK41

MWM UK31

 
3 Mapambo ya lugha Visawe

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma, kutumia na kuandika visawe huku akieleza maana -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

KS MFUTI

UK  44

MWM UK33

 
4 Sarufi Viunganishi

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma, kutumia na kuandika viunganishi vya chaguo na vya nyongaza kwa usahihi -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-Picha kiabuni KS MFUTI

UK  45

MWM UK34

 
5 Kuandika (insha) Lau ninekuwa waziri wa elimu

 

Kufikia mwisho wa kipindi mwanafunzi aweze kubuni na kuandika kwa hati zinazosomeka kulingana na kichwa alichopewa -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunz

-Kamusi

-Kamusi i

KS MFUTI

UK  46

MWM UK35

 
9 1 Kusikiliza na kuzungumza Ngonjera Kufikia mwisho wa kipindi mwanafunzi aweze kueleza maana ya ngojera, kukariri na kufafanua ujube -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kutambua

-kuunganisha

Kitabu cha wanafunzi KS MFUTI

UK 47

MWM UK37

 
2 Ufahamu Utenzi

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kueleza na kujibu  maswali ya ufahamu kwa usahihi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

-Picha kiabuni KS MFUTI

UK  5 6

MWM UK39

 
3 Mapambo ya lugha Nimino za makundi

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kueleza na kutumia  nomino za makundi katika sentensi kwa usahihi -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK  57

MWM UK40

 
4 Sarufi Viunganishi linganishi

 

Kufikia mwisho wa kipindi mwanafunzi aweze kueleza kubainisha na kutumia viunganishi linganishi kwa usahihi -kusoma

-kusikiliza

-kutambua

-kuzungumza

-kusoma

-kutumia

-kueleza

-kuuliza

-Picha kiabuni

-Kamusi

KS MFUTI

UK  17

MWM UK

 
5 Kuandika (insha) Utungaji mashairi Kufikia mwisho wa kipindi mwanafunzi aweze kuakifisha, kutunga shairi kuzingatia arudhi kulingana na kichwa na beti alizopewa -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

KS MFUTI

UK  58

MWM UK40

 
10 1 Kusikiliza na kuzungumza Msamiati wa mahakamani Kufikia mwisho wa kipindi mwanafunzi aweze kusoma nakutumia baadhi ya msamiati wa maneno wa mahakamani -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK  49

MWM UK

 
2 Ufahamu Usipoziba ufa

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kueleza na kujibu  maswali ya ufahamu kwa usahihi -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

-Kamusi

KS MFUTI

UK  19

MWM UK42

 
3 Mapambo ya lugha Methali Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kutambua na kutumia  methali na maana zake katika sentensi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-Picha kiabuni KS MFUTI

UK  59

MWM UK43

 
4 Sarufi Manano ya kutiria mkazo(takriri)

 

Kufikia mwisho wa kipindi mwanafunzi aweze kueleza maana yatakiririna kutumia maneno ya kutilia mkazo katika sentensi kwa usahihi -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

Kitabu cha wanafunzi KS MFUTI

UK  61

MWM UK45

 
5 Kuandika (insha) kumbumbu Kufikia mwisho wa kipindi mwanafunzi aweze kubainisha na kuandika insha ya kumbumbu -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK  62

MWM UK45

 
11 1 Kusikiliza na kuzungumza Mekoni

 

Kufikia mwisho wa kipindi mwanafunzi aweze kufasiri maneno ya picha, kujadili nakufafanua baadhi ya msamiati wa mekoni -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-Picha kiabuni KS MFUTI

UK  63

MWM UK46

 
2 Ufahamu Wageni mekoni

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kuandika nakujibu maswali ya ufahamu -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi

-Kamusi

KS MFUTI

UK  64

MWM UK47

 
3 Mapambo ya lugha Methali

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kutambua na kutumia  methali na maana zake katika sentensi -kusikiliza

-kutumia

-kueleza

-kuuliza

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

KS MFUTI

UK  76

MWM UK51

 
4 Sarufi Matumizi ya ‘si-’

 

Kufikia mwisho wa kipindi mwanafunzi aweze kutumia na kueleza matumizi sahihi ya ‘si –‘ -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK  77

MWM UK52

 
5 Kusikiliza na kuzungumza Vipindi vya redio na runinga

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusikiliza redio narununga nakujibu maswali yatayo ulizwa -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusikiliza

-kutumia

-kueleza

-kuuliza

Redio

runinga

-Kamusi

KS MFUTI

UK  78

MWM UK 54

 
12 1 Kuandika insha Jinsi ya kupika Kufikia mwisho wa kipindi mwanafunzi aweze kubuni na kuandika insha shahihiya maelezo jinsi ya kupika -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

KS MFUTI

UK  79

MWM UK55

 
2 Ufahamu Barua rasmi

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma, kueleza na kuandika barua rasmi kwa hati zinazosomeka na nadhifu -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK  79

MWM UK56

 
3 Mapambo ya lugha vitawe Kufikia mwisho wa kipindi mwanafunzi aweze kusoma, kutumia na kuandika vitawe huku akieleza maana -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

KS MFUTI

UK  78

MWM UK52

 
4 Sarufi Matumizi ya, ‘amba’ Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,  kutambua na kuondosha amba’ na kutumia ‘O’ rejeshi mahali pake katika sentensi kwa usahihi -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-Picha kiabuni

-Kamusi

KS MFUTI

UK 80

MWM UK58

 
5 Kusandika insha Barua rasmi Kufikia mwisho wa kipindi mwanafunzi aweze  kueleza na kuandika barua rasmi kwa hati zinazosomeka na nadhifu -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

KS MFUTI

UK 83

MWM UK60

 
13 MARUNDIO  
14 MTIHANI WA MWISHO WA MUHULA  

 

 

 

 

 

 

 

AZIMIO LA KAZI

DARASA LA NANE 2016

MUHULA WA II

 

 

 

 

               ASILIA

  1. KISWAHILI MFUTI
  2. Mwongozo wa Mwalimu
  3. Oxford
  4. Kamusi

 

 

 

 

 

JUMA KIPINDI FUNZO MADA MALENGO SHUGHULI ZA MWALIMU SHUGHULI ZA  MWANA FUNZI NYENZO ASILIA MAONI
1 KUFUNGUA SHULE NA MATAYARISHO  
2 1 Kusikiliza na kuzungumza/kongea Msamiati wa teknolojia Kufikia mwisho wa kipindi mwanafunzi aweze, kutumia baadhi ya msamiati wa teknolojia kwa usahihi -kusoma

-kusikiliza

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kutambua

-kuzungumza

-Picha kiabuni

-Kamusi

KS MFUTI

UK 88

MWM UK61

 
2 kusoma

Ufahamu

Uchungu wa mwana Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  ufahamu uchungu wa mwana na kujibu maswali ya ufahamu kwa usahihi -kueleza

-kutaja

-kuandika

-kuiga

-kueleza

-kutaja

-kuandika

-kuiga

-kadi

-picha

michoro

KS MFUTI

UK 89

MWM UK62

 
3 Maombo ya lugha Misemo Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kutamka na kutumia misemo kwa usahihi -kusoma

-kuandika

-kutamka

-kutumia

-kusoma

-kuandika

-kutamka

-kutumia

-kadi

-picha

michoro

-Kamusi

KS MFUTI

UK 89

MWM UK63

 
4 Sarufi Matumizi ya ‘na’ Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kubainisha na kutumia ‘na’ katika sentensi kwa namna mbalimbali na kwa usahihi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi KS MFUTI

UK90

MWM UK62

 
5 Kuandika (insha) Teknolojia Kufikia mwisho wa kipindi mwanafunzi aweze kuandika, kujadili  na kutumia  msamiati wa teknolojia kwa usahihi -kueleza

-kutaja

-kuandika

-kuiga

-kueleza

-kutaja

-kuandika

-kuiga

-Picha kiabuni

-Kamusi

KS MFUTI

UK91

MWM UK63

 
                   
3 1 Kuongea na kusikiliza Msamiati wa ukoo Kufikia mwisho wa kipindi mwanafunzi aweze kubainisha baadhi ya misamiatri ya ukoo. Na kuitumia kwa usahihi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

-kadi

-picha

michoro

KS MFUTI

UK91

MWM UK64

 
2 Ufahamu Sijafisha Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kujibu na kuandika maswali yaufahamu kwa usahihi kusoma

-kusikiliza

-kutambua

-kuzungumza

kusoma

-kusikiliza

-kutambua

-kuzungumza

-kadi

-picha

michoro

-Kamusi

KS MFUTI

UK92

MWM UK65

 
3 Mapambo ya lugha  methali Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kutambua na kutumia metheli zenye mizizi ‘ndi’ katika sentensi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi KS MFUTI

UK93

MWM UK66

 
4 Sarufi Matumizi ya ‘ndi’ Kufikia mwisho wa kipindi mwanafunzi aweze kusoma, kubainisha na kueleza matumizi ya ‘ndi’ katika ngeli zote -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-Picha kiabuni KS MFUTI

UK95

MWM UK67

 
5 Kuandika (insha) Insha ya maelezo Kufikia mwisho wa kipindi mwanafunzi aweze kumbuni, na kuandika insha akitumia maneno aliyopewa kwa usahihi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

-kadi

-picha

michoro

KS MFUTI

UK96

MWM UK68

 
4 1 Kusikiliza na kuzungumza Viwanda Kufikia mwisho wa kipindi mwanafunzi aweze kubainisha kutaja na kueleza aina za bidhaa/ vifaa vinavyo tengezwa katika viwanda -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK 96

MWM UK68

 
2 Ufahamu Kazi ya mkono haitupi Kufikia mwisho wa kipindi mwanafunzi aweze, kusoma kujadili na kujibu maswali ya ufahamu kwa usahihi kusoma

-kusikiliza

-kutambua

-kuzungumza

kusoma

-kusikiliza

-kutambua

-kuzungumza

-Picha kiabuni

-Kamusi

KS MFUTI

UK 104

MWM UK70

 
3 Mapabo ya lugha Methali zinazo fanana ki mawazo Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,  kutambua na kutumia methali zinazo fanana kimawazo     -kadi

-picha

michoro

KS MFUTI

UK104

MWM UK71

 
4 Sarufi Vielezi Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kueleza, kubainisha na kutumia vielezi katika sentensi kwa usahihi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK 107

MWM UK72

 
5 Kuandika  (insha) Mtungo Kufikia mwisho wa kipindi mwanafunzi aweze kumbuni, na kuandika isha akitumia maneno aliyopewa kwa usahihi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

-kadi

-picha

michoro

KS MFUTI

UK 108

MWM UK73

 
5 1 Kusikiliza na kuzungumza Matunda, miti na mimea Kufikia mwisho wa kipindi mwanafunzi aweze  kutambua , kuchora na kutumia msamiati wa matunda,miti namimea     -Picha kiabuni KS MFUTI

UK 109

MWM UK73

 
2 Ufahamu Mwadani wetu Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kubainisha na kujibu maswali ya ufahamu kwa usahihi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kadi

-picha

michoro

-Kamusi

KS MFUTI

UK 9

MWM UK

 
3 Mapambo ya lugha vitawe Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,  kutambua maana  na kutumia vitawe katika sentensi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi KS MFUTI

UK 110

MWM UK74

 
4 Sarufi Viulizi Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kueleza, kubainisha na kutumia viulizi katika sentensi kwa usahihi     -Picha kiabuni

-Kamusi

KS MFUTI

UK 11

MWM UK

 
5 Kuandika (insha) Miti Kufikia mwisho wa kipindi mwanafunzi aweze kuandika insha umuhimu wa miti ukijaza  mapengo ukitumia maneno uliyopewa -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK 11 1

MWM UK76

 
6 1 Kusikiliza na

kuzungumza

 

Vitabu vya maktaba Kufikia mwisho wa kipindi mwanafunzi aweze kueleza, kuorodhesha na kufafanua  na kuchangamkia kusoma vitabu vya maktaba -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha

-Kamusi wanafunzi

KS MFUTI

UK 112

MWM UK78

 
2 Ufahamu Tuzungukaje mbuyu? Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,  kutambua  na kujibu maswali yaufahamu na kujadili mafunzo katika ngojera hili -kusikiliza

-kutamka

-kutambua

-kuzungumza

-kusikiliza

-kutamka

-kutambua

-kuzungumza

-Picha kiabuni KS MFUTI

UK 113

MWM UK79

 
3 Mapambo ya lugha Vitendawili Kufikia mwisho wa kipindi mwanafunzi aweze kusoma, kutumia nakutega na kutegua vitendawili -kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kadi

-picha

michoro

-Kamusi

KS MFUTI

UK 114

MWM UK99

 
4 Sarufi Matumizi ya: ‘katika’ kwenye’ na ‘ni’

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kueleza, kubainisha na kutumia ‘katika’ kwenye’ na  ‘ni’ -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK 123

MWM UK81

 
5 Kuandika (insha) Mtungo Kufikia mwisho wa kipindi mwanafunzi aweze  kubunikuandika insha kwa kujaza mapengo ukitumia maneno uliyopewa -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

-Picha kiabuni

-Kamusi

KS MFUTI

UK 122

MWM UK82

 
7 MTIHANI WA KATI YA MUHULA  
8 1 Kusikiliza na kuzungumza Miti na mimea

 

Kufikia mwisho wa kipindi mwanafunzi aweze kubainisha, kueleza na kutumia msamiati wa wa miti na mimea kwa usahihi -kusoma

-kusikiliza

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK 129

MWM UK86

 
2 Ufahamu Mstahimilivu hula mbivu

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,  kujadili funzo,kutambua na kujibu  maswali ya ufahamu kwa usahihi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi KS MFUTI

UK 129

MWM UK87

 
3 Mapambo ya lugha Vitate

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma, kutumia na kuandika vitate  huku akieleza maana -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

KS MFUTI

UK  131

MWM UK88

 
4 Sarufi Ukubwa wa nomino

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kueleza, kutumia na kuandika ukumbwa wa nomino -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-Picha kiabuni KS MFUTI

UK  138

MWM UK90

 
5 Kuandika (insha) Wasifu (mekatilili)

 

Kufikia mwisho wa kipindi mwanafunzi aweze kubuni na kuandika insha kuhusu mutu mashuhuri kwa hati zinazosomeka kulingana na kichwa alichopewa -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi KS MFUTI

UK  138

MWM UK91

 
9 1 Kusikiliza na kuzungumza Majina ya kike na kiume Kufikia mwisho wa kipindi mwanafunzi aweze kueleza kumbainisha na kutumia  msamiati wa kike na kiume -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

Kitabu cha wanafunzi KS MFUTI

UK  139

MWM UK92

 
2 Ufahamu Mjadala: wanafaa kusoma pamoja

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma , kujadili ,kueleza na kujibu  maswali ya ufahamu kwa usahihi -kujadili

-kutumia

-kueleza

-kuuliza

-kujadili

-kutumia

-kueleza

-kuuliza

-Picha kiabuni KS MFUTI

UK  143

MWM UK92

 
3 Mapambo ya lugha Visawe

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kueleza maana na kutumia  visawe katika sentensi kwa usahihi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK  143

MWM UK93

 
4 Sarufi Hali ya udogo

 

Kufikia mwisho wa kipindi mwanafunzi aweze kueleza kubainisha na kutumia hali ya udogokutoka hali ya kawaida kwa usahihi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

-Picha kiabuni KS MFUTI

UK  144

MWM UK96

 
5 Kuandika (insha) mjadala Kufikia mwisho wa kipindi mwanafunzi aweze kuandika insha ya mjadala kulingana na kichwa ulicho pewa -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

KS MFUTI

UK 145

MWM UK97

 
10 1 Kusikiliza na kuzungumza Majina ya wafanyikazi mbalimbali Kufikia mwisho wa kipindi mwanafunzi aweze kusoma nakutumia baadhi ya msamiati wa majina ya wafanyi kazi mbalimbali -kusoma

-kusikiliza

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK  146

MWM UK98

 
2 Ufahamu Uhaba wa kazi

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kueleza na kujibu  maswali ya ufahamu kwa usahihi -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

KS MFUTI

UK  145

MWM UK97

 
3 Mapambo ya lugha Shairi : ‘nitafanya kazi ngani’ Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kukariri na kujadili funzu la shairi hili -kusoma

-kusikiliza

-kukariri

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kukariri

-kutambua

-kuzungumza

-Picha kiabuni KS MFUTI

UK  146

MWM UK97

 
4 Sarufi Usemi wa taarifa

 

Kufikia mwisho wa kipindi mwanafunzi aweze kueleza na kubainisha usemi wa taarifa na kubadilisha kwa usemi halisi -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

Kitabu cha wanafunzi KS MFUTI

UK  147

MWM UK98

 
5 Kuandika (insha) kazi Kufikia mwisho wa kipindi mwanafunzi aweze kubainisha na kuandika insha akitetea hoja kikamilifu -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK  147

MWM UK99

 
11 1 Kusikiliza na kuzungumza Mekoni

 

Kufikia mwisho wa kipindi mwanafunzi aweze kufasiri maneno ya picha, kujadili nakufafanua baadhi ya msamiati wa mekoni -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-Picha kiabuni KSMFUTI

UK  21

 
2 Ufahamu Wageni mekoni

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kuandika nakujibu maswali ya ufahamu -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi KS MFUTI

UK  22

 
3 Mapambo ya lugha Methali

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kutambua na kutumia  methali na maana zake katika sentensi -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

KS MFUTI

UK  23

MWM UK

 
4 Sarufi Matumizi ya ‘si-’

 

Kufikia mwisho wa kipindi mwanafunzi aweze kutumia na kueleza matumizi sahihi ya ‘si –‘ -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK  23

MWM UK

 
5 Kusikiliza na kuzungumza Vipindi vya redio na runinga

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusikiliza redio narununga nakujibu maswali yatayo ulizwa -kusikiliza

-kutumia

-kueleza

-kuuliza

-kusikiliza

-kutumia

-kueleza

-kuuliza

Redio

runinga

KS MFUTI

UK  24

MWM UK

 
12 1 Kuandika insha Jinsi ya kupika Kufikia mwisho wa kipindi mwanafunzi aweze kubuni na kuandika insha shahihiya maelezo jinsi ya kupika -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

KS MFUTI

UK  24

MWM UK

 
2 Ufahamu Barua rasmi

 

Kufikia mwisho wa kipindi mwanafunzi aweze kusoma, kueleza na kuandika barua rasmi kwa hati zinazosomeka na nadhifu -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK  25

MWM UK

 
3 Mapambo ya lugha vitawe Kufikia mwisho wa kipindi mwanafunzi aweze kusoma, kutumia na kuandika vitawe huku akieleza maana -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

KS MFUTI

UK  23

MWM UK

 
4 Sarufi Matumizi ya, ‘amba’ Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,  kutambua na kuondosha amba’ na kutumia ‘O’ rejeshi mahali pake katika sentensi kwa usahihi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-Picha kiabuni KS MFUTI

UK  25

MWM UK

 
5 Kusandika insha Barua rasmi Kufikia mwisho wa kipindi mwanafunzi aweze  kueleza na kuandika barua rasmi kwa hati zinazosomeka na nadhifu -kusoma

-kusikiliza

-kutambua

-kuunganisha

-kusoma

-kusikiliza

-kutambua

-kuunganisha

-kadi

-picha

michoro

KS MFUTI

UK  26

MWM UK

 
5                
13 JARIBIO LA PILI  
14 MTIHANI WA MWISHO WA MUHULA  

 

 

 

 

 

AZIMIO LA KAZI

DARASA LA NANE 2016

MUHULA WA II

 

 

 

 

               ASILIA

  1. KISWAHILI MFUTI
  2. Mwongozo wa Mwalimu
  3. Oxford
  4. Kamusi

 

 

 

 

 

JUMA KIPINDI FUNZO MADA MALENGO SHUGHULI ZA MWALIMU SHUGHULI ZA MWANAFUNZI NYENZO ASILIA MAONI
1 KUFUNGUA NA MATAYARISHO  
2 1 Kusikiliza na kuzungumza/kongea Mihadarati Kufikia mwisho wa kipindi mwanafunzi aweze, kutumia baadhi ya msamiati wa mihadarati na kufafanua adhari zake. -kusoma

-kusikiliza

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kutambua

-kuzungumza

-Picha kiabuni KS MFUTI

UK 154

MWM UK100

 
2 kusoma

Ufahamu

Sibaguami Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kujibu na kujadili ujube uliko katika msakala haya -kueleza

-kujadili

-kuandika

-kuiga

-kueleza

-kujadili

-kuandika

-kuiga

-kadi

-picha

michoro

KS MFUTI

UK 155

MWM UK101

 
3 Maombo ya lugha Misemo Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kubainisha na kutumia misemo katika sentensi -kusoma

-kuandika

-kutamka

-kutumia

-kusoma

-kuandika

-kutamka

-kutumia

-kadi

-picha

michoro

KS MFUTI

UK 156

MWM UK102

 
4 Sarufi Usemi halishi Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kubainisha usemi halisi na kubadilisha usemi wa taarifa kuwa usemi halisi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi KS MFUTI

UK157

MWM UK103

 
5 Kuandika (insha) Mazungumzo na methali Kufikia mwisho wa kipindi mwanafunzi aweze kuandika ya mazungumzo kutumia  methali kwa ushahihi -kueleza

-kutaja

-kuandika

-kuiga

-kueleza

-kutaja

-kuandika

-kuiga

-Picha kiabuni KS MFUTI

UK157

MWM UK104

 
                   
3 1 Kuongea na kusikiliza Mazingira Kufikia mwisho wa kipindi mwanafunzi aweze kujadili namna ya mazingira na njia za kukinga na kuzuia uharubifu -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

-kadi

-picha

michoro

KS MFUTI

UK4157

MWM UK103

 
2 Ufahamu Hotuba Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kuzingatia na kujibu maswali yaufahamu kusoma

-kusikiliza

-kutambua

-kuzungumza

kusoma

-kusikiliza

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK155

MWM UK104

 
3 Mapambo ya lugha Methali Kufikia mwisho wa kipindi mwanafunzi aweze kusoma ,  kutambua na kutumia  metheli katika sentensi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi KS MFUTI

UK156

MWM UK105

 
4 Sarufi Myambuliko wa vitenzi Kufikia mwisho wa kipindi mwanafunzi aweze kusoma, kubainisha na kutumia vitenzi katika sentensi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-Picha kiabuni KS MFUTI

UK154

MWM UK100

 
5 Kuandika (insha) Hotuba Kufikia mwisho wa kipindi mwanafunzi aweze kumbuni, na kuandika isha akitumia maneno aliyopewa kwa usahihi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

-kadi

-picha

michoro

KS MFUTI

UK156

MWM UK102

 
4 1 Kusikiliza na kuzungumza Ajira ya watoto Kufikia mwisho wa kipindi mwanafunzi aweze kufasiri ujumbe wa michoro, kueleza na kubainisha ubaya,hasara na madhara ya ajira ya watoto -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK 157

MWM UK103

 
2 Ufahamu Krismasi Kufikia mwisho wa kipindi mwanafunzi aweze, kusoma kutaja na kutumia msamiati mpya  kwa usahihi kusoma

-kusikiliza

-kutambua

-kuzungumza

kusoma

-kusikiliza

-kutambua

-kuzungumza

-Picha kiabuni KS MFUTI

UK 160

MWM UK105

 
3 Mapabo ya lugha Vitawe Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,  kutambua na kutumia vitawe katika sentensi     -kadi

-picha

michoro

KS MFUTI

UK 161

MWM UK105

 
4 Sarufi  ‘a’ unganifu Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kueleza, kubainisha na kutumia vihisishi katika sentensi kwa usahihi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

Kitabu cha wanafunzi KS MFUTI

UK 162

MWM UK159

 
5 Kuandika  (insha) Sherehe Kufikia mwisho wa kipindi mwanafunzi aweze kumbuni, na kuandika isha akitumia maneno aliyopewa kwa usahihi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

-kadi

-picha

michoro

KS MFUTI

UK 163

MWM UK 106

 
5 1 Jaribio la kwanza Dira Kufikia mwisho wa kipindi mwanafunzi aweze  kutambua , kuchora na kutumia msamiati wa dira kwa usahihi -kusoma

-kuchora

-kueleza

-kuuliza

-kusoma

-kuchora

-kueleza

-kuuliza

-Picha kiabuni KS MFUTI

UK 164

MWM UK107

 
2 Ufahamu Nyota njema Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kubainisha na kujibu maswali ya ufahamu kwa usahihi -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK 164-165

MWM UK108

 
3 Mapambo ya lugha Silabi- cha Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,  kutambua na kutumia methali katika sentensi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

Kitabu cha wanafunzi KS MFUTI

UK 165

MWM UK110

 
4 Sarufi Vihisishi Kufikia mwisho wa kipindi mwanafunzi aweze kusoma,kueleza, kubainisha na kutumia vihisishi katika sentensi kwa usahihi -kusoma

-kutumia

-kueleza

-kuuliza

-kusoma

-kutumia

-kueleza

-kuuliza

-Picha kiabuni KS MFUTI

UK 166

MWM UK112

 

 
5 Kuandika (insha) Barua ya kirafiki Kufikia mwisho wa kipindi mwanafunzi aweze kuandika insha kwa kujaza mapengo ukitumia maneno uliyopewa -kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kusoma

-kusikiliza

-kuandika

-kutambua

-kuzungumza

-kadi

-picha

michoro

KS MFUTI

UK 167

MWM UK112

 
6 1-5 Jaribio la kwanza Marudio Mwanafunzi aweze:

–        kujibu maswali aliyoulizwa

–        kukumbuka yale aliyo funzwa haponawali

–        kubuni na kuandika kwa hati ilyo someka

–        kustawisha ukakamavu wa kukabiliana na mtihani

-kueleza

-kuongoza kujibu maswali

-kusikiliza

-kuandika

 

 

-kueleza

-kuongoza kujibu maswali

-kusikiliza

-kuandika

 

 

-Kitabu cha mwana fuzi

-mitihani yajaribio

KS MFUTI

UK32-37

MWM UK115

 
                 
6   Jaribio la pili   Mwanafunzi aweze:

–        kujibu maswali aliyoulizwa

–        kukumbuka yale aliyo funzwa haponawali

–        kubuni na kuandika kwa hati ilyo someka

–        kustawisha ukakamavu wa kukabiliana na mtihani

-kueleza

-kuongoza kujibu maswali

-kusikiliza

-kuandika

 

-kueleza

-kuongoza kujibu maswali

-kusikiliza

-kuandika

 

Kitabu cha mwana fuzi

-mitihani yajaribio

KS MFUTI

UK 66-67

MWM UK25

 
7   Jaribio la tatu   Mwanafunzi aweze:

–        kujibu maswali aliyoulizwa

–        kukumbuka yale aliyo funzwa haponawali

–        kubuni na kuandika kwa hati ilyo someka

–        kustawisha ukakamavu wa kukabiliana na mtihani

-kueleza

-kuongoza kujibu maswali

-kusikiliza

-kuandika

 

-kueleza

-kuongoza kujibu maswali

-kusikiliza

-kuandika

 

Kitabu cha mwana fuzi

-mitihani yajaribio

KS MFUTI

UK 98-103

MWM UK48

 
8   Jaribio la nne   Mwanafunzi aweze:

–        kujibu maswali aliyoulizwa

–        kukumbuka yale aliyo funzwa haponawali

–        kubuni na kuandika kwa hati ilyo someka

–        kustawisha ukakamavu wa kukabiliana na mtihani

-kueleza

-kuongoza kujibu maswali

-kusikiliza

-kuandika

 

-kueleza

-kuongoza kujibu maswali

-kusikiliza

-kuandika

 

Kitabu cha mwana fuzi

-mitihani yajaribio

KS MFUTI

UK 132-137

MWM UK60

 
9 1 Jaribio la tano   Mwanafunzi aweze:

–        kujibu maswali aliyoulizwa

–        kukumbuka yale aliyo funzwa haponawali

–        kustawisha ukakamavu

-kueleza

-kuongoza kujibu maswali

-kusikiliza

-kuandika

-kueleza

-kuongoza kujibu maswali

-kusikiliza

-kuandika

 

Kitabu cha mwana fuzi

-mitihani yajaribio

KS MFUTI

UK 148-153

MWM UK80

 
10 1 Jaribio la sita   Mwanafunzi aweze:

–        kujibu maswali aliyoulizwa

–        kukumbuka yale aliyo funzwa haponawali

-kueleza

-kuongoza kujibu maswali

 

-kueleza

-kuongoza kujibu maswali

-kusikiliza

 

Kitabu cha mwana fuzi

-mitihani yajaribio

KS MFUTI

UK 168-172

MWM UK99

 
11 MTIHANI WA KCPE  

 

Grade 7 Computer Sciences Free CBC Schemes of Work {Updated Version)

2023 GRADE 7 LONGHORN COMPUTER SCIENCESCHEMES OF WORK TERM 3

Ref used:

Grade 7 Computer science Curriculum Design

Longhorn Computer Science Grade 7 (Teachers Guide and Learners Book)

NAME OF THE TEACHER                                                                SCHOOL                                             YEAR                         TERM            III       

Wk LSN strand Sub-strand Specific Learning Outcomes Key Inquiry Question(s) Learning Experiences Learning Resources Assessment Methods Ref  
1 1 COMPUTER NETWORKS Internet concepts

 

Internet as a resource that runs on a global network of computers

By the end of the lesson the learner should be able to:

Knowledge

a.      describe the internet as a resource that runs on a global network of computers

b.      explain benefits of internet in the immediate environment

Skill

c.      use IT devices to access the internet

Attitude

d.     appreciate the use of internet as a computer network resource.

1. Why do you use internet?

2. How do you connect to internet?

 

The learner is guided to:

• search for the meaning of the term internet and present to peers,

• debate on the benefits and challenges of the internet,

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

 

Longhorn Comp. Scie T.G Pg. 183-187

 

Longhorn Comp. Scie P.B Pg. 141-143

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  2   Challenges of the internet By the end of the lesson the learner should be able to:

Knowledge

a.      explain challenges of internet in the immediate environment

b.      explore ways of overcoming challenges of internet in the immediate environment.

Skill

c.      assess the internet connectivity in a nearby locality or school.

Attitude

d.     appreciate the use of internet as a computer network resource.

1. Why do you use internet?

2. How do you connect to internet?

 

The learner is guided to:

• in groups, discuss ways of

overcoming challenges of the internet in the immediate environment,

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 187-189

 

Longhorn Comp. Scie P.B Pg. 144-147

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  3   Basic requirements for internet connectivity By the end of the lesson the learner should be able to:

Knowledge

a.      identify basic requirements for internet connectivity

Skill

b.     observe pictures of the basic requirements for connecting to the internet.

Attitude

c.      appreciate the use of internet as a computer network resource.

1. Why do you use internet?

2. How do you connect to internet?

The learner is guided to:

• discuss the basic requirements for internet connectivity (Internet

Service Provider (ISP), Internet software, communication media,

communication device),

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

 

Longhorn Comp. Scie T.G Pg. 189

 

Longhorn Comp. Scie P.B Pg. 148-149

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
2 1   Connecting to the internet and using it By the end of the lesson the learner should be able to:

Knowledge

a.      identify the basic requirements for internet connectivity.

Skill

b.     connect to the internet to search for a topical issue.

Attitude

c.      appreciate the use of internet as a computer network resource.

1. Why do you use internet?

2. How do you connect to internet?

The learner is guided to:

• share experiences on interaction with the internet and list the services available,

• in turns select service available in the internet and use it to search for a

relevant topical issue,

• use the internet to search for a topical issue.

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 190-191

 

Longhorn Comp. Scie P.B Pg. 150

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  2   World Wide Web (WWW)

 

Importance of World wide web

By the end of the lesson the learner should be able to:

Knowledge

a.      explain the importance of WWW as used in computer networks

Skill

b.     make posters on WWW.

Attitude

c.      appreciate the use of WWW as a repository of information.

1. How do you access internet resources?

2. Why do you use a web browser?

 

The learner is guided to:

• use available learning resources to search for the meaning of the terms World Wide Web (WWW), web browsers, uniform resource locator (URL),

• in turns discuss examples of web browsers (explorer,

Firefox, Chrome, Netscape,

Opera, Safari)

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 192-195

 

Longhorn Comp. Scie P.B Pg. 151

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  3   Features of a web browser By the end of the lesson the learner should be able to:

Knowledge

a.      identify the features of a web browser

Skill

b.     Using IT devices locate the features of a web browser on the screen.

Attitude

c.      appreciate the use of WWW as a repository of information.

 

1. How do you access internet resources?

2. Why do you use a web browser?

The learner is guided to:

• launch and navigate a web

browser to identify its features,

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 195-196

 

Longhorn Comp. Scie P.B Pg. 152-153

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
3 1   Components of Uniform Resource Locator By the end of the lesson the learner should be able to:

Knowledge

a.      name the components of a URL

Skill

b.     describe the components of a uniform resource locator (URL) used to access resources in the internet.

Attitude

c.      appreciate the use of WWW as a repository of information.

1. How do you access internet resources?

2. Why do you use a web browser?

The learner is guided to:

• take turns to write URL format: protocol://hostname/other

information

• participate in giving examples of URL

• type a web resource Uniform Resource Locator (URL), and discuss its components,

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 196-197

 

Longhorn Comp. Scie P.B Pg. 154-155

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  2   Components of Uniform Resource Locator By the end of the lesson the learner should be able to:

Knowledge

a.      name the components of a URL

Skill

b.     describe the components of a uniform resource locator (URL) used to access resources in the internet.

Attitude

c.      appreciate the use of WWW as a repository of information.

1. How do you access internet resources?

2. Why do you use a web browser?

The learner is guided to:

• take turns to write URL format: protocol://hostname/other

information

• participate in giving examples of URL

• type a web resource Uniform Resource Locator (URL), and discuss its components,

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 196-197

 

Longhorn Comp. Scie P.B Pg. 154-155

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  3   Locating resources in the WWW using a web browser By the end of the lesson the learner should be able to:

Knowledge

a.      list the different types of web browsers.

Skill

b.     use a web browser to locate resources in the WWW

Attitude

c.      appreciate the use of WWW as a repository of information.

1. How do you access internet resources?

2. Why do you use a web browser?

The learner is guided to:

• take turns to demonstrate how web browsers work,

• practice using a web browser to locate relevant internet resources.

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 197-198

 

Longhorn Comp. Scie P.B Pg. 156

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
4 1 COMPUTER PROGRAMMING Computer programming concepts

 

 

Meaning of program and computer programming

By the end of the lesson the learner should be able to:

Knowledge

a.      define the terms program and computer programming

b.      explain the importance of programming as used in computing

Skill

c.      using IT devices locate a computer program

Attitude

d.     appreciate using computer programs in performing daily life activities.

1. Why do computers have programs?

2. How do you use computer programs?

 

The learner is guided to:

• use available learning resources to search for the meaning of the term

programming and programs,

• in groups discuss the importance of computer programs,

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 199-202

 

Longhorn Comp. Scie P.B Pg. 157-159

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  2   Application areas of computer programs By the end of the lesson the learner should be able to:

Knowledge

a.      identify areas where computer programs are used in daily life.

Skill

b.     Group computer applications according to functions.

Attitude

c.      appreciate using computer programs in performing daily life activities.

  The learner is guided to:

• take turns to discuss areas where computer programs are used in daily life and list them,

• share ideas on the use of

programming in daily life activities,

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 203

 

Longhorn Comp. Scie P.B Pg. 160-161

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  3   Launching and interacting with computer programs By the end of the lesson the learner should be able to:

Knowledge

a.      list the different types of computer programmes.

Skill

b.     launch and interact with a computer program for awareness.

Attitude

c.      appreciate using computer programs in performing daily life activities.

  The learner is guided to:

• share experience on performing daily life activities (playing computer games, listening to music, performing mathematical

operations, drawing objects, type text) using available computer programs accessories.

• in groups, start and interact with a computer program accessory such as, a computer game, calculator, paint, snipping tool, media player

and note notepad, accessories.

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

 

Longhorn Comp. Scie T.G Pg. 204-206

 

Longhorn Comp. Scie P.B Pg. 162-164

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
5 1   Visual programming concepts

 

Types of visual programming applications

By the end of the lesson the learner should be able to:

Knowledge

a.      identify types of visual programming applications for use

Skill

b.     demonstrate knowledge in visual programming.

Attitude

c.      appreciate navigating a visual programming application interface.

1. Why do you use visual? programming

applications?

2. How do you launch

visual programming

application?

The learner is guided to:

• use available resources to search for the meaning of the term visual programming,

• discuss and list examples of visual programming applications used in

computer programming,

• in groups discuss the procedure of launching a visual programming

application,

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 207-210

 

Longhorn Comp. Scie P.B Pg. 165

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  2   Launching a visual programming application By the end of the lesson the learner should be able to:

Knowledge

a.      explain the procedure of launching a visual programming application

Skill

b.     launch a visual programming application in a computer.

Attitude

c.      appreciate navigating a visual programming application interface.

1. Why do you use visual? programming

applications?

2. How do you launch

visual programming

application?

The learner is guided to:

• in groups, launch a visual

programming application such as Microsoft Make Code, Scratch, Code.org, Sprite box,

• share experiences on navigating the visual programming application interface with peers.

• consult a computer resource person to demonstrate how to

launch visual programming

applications used in computer programming,

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

 

Longhorn Comp. Scie T.G Pg. 210-211

 

Longhorn Comp. Scie P.B Pg. 166-167

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  3   Launching a visual programming application By the end of the lesson the learner should be able to:

Knowledge

a.      explain the procedure of launching a visual programming application

Skill

b.     launch a visual programming application in a computer.

Attitude

c.      appreciate navigating a visual programming application interface.

1. Why do you use visual? programming

applications?

2. How do you launch

visual programming

application?

The learner is guided to:

• in groups, launch a visual

programming application such as Microsoft Make Code, Scratch, Code.org, Sprite box,

• share experiences on navigating the visual programming application interface with peers.

• consult a computer resource person to demonstrate how to

launch visual programming

applications used in computer programming,

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 210-211

 

Longhorn Comp. Scie P.B Pg. 166-167

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
6 1   Visual programming features

 

Features of scratch

By the end of the lesson the learner should be able to:

Knowledge

a.      explore features of a visual programming application.

Skill

b.     match the functions of the features of visual programming application to their functions.

Attitude

c.      appreciate the application of the features of a visual programming application to create a sequence of instructions

1. Why is visual programming popular in introducing computer programming?2. How do you use visual programming application features?

 

The learner is guided to:

• in groups discuss the features of a visual programming application

• match the functions of the

features of a visual programming application to their functions

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

 

Longhorn Comp. Scie T.G Pg. 212-215

 

Longhorn Comp. Scie P.B Pg. 168

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  2   Features of scratch By the end of the lesson the learner should be able to:

Knowledge

a.      explore features of a visual programming application.

Skill

b.     match the functions of the features of visual programming application to their functions.

Attitude

c.      appreciate the application of the features of a visual programming application to create a sequence of instructions

1. Why is visual programming popular in introducing computer programming?2. How do you use visual programming application features? The learner is guided to:

• in groups discuss the features of a visual programming application

• match the functions of the

features of a visual programming application to their functions

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 212-215

 

Longhorn Comp. Scie P.B Pg. 168

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  3   Functions of the features of scratch By the end of the lesson the learner should be able to:

Knowledge

a.      relate the features of a visual programming application to their function

Skill

b.      match the functions of the features of visual programming application to their functions.

Attitude

c.      appreciate the application of the features of a visual programming application to create a sequence of instructions

1. Why is visual programming popular in introducing computer programming?2. How do you use visual programming application features? The learner is guided to:

• discuss the functions of the

features of a visual programming application

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

 

Longhorn Comp. Scie T.G Pg. 215

 

Longhorn Comp. Scie P.B Pg. 169

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
7 1   Functions of the features of scratch By the end of the lesson the learner should be able to:

Knowledge

a.      relate the features of a visual programming application to their function

Skill

b.      match the functions of the features of visual programming application to their functions.

Attitude

c.      appreciate the application of the features of a visual programming application to create a sequence of instructions

1. Why is visual programming popular in introducing computer programming?2. How do you use visual programming application features? The learner is guided to:

• discuss the functions of the

features of a visual programming application

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

 

Longhorn Comp. Scie T.G Pg. 215

 

Longhorn Comp. Scie P.B Pg. 169

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  2   Terminologies used in visual programming applications By the end of the lesson the learner should be able to:

Knowledge

a.      describe terminologies used in a visual programming application

Skill

b.     demonstrate the use of Visual programming terms.

Attitude

c.      appreciate the application of the features of a visual programming application to create a sequence of instructions

1. Why is visual programming popular in introducing computer programming?2. How do you use visual programming application features? The learner is guided to:

• in turns discuss and demonstrate the use of visual programming terms (reserved words, syntax, variables, input output statements, control structures,

variable declarations).

 

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 216

 

Longhorn Comp. Scie P.B Pg. 172-175

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  3   Using scratch to create a sequence of instructions By the end of the lesson the learner should be able to:

Knowledge

a.      name the features of the scratch program.

Skill

b.     use the features of a visual programming application to create a sequence of instructions.

Attitude

c.      appreciate the application of the features of a visual programming application to create a sequence of instructions

1. Why is visual programming popular in introducing computer programming?2. How do you use visual programming application features? The learner is guided to:

• in groups, create a sequence of actions using the features of a visual programming application (animations, sound)

• share experience on the use of the features of a visual

programming application

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

 

Longhorn Comp. Scie T.G Pg. 218-220

 

Longhorn Comp. Scie P.B Pg. 176-179

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
8 1   Using scratch to create a sequence of instructions By the end of the lesson the learner should be able to:

Knowledge

a.      name the features of the scratch program.

Skill

b.     use the features of a visual programming application to create a sequence of instructions.

Attitude

c.      appreciate the application of the features of a visual programming application to create a sequence of instructions

1. Why is visual programming popular in introducing computer programming?2. How do you use visual programming application features? The learner is guided to:

• in groups, create a sequence of actions using the features of a visual programming application (animations, sound)

• share experience on the use of the features of a visual

programming application

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

 

Longhorn Comp. Scie T.G Pg. 218-220

 

Longhorn Comp. Scie P.B Pg. 176-179

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  2   Using scratch to create a sequence of instructions By the end of the lesson the learner should be able to:

Knowledge

a.      name the features of the scratch program.

Skill

b.     use the features of a visual programming application to create a sequence of instructions.

Attitude

c.      appreciate the application of the features of a visual programming application to create a sequence of instructions

1. Why is visual programming popular in introducing computer programming?2. How do you use visual programming application features? The learner is guided to:

• in groups, create a sequence of actions using the features of a visual programming application (animations, sound)

• share experience on the use of the features of a visual

programming application

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

Longhorn Comp. Scie T.G Pg. 218-220

 

Longhorn Comp. Scie P.B Pg. 176-179

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
  3   Using scratch to create a sequence of instructions By the end of the lesson the learner should be able to:

Knowledge

a.      name the features of the scratch program.

Skill

b.     use the features of a visual programming application to create a sequence of instructions.

Attitude

c.      appreciate the application of the features of a visual programming application to create a sequence of instructions

1. Why is visual programming popular in introducing computer programming?2. How do you use visual programming application features? The learner is guided to:

• in groups, create a sequence of actions using the features of a visual programming application (animations, sound)

• share experience on the use of the features of a visual

programming application

Digital devices, reference materials, productivity tools, compute software (OS, Utility software and Application programs), computer hardware, manila papers, Internet, video, audio clips

 

Longhorn Comp. Scie T.G Pg. 218-220

 

Longhorn Comp. Scie P.B Pg. 176-179

Rating scales, rubrics, questionnaires, projects, journals, portfolios, oral questions, aural questions

 

   
9 END YEAR ASSESSMENT/CLOSING

 

 

Agriculture KCSE Finest Revision Resources (Questions and Answers)

AGRICULTURE  I

 

SECTION A. 

 

  1. State two uses of a rotavator. 1mk
  2. List two methods of controlling coffee berry disease 1mk
  3. Identify four sources of agricultural information 2mks
  4. State two ways of controlling Rinderpest disease 1mk
  5. Mention four signs of infertility in cows 2mks
  6. How is cannibalism controlled in poultry 2mks
  7. List four maintenance of ox-cart 2mks
  8. Mention four common fodder crops grown in Kenya       2mks
  9. Name two diseases that attack cassava in the field 1mk
  10. Identify six physical measures of pest control in crops 3mks
  11. Differentiate between grass strips and trash lines 1mk
  12. Name two categories of land tenure 1mk
  13. Give four reasons why rabbits should be fed on balanced feeds 2mks
  14. List four causes of poor quality concrete 2mks
  15. Mention four reasons for maintaining farm buildings 2mks
  16. Identify four ways of reducing the problem of ewes disowning their lambs                   2mks
  17. State four advantages of using certified seeds from Kenya seed company             2mks
  18. State two symptoms of damping off disease in cabbages             1mk
  19. Name two methods of harvesting rice 1mk
  20. Mention two ways of classifying pests in crop production 1mk
  21. Name the dual purpose sheep 1mk
  22. Name the young of a donkey 1mk
  23. Name two primary hosts for liver flukes 1mk
  24. State one use of a tag applicator 1mk

 

SECTION B.

 

  1. a. State four effects of excess Nitrogen in crops 4mks
  2. List six roles of organic matter in the soil 6mks
  3. Mention three characteristics of phosphatic fertilisers 3mks
  4. a. State three ways of collecting semen from bulls for artificial insemination          3mks
  5. List six ways of carrying out pregnancy diagnosis in cows                                  6mks
  6. Give four reasons for conserving pastures                                                          4mks
  7. a. Name three tomatoes for fresh market 3mks
  8. State five reasons for pruning in crops 5mks
  9. i. Mention four factors to be considered when siting grain storage structures 4mks
  10. Differentiate between seed rate and plant population   2 mks

 

SECTION C.

 

  1. Explain field production of sweet potatoes from field preparation upto harvesting. 20mks.
  2. a. Explain management practices that ensure maximum production of eggs in poultry 12mks
  3. Explain management practices a beef farmer would carry out to minimise effects of low

rainfall                                                                                                              8 mks

  1. a. Mention five characteristics of pyrethrums                                                         5mks
  2. Explain environmental factors that affect effectiveness of agro-chemicals in crop

production.                                                                                                        5mks

  1. i. Differentiate between land consolidation and land fragmentation and settlement. 3mks.
  2. Explain objectives of land reforms.                                                               7mks.

 

AGRICULTURE I

MARKING SCHEME.

 

SECTION A.

 

  1. Uses of a rotavator
  • Refining seed bed / secondary tillage
  • Cultivation in clay soil / weeding in waterlogged soils.

2 x ½ = 1 mark.

 

  1. Control of C.B.D.
  • Open pruning
  • Plant resistant varieties e.g Ruiru II
  • Regular spraying with appropriate fungicides / copper fungicides / captafol

2 x ½ = 1 mark.

 

  1. Sources of agri – information.
  • Agricultural research stations
  • Neighbouring farms
  • Agricultural field days
  • Agricultural shows
  • Mass media eg. T.V, Radios
  • Agriculture extension officers
  • Farmers training centres
  • Agricultural training and education institution
  • Chiefs baraza.

4 x ½ = 2 marks.

 

  1. Control of Rinderpest.
  • Regular Vaccination
  • Separate sick and healthy ones
  • Imposing quarantine
  • Kill affected animals and proper disposal
  • Notify authorities of an outbreak.

2 x ½ = 1 mark.

 

  1. Signs of infertility.
  • A cow does not show heat signs
  • Abnormal discharge from vulva
  • Prolonged heat period
  • Irregular heat intervals / too short or too long heat intervals.

4 x ½ = 2 marks.

 

  1. Control of cannibalism.
  • Avoid bright light in brooder
  • Keep birds busy / hang green leaves
  • Give balanced feeds
  • Control external parasites / fleas
  • Debeak hens which peck others
  • Keep birds according to age-groups
  • Avoid overcrowding of birds / provide enough space to the birds.

4 x ½ = 2 marks.

 

 

  1. Maintenance of Ox-cart.
  • Lubricate moving parts regularly
  • Check tyre pressure and adjust accordingly
  • Clean it after use
  • Repair / replace any worn out / broken parts eg. yoke
  • Proper storage in a shed. 4 x ½ = 2 marks.

 

  1. Common fodder crops
  • Napier grass / elephant grass
  • Guatemala grass
  • Edible land
  • Kales
  • Manigolds
  • Lucerne / alfafa
  • Sorgum 4 x ½ = 2 marks.
  1. Cassava diseases.
  • Cassava mosaic
  • Brown streak
  • Bacterial bright. 2 x ½ = 1 mark.
  1. Physical measures of pest control.
  • Use of lethal temperature
  • Drying the grains
  • Irrigation / flooding the field
  • Suffocation / use of airtight stores
  • Physical destruction / trapping / picking and killing
  • Use of electromagnetic radiation.

6 x ½ = 3 marks.

  1. Grass strips uncultivated strips of grass left between cultivated strips to control soil erosion while trash lines are heaps of cup residues placed along contours in the cup field to reduce soil erosion.

1 x 2 = 2mks (mark as a whole)

  1. Categories of land tenure.
  • Collective land tenure
  • Communal land tenure. 2mks
  1. Reasons for balanced feeding.
  • For faster and quick growth
  • For earlier maturity
  • For good quality products
  • For increased resistance against diseases
  • For increased feed conversion

4 x ½ = 2 marks.

  1. Causes of poor quality concrete.
  • Less water to harden it used.
  • Too much sand / too little cement / ratio of sand to cement mixture not appropriate.
  • Premature drying of concrete.
  • Size of individual aggregates / coarse sand.

x ½ = 2 marks.

  1. Reasons for maintaining farm structures.
  • To last longer / for durability
  • For easy cleaning
  • For proper ventilation / to avoid dampness / too much heat
  • To reduce maintenance cost
  • To be more efficient in use
  • To be more secured / for security.

x ½ = 2 marks.

  1. Reducing disowning of lambs.
  • Ensure the ewes recognises its lamb soon after lambing.
  • Proper steaming of ewe before lambing.
  • Use of lambing pens
  • Mastitis control / dry ewe therapy
  • Blind folding of ewes to activate maternal instinct.

x ½ = 2 marks.

  1. Advantages of certifield seeds.
  • Free from pests and diseases and weeds
  • Viable / uniform germination
  • Vigorous growth / faster maturity
  • High yields
  • High quality product
  • True to type / not contaminated with other seeds
  • Reduces cost of treatment / treated with chemical.

x ½ = 2 marks.

  1. Symptoms of damping off
  • Falling of seedlings / withering and death of seedlings
  • Cobweb like black mass of fungi making a ring on the stem base.

x ½ = 1 mark.

  1. Methods of harvesting rice.
  • Use of sickles
  • Use of combine harvesters.

x ½ = 1 mark.

  1. Classification of pests.
  • According to where they are found or attack the crop produce
  • Nature of damage caused
  • Nature of mouthparts
  • Part of the plant damaged

x ½ = 1 mark.

  1. Dual purpose sheep.
  • Dorper
  • Harmpshire doam
  • Dorset horn

x ½ = 1 mark.

  1. Young of donkey

– foal

1 x 1 = 1mark

  1. Primary hosts for liverflukes.
  • Cattle
  • Sheep
  • Goats

1 x ½ = 1 mark.

  1. Use of a tag applicator
  • To cut and insert a tag in the ear of an animal for identification purposes.

1 x 1 =  1 mark

 

SECTION B.

 

  1. Effects of excess nitrogen.
  • Burning / scorching of the leaves
  • Weak stems and fruits
  • Delayed maturity
  • Excess succulency and crop lean or fall / lodging of crops

4 x 1 = 4mks.

 

  1. Role of organic matter
  • Binds soil particles so improve soil structure and texture
  • Reduces leaching and soil erosion
  • Buffers soil PH
  • Reduces toxicity of plant poisons
  • Provides food and shelter to micro-organisms
  • Makes phosphorous more available.
  • Increases water holding capacity of soil and water infiltration
  • Makes soils warm as it is dark.

1 x 6 =  6marks.

  1. Characteristics of phosphatic part.
  • Slightly soluble in water
  • Have a long residual effect
  • Not liable to leaching
  • Slight scorching / burning effect.

1 x 3 = 3 marks.

  1. a. Collecting semens.
  • Use of artificial vagina
  • Recovery of semen from vagina of females soon after natural service
  • By electrical stimulation of a bull to ejaculate.

1 x 3 = 3mks.

  1. Pregnancy diagnosis.
  • Noting absence of heat period after service
  • Decline in milk yield
  • Thick secretion from cervix
  • Rise in vaginal temperature above normal
  • Thick honey like secretion in the teat
  • Feel method / apply pressure on right flank to the foetus.
  • Rectal palpation / touching the rectism and feeling minute movements.
  • Check specific gravity and PH of cervical means change drastically.

1 x 6 = 6mks.

  1. Reasons for conserving forages.
  • To provide feed in dry season
  • To ensure feeding of animals throughout the year
  • To conserve forage for sale
  • For better and efficient utilisation of land.

1 x 4 = 4 marks.

27.a.  Fresh market tomatoes

  • Marglobe
  • Money maker
  • Hotset
  • Ponderosa
  • Super marmande
  • Early beauty
  • Hundred fold
  • Best of all.

 

  1. Reasons for pruning.
  • To maintain appropriate crop / leaf ration for maximum yields
  • To attain a regular cropping / to increase productive rate
  • To control over breaking and regulate cropping
  • To facilitate air circulation and light penetration.

1 x 5 = 5mks

 

  1. i.  Siting grain stores.
  • Accessibility / nearness to road
  • Security
  • Space for future expansion
  • Topography of the place
  • Direction of prevailing wind
  • Drainage of the area
  • Nearness to power source
  • Distance from crop field.

1 x 4 = 4mks

  1. Seed rate is the quantity of seeds planted in a given area of land while plant population

is the number of plants lawns growing in a given area of land after germination.

1 X 2 = 2 marks.

                        (mark as a whole)

 

SECTION C.

 

  1. Production of sweet potatoes.
  2. Field preparation
  • Should prepare during dry season
  • Should be prepared thoroughly
  • Clear the vegetation using appropriate tools eg. pangas, slashers
  • Dig / plough the field to uproot all needs.
  • Break the dods to medium tilth
  • Dig ridges spaced 90 – 150cm apart and add some organic manure

1 x 5 = 5 mks.

 

  1. Planting
  • Select vines / stem cuttings from the tips of vines of mature plants
  • The cuttings are about 25 – 90cm long
  • Plant when there is enough moisture in the soil / soon after onset of rains
  • Cover the vines up to 2/3 of their total length
  • Plant at a spacing of 30 – 60cm along the ridges between vines
  • Healthy vines should be used.

1 x 6 = 6 mks.

 

  • Field operations.
  • Uproot the weeds using a panga / forked jembes
  • Avoid cutting the roots
  • Earth up the soil on the vines to encourage more rooting and tuber expansion.
  • Uproot weeds by hands after crop establishment/ after 2 months in the field.
  • Control moles by trapping, also porcupines and squirrels
  • Spray with appropriate pesticide / endusulfan / fenithism to control sweet potato weevils
  • Control mites using appropriate chemicals to control spread of sweet potato virus B disease.

1 x 6 = 6mks.

  1. Harvesting
  • They are ready for harvesting 4 –5 months after planting
  • Large root tubers cause the ground to crack indicating readiness
  • Harvesting for food is done piecemeal using a sharpened stick or forked jembes.
  • Complete harvesting is done when the root-tubers are to be marketed.

1x 3 = 3mks.

 

 

  1. a. Maximum production of eggs.
  • Provide enough space / avoid overcrowding
  • Proper feeding on layers mash / ensure balanced feeding
  • Provide sand / grit to aid digestion
  • Provide clean rest boxes / clean shelters / observe hygiene
  • Provide enough laying boxes
  • Vaccinate against diseases / new castle / fowl typhoid.
  • Cull poor layers.
  • Frequent egg collection.
  • Control external parasites with appropriate pesticides
  • Avoid disturbance / proper handling
  • Gradual change of routine / feeds
  • Protection from extreme weather / No cold or hot conditions
  • Treat sick birds with appropriate drugs
  • Isolate sick birds from healthy ones
  • Administer coccidiostale in feed or water
  • Debeak egg eaters

1 x 12 = 12 marks.

 

  1. Management done by a beef farmer.
  • Conservation of forage when excess as hay / silage / standing forage
  • Paddocking of pastures / rotational grazing
  • Reseeding of pastures at beginning of the rains for maximum production
  • Water collection and storage / construct dams / ponds / boreholes to ensure constant supply.
  • Provide supplementary feeds/ concentrates during scarcity
  • Irrigation of pastures to increase yield during dry season
  • Planting pasture species adaptable to dry and conditions resistant to draught.
  • Select breeds of beef cattle resistant to dry conditions/ hardy cattle.

(Award for stating correct practice)

Award for practice for explanation)

1 x 8 = 8 mks.

 

  1. a. Characteristics of pyrethrums.
  • Act as nerve poisons
  • Have repellant effect
  • Have limited persistance
  • Are safer to the user
  • Have a rapid knock down effect.

1 x 5 = 5mks.

 

  1. Environmental influence on agro-chemicals.
  • Wind – men blow the chemical away from intended crops
  • Rain – may dilute or wash away the chemicals
  • Soils – some may absorb and retain more chemicals than others.
  • Light – may decompose some chemicals.
  • Temperature – increases translocation and hence absorption of chemicals.

 = 5mks.

 

  1. i. Land consolidation is putting together under one holding different
  • Land fragmentation is a situation where an individual farmer owns many separate pieces of land scattered over a wide area.
  • Settlement refers to occupation of land which was previously uninhabited

(mark as a whole)    1 x 3 = 3mks.

 

  1. Objectives of land reforms
  • To increase agricultural output through properland use
  • To orientates agricultural production to meet market demands.
  • To enhance efficient utilisation of land
  • To put idle land to use
  • To encourage commercial instead of subsistence production.
  • To encourage conservation and improvement of land and its resources
  • To settle the landless and ease population density pressure in some areas.
  • To create self employment.

1 x 7 = 7mks.

 

AGRICULUTURE II

 

Section A .

 

  1. What benefit do pastoralist farmers attain from camels? (1 mark)
  2. (i) State FOUR factors influencing soil formation                                               (4 marks)

(ii)     State the importance of the colour of a soil in soil forming process                 (2 marks)

  1. State FOUR methods that can be used to raise production in a group of sows (2 marks)
  2. State FOUR advantages of free-range system in poultry production (2 marks)
  3. Give FOUR differences between indigenous and exotic cattle (2 marks)
  4. State TWO advantages of rolling in land preparation (1 mark)
  5. Apart from bacteria and fungi name TWO other causes of diseases in crops (1 mark)
  6. (i) Name the primary host of tapeworm                                                               (1 mark)

(ii)     List TWO internal parasites that attack sheep.                                                 (1 mark)

  1. (i) State TWO ways of controlling Nematodes                                                      (1 mark)

(ii)     Give TWO pests that are disease vectors in crops                                             (1 mark)

  1. Mention FOUR farm structures which may be found in a mixed farm. (2 marks)
  2. What FOUR factors would one consider when choosing feedstuffs for preparing a livestock

ration?                                                                                                                    (2 marks)

  1. Why is it not necessary to have grass pasture as a livestock feed in pig production? (1 mark)
  2. State FOUR stages of curing hides and skins. (2 marks)
  3. State the role of Isthmus in the reproductive system of a layer (1 mark)
  4. (i) Name the crop usually attacked by the Zebra disease.                                    (½  mark)

(ii)     What is decortication?                                                                                    (1 mark)

  1. State THREE importance of lime in crop production. (11/2 marks)
  2. Name TWO characteristics that make Katumani maize variety recommended for marginal

areas.                                                                                                                      (1 mark)

  1. State FOUR ways of preventing swarming in bees. (2 marks)
  2. State FOUR factors that contribute to the competitive ability of weeds over cultivated crops. (2 marks)
  3. Mention any FOUR factors that affect the quality of hay. (2 marks)
  4. How can a farmer improve soil PH in his farm?                                                      (1 mark)
  5. State ONE reason for foot trimming in livestock. (1 mark)
  6. State FOUR characteristics of a good wood preservative. (2 marks)

Section B

 

  1. Mention THREE methods of classifications of herbicides. (3 marks)
  2. State FOUR factors considered when selecting a site for a vegetable nursery bed. (4 marks)

 

(i)      Name the class of animal represented by the above digestive system.               (1 mark)

(ii)     Label the parts indicated by the letters.                                                          (4 marks)

P     ___________________________

Q    ___________________________

R    ___________________________

S    ___________________________

 

(iii)    State one function of P and S                                                                        (2 marks)

(iv)    Give ONE reason why rabbits and donkeys are able to digest cellulose material.         (1 mark)

  1. Give a benefit for feeding colostrum to a newly born calf. (1 mark)
  2. Name THREE soil fractions. (3 marks)
  3. (a) State FOUR activities carried out on a site before laying a farm structure.       (2 marks)

(b)     State TWO methods commonly used in wood treatment.                               (2 marks)

(c)      Give FOUR maintenance practices done on live fence.                                   (2 marks)

(d)     Give ONE disadvantage of using stones for building.                                       (1 mark)

  1. (a) State TWO advantages of chemical control on crop pests.                               (2 marks)

(b)     State FOUR effects of plant diseases on crop production.                                (4 marks)

(c)      List TWO diseases and TWO pests that attack bananas in the field.                (2 marks)

  1. (a) Explain the difference between risk and uncertainty.                                       (1 mark)

(b)     Give FOUR ways in which farmers adjust to risk and uncertainties.                (2 marks)

(c)      State FOUR crucial questions a farmer would be trying to answer when preparing
a  partial budget.                                                                                           (2 marks)

 

Section C

 

32.(a) Describe the production of dry beans on a piece of land that has been follow under the                    following headings.

(i)      Ecological requirement                                                                                  (5 marks)

(ii)     Land preparation and planting.                                                                      (5 marks)

(iii)     Field management                                                                                        (5 marks)

(b) (i)       Give THREE causes of Blossom end rot?                                                      (3 marks)

(ii)     What is Blossom end Rot?                                                                             (2 marks)

  1. (a) State TWO types of fences.                                                                           (2 marks)

(b)     Give the advantages of hedges                                                                       (3 marks)

(c)      Outline the maintenance of the fence.                                                           (3 marks)

(d)     What information is found in calving records?                                               (3 marks)

(e)      What are aspects of wind that affect crop production?                                    (3 marks)

(f)      Briefly describe the effect of parasitism in livestock production.                     (6 marks)

 

 

 

34.(a) What characteristics make cassava an important food crop in most parts of Kenya?           (14 marks)

(b)   Name TWO diseases of cassava.                                                                           (2 marks)

(c)   State TWO symptoms of each disease named above.                                          (4 marks)

 

agRICULTURE II

MARKING SCHEME

 

  1. * Milk

*        Fur

*        Meat

*        Manure

*       Transport

*       Hide                                                                                                        (2 x1/2  =  1)

2.(i)   *        Topography / drainage

*        Living organism / vegetation

*        Climate – man – time

*        Parent rock  (4mks)

(ii)     *        Influence soil temperature whereby dark colour soils absorb and retain more
heat hence activities and survival of microbes.                                       (2 mks)

  1. * Proper feeding

*        Proper disease and pest control

*        Proper breeding

*        Proper control of worms

  1. * Cannibalism and egg eating are reduced

*        No need to provide grit as birds pick it from the soil

*        Less feed used

*        Manure is evenly spread to the runs, this helps vegetation to regenerate.     (4×1/2 = 2)

  1. Indigenous Exotic

(i).  Have hump                                          (i).  No hump

(ii). Resist high temperatures                      (ii)  Cannot resist high temperatures

(iii) Have dew lap                                        (iii) No no dew lap / small

(iv) Have high tropical diseases and             (iv) Have low resist tropical diseases and pests

pests

(v)  Can walk for long distance in                (v)  Cannot walk for long distance in search of
pasture search of pasture and water.              Pasture and water.

(vi) Have long calving interval                    (vi) Have short calving interval      (4 x 1/2  =  2)

 

  1. * To level the soil

*        To turn the soil                                                                                       (2 x 1/2  =  1)

  1. * Virus

*        Nutritional                                                                                             (2 x 1/2  =  1)

  1. (i) Man

(ii) –   Tapeworm

–  Liverfluke                                                                                               (2 x 1/2  =  1)

  1. (i) *        Soil fumigation

*        Use of crop rotation                                                                       (2 x 1/2  =  1)

(ii)     *        Aphid

*        Whitefly                                                                                       (2 x 1/2  =  1)

  1. * Crop stores

*        Crushes

*        Beehive / Breeding structures

*        Homestead

*        Fences

*        Silos

*        Cribs                                                                                                      (4 x 1/2  =  2)

  1. * Cost

*        Availability of feeds

*        Palatability

*        Physiological status / health

  1. * Because pigs are monogastric hence cannot digest grass / pasture.                        (1mk)
  2. * Washing

*        Fleshing

*        Trimming

*        Salting / drying                                                                                       (4 x 1/2  =  2)

  1. * Shell membrane is added

*        Addition of mineral, water and thin outer albumen.                                            (1mk)

  1. (i) Sisal                                                                                                                (1/2mk)

(ii)     Removal of the fleshy tissues from the fibres of sisal.                                          (1mk)

  1. * To increase the soil pH.

*        To supply soil nutrients

*        Improve soil structure                                                                     ( 3 x 1/2  =  11/2  )

  1. * Require little rainfall / precipitation

*        Mature faster                                                                                           (2 x 1/2  = 1)

  1. * Giving bees water during dry conditions

*        Controlling pests and diseases

*        Proper harvesting of honey to avoid killing bees

*        Siting  beehive away from the public road.                                              (4 x 1/2  =  2)

  1. * Heavy feeders

*        Produce a lot of seeds

*        Allotophetic

*        Smoothening effect

*        Some are parasitic                                                                                   (4 x 1/2  =  2)

  1. * Stage of growth at which forage is harvested.

*        Leaf content of the forage material

*        Method of handling and curing the material

*        Form in which material is fed to the animal

*        Species of forage used

*        Amount of foreign material in forage.                                                     (4 x 1/2  =  2)

  1. * Application of lime

*        Use of acid fertilizers                                                                                          (1mk)

  1. * To prevent foot rot.                                                                                            (1mk)

 

  1. * Poisonous to destructive agent

*        Permanent / stay long

*        Good penetration

*        Harmless to wood and metal fastener                                                       (4 x 1/2  =  2)

 

Section B.

 

  1. * Form e.g. granular / powder                                                                               (3mks)

*        Time of application e.g. pre-post plant

*        Mode of action e.g. systemmatic / non-systemmatic

 

  1. * Sheltered area protected from strong wind

*        Easily drained area

*        Near source of water

*        Should be free from stumps and roots

*        Should be fertile and well drained soil                                                          (4 x 1 = 4)

  1. (i) Polygastric / Ruminant animal                                                                            (1mk)

(ii)     P – Rumen

Q – Recticulum

R – Omasum

S – Absoption                                                                                                   (4 mks)

(iii)    P – Digestion of cellulose

S – Abomsm of water                                                                                 (2 x 1  =  2)

(iv)    * They digest in the ceaca                                                                                   (1mk)

  1. * Impart immunity to a newly born calf.

*        Has antibodies that enable the calf to resist diseases

*        Clean the bowel

*        Contain vitamins any                                                                           (Any 1 x 1 = 1)

  1. * Sand

*        Silt

*        Clay                                                                                                          (3 x 1  =  3)

29.(a) *        Clearing of bushes / grass

*        Levelling

*        Water drainage system around the structure to be made.                          (4 x 1/2  =  2)

(b)   *        Painting with old engine oil

*        Soaking in dieldrin

*        Applying chemicals using pressure

*        Charring

*        End diffusion / sap displacement / sap stream                                          (4 x 1/2  =  2)

(c)   *        Prunning

*        Gapping

*        Weeding

*        Controlling pests and diseases                                                                 (4 x 1/2  =  2)

(d) *        Expensive to transport / buy

*        Requires labour to shape them

*        Requires skills / techniques when laying                                                      (1 x 1  = 1)

30.(a) *        It is faster in pest control than other methods

*        The results of pesticides can be predicted.

*        More effective                                                                                           (1 x 2  =  2)

(b)  *        Lower yields / quantity

*        Lower quality of crop production

*        Causing poisoning to consumers

*        Increase production costs                                                                        (4 x 1/2  =  2)

(c)    *        Banana diseases – Cigar end rot, Panama disease, Sigatoka                     ( 1 /2  x 2 =  1)

*        Banana pests – Banana weavil, Banana thrips, Nematodes                       ( 2 x 1 /2 =  1)

31.(a)     Risk is the difference between the expected and the actual outcome and it could be
estimated while uncertainty is a situation in which the outcome cannot be estimated .
( 1mk)

(b) *        Diversification

*        Selecting a more reliable enterprise

*        Insurance

*        Guranteed prices

*        Input rationing                                                                                      ( 4 x 1 /2 =  2)

(c) *        What extra cost the farmer is going to incur

*        What revenue is foregone as at the proposal

*        What extra revenue is to be earned

*        What costs are saved as a result of the proposal                                       ( 4 x 1 /2 =  2)

 

Section C.

 

BEANS

32.(a) (i) Ecological requirements

*        Well drained loam soils

*        Beans require moist soils through out growing period

*        Moderate rainfall

*        No rain at harvesting time

*        Irrigation can be done                                                                                   (5 x 1 – 5)

(ii)Land Preparation

*        Clear land before onset of rains

*        Cultivate land to required tilth not so fine (medium tilth)

*        Seeds should be dried before planting – select seeds for planting

*        Plant at the onset of rains

*        Planting is done by placing 2 – 4 seeds per hole

*        Spacing (30 x 15)cm

*        Apply Diammoniun phosphate at rate of 200kg/ha before planting.                  ( 5×1=5)

(iii)    Field Management

*        Weeding done before flowering

*        Weeding done when soil is dry

*        Hand weeding is done

*        During dry months irrigate land

*        Use furrow irrrigation

*        Control of diseases such as Bacterial Halo/blight and Anthracose, planting                                                         resistant varieties.

*        Pest control by use of Benomyl Copper fungicide or mancazeb

*        Pests include bean aphid; bean bruchids; Spotted borer, American
bollworm, Beanfly, Golden ring moth                                                           ( 5×1=5)

(b)(i)  *        Too much Nitrogen

*        Too dry and too wet conditions

*        Calcium deficiency                                                                                          ( 3mks)

 

(ii)     Physiological diseases that occur when tomato plants are exposed to too dry and                                    too wet conditions, calcium deficiency                                                    ( 2mks)

33.(a) (i) Live fence e.g. (a) Hedges and growing trees (b) Electric fence

(ii)     Dead fence e.g.

*        Post and wire i.e. barbed wire, plain wire

*        Post woven wire rail fence

*        Wall fence e.g. stone break

*        Trench fence                                                                                                    ( 2mks)

(b)     *        Has more aesthetic value i.e natural

*        Act as a windbreak and controls soil erosion

*        It is easy and cheaper to establish

*        May be a source of fodder to livestock

*        It can be a source of firewood.                                                                          ( 3mks)

(c)     *        Replace any broken / rotten post

*        Any loose fence wire should be tightened appropriately

*        Trim the hedge and any gaps filled, the fencing posts should be treated first with                            preservatives such as old engine oil, creosate or charred.                                    ( 3mks)

(d)     *        Sex of the calve

*        Date of birth

*        Breed of the mother / dam

*        Weight

*        Breed of size                                                                                    ( any given 3mks)

(e)     *        Strength

*        Direction

*        Humidity    ( 3 x 1 = 3)

(f)      *        Cause wounds

*        Low quality

*        Cause irritation

*        Spread diseases

*        Reduce quality of product

*        Increase cost of production                                                                         ( 6 x 1 = 6)

  1. (a) *        It is drought resistant

*        Gives good yield in poor soils

*        Require less labour

*        Sheds its leaves during dry season thus reduce water evaporation

*        Has low nutrient requirement

*        Resistant to certain pests and diseases

*        Can be used as flour and boiled                                                                 ( 7 x 2 = 14)

(b)     Diseases of cassava

(i)      Cassava mosaic                                                                                                 (1mks)

Symptoms

         *        Leaves turn yellow

*        Reduced yield

*        Stunted growth

*        Distorted leaves                                                                                                (2mks)

(ii)     Brown streak                                                                                                    (1mks)

Symptoms

*        Old leaves develop yellow buds at veins

*        Brown patches on tubers                                                                                   (2mks)

(iii)    Bacterial blight                                                                                                 (1mks)

Symptoms

*        Leaves develop brown patches

*        Shoot tips turn brown

*                   Wilting of leaves that easily fall off            (2mks)

 

AGRICULTURE III

PART I

 

SECTION A:

 

  1. What is a forage crop? (1 mk)
  2. List four features of a fish pond. (4 mks)
  3. State two disadvantages of persistent use of pesticides in crop production. (2 mks)
  4. State two reasons for seasoning timber before use. (2 mks)
  5. State four disadvantages of natural mating as a method of breeding in dairy cattle. (4 mks)
  6. Why should smoke be used during harvesting of honey. (2 mks)
  7. State four functions of potassium in plant growth. (4 mks)
  8. State four routine management practices that should be carried out in sheep production

(4 mks)

  1. State four ways of improving productivity of farm labour. (4 mks)
  2. If you are told a newly released Kitale hybrid maize seed is H823. What does the 3 digits stand for? (1 ½ mks)
  3. (a) Why are insecticides made from pyrethrum recommendable than synthetic insecticides?

(b)  (i)Pyrethrum is propagated by which means?                                                     (1 mk)

(ii) What is cutting back in pyrethrum growing?                                                            (1 mk)

  1. How does a vaccine work in the body of an animal? (2 mks)
  2. State one characteristic of organic matter that enables it to perform each of the following roles in the soil.

(i) Improvement of soil structure.                                                                            (2 mks)

(ii) Improvement of water holding capacity.                                                             (2 mks)

  1. State the symptoms of attack by the sweet potato weevils. (2 mks)
  2. Name two tools that may be used to dock lambs. (2 mks)

 

SECTION B:

 

  1. Give four reasons why the feeding of colostrum is important in the rearing of piglets. (2 mks)
  2. Give two reasons why green manures are not commonly used by small scale farmers. (2 mks)
  3. When is opportunity cost said to be zero? (1 mk)
  4. State two reasons for proper record keeping. (2 mks)
  5. State four reasons why zero grazing is becoming increasingly popular in small scale farming
    in Kenya. (2 mks)
  6. Name four methods of land reclamation. (2 mks)
  7. State four components of cattle dip. (4 mks)
  8. State two functions of ventilation in an animal house. (2 mks)
  9. Mention four important crush practices in a dairy farm. (2 mks)
  10. (a) Differentiate between the following. (3 mks)             (i) Furrowing and farrowing.

(ii) Candling and kindling.

(iii) Undersowing and oversowing.

  1. Mention four problems associated with soil erosion. (2 mks)
  2. Name three sources of water in the farm. 2½ mks)
  3. (a) Give two factors that may lead to longer calving interval in a dairy cow. (2 mks)

(b) State three benefits of giving a pregnant dairy cow a special diet for 6 weeks

before calving.                                                                                                   (2½ mks)

(c)  List four signs of infertility in dairy cows.                                                         (2 mks)

  1. (a) List three problems that are faced by farmers who practice mono-culture. (3 mks)

(b)  What is meant by seed inoculation?                                                                  (1 mk)

  1. State the main reason why sharp corners should be avoided in a brooder for chicks. (1 mk)
  2. (a) Apart from reducing friction, what other functions does lubrication perform in farm

machinery? (List four).                                                                                                (2 mks)

(b)  Give four ways through which a farmer may improve production efficiency without necessarily incurring extra cost.                                                                         (2 mks)

 

SECTION C:

 

32.(a)i.  How is the nitrogen element lost from the soil?                                    (5 mks)

  1. Describe the method  by which nitrogen may be restored to the soil.      (8 mks)

(b) (i) Describe how high quality farm yard manure could be made and stored.   (5 mks)

(ii) Outline the factors which influence the quality of farmyard manure.      (2 mks)

  1. (a) List the factors that could make meat unfit for eating by man.              (4 mks)

(b)  Outline the procedure for clean milk production.                                               (10mks)

(c)  Why are goats suited to most parts of Kenya?                                        (6 mks)

  1. (a) What are the effects of endo-parasites in livestock? (5 mks)
  • With the aid of diagrams describe briefly the life cycle of liverfluke. (5 mks)
  • Describe foot and mouth disease under the following headings:

(i) Cause.                                                                                              (1 mk)

(ii) Symptoms.                                                                                      (2 mks)

(iii) Control measures.                                                                          (2 mks)

(d)  State five signs of a good layer among a flock of hens.                           (5 mks)

 

 

AGRICULTURE III

MARKING SCHEME

 

  1. Plant which either grows naturally or cultivated and used for feeding livestoc (1 mk)
  2. (i) Inlet                                         (iii)  Pool/tank
  • Spillway (iv)  Fence
  • Outlet (any 4×1 = 4 mks)
  1. – Pesticides kill beneficial insects.
  • Some pests build resistance to pesticides.
  • Pesticides have residual effects.
  1. – To avoid warping.
  • To reduce attack by pest/fungi.
  1. – There is no control in breeding.
  • Can cause spread of vaginitis disease.
  • Wastage of sperms.
  • Sperms can be used in remote areas. (4×1 = 4mks)
  1. To make bees less aggressive/less active. (2 mks)
  2. Mark any relevant answer correct.
  3. – The feet should be trimmed regularly.
  • Dipping should be regular, but should be stopped one month before lambing.
  • Sheep should be vaccinated regularly.
  • Sheep should be dewormed regularly. (4×1 = 4 mks)
  1. – Giving incentives.
  • Training labour.
  • Farm mechanization.
  • Labour supervision.         (any 4×1 = 4 mks)
  1. 8 = Altitude.

2 = Serial number.

3 = Number of crossing.                                                                              (1 ½ mks)

  1. (a) Easily broken down. (1 mk)

(b)  (i) Splits                                                                                               (1 mk)

(ii) Removal of old stems upto the level of foliage.                                 (1 mk)

  1. A vaccine induce temporary antibodies which initiate the formation of antigen in balance to await the anticipated antibodies of the diseases to attack. (2 mks)
  2. (i) Decomposing enhancing the breaking down of particles of soil thus the soil structure is

improved.                                                                                             (2 mks)

  • The organic matter decomposes to form the elements that are incorporated in soil, thus increasing capilarity adhesiveness of H2
  1. – Larvae

–     Tubers become discolored .                                                       (any two 2×1 = 2 mks)

  1. – Scalpel.

–     Docking knife.

 

SECTION B:

  1. – Impart immunity.
  • Contains necessary nutrients.
  • Faster growth.
  • Facilitate metabolism. (4 x ½ = 2 mks)
  1. – Crops are harvested when green matter has been withered.

–   It is expensive.                                                                          (2×1 = 2 mks)

  1. When there is no choice to be made. ( 1mk )

 

 

  1. – To follow good farm plant.

–   To assist in credit acquiring.                                                      (2×1 = 2 mks)

  1. – Mark any correct relevant answer.
  2. – Drainage.

–     Tse-tse fly control.

–     Reafforestation.

–     Afforestation.                                                                                       (4 x ½ = 2 mks)

22.(a)-  Collecting pen.

  • Dipping tank.
  • Drainage race.
  • Entrance race.
  • Waste pit/tank.
  • Water source/tank. (4×1 = 4 mks)

(b)  –     Remove mud from hooves/clean hooves.                                                           ( ½ mk)

  1. – Allows efficient air circulation in the house.
  • Prevents inside of the house from becoming humid.
  • Controls temperature in the house. (2×1 = 2 mks)
  1. – Milking.
  • Isolation
  • (4 x ½ = 2 mks)

25.(a)-  Cutting/making ditches/channels on farm.

–     Giving birth in pigs.                                                                              (1×1 = 1 mk)

(b)  –     Is observation of the inner part of the egg against strong light.

–     Is the giving birth in rabbits.                                                                 (1 x 1 = 1 mk)

(c)  –      Less seed rate applied/establishment of pasture under an already growing crop/nurse crop

–   More seeds rate applied/establishment of a pasture legume/grass on an existing grass

pasture.                                                                                                                                                         (1 x 1 = mk)

  1. – Fertility is reduced.
  • Useful micro-organisms are washed.
  • Dams are silted.
  • Plants roots exposed.
  • (Any 4 x ½ = 2 mks)
  1. – Dams/reservoirs.
  • Tanks
  • Rivers/sea
  • (Any 3 x ½ = 2 ½ mks)

28.(a)-  High milk production.

  • Irregular heat signs.
  • Poor nutrition.
  • Poor health.
  • Incorrect timing of service. (2×1 = 2 mks)

(b)  –     High quality colostrum.

  • Strong heavy and health calf is obtained.
  • Results in high milk yield. Gives cow enough energy for calving.
  • To a customs the animals to concentrates feed.
  • To stimulate development. (Any3 x ½ =2½ mks)

(c)  –     Absence of estrous.

  • Prolonged estrous.
  • Irregular heat intervals.
  • Conception failure after service.
  • Abnormal discharge from vulva. (Any 4x½ = 2 mks)

29.(a)-  Build up of pest or diseases.

  • Breakdown of soil structure.
  • Soil erosion may be a problem.
  • Loss of soil fertility. (3×1 = 3 mks)
  • Introducing a suitable strain of nitrogen fixing bacteria to legume seeds. (1 mk)

 

  1. To avoid suffocation due to overcrowding/kindling of chicks in corners. (1 mk)

 

31.(a)-  Cushions/prevents rubbing together of moving parts in machinery.

  • To prevent rusting.
  • Absorbs heat and traps pieces of metal, which comes from moving surfaces.
  • Improves work efficiency machines.
  • Reduces rate of rear and lear. (any 4x½ = 2 mks)

(b)-   Use of improved or modern farming methods.

  • Improved farm management.
  • Mechanization of farm operations.
  • Efficiency in use of labour.
  • Select enterprises that can bring highest return.
  • Organizing marketing activities to realize highest price. (any 4x½ = 2 mks)

32.(a)-  Denitrification.

  • Soil erosion.
  • Crop absorption. (any 5×1 = 5 mks)

(b)  –     Nirtogen fixation by root nodules bacteria and free living bacteria.

  • Application of inorganic manure e.g. guano manure.
  • Crop rotation/planting of legumes.

(c)(i)-   Plant material is placed on concrete floor indoors.

  • Animals defecate on it and mix it with urine and dung .
  • Cover the heap with soil or polythene sheet to prevent leaching.
  • Consolidate heap to prevent entry of water.
  • Allow it to dry completely before being used. (1×5 = 5 mks)

(ii)

  • Type of bedding used.
  • Age of animal.
  • Type of feed given to animal.
  • Type of animal from which dung is obtained. (4 x ½ @ = 2 mks)

33.(a)-  Animal infested by zoonotic diseases animal infested by par.

  • Animal found dead due to unknown cause.
  • Uninspected meat by environmental health authorities.
  • Contamination e.g. feaces, flies.

(b)  –     Ensure that the cows are clean and healthy.

  • Milking equipment this should be seamless to make them easy to clean – they should be cleaned thoroughly after each milking.
  • Milking parlour/shed.
  • Situated in dust free environment.
  • The floor should be made of concrete to facilitate cleaning.
  • Cleaned thoroughly after milking.
  • Milking jelly to avoid cracking.

 

 

(iv)  –     Milker – should be healthy i.e. free from zoonoses.

–     Should be clean always.

  • Should have white coat to help in detecting dirt.
  • Should have short finger nails.
  • Wash his hand well with soap before milking.

(v)  –     Best for mastitis before milking. Using a strip cup, animals having mastitis should be milked last to avoid spread of diseases.

(vi) –     Milk storage and handling.

  • Cool milk soon after milking and store it.
  • A cool place to reduce bacterial growth and multiplication.
  • Filter and cover milk to.
  • Remove and keep away dirt resepctively.(1mk each points x 5 = 5mks

1mk for explanation x 5 =5mks)            =10 mks)

(c)  –     Goats are tolerant.

  • Goats are drought resistant.
  • They eat variety of vegetation.
  • They are tolerant to heat and high temperature. (Any 3 points x 2mks = 6 mks)
  1. (a) – Damage to animal organs.
  • Obstruction in the alimentary canal.
  • Parasites eat food intended for use by livestock.
  • Suck blood.
  • Cause irritation. (Any 5×1 = 5 mks)

 

(b)

Developing

Embryos

 

 

Egg in feaces                                              larval

Forms

In snail

 

 

 

Animal eats                                                cyst

Encycited

Cercarian

Drawing 2 ½ mks

                                                                  Description 2 ½ mks

(c) Foot and mouth:

Causes – virus type A, C and D.             (1 mk)

 

Symptoms:

  • Blisters/wounds appear on the mouth and feet.
  • Tongue ,lips, and gums are inflamed.
  • Lesions appear between the skin and hoof.
  • Profuse salivation.
  • Animal weak and thin.
  • Drops in milk production. (Any 2×1 = 2 mks)

      Control:

  • Kill animal affected.
  • Nurse animals with wound by use of antibiotics. (Any 2×1 = 2 mks)

 

 

 

(d)  –    Signs of good  layers:

  • Combs and wattles are large, warm and waxy and red.
  • Bright orange and alert eyes.
  • Pale beak.
  • The rent is oval, moist, reddish in colour and active.
  • Abdomen is soft, pliable and wide.
  • The space between keel and pelvic bone is wide and can fit 3-4 fingers.
  • Alert and active.
  • Dry and rugged feathers.
  • Moulting starts late.
  • Shanks are pale.
  • Brooding is rare. (Any 5×1 = 5 mks)

 

AGRICULTURE IV

 

SECTION A.

 

  1. State four factors that should be considered when selecting materials for constructing a dairy shed. (2mks)
  2. Give two reasons for raising Kale seedlings in the nursery before translocating (2mks)
  3. State two conditions that may lead to sub-division of agricultural land (1mk)
  4. State four reasons why piglets should be weighed immediately after farrowing and weaning

(2mks)

  1. List four characteristics of clay soil (2mks)
  2. Give two reasons why sorghum should be harvested, for making silage just before flowering stage (2mks)
  3. (a) What is the duration of oestrus cycle in a cow. (1mk)
    (b) If  a dairy cow is noticed to be showing first signs of heat at 6.00am. What time should it

be inseminated                                                                                                               (1mk)

  1. State three advantages of chilting in Irish potato seeds (3mks)
  2. State four tests a farmer would carry out in determining the quality of a fresh egg. (2mks)
  3. Differentiate between Dams and well as used in water supply (2mks)
  4. State any two characteristics symptoms of viral diseases in plants (1mk)
  5. State four stages of curing hides (2mks)
  6. Give two factors that influence the time of first defoliation in newly established pastures

(2mks)

  1. Outline two effects of adding organic matter to sandy soil
  2. State four characteristics of horizon ‘A’ of a typical soil profile                   (4mks
  3. Why should a farmer plant grass around a fish pond. (2mks)
  4. Give three causes of post- harvest losses of maize to the farm. (3mks)
  5. Give two characteristics of the head of a corriedale sheep breed. (2mks)
  6. Mention four pests that may cause reduction in honey yield from top bar hive. (2mks)
  7. Give three danger signs indicating breeding problems in a Friesian cow.

SECTION B

  1. Give four factors that determine the pyrethrin content in pyrethrum (4mks)
  2. Why do cereal crops dominate arable land. (3mks)
  3. (a) List four essential components of zero grazing unit. (2mks)
    (b) State four reasons for practising zero grazing system.                                                 (2mks
    (c) Give two reasons for the two month dry period, the cow, requires before parturation.(2mks)
  4. State three ways of increasing efficiency in farming.                                                       (3mks)
  5. (a) State two reasons for maintaining livestock in good health.                             (2mks)
    (b) Name two noticeable diseases of cattle.                                                                      (2mks)
    (c) State four measures that should be taken to prevent an out break of Newcastle disease in

poultry                                                                                                                         (2mks)

  1. (a) Differentiate between market and marketing. (2mks)
    (b) What is an imperfect market.                                                                                        (1mk)
    (c) How will the price of mangoes in the short run, be affected if the quantity of mangoes supplied in a market is increased?                                                                                     (1mk)
  2. State any three advantages of vegetative propagation. (3mks)
  3. State three factors which would be considered when siting a compost pit. (3mks)
  4. State four factors that should be considered when planning a lay out of mixed farm. (2mks)
  5. (a) Name any three roofing materials commonly used. (3mks)
    (b) Give three disadvantages of using wood fuel as a source of power.
  6. (c) Name three examples of one point hitch implements drawn by the tractors. (3mks)

SECTION C

  1. (i) Describe cattle Trypanosomiasis under the following sub headings.
  2. (a) Cause and Transmission. (1mk)
    (b) Symptoms.                                                                                                                     (5mks)
    (c) Control                                                                                                                           (3mks)
    (ii) What are  the causes of infertility in cows.                                                                   (4mks)
    (iii) Describe the management of rabbits between mating and weaning.                           (3mks)
  3. Describe the management practices that should be carried on a low yielding herd of dairy cattle to make it more productive.                                                                                     (2mks)
  4. (a) What causes losses of maize during storage. (4mks)
    (b) What practices should be carried out to minimise losses of maize during storage?     (6mks)
    (c) What are the functions of the National Cereals and Produce board in the marketing of

maize?                                                                                                                         (10mks)

AGRICULTURE IV

MARKING SCHEME.

 

  1. – The cost
    – Durability
    – Workability
    – Easy to clean
    – Availability
    (4 x ½ = 2mks)

 

  1. To increase rate of germination
    – To nurse the seedlings before transplanting so that they have good establishment
    (2 x ½ = 1mk)

 

  1. – Selling of land
    – Population increase/Pressure on land
    (2 x ½ = 1mk)

 

  1. – To determine the food to be eaten
    – Check whether the piglets are healthy
    – To keep proper records
    – To have proper management
    (4 x ½ = 2mks)

 

 

  1. – Poorly drained
    – Poorly aerated
    – Fine texture
    – Has higher amount of nutrients
    (4 x ½ = 2mks)

 

  1. – High nutrient value
    – Succulent hence palatable
    – Higher decomposition
    (2 x 1 = 2mks)

 

  1. (a) 19hrs (1mk)
    (b) 6.00 am + 10 hrs = 4.00 pm (1mk)

 

  1. – Easy establishment
    – Early maturity
    – Easy germination

 

  1. – Candling
    – Water method
    – Opening the egg
    – Observation
    (4 x ½ = 2mks)

 

  1. A dam is a wall build across a river to prevent flow of water to form a lake. While a well is a wall/stamped across water. (2mks)

 

  1. – Swelling
    – Mottling of leaves
    – Chlorosis of leaves

 

  1. – Washing
    – Fleshing
    – Salting/Drying
    (4 x ½ = 2mks)

 

  1. – The rate of growth of forage
    – The type of animal to be fed
    (2 x 1 = 2mks)

 

  1. – Moisture content of Soil
    – Tilth of soil
    (2 x 1 = 2mks)

 

  1. – Tap roots are formed here
    – Little humus
    – Slightly compact/Yellowish
    – Active micro – organisms
    (4 x 1 = 4mks)

 

  1. – To prevent soil erosion/forming the soil
    – Attract insects to the pond for fish to feed on.
    (2 x 1 = 2mks)

 

 

  1. – Improper or incomplete drying of grains
    – Wet heating (Placing in sacks with the maize on the floor)
    – Drainage by pests e.g rodents/weavils
    (3 x 1 = 3mks)

 

  1. – Harmless
    – Well wooled but free from wool blindness
    (2 x 1 = 2mks)

 

  1. – Ants – birds
    – Wax – moth – Bee louse
    – Hive beetle – Death head hank moth
    – Pirate wasps
    (4 x ½ = 2mks)

 

  1. – Lack of heat
    – Abnormal discharge from the reproduction track
    – Irregular heat intervals
    – Abortion
    – Prolonged heat period
    (3 x 1 = 3mks)

 

  1. – Part of the plant
    – Stage of flower development
    – Genetic constitution
    – Age of the plant
    – Handling/Care during picking drying and dispatch
    (4 x 1 = 4mks)

 

  1. – They are the most important source of carbohydrate
    – Harvesting of cereals is easier
    – Storage and transportation are easier
    – Cereals are widely adapted to various ecological zones
    (3 x 1 = 3mks)

 

  1. (a) – Sleeping cubicles
    – Milking place
    –  Feeding and watering troughs
    –  Calf pens
    –  Loafing/exercise area
    –  Store

 

– High milk production is obtained
– Allows higher stocking rate
– Animals make maximum use of fodder without wastage
– Rapid accumulation of manure
– Minimises outbreak of diseases
– Avoids overgrazing
(4 x ½ = 2mks)

(c)
– To obtain a strong, heavy and healthy calf at birth
– To increase milk production in the next lactation
– To allow accumulation of body reserves used in the formation of colostrum

  1. – Select proper/more paying enterprise
    – Proper crop husbandry practices
    – Proper livestock husbandry
    – Mechanisation
    – Adopt new techniques
    – Timeliness of farm operations
    (3 x 1 = 3mks)

 

  1. (a)
    – Increase quantity of livestock product
    – To increase quality of livestock product
    – To increase profit level
    – To prevent the spread of zoonotic diseases
    – To increase productive life of livestock
    (2 x 1 = 2mks)


(b)
(i) Foot and Mouth disease
(ii) Anthrax
(iii) Rinder pest
(iv) Lampiy sick disease
(v) Rabies
(2 x 1 = 2mks)

(c)
– Vaccinate birds at regular interval
– Isolate affected birds/destroy affected birds
– Disinfect the house before bringing in new stock
– Impose quaratine in the farm
– Ensure proper farm hygiene
(4 x ½ mks = 2mks)

 

  1. (a) – Market is a place where buyers and sellers meet to buy and sell goods and services,

where as marketing is the performance of business activities that direct the flow of goods

and services from producers to consumers                                                       (2mks)
(b) An imperfect market is a situation in which some buyers, some sellers or both have limited

knowledge of goods and services offered for sale at various forces/few buyers and sellers

/different, goods in to the market.

(c) The Price of mangoes go down.        (1mk)

 

  1. – Easy establishment of plant
    – Maintains the present genetics
    – Seedless plants can be propagated
    (3 x 1= 3mks)

 

  1. – Proximity to the farm/Crop field
    – Accessibility
    – Topography/Soil drainage
    – Soil type
    (3 x 1 = 3mks)

 

  1. – Size of the land
    – Topography of the land
    – Direction of the wind
    – Soil fertility
    – Capital availability
    (4 x ½ = 2mks)

 

  1. (a)
    – Corrugated iron sheets
    – Tiles
    – Asbestors
    – Alluminum sheets
    – Thatch/Palm leaves
    – Timber/Wood shingles
    (3 x 1 = 3mks)

    (b)
    – Leads to environmental destruction
    – Air pollution
    – Exhaustible
    – Limited uses
    – Cannot be regulated/uneconomical
    (3 x 1 = 3mks)

 

  • – Trailers
    – Heavy harrow
    – Planters (3 x 1 = 3mks)

 

  1. (a) Caused organism protozoa (1mk)
    (b) Symptoms
    –   High temperature or fever
    –  Dullness
    –  Animal looses appetite
    –   Body becomes very weak
    –   Lachrimation
    –   Diarrhoea
    –   Milk production decreases
    –   Loss of hair at Tail end
    –   Animal has anaemia
    –   Abortion may occur in pregnant females

 

  • Control
    – Treat the animal with typanoccidal drug
    – Control tsetse flies by bush clearing spraying
    – Confinement of game animals in game parks

(ii) –  Causes  infertility in cows
–  Damaged uterus caused by abortion
– Infection such as viginitis, brucel lossis;
–  Retained placenta
– Blocked fallopian tube as a result of infection
– Lack of essential nutrients like vitamin E
– Frematin: a heifer born twin with a bull is 90% infertile

 

Mating
–  Young females should be mated when they are 6 – 7 months old
–  Old females should be mated after kindling
–  Take the doe to the bucks  hutch for mating
–  Have one back to give does

Preparation for Kindling
   –  Gestation period 31 days (one month)
–  Clean and put in  a nest box  nestling materials four days to kindling
–   Place the box in the darkest corner of the hutch since the doe likes kindling where it is

dark

 

Rearing the Kindles
–  Regularly check the nest to remove sick, weak and dead babies
–  Check daily to ensure that all the babies feed well
–  Feed kindles on mothers milk for the 2 ½ – 3 weeks
–  They allowed to stay with mother until 8th week when they are weaned
(8 x 1 = 8mks)

 

  1. – Proper selection of the herd
    – Proper disease control
    – Proper pest control
    – Proper housing
    – Proper feeding
    – Spraying animals against external parasites
    – Proper recording of management activities
    – Cross breeding up grading the herd
    – Proper serring the dam
    – Proper use of good sire
    – Deworming against internal parasites
    (stating 10 x 1 = 10mks, Explanation 10 x 1 = 10mks)

 

  1. (a)
    – Poor drying
    – Damage by pests
    – Poor processing
    – Wet heating
    (4 x 1 = 4mks)

 

  1. – Drying
    – Control
    – Proper processing
    – Dusting
    – Storage in ventilated stores
    – Store away from wet conditions
    (6 x 1 = 6mks)

 

  1. (i) Buy maize when harvest is high
    (ii) Store maize
    (iii)  Sell maize when there is shortage
    (iv)  Import maize
    (v)  Export maize
    (vi)  Dispose maize
    (vii)  Advice the Minister on the proper production of maize
    (ix)  Control prices of maize in the market
    (10 x 1 = 10mks)

 

 

 

AGRICULTURE V

SECTION A.

 

  1. List four advantages of individual owner tenure system (2mks)
  2. State two ways to show how check dams reduce soils erosion (1mks)
  3. Identify four factors that contribute to competitive ability of weeds (1mk)
  4. Mention four ways of classifying herbicides (2mks)
  5. List two ways of controlling smut disease in the field. (1mk)

(a) Name any two pests that attack bean pods in the field                                           (1mk)
(b) Give four examples of French beans/Green beans                                                  (2mks)

  1. What four factors should a farmer consider for effective control of pests in the field (2mks)
  2. List four ways of increasing carrying capacity of pastures (2mks)

(a) Name any four livestock attacked by trypanosomiasis disease                                 (2mks)
(b) What organism causes Gall sickness disease in livestock                                        (1mk)

  1. State four signs of heat in a heifer (2mks)
  2. List four factors that determine the number of times a seed bed is harrowed (2mks)
  3. Name four systems of irrigation (2mks)
  4. Name any two examples of dual purpose breeds of cattle (1mk)
  5. State any two factors that influence the rate of leaching in soils (1mk)
  6. Give any four information contained in a feeding record in pig production (2mks)
  7. State four ways of modifying temperature in crop production (2mks)
  8. Name four methods of vegetative progagation (2mks)
  9. Give two reasons why ewes disown their lambs soon after lambing (1mk)
  10. What four factors determine the type of fence to construct in the farm (2mk)
  11. List any two deficiency symptoms of nitrogen in crops (1mk)
  12. List any four examples of marking tools in a workshop   (2mks)

(a) Name the system of breeding where low grade female is mated to pure bred sire    (1mk)
(b) Give two classes of concentrate feedstuffs                                                                          (1mk)

SECTION B

  1. (a) State four signs of attack by Rinderpest disease (2mks)
    (b) List three disease predisposing factors in livestock                                               (3mks)
    (c) (i) State four disadvantages of animal drawn implements                                     (2mks)
    (ii) List two advantages of artificial incubation of eggs                                         (1mk)
  2. (a) Define the term pest in agriculture (1mk)
    (b) State four physical measures of controlling pests in the field                               (4mks)
    (c) (i) Name two common viral diseases in cassava                                                   (2mks)
    (ii) List two physiological disorders in crops                                                             (1mk)
  3. (a) Give three uses of cement in the construction of farm structures (3mks)
    (b) What four factors should be considered when siting a nursery bed                      (2mks)
    (c) (i) Mention two farm structures used for water storage in the farm                     (1mk)
    (ii) List four disadvantages of a cattle dip                                                           (2mks)
  4. (a) State one function of each one of the following parts of an ox-plough
    (i)  Main beam
    (ii) Frog
    (iii) Share
    (iv) Hake                                                                                                                (2mks)
    (b) Identify three maintenance practices done on the ox-plough                               (3mks)
    (c) List three operational differences between a mould board plough and a disc plough                                                                                                                                      (3mks)

27.(a) State three important characteristics of Hereford breed that makes it particularly suitable

for beef production                                                                                              (3mks)

 

(b) What three major character expressions would you look for in the performance records of a

dairy cow for breeding                                                                                          (3mks)
(c) Give the period of sexual maturity in months in each of the following livestock      (2mks)
(i) Sow
(ii) Ewe

28.(a) Explain ways of improving the quality and quantity of crop yields.                      (13mks)
(b) Explain the effect of soil PH on crop production                                                 (7mks)

  1. Describe production of maize for dry grains under the following sub-heading (20mks)
    (i) Land preparation
    (ii) Planting
    (iii) Field operations                                                                                               (5mks)
    (iv) Harvesting                                                                                                      (3mks)
  2. (a) Describe fowl pox disease under following sub-headings
    (i) Animals attacked                                                                                              (1mk)
    (ii) Causal organism                                                                                               (1mk)
    (iii) Symptoms of attack by diphtheritic and cutaneous types of fowl pox                 (8mks)
    (iv) Control measures                                                                                            (2mks)
    (b) Explain factors that influence vices in poultry                                                   (10mks)

 

AGRICULTURE  V

MARKING SCHEME

 

  1. – Title deed used to secure loan credit
    – Independent production plan
    – Accessibility to agricultural advice
    – Earn income by leasing/renting land
    – High production of quality produce
    – Proper supervision of land                                                                        (4 x ½ = 2mks)
  2. – Trap sediments/soil
    – Slow down the speed of run off
    – Reduce volume of run off (2 x ½ = 1mk)
  3. – Produce large quantities of seeds
    – Seeds have a long viability
    – Propagate vegetatively
    – Extensive root system
    – Hardy/survive adverse weather and soil condition
    – Have a short life cycle
    – Gross feeders/heavy feeders (4 x ½ = 2mks)
  4. – Mode of action
    – Time of application
    – Selectivity
    – Formation/physical form of herbicide (4 x ½ = 2mks)
  5. – Rogueing/uproot and burn infected crop (reject rogueing alone)
    – Crop rotation
    – Plant resistant varieties of maize
    – Plant certified seeds
    – Avoid application of infected organic manure (2 x ½ = 1mk)

6.a. – American bollworm (reject bollworm alone)
– Flower thrips                                                                                             ( 2x ½ = 1mk)
(b) – Primeur
– Long tom
– Saxa
– Master Piece
– Monel                                                                                                 (4 x ½ = 2mks)

  1. – Nature of crop damage caused
    – Biology of pest/reproduction of pest/feeding habits
    – Weather conditions favouring pest attack
    – Whether pest has natural enemies
    – Population level of pest (4 x ½ = 2mks)
  2. – Irrigation
    – Application of manures / fertilisers
    – weed control
    – Control of pests
    – defoliation at right growth stage (4 x ½ = 2mks)

9.(a) – Cattle (reject cow or bull alone)
– Sheep
– Goats
– Pigs                                                                                                      (4 x ½ = 2mks)

(b)   –  Protozoa – Anaplasma marginale                                                           (1mk)

  1. – Reddening and swelling of vulva
    – Heifer stands still to be mounted by others and also mounts other cows
    – Clear slimy mucus discharge comes out of vulva and sticks on the tail
    – rise in body temperature/above 39oC
    – Heifer is restless and loses appetite
    – constant bellowing/mooing
    (reject slight drop in milk production) (4 x ½ = 2mks)
  2. – Type of crop to be planted/size f planting material
    – Initial condition of land
    – Time available before sowing
    – Topography of land (4 x ½ = 2mks)
  3. – Surface/flood/furrow/basin irrigation
    – Overhead/Sprinkler
    – Drip/Trickle irrigation
    – Sub- surface (4 x ½ =  2mks)
  4. – Sahiwal
    – Red poll
    – Simmental (2 x ½ = 1mk)
  5. – Soil type
    – Vegetation cover
    – Rainfall amount and intensity
    – Topography of land/slope of land (2 x ½ = 1mk)
  6. – Date
    – Number of pigs

–  Amount of feed received
–  Amount of feed used
–   Type of feed
–   Balance in stock
–   Remarks

  1. – Mulching
  • Shading crops
  • Irrigation / watering
  • Pruning (4 x ½ = 2mks)
  1. – Grafting
    – Budding/bud-grafing
    – Layering
    – Use of storage organs
    – Use of stem cuttings                                                                               (4 x ½ = 2mks)
  2. – Painful udder
    – Poor maternal instinct
    – Too weak lambs
    – Low milk production (4 x ½ = 2mks)
  3. – Local availability of materials
    – skills available
    – Purpose of the fence- climate of the area
    – Topography of land
    – Capital available/cost of the fence
    – Preference of the farmer (4 x ½ = 2mks)
  4. – Leaf chlorosis/yellowing of leaves
    – Stunted growth
    – Brown leaves and premature leaf fall
    – Retarded flowering and fruiting
    – Retarded flowering and fruiting (2 x ½ = 1mk)
  5. – Scriber
    – Knives
    – Divider
    – Pencils
    – Marking gauge/mortise gauge
    – Punches/centre punch/dot punch/nail punch (4 x ½ = 2mks)
  6. a. – Grading up/ upgrading                                                                      (1mk)
    (b) – Energy/carbohydrate concentrate
    – Protein concentrate

23.(a) – High body temperature/high fever
–   Starring coat
–   Discharges in the mouth and nose
–   Watery eyes
–   Diarrhoea and dysentery
–   Red mucus membrane of the mouth, nose
–   Emaciation
–   Grinding of teeth
–   Death in 2 – 10 days in acute cases but may live for 3 or more weeks in less acute cases
(4 x ½ = 2mks)

(b) – Age of animal
– Sex of the animal
– Colour of the animal
– Body conformation
– Change of climate
– Environment
– Size of the herd/flock
– Body conditions/fatigue/weakness
– Animal movements

(c) (i) – More tedious to the operator
– Requires more than one person
– Slower/work output less
– Animal tires quickly
– Attack by diseases makes it difficult to use the animals
– A piece of land has to be set aside to grow fodder crop or develop pasture for animals         (4 x ½ = 2mks)

(ii) – Many chicks can be hatched at ones time
– It is possible to plan when to hatch chicks                                               (2 x ½ =1mk)

  1. (a) Organisms that causes damage to plants either directly or indirectly by introducing

organisms which cause disease.                                                             (1mk)

(b) – Use of lethal temperature
– Drying
– Irrigation/flooding the field
– Suffocation
– Physical destruction
– Use of physical  barriers
– Use of electro magnetic radiation                                                            (4 x1 = 4mks)

(c)(i)– Cassava mosiac
–Brown streak                                                                                           (2 x 1 = 2mks)

(ii) – Hot and cold disease in coffee
– Leaf wrinkle
– Elgon dieback in coffee
– Blossom end-rot in tomatoes
– Thick necked condition in onions                                                            (2 x ½ = 1mk)

25.(a) – Making mortar
–  Making concrete
–  Making concrete blocks
–  Making anil                                                                                           (3 x 1 = 3mk)

(b) –   Nearness to water source
–  Slope of the land/levelness of the area
–  Drainage of soil
–  Nearness to mainfield
–  Freedom from pests and diseases/weeds
–  Shelter from winds                                                                                (4 x ½ = 2mks)

(c)(i) – Ponds
– Dams/weirs
– Storage tanks                                                                                          (2 x ½ = 1mk)

(ii)  – Expensive/high initial cost
– High labour demand/easy transmission of diseases
– Difficult to maintain the strength/ concentration of dipwash                    (4 x ½ = 2mks)

26.(a)(i) For attachment of all parts of the plough
(ii) For attachment of the mouldboard, share and landslide
(iii) Cuts the furrow slice horizontally
(iv) For attaching depth rod that can be adjusted before and also during ploughing
(4 x ½ = 2mks)

(b) – Lubricate land wheel bearing
– Repair/replace wornout shares
– Tighten loose bolts and nuts
– For long storage with old engine oil to prevent rusting
– Store properly under a shed (reject shade)   ( 3 x 1= 3mks)

(c) Mouldboard plough                                      Disc plough
(i) Not used in a field with obstacles/            (i) Used in a field with obstacles /rolls over
cannot ride over stones, roots                         roots, stones
(ii) Inverts furrow slice completely/leaves      (ii) Does not invert furrow slice completely/

a clean field                                                  leaves a rough field
(iii) Operates at uniform depth/share             (iii) Cuts at varying depths as it rides over

             furrows same depth once set                             obstacles
(iv) Rigid, easily broken by obstacles             (iv) Not easily broken by obstacles as it rides

                                                                                        over them
(v) Requires more tractors power to pull        (v) Requires less tractor power to pull it when
when operating                                                operating
( 3 x 1 = 3mks)

27.(a) Hereford
–   Fairly large body/weighs 80 – 1000 kg
–  Good depth and width/compact body
–  Fast growth and reach market weight early
–  Efficient food converter into flesh                                                          (3 x 1=3mks)

(b) –  Character expressions
– Ability to milk/easy milk let-down
–  Quantity of milk produced
–  Butterfat content /quality of milk
–  Number of calves born by the cow
–  Ability to resist diseases
–  Weight gain/growth rate of calf born by the cow                                     (3 x 1 = 3mks)

(c)  (i) Sow – 4 – 6months
(ii) Ewe – 8 – 12 months                                                                           ( 2 x 1 = 2mks)

28.a – Plant appropriate feeds depending on ecology
– Planting healthy seeds/materials/certified seeds
– Early planting
– Proper seed bed preparation/correct tilth
– Weed control
– Disease control
– Pests control
– Soil and water conservation/terracing
– plant at correct spacing
– Proper seed rate
– Application of inorganic fertilisers
– Irrigation watering
– Gapping
– Thinning
– Crop rotation
– Timely harvesting                                      ( 13 x 1 = 13mks)
(Award a full mark for a well explained point)

(b) – Influences availability of soil nutrients
–Influences activity of soil micro-organisms/nitrifying and nitrogen fixing bacteria
– Influences balance of different micro-organisms by influencing their competitive ability
– Influence presence of disease organisms/fungus and bacteria
– Influence soil borne pests e.g nematodes in acidic soils
– Influences the type of fertiliser to be applied whether acidic or alkaline
– Influences concentration of iron and Aluminium to injurious or toxic levels to crops
( 7 x 1 = 7mks)

  1. (i). Land Preparation
    – Prepare the land early during the dry season to allow stubble to rot
    – Clear the vegetation to ease tillage
    – Plough/dig deep to uproot all weeds
    – Break soil clods/harrow the field to medium tilth
    – Dig terraces if the land is slopy/carry out soil and water conservation measures
    (4 x 1= 4mks)
    (ii) Planting
        
    – Do dry planting in areas with short rainy seasons
    – Use certified maize seeds suitable to ecological conditions
    – Dig holes 2.5 cm to 10cm deep depending on soil moisture
    – Space holes 23 – 30 cm x 75 – 90 cm depending on variety of maize and soil fertility/soil

moisture
–  Apply 1 tablespoonful/DSP fertiliser per hole/100 – 150 kg DSP per hectare
–  Mix the fertiliser with the soil to prevent seed scorching
–  Apply well decomposed organic manure
–  Place 1 or 2 seeds per hole and cover with the soil                                     ( 8 x 1= 8mks)

(iii) Field operation
– Thinning soon after germination when soil is moist
– Gapping/replanting soonest for uniform growth
– Uproot with a jembe/panga or use chemical herbicides eg, MCPA, 2, 4- D
– Control diseases using appropriate chemicals/uproot and burn infected crop
– Control pests using appropriate pesticides e.g maize stalk borer use dipterex/uproot and burn

infected crop
– Irrigate during dry weather                                                                       (5 x 1 = 5mk)

 

(iv) Harvesting
– Start 4 – 5 months after sowing depending in variety and altitude
– Done during dry weather
– Harvest when leaves and cobs dry
– Cut the stalks, remove the cobs or open the ear, break the cobs and then cut the stalks
– Put in the sun to dry, snell, winnow, dry the grains to include 12 – 13%, treat with actelic

dust before storage                                                                                (3 x 1 = 3mks)

30.(a) Fowl pox disease

(i) – Chicken
– Turkeys
– Pigeons
– Other birds
(Poultry alone is correct)                                                                              (1 x 1= 1mks)

(ii) – Virus                                                                                                       ( 1 x 1 = 1mk)

(iii) – Symptoms of attack
Dipheritheritic type
– Yellowish, raised lesions inside throat and mouth mucus membrane
– Difficulty in breathing & swallowing
– Watery discharges from the eyes and nose at early stages
– Loss of appetite
– Dullness
– Emaciation and death may occur                                                                (5 x 1 = 5mks)

(iv) Cutaneous type
– Injuries/lessions on combs, wattles/lessions at first are small and greyish white and later
turn yellowish brown and bigger in size
– Lessions on the, vent, feet and under the wings
– Loss of appetite, emaciation and death                                                        (3 x 1 = 3mks)

(iv) Remove affected birds and kill
– Vaccinate healthy birds                                                                              ( 2 x 1= 2mk)

(b) – Presence of broken/soft shelled eggs
– Bright lights in the nests
– Idleness
– Inadequate nest/birds lay on floor
– Incorrect feeding/lack of minerals/unbalanced feeding
– Overcrowding of birds
– presence of external parasites on combs wattles e.g. fleas
– Prolapse cloaca does not retract after the hen has laid
– Introduction of new birds in a flock causing fighting
– Keeping birds of different ages together/no age grouping
– Bright light in the brooder                         ( 10 x 1 = 10mks)

AGRICULTURE VI

SECTION A :

 

  1. State four practices which destroy soil structure. (2mks)
  2. Give the function of a creep area in a pigsty.                                                          (2mks)
  3. a) State one use of each of the following farm tools:
  4. i) a mason trowel.                                                                                                   (1mk)
  5. ii) a pair of tin snips.                                                                                               (1mk)
  6. b)   Which tool would be required for each of the following operations?
  7. i)    Cutting wool from sheep.                                                                            (1mk)
  8. ii)   Castrating piglets.                                                                                       (1mk)
  9. Give four precautions you would take when harvesting to ensure that cotton picked is of high

quality                                                                                                                   (2mks)

  1. List four factors that influence herbicidal selectivity and effectiveness in weed control.                                                                                                                                     (2mks)
  2. a) Give four advantages of a tractor in farm mechanisation.                                         (2mks)
  3. b)   Outline two limitations of tractor power.                                                                 (2mks)
  4. Give the name of symbiotic bacteria which fixes nitrogen in the root nodules of leguminous

plants                                                                                                                          (1mk)

  1. Give two reasons why a rabbit may disown its young ones. (2mks)
  2. a) State four types of risks and uncertainities.                                                        (3mks)
  3. b)   Outline how the government helps farmers to overcome risk and uncertainities. (3mks)
  4. a) In which ionic form is the element sulphur absorbed by plants. (1mk)
  5. b) i) State any three deficiency symptoms of nitrogen in crops. (3mks)
  6. ii) State two sources of Nitrogen in the soil for plants. (2mks)
  7. a) Define the following terms:
  8. i) Forage crop.      (1mk)                           ii) An Apiary.   (1mk)
  9. b) How do you ensure proper forage utilization in livestock production .                 (2mks)
  10. a) What is leaching. (1mk)
  11. b)   Name two factors which increases the rate of leaching.
  12. a) State one important role of the hormone testerone in male livestock. ( ½ mk)
  13. b) State three important objectives of steaming up in livestock .                    (1 ½ mks)

 

SECTION  B.

 

  1. State four factors that determine siting of a farm structure. (2mks)
  2. State any four uses of crush in a farm. (2mks)
  3. a) Mr. Akugo wishes to fence a straight fence 4.40 m long bourdering the school farm. Find out how many cedar posts he will require if the distance from one post to another is 2m (2mks)
  4. b) Mr. Akugo Applied 150kg N.P.K 25:20:15 to his one hectare of tobacco in Osogo area.         Calculate how many kg of each of the fertilizer element he applied.                         (3mks)
  5. a) State two disadvantages of serving Fresian heifers when they are less than 18 months old

(2mks)

  1. b) List three factors that make it possible for a camel to survive in arid and semi-arid areas.

(3mks)

  1. a) What factors contribute to the success of a co-operative society. (3mks)
  2. b) State four main functions of marketing boards.                                                  (2mks)
  3. c) Name two channels through which marketing boards buy produce from farmers.      (2mks)
  4. Distinguish between passive and active acquired immunity in livestock health. (2mks)
  5. Give one characteristic symptoms of attack by nematodes on roots of tomatoes. (1mk)
  6. a) State three ways of increasing efficiency in farming. (3mks)
  7. b)   List four factors which effect the profitability of egg production enterprise.        (4mks)
  8. a) Give two methods that can be used to improve local breeds of livestock.       (2mks)
  9. b)   State four merits of using AI in livestock breeding .                                           (4mks)
  10. State six reasons why farmers are adviced to practice mixed farming. (3mks)

 

SECTION  C.

 

24.a) Describe the management practices involved in rearing of a day-old chicks upto the age of 8

weeks                                                                                                                  (16mks)

  1. b) Explain the management practices livestock farmers should adopt to reduce the problem of feed shortage during drought. (4mks)
  2. The transaction below shows Mrs.Odhiambo’s financial position in farm business for the year 1997. Use this information to answer the question that follow:

–     Milk sales                             –   sh 8,000.00

–     Purchase of farm tools                 –   sh 1,000.00

–     Sales of goats                               –   sh    500.00

–     Constraction of zero grazing unit- sh 10,000.00

–     Closing valuation                         – sh 16,000.00

–     Depreciation of machinery          –  sh      800.00

–     Interest payable                            –  sh     750.00

–     Purchase of pesticides                  –   sh    300.00

–     Veterinary bills                             –  sh     400.00

–     Sales of cabbage                           –  sh     750.00

–     Wages                                           –  sh   4800.00

–     Sales of one heifer                        –  sh   9400.00

–     Opening valuation                        – s h12,000.00

–     Tea sales                                       –  sh  4,700.00

  1. i) Prepare the profit and loss account of Mrs.Odhiambo’s farm.       (15mks)
  2. ii) What percentage profit or loss did Mrs. Odhiambo make during the year 1997.     (5mks)

26  a)   State the principles involved in planning a crop rotation programme.                 (6mks)

  1. b)   Describe field management praticises involved in coffee production.                  (24mks)

 

AGRICULTURE VI

MARKING SCHEME

 

  1. – Over cultivation

–   Burning of land

–  Monocropping / monoculture / failure to practice crop rotation

–   Wriking soil when it is wet

–  Repeating use of heavy machinery for cultivation

  1. – Where special feed is provided to piglets (creep feed).

– Heat / warmth is provided to piglets to prevent chilling.

–  Prevent mother pig from crashing the piglets.

–  Ultraviolet rays from bulb assist in synthesis of vitamin D under the skin.

3.a)i) For applying mortar / concrete on walls / floors.    ii) For cutting metals.

b)i) Pair of shears.      ii) A scapel.

  1. – Picking should be done every week to ensure no foreign materials.

– Avoiding picking when it is wet.

–  Picking is done manually.

  1. – Active ingredient.

–  The type of herbicide.

–  Time of application.

6.a)- Quick rate of work

– Better burial of weeds during ploughing

– Efficient work

– Less labour demanding

– Can be used to transport farm produce

– Operation done in line

– Does ploughing, planting and harvesting

– P.T.O shaft can be used for irrigation, sprinkling water etc.

  1. b) – Can only work in flat surfaces

–  Needs frequent repair and maintenance

–  Initial capital for purchase is high

–  Requires technical know how to operate

–   Only confined to large scale farms

 

  1. – Rhizobium
  2. – Poor feeding of the mother rabbit.

– Touching kindles with strong / bad smelling substances.

– When the doe is frightened or disturbed after kindling.

9.a) – Weather changes.

–  Out break of pests and diseases.

–  Natural calamities e.g earth quakes.

–  Obsolenscence (become out of date) i.e machinery.

–  New production techniques.

– Changing prices of commodities.

–  Low or high yields of production.

–  Theft cases – sickness / injury.

–  Fire out breaks death.

  1. b) – Weather forecast.

–  Provision of extension service and advice to farmers.

–  Adjusting future prices of commodities from the farm and giving farmers in advances.

–  Stabilization of prices of certain farm produce so that they don’t flactuate.

–  Provision of loans to farmers to improve production.

–  Provision of research on animals and plants e.g hybrids diseases etc.

–  Provision and subsidisation of agricultural inputs.

  1. a) SO4-2 (Sulphate ion).
  2. b) i) – Yellowish green colour / chlorosis.

–  Stunted growth.

–   Premature drying of leaves.

–  Defoliation / falling of leaves.

  1. ii) – Inorganic manure

–  Organic manure

–  Nitrogen fixation

–  Fresh organic matter

11.a) i) – Is any plant established naturally or artificially for feeding livestock.

(ii) – Site for beehive / bee keeping.

  1. b) – Better forage management.

– Use of high yielding / performing animal.

–  Conservation of forage.

  1. a) – Loss of plant nutrients through percolation into deeper layer of soil by water.
  2. b) – High rainfall / excess irrigation water.

–   Solubility of mineral salts.

–   Rate of soil structure / infiltration / texture drainage.

13.a) – Necessitates sperm production.

b)- Provide sufficient nourishment for the development of the foetus and the expectant mother.

–   Helps to prevent pregnancy to xermical (twin lamb disease).

–   Result in strong healthy lamb at birth.

–   Promotes udder development and better milk following lambing.

–   Enhances muscle reserve build up in ewes.

 

SECTION B

 

  1. – Topography

–  Soil

–  Direction of wind

–  Relationship to other structures

–  Accessibility

–  Size of the farm

 

 

 

  1. – Used when castrating male animals.

–  Used when carrying out livestock treatment.

–  Used to restrain animals during artificial insemination.

–  Used to restrain animals during weighing.

16.a) – 3 cedar posts.

  1. b) – N.P.K total ratio = 60

–  Nitrogen  = 62.50 Kg

  • Phosphate = 50Kg
  • Potassium = 37.5Kg

17.a) – Stunted growth

– Low milk yields

  1. b) – Presence of hump where fats are stored

–  Presence of dehydrating tissues

–  Resistance to drought

–   Ability to drink large amounts of water

18.a) – Sound economic base / finance

–  Loyalty of members

–  Good leadership

–  Effective control of funds

–   Education of members on their rights and obligations

 

  1. b) – Carry out efficient marketing of farm products

–  Provide loans to farmers

–  Keep prices stable

–  Control quality and quantity of products

19.- Passive immunity is one passed from the mother to the child while active immunity is

acquired after a previous developing antibodies against the disease.

  1. – Presence of root knots.

–  Wilting.

21.a) – Select proper / more paying enterprises.

–   Proper crop husbandry practices.

–   Improve farm management.

–   Mechanization of farm operation

–   Adopt new technique of production

–   Time line of farm operation

  1. b) – Laying percentage

–  Price of culled birds

–  Replacement cost

–  Casual labour

22.a) –  Cross breading

–   Up – grading.

  1. b) – Eliminate the risk of expenses.

–  It is possible to use sire on smaller cows without any physical injury.

–  Permits the use of proven sires beyond the economy lease of an average farmer.

–  It makes it easy to control in breeding.

  1. – To maintain soil fertility.

– To reduce / spread risks

–  To maximise revenue / incure

–  To obtain balance diet for the family

–  Crop residues are feed to livestock

–  Income is obtained through out the year

 

 

 

 

SECTION C.

 

24.a) – Litter should be provided for to provide warmth as well as absorbing moisture.

–  Fresh air – Holes for ventilation should be made on the walls of the brooder to allow gaseous exchange.

–   Heat source – A wire should be around heat source to prevent chicks from burning.

–   The correct temperatures should be:

32 – 35oC 1st week

29.7 – 32.2oC 2nd week

26.6 – 29.7oC 3rd week

–     Source of heat should be removed in the fourth week gradually to avoid stress.

–     Light – provide light in the brooder for the chicks to see food (dim light).

–     Feeders – provide clean feeders which should be cleaned every morning.

–     Waterers – provide waters which should clean have pointed tips to avoid water being dirty.

–     Avoid sharp corners as these encourages overcrowding causing suffocation.

–     Feed chick and duck marsh at recommended ratio depending on the number of chicks.

–     Vaccinate chicks with Newcastle vaccine.

  1. b) – Conserve excess feed during plenty in form of silage or hay.

–     Ensure selective stock disposal when drought is approaching.

–     Provide crop residues to livestock.

–     Irrigation of pastures.

–     Give supplementary feed to livestock.

–     Plant drought tolerant fodder crops.

  1. – Profit and loss account for Mrs Odhiambo’s farm for 1997.

–     Profit and Loss account.

–     Profit and loss of Mrs Odhiambos farm for 1997

 

Sales and reciepts                                Purchases and expenses

Shs   cts                                                        Shs

Closing valuation                 16,000 00  Opening valuation                    12,000

Milk Sales                    8,000  00          Purchases of farm tools                1000

Sales of goats                500 00              Consts of zero unit                   10,000

Sales of cabbages          750 00              Depr of machinery                         800

Sales of heifer                 9,400 00         Interest payable                             750

Sales of tea                   4,700 00          Purchase of pesticides                    300

Veterinary bills                              400

Wages                                        4,800

                                                            Profit                                          9,300

39,350                                                                39,350

 

Mrs Odhiambo started her operations at a farm value Shs. 12,000 and therefore this is the base. The farm made a profit of Shs 9,300. The percentage profit is therefore

 9,300   x  100 = 77.5%

12,000

26.a) – Shallow rooters should alternate with deep rooters.

–     Crop attacked by the same pests and disease should not follow each other.

–     Crops with high nutrient requirement should come first in a newly ploughed land.

–     Legumes should be included in the programme to increase nitrogen content of soil.

–     Fallow period / grass should be included in the rotation to build soil structure.

–     Crops which are hard to weed should alternate with those that are easy to weed.

  1. b) –    Proper weed control to reduce competition for nutrients and water.

–   To improve coffee yield and quality.

 

 

Methods

  • Cultural e.g mulching
  • Mechanical e.g slashing
  • Chemical use of herbicides
  • Mulching

–     Mulch should not be in contact with coffee stem

Reason:

  • Preserve soil moistures
  • Suppress weeds
  • Supplies nutrients when it decomposes
  • Reduce soil erosion

 

      Pest s and diseases

Coffee pests:

  • Leaf miner
  • Control spray with recommended insecticide
  • Antesia bug
  • Proper pruning
  • Spray with recommended pesticide

 

Coffee disease:

  • CBD
  • Proper prunning

–     Planting resistant cultivers

  • Spray with recommended fungicide

 

Coffee leaf rust:

–     Plant resistant varieties

–     Time by application of fungicide

 

         Prunning

  • Single stem
  • Multiple stem
  • Change of cycle

 

         Reasons For Prunning

  • To improve yields and quality of coffee
  • To assist in disease and pest control
  • To open the crop to light and air
  • To avoid die-back of primaries and roots

 

Fertilizer and manure applications:

–     Phosphate fertilizers – Applied at planting time to supply phosphorous for proper root growth and development

 

–     Nitrogenous fertilizer – The crop should be top dressed when soil is moist.

Reason:- To supply nitrogen required for vegetation growth, flowering and fruiting.

 

–     Manure(compost) application:- Apply in planting holes.

Reason: – To improve soil structure.

– To supply nutrients.

–  To increase activity of micro-organism.

 

 

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Business Form one to four Finest Revision Resource

TOP STUDENT

REVISION

BUSINESS STUDIES

 

 

 

By; Charles Tole

 

 

 

 

 

 

 

 

 

8 Sample Papers Complete with their Marking Schemes

 

 

 

 

 

 

 

 

 

 

Produced By;

Idealprice Consultants,

P.O. Box 15339-00100,

Nairobi.

 

© 2008

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

TEST PAPER I

 

  1. Mention four limitations of advertising as a means of sales promotion.                      (4mks)
  2. The capital structure of a company is divided. Name four such divisions.   (4mk)
  3. List down four features of co-operative society. (4mks)
  4. Indicate the account in which the following entries as made. (4mks)

 

Entry Account
Bad debts  
Carriage inwards  
Returns inwards  
Discounts given  
Gross profit  
Purchases  
Drawings  
Pre-paid rent  

 

  1. Highlight four ways in which Insurance companies make profits.               (4mks)
  2. The major objective in commerce is to maximize profits. Highlight four ways in

which businesses can achieve this important objective.                                                              (4mks)

  1. Use the following information to Calculate the required

Sundry debtors                 52,000

Stock                                   5,000

Cash at bank                   100,000

Capital                              40,000

Net profit                          16,000

Current liabilities             82,000

Calculate:

  • Current ratio.       (3mks)
  • Quick ratio. (2mks)

 

  1. Mention four types of wholesalers.       (4mks)
  2. List down four machines used in the mail nom and their use.       (4mks)

 

MACHINE USE
(i)  
(ii)  
(iii)  
(iv)  

 

  1. The law of demand applies ‘ceteris penbus’ all other factors held constant ). Highlight

four circumstances under which this law may not hold.                                                             (4mks)

  1. As the transport manager of a busy company. Highlight four factors you would consider

before choosing a mode of transport for your goods and services.                                (4mks)

 

 

 

 

 

  1. The following balances were extracted from the books of Beauty Care Enterprises on

31st of March 1997.

Shs.

Fixed Assets                    42,000

Stock                                48,000

Debtors                             50,000

Bank overdraft                 25,000

Creditors                          40,000

Cash in hand                      2,000

 

During the month of April  the following transactions took place.

  • Bought Motor Vehicle on credit valued at kshs. 80,000 from General motors.
  • With drew Sh. 28,000 for personal use.

Required: Prepare a balance sheet as at  April 30th 1997.                                          (5mks)

 

  1. Give four reasons why an office may prefer storing information in Micro-films instead

of files.                                                                                                                                (4mks)

  1. Name four documents used in home trade.                      (4mks)
  2. Mention four reasons to justify that the Central Bank is not a Commercial Bank.    (4mks)
  3. Give four reasons why people prefer using a cheque to cash when effecting payment. (4mks)
  4. The following information relates to Sis Mis traders for the year ended 31st Dec 1996.

Opening stock

Closing stock turnover 8 times

Sales for the year 293,000

            Determine:

  • Cost of sales
  • Purchases
  • Gross profit or Gross loss.  (5mks)
  1. Give four advantages of personal selling as a means of sales promotion.  (4mks)
  2. Mention four of the contents in articles of association. (4mks)
  3. International trade improves the economic welfare of a country. The recent opening

of the East African co-operation seeks to diversify International trade. Highlight four

factors that may hinder this good intention.                                                                    (4mks)

  1. The following balances relates to a business for the year ended 31 Dec 1990.

Capital as at 31 December 1990          170,000

Drawings for the year                               5,000

Net profit for the year                              5,000

Extra capital introduced                         52,000

Purchases                                                            20,000

Determine the amount of capital as at 1 Jan. 1990                                                       (3mks)

(ii) Net profit ratio.                                                                                                          (2mks)

 

  1. Give four functions of commercial attaches.             (4mks)
  2. Mention four investment services offered by the Kenya Posts and Telecommunications.

(4mks)

  1. Name four file covers used in an office.       (4mks)

 

 

 

 

TEST PAPER I

MARKING SCHEME

 

  1. – Its Expensive
  • It leads to unnecessary competition
  • It doesn’t lead to sales increases it turns out to be a waste
  • Most advertisements are brief thus they may be misinterpreted.

 

  1. – Called – up capital
  • Paid – up capital
  • Registered capital
  • Issued capital

 

  1. – Voluntary Association – Service motto – serve its members
  • Democratic management
  • Division of profits
  • State control

 

  1. – Bad debts – profits & loss A/C
  • Carriage inwards – Trading account
  • Returns inwards –
  • Discounts given – profit and loss
  • Gross profit – profit and loss
  • Purchases – Trading account
  • Drawings – Balance sheet
  • Prepaid rent – profit & loss

 

  1. – By investing their funds
  • From differences in claims
  • From forfeited funds by policy holders
  • By charging those who bleach the contract of the Policy – a certain amount is deducted when one quits the policy before agreed time.

 

  1. – Reducing prices to increase sales
  • Reduce the cost of production
  • Increase prices of products
  • Expand production and market

 

7.

Current ratio =  Current Assets

Current liabilities

Current Assets:                                   Current liabilities = 82,000

Debtors           52,000

Stock                 5,000                         Current ratio   = 157000

Cash at bank  100,000                                                                 82000

= 1.914

  • Quick ratio = Current assets – Stock = 152,000        = 1.853

                                               Current liabilities                   82,000

 

 

  1. – Truck wholesalers
  • Regional wholesalers
  • General wholesalers
  • Specialized wholesalers
  • Nationwide wholesalers
  • Cash and carry wholesalers

 

  1. – Letter openers – opening mail
  • Franking machines – stamping letters
  • Parcel scales – weighing parcels before dispatch
  • Shredding machine – Shredding unwanted paper
  • Guillotine machine – trimming papers to the required size
  • Stapling machine – a fixing pages together

 

  1. – When the prices of other products change
  • Tastes and preferences of individuals change
  • When products don’t have substances.
  • Incase of primary products that must be consumed
  • Incases of monopolies who are the only producers.

 

  1. – Cost of the transport mode.
  • Distance to the covered
  • Type of goods to be transported
  • Modes of transport available
  • Quantity of goods to be transported.

 

12.

Beauty Care

                     Balance Sheet

                          As 31st March 1997

 

Capital                                    77,000             Fixed Assets        42,000

Less drawings             28,000             Motor Vehicle                  80,000

Net capital                  49,000                                                          122,000

Loan                            28,000

77,000             Current Assets

Current liabilities       25,000             Stock                     48,000

                        Bank overdraft            40,000             Debtors                 50,000

Creditors 1                  80,000             Cash in hand           2,000   100,000

Creditors 2                222,000                                                         222,000

 

13 .-    They are more confidential than files

  • More durable compared to files
  • They take less space
  • Easy to move about and convenient to carry.

 

  1. – Letter of inquiry
  • Catalogue
  • Proforma Invoice
  • Delivery note
  • Promising note
  • Credit note
  • Debit note

 

  1. – Does not rent money to individuals
  • Doesn’t receive deposits from individuals neither does it run accounts for individuals.
  • Not an agent to the stock exchange
  • Does not discount bills of exchange
  • Doesn’t offer money transfer services for individuals.

 

  1. – Cheque is more secure than cash
  • It also acts as evidence of payment
  • Its convenient instead carrying large sums of money
  • Saves ones time as they may not have to go withdraw money to make payment

 

17.

Rate of stock turn – over                                                        Gross loss

Cost of sales                                                                       = Sales – Cost of sales

Average stock                                                                     = 293000 – 464000

8 x average stock) = cost of sales.                                         = 171,000

8 x (opening stock + closing stock) = cost of sales

2

8 x (52000 + 64000) = C.O.S

2

8 x 58000  =  C.O.S

464,000     =  C.O.S

 

  • Purchases + Opening Stock – Closing Stock = C.O.S

464,000 – 52,000 + 64,000  =  476,000

 

  1. – Personal contact with the customers
  • Immediate feedback on customers feelings about a certain product
  • Salesman goes to the particular group targeted.
  • Salesman can persuade and convince people to buy.

 

  1. – Classes and rights of shareholders.
  • Issue and transfer of shares
  • How alterations in the capital could be treated
  • General meeting procedures & sorting rights
  • Qualifications , Duties , and powers of directors
  • Auditing of books
  • Borrowing , dividends and reserve policies.

 

  1. – Some countries may feel they aren’t benefiting much
  • Political differences
  • Instability in neighbouring countries
  • Influx of refugees
  • Different currencies used.

 

 

 

 

21.

Capital owned at the end of the year

= initial capital + Net profit – Drawings – Additional capital

= 170,000 + 5,000 – 100,000 + 52,000

= 222,000 – 105,000

= 122,000

 

  • Net profit ratio =  Net profit x 100

sales

Sales   =  Purchases + Net profit

= 20,000 + 5,000

= 25,000

 

Net profit ratio  =  5,000 x 100          = 20%

25,000

 

22  .-   Search for Kenyan markets abroad

  • Inform businesses abroad of Kenyan goods
  • Negotiate to expand Kenyan Market abroad
  • Liase with the ministry of Commerce to educate Kenyan producers/manufactures.

 

  1. – Sell bearer and premium bonds
  • Operate savings accounts
  • Operate SAYE (Save As You Earn) scheme
  • Have fixed deposit facilities

 

  1. – Box files
  • Flat files
  • Lever arch files
  • Loose leaf binders
  • Ring binders

 

 

TEST PAPER  II

 

  1. (a) Marketing boards are producer organizations set up to encourage and control the

selling of agricultural products. Discuss five functions of these organizations.

(10mks)

(b)  Discuss five differences between a bank statement and a bank pass book.                   (10mks)

 

  1. (a) Despite major development in the transport sector. Many people are still using

carts (both those driven by animals and those pushed by human beings).

Discuss five reasons to show why this type of transport is still preferable to others.

(10mks)

  • Money evolved through several stages to get to the currency notes and coins we

have today. However “ Modern – day trade may compare negatively with barter

trade”. Discuss this statement giving it your support.                                             (10mks)

 

 

  1. (a) A business enterprise should keep a budget other than the other statements it

keeps.   Discuss reasons why this activity is important.                                           (10mks)

  • You are the sales manager of ‘speed’ Industries Ltd. that manufactures tyres.

You are faced with the decision to choose an appropriate channel to distribute the

product.  Discuss the factors you consider in your choice.                                 (10mks)

 

  1. (a) Mail order business is whereby customers place their orders through mail  and goods

are also send through mail. This business  is however not popular in Kenya.

Discuss five reasons to encourage your colleagues to start this type of business.             (10mks)

  • Discuss the factors that influence the choice of a filing system in an organization. (10mks)
  1. (a) Briefly explain the following terms as used in Capital structure of a company.
  • Registered capital.        (2½ mks)
  • Called – up capital. (2½ mks)
  • Issued capital. (2½ mks)
  • Paid – up capital. (2½ mks)

 

  • Specialization is the separation of jobs, activities and processes so that every individual concentrates on what they can do best. Name and explain four major such divisions.

(12mks)

 

  1. (a) Discuss five benefits that the Kenyan government gets from Binded Warehouses. (10mks)
  • Mwachai BINA’s books disclose the following information for the period ended

31 Dec. 1996.

 

Opening Stock                        21000  
Purchases                                40000  
Closing Stock                          10000  
Sales                                        82000  
Returns inwards                        2000  
Motor Vehicle                         10000  
Rent                                           3000  
Sales expenses                         18000  
Discount received                      3000  

 

Required: Draw up Mwachai’s trading profit and loss  account for the period ended.     (10mks)

 

 

TEST PAPER  II

MARKING SCHEME

 

  1. (a) –    Encourage and control marketing of agricultural products by purchasing them from

farmers.

  • Encourage stable prices and incomes by keeping prices fairly fixed and store
  • unsold goods to act as buffer stock.
  • Offer loans for farm input purchasing.
  • Help government to regulate, license and impure major export crops e.g. tea.
  • Provision of a wide range of experts dealing with agricultural products
  • They do research work to improve yield.

 

 

  • – A bank statement is issued after an account holder has requested for it whereas a bank pan

book is issued after one opens an account.

  • Pass book can be used to withdraw money whereas a statement cannot.
  • A bank pass book contains all the particulars of a person e.g. a passport size. Whereas a bank statement doesn’t have
  • A bank statement is seasonal that is its issued from time to time whereas a pass book is more permanent.
  • A bank statement shows only the transactions that were undertaken in a certain period of time but a passbook may contain all the transactions since the account was opened.
  • A bank statement may be given to anyone who holds either a savings or current account but pass book may be given only to those with savings account where cheque facilities aren’t available.

 

  1. (a) –     Carts are cheap to buy and maintain
  • They are cheap and negotiable
  • They are flexible – can be used where there aren’t any roads
  • Convenient for short distances
  • Can deliver at the doorsteps
  • Readily available.

 

(b) –   Trading procedures have become complicated

  • Increased fraud and cheating over money
  • Trade has become risky due to theft and carrying of large sums of money
  • Trade has become expensive where one has to bank money, pay interest for loans etc.
  • It has been dominated by only a few leaving out the majority.

 

  1. (a) – Business is done to compare the actual results with the budgeted plans.
  • Aids in communication e.g. employees are fully aware of tasks to undertake within a certain period of time i.e. time budget.
  • Increase efficiency as everyone want to accomplish what is expected of them.
  • Evaluation of performance to know what was achieved according to the budget

and what was not and why.

  • Avoid time and money wastage as each is accounted for at the end of the period.
  • Know output of employees as each departments achievement are seen.

 

  • – Cost – choose channels which are not too expensive so that profits aren’t reduced.
  • Nature of my produce – Requires middlemen with tyre knowledge.
  • Size or Nature of market – if the market is large enough around the factories

the short channels are chosen.

  • Marketing risks – to minimize these then middlemen will be involved
  • Government policy – If there is any directive according to the state law as to who should sell tyres.
  • Competition – If we are forcing steep competition then its advisable to market the goods ourselves to take care of our customers and convince them to buy.

 

4 . (a)  –    It is cheap as its not necessary to maintain expensive showrooms.

  • Doesn’t require a fleet of transport vehicles
  • Salesmen may not be required and few employees can handle the mail.
  • Distributor may contain complete control over the products on his own.
  • Goods sold for cash checks on bad debts
  • Can operate with little capital as its unnecessary to have huge stocks of goods.

(b) –     Number of files & documents to be kept

  • Simplicity – to be understand easily by new employees
  • Flexible to accommodate more information without having to change existing documents .
  • Easy reference and minimum risk of misfilling.
  • Documents shouldn’t be filed more than once in different subjects.
  • Must be affordable.

 

  1. (a) Registered capital – Maximum amount of capital a company expects to raise from

its shares and its also the amount stated in the Articles of Association.

 

Called – up capital – once the shares have been put to the public for subscription they may be called upon to pay for either all the shares or only a certain amount. The total of the amount that will be paid is called – up capital.

 

Issued capital – Although the company has registered capital it may decide to put only a portion of these to the  public to start with. The total amount released then is issued – capital.

 

Paid – up capital – The total amount received from subscribers by the company out of the called – up capital.

 

(b)    Trade divisions (professions division)

A person concentrates on the professional trade or service e.g. teaching , fishing.

 

Process division – Although in the same trade different people have their work being divided into a number of stages, activities or processes each being undertaken by a different group of people.

 

Regional or Territorial specialization – Two or more firms on the same product to complete the production process e.g. firm may undertake the processing of cotton whereas as B does the  packaging.

 

Nationalized specialization – Based on soils , climates , skills and costs of production. Applied in International trade whereas a country concentrates on the production of a particular commodity or supply of service e.g. Kenya is known for coffee production.

(4 x 2 ½) = 10 mks

 

6   (a)  –     Collecting of duties which the government revenue.

  • Checks on imports – Quantity and equality
  • Checks on illegal trade e.g. drugs / firearms
  • Use them to compile statistical data or imports.
  • Use them to keep confiscated goods that would later be government property.

 

 

 

 

 

 

 

 

 

 

(b)

            MWACHAI’S BINA

                    Trading profit and loss account for the year ended Dec. 31 1996

 

 

Opening Stock                        21,000                         Sales               82,000

Purchases                    40,000                         Returns             2,000

G.A.S                          61,000                         inwards

Less Closing Stock     10,000

Cost of sales               51,000

Gross profit                29,000

80,000                                                 80,000

Rent                              3,000                         Gross profit                            31,000

Sales expense             18,000                              b/f

Net profit                    13,000                         Discount received                     3,000

34,000                                                                         34,000

 

20 times ½ = 10mks

 

TEST PAPER III

 

  1. Distinguish the following terms:       6mks
  2. Free goods and economic goods
  3. Consumer goods and producer goods
  4. Business wealth and personal wealth
  5. State four advantages of a bonded warehouse       4mks
  6. Name the factor of production that each of the following resources relate to.       5mks

Resource                                                   Factor of production

  1. Teacher ______________________________________
  2. Computer ______________________________________
  3. Water ______________________________________
  4. Sole proprietor ______________________________________
  5. Forest ______________________________________

 

  1. Mention any four methods used for sending money through the post office.       4mks
  2. Using the book keeping equation, indicate the correct figure in the empty spaces provided.

 

Assets Capital Liabilities.
180,000 50,000  
200,000   90,000
  45,000 38,000
150,000   50,000

 

 

 

 

 

 

  1. The following information was drafted from the books of Rayrose college on 31st Jan 2002.

Item                                         Amount (Kshs)

Opening stock                         400,000

Sales                                       700,000

Closing stock                          10,000

Purchases                                150,000

Sales returns                           20,000

Purchase returns                     50,000

Salaries and wages                 80,000

Rent                                        70,000

Income                                                80,000

 

Calculate:

  1. Gross Profit            4mks
  2. Net profit            4mks
  3. Mark-up percentage            2mks

 

  1. State any four conditions under which a proforma invoice may be used.
  2. Highlight any four services rendered by retailers to consumers.       4mks
  3. Outline any four functions of the Kenya External Trade Authority by (K.E.T.A.) 4mks
  4. Mention four limitations of advertising as a means of sales promotion       4mks

 

  1. As the transport manager of a busy company highlight four factors you would consider

before choosing a mode of transport for your goods and services.                                4mks

  1. “Most secondary schools maintain that students should pay fees with cheques and not cash” Account for this statement.                                                                                         4mks
  2. Indicate the account in which the following entries are made.

Entry

Account
Carriage inwards  
Carriage outwards  
Incomes receivable  
Return inwards expenses.  

 

  1. List four features of co-operative societies.       4mks
  2. Outline any four barriers that hinder effective communication.       4mks
  3. Highlight any four problems or challenges that businesses would face due to adoption

of the decentralization policy.                                                                                   4mks

  1. Give any four benefits that a farmer would reap from being a member of a producer cooperative.       4mks
  2. Outline any four features of multiple shops.       4mks
  3. With the aid of a tree diagram, show the main divisions of trade.       3mks
  4. What is meant by the following abbreviations as used in the study of commerce. 4mks
  5. O. R __________________________________________________________
  6. O . M . O __________________________________________________________
  • I . F ___________________________________________________________
  1. W. O ___________________________________________________________

 

 

  1. State any four methods that the government of Kenya uses to protect its consumers. 4mks
  2. Give any three characteristics of money as a medium of exchange       3mks
  3. What is a ‘bill of exchange’       2mks
  4. Highlight any three demerits of the barter system of trade.       3mks

 

TEST PAPER  III

MARKING SCHEME

 

  1. Distinguishing Concepts.
  2. a) Free goods                 –  Those things which are given free by nature eg air.  Have utility but

no economic value.

Economic goods          –  Those things which are scarce when compared to human wants and

human effort required to obtain them.  Have utility and economic

value (price)

  1. b) Consumer goods        –  Goods meant for direct consumption and they satisfy economic

wants may be  perishable or durable.

Produce goods           –  Goods that help in further production of consumer goods.  They

assist in  production eg machinery.

  1. c) Business Wealth –  Goods of value (assets) that belong to a business eg. Premises.

Personal Wealth        –  Consists of items that belong to an individual person eg clothes,

TV,   etc.                                         1 x 6

 

  1. Advantages of a bonded warehouse.
  • Goods can be prepared for sale while in bond
  • If goods are sold on bond, the duty passes to the buyer
  • The owner can look for market first before paying the duty
  • Some goods lose weight so the duty paid becomes lower if its based on weight
  • Ensures that no duty is evaded since the goods cant be released without payment
  • Enables the government to check on prohibited goods and smuggling.

Any other relevant point.

 

  1. Resource Factor of production
  2. a) Teacher                                     Labour
  3. b) computer Capital
  4. c) Water Land
  5. d) Sole proprietor Enterprenuer
  6. e) Forest Land                            1 x 4

 

  1. Methods of sending money through post office.
  • Telegraphic money order
  • Ordinary money order
  • Registered mail
  • Postal order. 1 x 4

 

 

 

 

 

 

Assets

Capital

Liabilities
180,000 50,000 130,000
200,000 110,000 90,000
83,000 45,000 38,000
150,000 100,000 50,000

1 x 4

  1. a) Calculating gross profit

 

Rayrose College

Trading account

 For the year ended 31st Jan 2001

 

Kshs.                                                                                                         Kshs.

Opening stock                                 400,000                                Sales                     700,000

Add purchases      130,000                                   Less sales returns                                20,000

Less purchases       50,000               100,000                    Net sales                                 680,000    

500,000

Less closing stock                                      10,000

Cost of goods sold                                    490,000

Gross profit c d                            190,000

            680,000                                                                 680,000

Gross Profit b d           190,000

   4mks

Calculating Net profit.

Net profit   = (Gross profit + income) – expenses

Income   =  80,000

Expenses         = Salaries & wages + rent

=  80,000 + 70,000

=   130,000

hence N.P        = ( 190,000 + 80,000)  – 150,000

=   270,000  –  150,000

kshs. 120,000                                                                4mks

 

  1. Calculating mark-up percentage.

Mark – up       =     Gross profi     x 100

Cost of goods sold

190,000    x 100

490,000

=  38.8%                                                                                 2mks

 

  1. When a proforma invoice may be used.
  • When the seller wants to request for payment before goods are delivered
  • When the seller does not want to give credit to the buyer
  • Issue to an agent who sells goods on behalf of the seller
  • Used by importers to get customs clearance before the goods are sent

 

 

 

  • Serve the purpose of a quotation
  • Sent to the buyer to show him what he would have to pay if he bought goods sent on approval. 1 x 4

 

  1. Services rendered by retailers to consumers
  • Selling goods to consumers in affordable units
  • Display goods to enable consumers know which alternatives exist.
  • Advise the consumers on use. Servicing and maintenance of goods
  • Stock a large variety from different manufacturers as per the needs of consumers
  • Provide credit facilities to worthy customers
  • Offers after sales services on technical goods to the consumers

Any other relevant point.                   1 x 4

 

  1. Functions of K.E.T.A.
  • Marketing of Kenyan products in foreign markets
  • Advise manufacturers on how to improve the quality of their products in order to get bigger markets overseas.
  • Providing training facilities on export matters
  • Arrange trade fairs and exhibitions in foreign countries to introduce Kenya products in foreign countries.
  • Publishing information on the possibilities of the expansion of exports
  • Encourage the production and marketing of handicrafts
  • Provide trade facilities and negotiate with transport companies on behalf of exporters.

1 x 4

 

  1. Limitations of Advertising.
  • Its expensive to the producer who may have to go through advertising agencies.
  • May be misleading or deceitful since they create unproductive competition.
  • The consumer is penalised in terms of prices since advertising costs are passed over to him
  • Creates mechanical buying particularly for permanently advertised products
  • Some advertisements are nasty and dirty hence leads to decay of good morals

Any other relevant point.                   1 x 4

 

  1. Factors to consider when choosing mode of transport.
  • The cost of transport
  • Speed and time taken
  • Nature of goods
  • Flexibility
  • Packing requirements
  • Distance to be covered
  • Insurance risks. 1 x 4

 

  1. Advantages of cheques.
  • Secure since if stolen they can be traced
  • Convenient to carry than coins and notes
  • Payment can be made without the need to travel
  • Acts as evidence of payment.

                                   Any other relevant point.                   1 x 4

 

Entry Account
Carriage inwards Trading account
Carriage outwards Profit and loss
Incomes receivable Profit and loss
Return inwards Trading account

 

  1. Features of cooperatives.
  • Voluntary Association ie open to all
  • Service motto ie serves its members and that profit is not the only motive
  • One for all and all for one ie self help through mutual help
  • Democratic management
  • Division of profit is done on the basis of service rendered
  • State control ie state exercises chose control. 1 x 4

 

  1. Barriers to effective communication
  • The language used
  • The medium used
  • Willingness of the receiver to listen
  • Noise
  • Attitudes of the communicators
  • Poor planning
  • Improper timing Any other relevant point.                   1 x 4

 

  1. Problems of decentralization.
  • Lack of trained personnel
  • Lack of basic infrastructures such as roads
  • Basic services like banking may not be found in all areas
  • Inadequate market
  • Certain benefits eg entertainment may not be found in the decentralized industries.

Any other relevant point.             1 x 4

 

  1. Benefits from co-operatives.
  • Fair price for his her produce
  • Reduced costs of transport, processing and marketing
  • Get ideas regarding quality improvement, better methods of cultivation, etc
  • Secure credit on reasonable terms than an individual could
  • Purchasing farm equipment’s at reasonable prices
  • A farmer is assumed of a stable price for his produce. 1 x 4

 

  1. Features of multiple shops.
  • Uniformity of the premises, external shop front, internal layout, etc
  • Control is vested in the central organisation
  • Buying if centralised and selling is decentralized through several branches.
  • Deal only in a particular set of goods
  • Sales are strictly cash
  • Selling prices are fixed by the central organisation.

Any other relevant point.             1 x 4

 

 

  1. Divisions of trade.

Trade

 

 

 

Home trade                             Foreign trade

 

 

 

Retail          Wholesales                     Import             Export

 

  1. Meaning of abbreviations.
  2. i) O.R –  Free On Rail
  3. ii) N. O. – Or Nearest Offer

iii)  C. I. F       –  Cost, Insurance and Freight

  1. iv) W. O –  Cash With Order.

 

  1. Government methods of consumer protection.
  • Setting up standards eg Kenya Bureau of Standards
  • Price control
  • Licences
  • Food and Drugs Acts.
  • Sale of Goods Act.
  • Public Heath Act.

 

  1. Characteristics of money
  • Acceptability
  • Portability
  • Divisibility
  • Durability
  • Homogeneity
  • Stability
  • Scarcity
  • Cognizability 1 x 3

 

  1. A bill of exchange is an unconditional order from one party to the other to pay on demand or at a stated future date the sum on the bill. It must be signed by the drawer and be accepted by the drawee.                                                                                                         2mks

 

  1. Demerits of barter trade.
  • Lack of double coincidence of wants
  • Lack of standard measure of value
  • Indivisibility of some products
  • Bulkiness of some products
  • Lack of storage facilities. 1 x 3

 

 

 

 

 

 

 

 

 

TEST PAPER  IV

 

  1. a) Explain the role played by the Jua Kali sector in the development of Kenya today            10mks
  2. b) Explain the problems experienced by cooperative societies.                                          10mks

 

  1. a) Discuss the factors that may influence the growth of a business unit.                                   10mks
  2. b) Discuss the factors that you would consider when choosing the mode of transport. 10mks
  3. a) The following information relates to demand and supply schedule for good.

Price per unit (Kshs.)             Quantity demanded                Quantity supplied.

70                                100                                          1000

60                                500                                          800

50                                500                                          400

30                                800                                          400

20                                900                                          300

10                                1000                                        100

  1. On the graph paper provided, draw the demand and supply curve on the same graph

and determine equilibrium price and quantity.                                                  10mks

  1. What would happen if the price of good X was increased from shs. 50 to shs. 70.                                                                                                         2mks
  • What would happen if the price of good X was decreased from shs. 50 to shs. 10.                                                                                                              2mks
  1. b) Outline the disadvantages if sole proprietorship.             6mks

 

  1. a)  Explain the differences between private limited company and public limited

company.                                                                                                       10mks

  1. b) Explain the role played by warehousing as an aid to trade among others.            10mks
  1. a)  Explain the problems that  Kenya experiences due to concentration of industries

in the urban centres.                                                                                             10mks

  1. Discuss the main functions of the accounts department in any business enterprise.

10mks

  1. a)  Discuss the main factors that influence the demand for a particular good or service.

10mks

  1. b) Explain the factors to consider when choosing means of communication.            10mks

 

 

TEST PAPER  IV

MARKING SCHEME.

 

  1. a) Role played by the Jua Kali sector in the development of Kenya today
  • Source of income to individuals
  • Source of employment both to owners and others
  • Provide market for raw materials produced in rural areas.
  • Earns foreign exchange eg curios and sold abroad
  • Government gains revenue from taxes
  • Provide goods at affordable prices to the locals
  • Provides investment opportunities
  • Promotes communal development as evidenced by women groups.

Any other relevant point                                2 x 5

 

   

 

  1. b) Problems experienced by cooperative societies.
  • Limited supply of goods and services since they concentrate on pure or main goods for export
  • Managerial problems due to inefficient committees, who lack appropriate education and relevant experience.
  • Financial problems due to embezzlement of funds by leaders.
  • Scattered population and poor basic infrastructures such as roads and other communication network.
  • Political interference since the government does not let the co-operatives to operate independently
  • Corruption particularly in recruitment of personnel which leads to inefficiency.

2 x 5

 

  1. a) Factors that may influence the growth of a business unit.
  • Availability of sufficient capital
  • Availability of market for the product
  • Favourable legal factors
  • Mature of product service
  • Minimal costs of production leading to high profits
  • Availability of transport and communication facilities.

Any other relevant point   2 x 5

 

  1. b) Factors that you would consider when choosing the mode of transport.
  • Cost freight rate:  Amount of money paid to transport companies and should be maintained at lowest levels possible.
  • Speed and time: Time required to transport goods from source to destination.  Fast moving goods need a means of transport that is fast e.g newspapers.
  • Nature of goods: Bulky goods are transported by sea or rail while valuable and perishable goods are transported by road or air.
  • Flexibility: Ability of a mode of transport to reach its destination.  Road transport is the most flexible.
  • Distance: The length of the journey for which goods have to be transported.  Short journeys are more economical by road.
  • Packing requirements: This is how goods have to be packed to meet a particular mode of transport.  Special packing might add to the cost and become expensive.
  • Convenience: This is the certainity that goods will reach the expected destination in required form.
  • Availability: This deals with how possible it is to get a certain mode of transport.

2 x 5.)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3)  Combined dd ss Curve for good X

 

 

 

70

60

   50

40

30

20

10

100      200      300      400      500      600      700      800      900      1000

Qty demanded  supplied.

 

E . P    = 50 shs.

E . Q    =  500

 

  1. i) From 50 to 70 shs.            =  at 50 sh:  Qd           =  500QS         =  500

at 70 shs:  QD  =  100 Q5  – 1000

Excess supply of 900 units

Supply >  dd by 900 units

Due to price increase.

 

  1. ii) From 50 to 10 =  at 50   –  QD   =   500  QS   =  500

at   10 sh   =  Qs   =  100  Qd   =  1000

shortage of 900 units

demand increase from 500 to 1000 units

excess demand but these is a shortage

due to price decrease.                                 1 x 6

 

  1. b) Disadvantages if sole proprietorship.
  • Meets the losses of the business alone
  • Unlimited liability
  • Insufficient capital for expansion
  • Inability to carry out research
  • Jack of all trade (over working)
  • In case of death, the business collapses
  • No economies of scale. 1 x 6.

 

 

 

 

 

 

 

  1. a) Differences between private limited company and public limited company.

 

Private ltd Co. Public Ltd Co.
–       Membership is from 20 and limited to 50

–       No transfer of shares

–       Commences trading after receiving a certificate of registration

–       The final accounts are private and confidential.

–       The directors do not need to be specified on formation

–       Aim is to have limited liabilities and safeguard personal properties

–       They comprise family business owned by friends

–       They are not quoted in the Nairobi stock Exchange.

–        Membership is from 7 and there is no limit

–        There is transfer of shares.

–        Commences trading only after receiving a certificate of incorporation

–        The final accounts have to be audited by a certified public accountant.

–        A list of directors and their portfolios have to be stated on formation.

–        Aim is to raise required capital from the public

–        The only relationship is that of business.

–        Quoted in the Nairobi stock Exchange.

 

1 x 10

  1. Role played by warehousing as an aid to trade among others.
  • Preserves surplus, stabilizes prices and helps to save huge quantities of good which would otherwise be wasted.
  • Goods can be prepared for sale while in the warehouse in terms of sorting, weighing, packing etc
  • Provide temporary accommodation of goods for producers with no sufficient room for storage.
  • Acts as a reservoir enabling trade to continue throughout the year.
  • Deferred payment on highly taxed goods enables the importer to look for a market for the good
  • Enables the government to get revenue through duty paid. 2 x 5

 

  1. a) Problems that a Kenya experiences due to concentration of industries in the urban

centres.            

  • Rural urban influx
  • Increase in criminology
  • Increased environmental pollution
  • Unbalanced regional development
  • Increase in unemployment in urban areas
  • Development of slums
  • High cost of living in urban areas
  • Inadequate social goods like schools
  • Destruction of road network
  • High dependence ratio if the rural folks on those who work in town.

Any other relevant point.             2 x 5

 

  1. Functions of the accounts department in any business enterprise.
  • Maintain the accounting records
  • Prepare and send invoices to the customers
  • Prepare the budget and plan the activities of the organisation
  • Receive money and make payments on behalf of the organisation
  • Calculate and make payments on behalf of the organisation to the employees

 

  • Prepare final accounts at the end of the year.
  • Analyse the financial position of the business
  • Carrying out internal audit. 2 x 5

 

  1. a) Factors that influence the demand for a particular good or service.
  • Own price of the commodity
  • Price of related goods ie substitutes and complements
  • Income levels of consumers
  • Tastes, fashions and preferences
  • Weather conditions eg demand for umbrellas and gumboots increases during the rainy season.
  • Government policy.
  • Expected changes in price of goods. 2 x 5

 

  1. b) Factors to consider when choosing means of communication.
  • Speed: When the message is urgent, telex, telephone or telegram is the best
  • Cost: Refers to expenses incurred in sending a message.  Generally, the letter is cheaper to send messages.
  • Confidentiality: The ability of the message to reach to the intended person. Letters are better than telephone for confidence purposes.
  • Distance: Represents the geographical gap between the sender and the recipient
  • Evidence: Refers to the record of the message conveyed.  All means of written communication provide evidence.
  • Reliability: The assurance  certainity that the message will reach the intended recipient, at the intended time, place and the right form.  Face to face is the best for this purpose.

2 x 5

 

 

TEST PAPER V

 

  1. Distinguish the following pair of concepts (6mks)
  • Transport and communication
  • Warehouse and warehousing.
  • Cheque and bankdraft
  1. Explain the role played by transport as an aids to trade            (4mks)
  2. Outline any four circumstances under which a manufacturer would sell goods directly to

consumers                                                                                                                          (4mks)

  1. Highlight any four reasons why a government may set up parastatals to supply essential

goods to its people                                                                                                             (4mks)

  1. Mention any four factors that account for the high demand for cellphones/mobile phones in

Kenya today                                                                                                                     (4mks)

  1. State any four characteristics of human wants            (4mks)
  2. The table below shows various forms of production. Indicate using a tick the correct

classification in the appropriate column                                                                         (5mks)

Form of production

 

Primary

Secondary Tertiary
Transportation      
Farming      
Food processing      
Retailing      
Lumbering      

 

  1. Outline four sources of capital to co-operative societies (4mks)
  2. Mr Ryan had he following assets and liabilities on 30th April 2002.

Kshs

Cash at bank                                      23,100

Cash in hand                            3,250

Debtors                                  45,300

Stock                                      48,200

Furniture                                 11,600

Motor-vehicle                        65,000

Creditors                                56,000

            Bank loan                               25,000

 

  1. a) Calculate his capital              (4mks)
  2. Outline any four services rendered by a public warehouse              (4mks)
  3. Mention four ways in which commercial attaches in our foreign embassies help out our

traders                                                                                                                                (4mks)

  1. Explain any three parties to a cheque  (3mks)
  2. Outline any four essentials of a warehouse (4mks)
  3. Highlight any four forms of after-sales service that may be extended to a consumer. (4mks)
  4. Outline any three qualities of an office worker             (3mks)
  5. Outline any four ways though which co-operatives help Agricultural development in

Kenya.                                                                                                                                 (4mks)

 

  1. The following balances were extracted from the books of Abdi Traders.

Kshs

Opening stock                         22000

Closing stock                          26000

Purchases                                349000

Expenses                                 120000

Sales                                       500000

Capital                                    200000

 

Calculate :

(a)  Margin percentage                                                                                             (3mks)

(b) Stock turn over                                                                                                          (3mks)

(c) Cost of sales                                                                                                                   (2mks)

 

 

 

  1. List any four principles of insurance  (3mks)
  2. Highlight any four reasons why international trade is said to be of great importance to a

developing country                                                                                                           (4mks)

  1. Outline any four factors that have promoted the growth of commerce over the years (4mks)
  2. Account for the popularity of supermarkets in Kenya today                               (4mks)
  3. Why in your opinion do you think oral communication lost popularity in Kenya today.

(4mks)

  1. Other than commercial banks, mention any other four financial institutions that provide

financial services in Kenya                                                                                              (4mks)

  1. Name any three factors of production and their respective rewards (3mks)

 

TEST PAPER V

MARKING SCHEME

 

  1. Distinguishing concepts.
  • Transport – The process of moving goods or people from one place to another.
  • Communication – The process of disseminating information /message from one person to another.
  • Warehouse – Is a building where goods are received and stored prior to being despatched to the people who require them
  • Warehousing – Is the process of receiving goods into warehouses, storing them and dispatching them to the users
  • Cheque – Is an order by a depositor/ drawer to the bank/drawee to make a specific payment to the person named /payee
  • Bank draft- A cheque drawn by a bank in favour of a married person and serves as a guarantee of payment .                       (1 x 6  = max 6 marks)

 

  1. Role of transport
  • Facilitates movement of materials from their source places to the factories/industries
  • Movement of finished goods from factories to consumers
  • Increases utility of goods and services
  • Enhances movement of bulky goods
  • Avoids wastage since surpluses are transferred to areas where they are scarce
  • Makes possible exploitation of resources

Any other relevant point                                ( 1 x 4 = max 4marks)

 

  1. When a manufacturer sells directly to consumers
  • When goods require sufficient after-sales services
  • When goods are very expensive e.g. motor cars.
  • When goods are branded and widely advertised
  • When goods are highly fashionable with high profit margin
  • When goods are highly fragile e.g. glass materials
  • When goods are highly perishable e.g. fruits and vegetables .
  • When the manufacturer wants to exercise some control during the distribution process. Any other relevant point ( 1 x 4 = max 4 marks)

 

 

 

 

  1. Why government sets up parastatals to provide essential services
  • Prevent exploitation of the public by private businessmen
  • Provide goods and services that private enterpreneurs are unwilling to venture due to low profits involved
  • Provide large amount of capital(initial) where private organisation are not able to finance
  • Provide essential goods and services that are too sensitive to be left in the hands of a private sector
  • Stimulate economic development of the country by providing the essential social overhead capital
  • Establish monopolies in areas of interest where high profits are involved

(any other relevant point ) ( 1×4 = max 4 marks )

 

  1. Factors accounting for high demand of cellphones
  • They are highly portable
  • Convenient since the owner can be reached anywhere and anytime
  • They are cheaper since the cost of calling a trunk call or otherwise is the same .
  • They are time saving since one does not have to travel to receive or make a call
  • They are easier and takes less time to install
  • Serves as a diary since telephone numbers can be stored

(any other relevant point) 1×4  max = 4 marks

 

  1. Characteristics of human wants
  • They are insatiable
  • They are habit forming
  • They are competitive
  • They are recurring
  • Particular wants can be satisfied

(Any other relevant point) 1 x4 max = 4marks

 

 

Form of production

 Primary
 

Secondary

 

Tertiary

Transportation              √
Farming         √    
Food processing            √  
Retailing    
Lumbering    

1 x 5 max 5 marks

  1. Sources of capital to co-operatives
  • Membership contributions
  • Retained profits
  • Credit facilities
  • Entrance fees
  • Surplus and reserves                             (1 x 4 max 4marks)

 

  1. Capital = Assets – Liabilities

Assets = ( Cash at bank + cash in hand +debtors + stock + Furniture + motor-vehicle)

(23100 + 3250 + 45300 + 48200 + 11600 + 65000)

= 196450

Liabilities = Creditors +Bank loan

= 56100+ 25000

= 81000

Hence capital = 196450-81000

=  115350

 

  1. Services rendered by a public warehouse
  • Provide storage facilities
  • Serves as a place for preparation of the goods for subsequent use by a change in their condition e.g. ripening of bananas
  • Renders a warrant which is a document of title issued to the merchant owning the goods
  • Offers protection by insurance against loss by fire, theft, etc
  • Prepares sample for report and market usage

( any other relevant point ) 1x 4 =  max 4

 

  1. Role of commercial attaches
  • Explore new markets for more export opportunities
  • Study and analyse markets for exports for their home countries
  • Research and keep data for new markets
  • Publish and advertise their countries exports
  • Select buyers agents, agents and distributors of the home country’s exports
  • Inform traders in their home countries of the standards required for exports .
  • Organise visits to trade fairs and exhibitions for businessmen from their home countries. (any other relevant point ) 1 x 4 = max 4 points

 

  1. Parties to a cheque.
  • The drawer – the account holder who drafts the cheque
  • The drawee- The bank that is to make payment.
  • The payee – The named person who is to receive payments

  (½  x 6  = 3marks)

 

  1. Essentials of a warehouse
  • Ideal location where goods are produced in greater quantity
  • The building must be suitable and of large size appropriate to store goods efficiently
  • Properly equipped with necessary facilities to store goods for a longer time without getting spoilt
  • Efficient and well trained staff with ability to protect goods against damages by fire, water or bad weather.
  • Should have efficient transport system to enhance movement of goods into and out of the warehouse.
  • Should have adequate protection measures

(any other relevant point) 1 x 4  = max  4 marks.

 

  1. Forms of after-sales services
  • Free of charge repair for specific period of time
  • Cleaning or servicing of the equipment at regular intervals free at very reasonable rates.
  • Provide technical advice regarding he use of equipment if the customer has some problem.
  • Ensure availability of spares

(Any other relevant point ) 1 x 4 = 4 marks)

 

 

  1. Qualities of an office worker
  • Personal attribute e.g. neatness, cleanliness.
  • Office etiquette e.g. respect courtesy
  • Personal knowledge and skills

 

  1. 16. How co-operatives help Agricultural development
  • Help farmers get better prices for their produce
  • Assist farmers seeds free or at reduced prices
  • Fertilisers, farming tools and implements are provided at subsidised rates
  • Offer useful advice on farming
  • They sometimes grant loans to farmers for development .
  • Provide a channel of distribution of loans offered by the government to relevant members. (any other relevant point) 1 x 4 = max 4 mks)

 

  1. Margin percentage

Gross profit   x 100

Sales

Gross profit = Sales – cost of goods sold(Opening +Purchases – closing stock

=  500000 – (22000+349000-26000)

=  500000- 397000

= 103000

 

Hence margin  103,000  x 100

500000

= 20.6%

 

(b) Stock turn – over

= Cost of sales

                        Average of stock

Cost of sales = Opening stock + Purchases – Closing stock

Average stock = Opening stock + Closing stock

2

Hence Turn-over  = 22000 + 349000  -26000

22000 +26000

2

103000

24000

= 4.3                                    ( 1 x 3) =  3marks

 

  1. c) Cost of sales

= Opening stock + Purchases

= 22000 +349000

= 371000                                             (1 x 2 ) = 2marks

 

  1. Principles of insurance
  • utmost good-faith uberimae or fidei
  • Insurable interest
  • Subrogation
  • Proximate cause
  • Indemnity
  • Contribution/average clause
  • Pooling risks
  • Insurable risks

 

  1. Importance of international trade
  • Individual are able to get a variety of goods which they do not produce themselves.
  • Encourages specialisation by countries
  • Encourages interdependence hence international peace and understanding
  • Encourages a country to dispose off its surplus produce
  • Earning of foreign exchange
  • The government earns income from tariffs and other duties

(Any other relevant point) 1×4 = max 4 marks)

 

  1. Factors that promoted growth of commerce
  • Development of money
  • Diversification of resources
  • Specialisation and division of labour
  • Government involvement in trade activities
  • Development of infrastructures
  • Development of auxiliary

 (any other relevant point ) 1 x 4 = 4 marks )

 

  1. Why supermarkets are popular.
  • Offer fair prices to their customers since they buy in bulk hence get huge discounts
  • Saves time for customers since they get what they want under roof.
  • Employs few attendants hence operate at low cost.
  • Impulse buying leads to more sales since customers have access to goods for easy choice
  • They do not sale credit hence avoid the problem of bad debts

Any other relevant point ( 1 x 4)

 

  1. Why oral communication has lost popularity.
  • It’s expensive in long distance
  • its not suitable for confidential messages.
  • It’s open to exaggeration where favours are expected
  • No record keeping making reference difficult
  • It can be time wasting.
  • It may be treated lightly if not properly planned

 

  1. Financial institutions providing financial services.
  • Merchant bank
  • Hire purchase finance companies.
  • Building societies.
  • Insurance companies and pension funds
  • Post office savings bank. Any other relevant point  1×4

 

  1. Factor  Reward

–  Land                                rent

–  labour                             wages

–  Capital                            interest

–  Enterprenuership                        profit                                       ½ x 6

 

 

TEST PAPER  VI

  1. a) Explain six reasons why a businessman would choose to operate a current account instead

of a savings account.                                                                                                    12mks.

  1. b) Highlight any four circumstances under which a businessman would communicate to the

customers by letters rather than telephone.                                                                 8mks.

  1. a) Discuss the main challenges/problems that face the small scale industry in Kenya today                                                                                                                                                  10 mks
  2. Explain the factors that a seller should take into account when choosing the best medium

for advertising his products                                                                                   10mks

  1. a) Explain the benefits that Kenya is likely to enjoy for being a member of the common

market for East and Southern Africa (COMESA)

  1. b) Clearly describe the services rendered by wholesalers to growers, producers and

manufacturers                                                                                                                   10mks

  1. a) Clearly describe how the central bank limits the credit creation of the commercial banks.                                                                                                                                                 10mks
  2. b) Explain the role played by marketing boards             10mks

 

  1. (a) The following information relates to African Retail Traders for the year ended

31st  Dec 1999

Gross profit                              600000

Opening stock                           285000

Sales                                       2400000

Purchases                                1830000

Expenses                                 360000

Net profit                                240000

Closing stock                          315000

Fixed assets                            400000

Debtors                                   900000

Cash at bank                             35000

Current liabilities                   438000

a)  Calculate

  1. Cost of goods sold
  2. Mark –up percentage
  3. Debtors ratio
  4. Current ratio     12mks
  5. b) Outline any four reasons that may lead to dissolution of a partnership business             8mks

 

6 .  a) What arguments can be advanced for state ownership of business understandings                10mks

  1. b) Write short notes on the following terms as used in insurance
  2. Indemnity
  3. Utmost good faith
  • Proximate cause
  1. Insurable interest
  2. Subrogation                                                                                                 10mks

 

 

 

 

 

 

TEST PAPER  VI

MARKING SCHEME

 

  1. a) Why a businessman would choose to operate a current account
  • Excellent means of settlement of accounts speedily and economically avoiding the use of the legal tender
  • Can be used for direct receipt of, regular, sums by direct credit instead of legal tender
  • Withdrawals can be effected on demand without notice
  • It is a basis for obtaining loan funds or of overdraft facilities
  • Provide means to effect regular payment at stated intervals of time by standing order eg rent.
  • Safety, reliability and ready availability of funds kept in this way.
  • Provides periodic bank statements issued which schedule all bank transactions completed for the period. Any other relevant point  =  2 x 6

 

  1. b) Circumstances when a businessman would communicate to customers with letters than

            telephone.

  • When permanent records for future reference are required
  • When communication needs to be highly confidential
  • When communication need to reach many people at a time when he/ she needs to accommodate complex messages such as tables and graphs for illustrations.
  • When the customer is far away and it may be two expensive to call.

Any other relevant point  =  2 x 4

 

  1. a) Problems facing small scale industry.
  • Lack of sufficient capital
  • Lack of proper market information
  • Poor management due to low education levels
  • Lack of credit facilities due to lack of proper securities
  • Poor financial control since most owners do not know how to keep clean records.
  • Lack of tenacity / persistence since most owners expect returns overnight.

Any other relevant point  =  2 x 6

 

  1. Factors to consider when choosing the best medium for advertising.
  • The ‘reach’ of the media ie the proportion of the target audience that can be contacted.
  • The relative cost per unit cost should be low
  • The impact of the media ie the extent to which the message is effectively taken in.
  • The permanence of advertising
  • The selectivity of the media.
  • The product being advertised. Any other relevant point  =  2 x 5

 

  1. a) Benefits that Kenya reaps from COMESA.
  • High quality products
  • Fairly priced products
  • Wide variety of products
  • Access to a wider market.
  • Wider consumer choices
  • International peace and understanding. Any other relevant point  =  2 x 5

 

 

 

  1. Services rendered by wholesalers to growers, producers & manufacturers.
  • Market to purchase all their products, providing warehousing in preparation for subsequent distribution to retailers.
  • Provision of transport services: often packing for despatches of the goods:  prompt settlement:  display and advertising of the product.
  • Link with retailers and consumers’ demands to facilitate production in anticipation of demand and with fore knowledge of changes in consumers’ selection, taste and requirements.
  • Relieves the suppliers off the problem of finding diverse markets, finance and storage problems.
  • Provides for further services necessary in preparation for retail marketing eg. Grading, blending, refining, etc
  • Fulfils the manufacturers’ needs for a steady supply of raw materials or semi – finished goods to ensure continuity of his production lines.

Any other relevant point  =  2 x 5

 

  1. a) How Central Bank limits credit creation of Commercial Banks.
  • Open market operations
  • Bank Rate Policy
  • Margin Requirement
  • Selective Credit Control
  • Rationing of Credit.
  • Direct Action, Moral persuasion and publicity. 2 x 5

 

  1. Role of marketing boards.
  • Buying and selling produce at fair prices
  • Control of production
  • Collection of produce and transportation
  • Storage of produce
  • Financing functions
  • Research and advisory services
  • Advise the government on matters related to the produce they handle
  • Promote interests of the farmers / growers by supplying fertilizers, seeds etc to enhance productivity and good farming.
  • High quality products
  • Fairly priced products
  • Wide variety of products
  • Access to a wider market.
  • Wider consumer choices
  • International peace and understanding. Any other relevant point  =  2 x 5

 

  1. a) Cost of goods sold.

=  opening stock  + purchases – closing stock

=  285,000 + 1,830,000 – 315,000

= 1,800,000

 

 

 

 

 

 

  1. Mark-up percentage

=          Gross profit    x 100

Cost of sales

=    600,000     x 100

1,800,000

=  33 1/3 %

 

  1. Debtors ratio

=  Debtors     x 365

sales

=   9 00,000     x 365

2,400,000

=  136.9

=  137days.

 

  1. Current ratio.

=           Current Assets

Current Liabilities

=   Stock + Debtors + Cash at bank

Current Liabilities

 

315,000 + 900,000 + 35,000

438,000

=     1,250,000

438,000          =   3.08  : 1   +  3 . 1 : 1.                            1 x 3

 

  1. Reasons that may lead to dissolution of partnership.
  • Completion of the intended purpose
  • Loss of one partners
  • Request of dissolution by one partner
  • A change in the country’s law disqualifying the business
  • A court order
  • Insolvency of the business. Any other relevant point  =  2 x 4

 

  1. a) Arguments for state ownership of business undertakings.
  • They promote general welfare of the public by providing social goods such as medical facilities, sewerage, etc
  • Avoid wastage and inefficiency since some establishments are nationalised to avoid duplication.
  • They encourage private investment in activities which require heavy capital but take long to bring returns.
  • They provide services to the public at fair prices since the interest of the public is the main consideration.
  • Encourages democratic control through the state and local authority and profits are not to a limited number of shareholders.

Any other relevant point  =  2 x 5

 

 

 

 

 

  1. Short notes:
  2. Indemnity : A principle of insurance which states that an individual should only be compensated upto the actual monetary loss caused by the event insured against.  The maximum one can get is the value of the property insured when there is total loss caused by the risk insured against.  Hence the insured should never benefit / gain.
  3. Utmost good faith – this principle requires that the insured person must disclose all the material facts about the insured property to the insurance company.
  • Proximate cause – This principle states that for an insurer to be liable, the loss must be dominantly and effectively resulting from a risk for which the insurance policy was taken out.
  1. Insurable interest –  This principle states that an insurance claim cannot be valid unless an insured person can prove that he has suffered a financial loss because the insured event has occurred.  The insured person must not gain from the occurrence of the insured event hence should only insure upto the amount that you can claim when the event occurs.
  2. Subrogation – This principle maintains that after the one insurer has indemnified the insured, the rights in the remains of the damaged proper by pass on to the insurer.                        2 x 5

 

 

TEST PAPER  VII

 

  1. Highlight any four  problems that a country may experience when it has decentralized its industries.                                                                                                                          4mks
  2. Highlight any four characteristics of money that enables it to serve its purpose efficiently.                               4mks
  3. Explain any four problems that Kenya would experience from participating in international trade.                   4mks
  4. Outline any four importance’s of advertising as a sales promotion method.       4mks
  5. Highlight any four ways in which insurance companies make profits.       4mks
  6. Mention four of the contents in the articles of association.       4mks
  7. In the spaces provided, name the type of business organisation described by each of the following features.       4mks

Description                                                                 Business organisation

  1. Each member has only one vote             ………………………………………
  2. The business enjoys perpetual succession             ………………………………………
  3. Each member may act on behalf of the business ………………………………………
  4. Formed by an act of parliament             ………………………………………

 

  1. Identify four services that can be conveniently paid for using standing orders (bank orders)                   4mks
  2. Give any four reasons why the Kenyan government is actively involved in licencing

business activities.                                                                                                                       4mks

  1. Outline any four main principles of investment       4mks
  2. Explain the meaning of the following terms.       4mks
  3. Par value
  4. Blue chips
  • Cum right or Ex-Right
  1. Tilt Edged securities

 

  1. The following is a summary of transactions for Ryan Traders for the month of April 2002.

2nd April    –  Commenced business with shs. 12,000 in cash and shs. 150,000 at the bank

3rd  April    –  Made cash sales amounting to shs. 6,000

5th April    –  Received cash shs. 16,000 from a debtor

12th April  –  Paid for motor expenses shs. 7,000 in cash

13th April  –  Drew shs. 30,000 from bank for office use

15th April  –  Paid salaries in cash amounting to shs. 15,000

17th April  –  Bought furniture worth shs. 18,000 paying by cheque.

18th April  –  Paid expenses amounting to shs. 15,000 by cash

19th April  –  Cash sales of shs. 40,000 received by cheque

23rd April  –  Paid shs. 20,000 cash into the bank

25th April  –  Paid a creditor shs. 16,000 by cheque

 

You are required to prepare a two-column cash book for Ryan Traders for the month of April 2002.                                                                                                                             8mks

  1. Highlight any four reasons why the Kenyan government finds it necessary to train businessmen.                   4mks
  2. The following balances were extracted from the books of A & B Co. as at 1st April 2002.

Capital      shs. 200,000

Fixed Assets  shs. 180,000

Current Assets      shs. 40,000

Current Liabilities shs. 20,000

 

During the month, the following took place.

  1. Goods worth 50,000 were taken away from the business for personal use.
  2. Net profit of shs. 80,000 was realized.

Calculate capital as at 30th April 2002.                                                               4mks

 

  1. Other than advertising, mention any other four methods of sales promotion.       4mks
  2. Highlight any four reasons why most businessmen do not prefer transferring their goods

via canals.                                                                                                                                    4mks

  1. Alex is an employee of an advertising firm as a sales executive. He receives a salary of

Kshs. 10,000 plus commission of the following basis:-

On the first sales of shs. 100,000 – 2%

On the next shs. 200,000              – 5%

On the balance of sales                – 10%

He sold an advertising space worth shs. 450,000 during the month of January 2002.

Calculate his total income for the month.                                                                             4mks

  1. Highlight any four methods that the government can use to influence the quantity of a commodity supplied.       4mks
  2. Indicate the type of resource represented by the following.       4mks

Type of Resource

 

Forest                                      …………………………………………………………….

Road                                        …………………………………………………………….

Doctor                                     …………………………………………………………….

Wildlife                                  …………………………………………………………….

 

 

 

 

  1. Outline any four methods of determining prices in the market.
  2. Write short notes on the following types of partners.       4mks
  3. Active Partners
  4. Minor partners
  5. Quasi partner
  6. Outgoing partner
  7. “The office performs a fundamental role in any business organisation.”  Account for this statement.                                                                                                                        4mks
  8. Outline any four factors that an investor should consider before setting up a small scale

retail shop in a town.                                                                                                        4mks

 

 

TEST PAPER  VII

MARKING SCHEME

 

  1. Problems of decentralization.
  • Spread of industrial pollution
  • Strain / over – utilization of a country’s natural resources
  • Certain special services may not be available in the rural areas
  • Incentive offered by the government to the firm is a burden to the tax payer
  • Decentralized firms may not enjoy benefits that other firms realize in urban areas.

Any other relevant point                    any four pts 1mks each

 

  1. Characteristics of money.
  • Portability
  • Homogeneity
  • Divisibility
  • Scarcity
  • Cognizability
  • Durability
  • Acceptability
  • 1 x 4mks

 

  1. – Inferior products may be imported.
  • Over dependence
  • Balance of payment problems
  • Loss of cultural values and prestige
  • Moral decay
  • Lack of cultural values and prestige
  • Moral decay
  • Lack of development skills for countries that specialize in production of primary products.

Any other relevant point                    any four pts 1mks each

 

  1. Importance of advertising.
  • Informs the prospective customers that a certain product has been placed in the market
  • Help consumers to make a more informed choice on the goods they intend to buy
  • Increases the volume of sales hence realization of higher profit levels
  • Acts as an aid to product identification since an advert clearly shows the brandname and colour of the product.
  • Assists in planning production as it plans a significant role in reducing sales fluctuations
  • Helps to inculcate customers goodwill particularly the reminder advert which keeps customers aware that the product is still in the market.

Any other relevant point                    1 x 4

 

  1. How insurance companies make profits
  • Investing their funds
  • Differences in claims
  • Forefeited funds by policy holders
  • By taking whatever remains after compensation eg remains of a car.

Any other relevant point                    1 x 4 mks

 

  1. Contents of the articles of association
  • Classes and rights of shareholders
  • The issue and transfer of shares
  • Methods of dealing with any alterations on the capital
  • Qualifications, duties and powers of directors
  • Borrowing, dividend and reserve policies
  • Rules regarding preparation and auditing of accounts.

Any other relevant point                    1 x 4 mks

 

  1. Description Business Organisation
a)     Each member has only one vote

b)     Business enjoys perpetual succession

c)     Each member may act on behalf of the business

d)     Formed by an act of parliament

Co-operative

Soleproprietorship

 

Partnership

Parastatal

 

  1. Services that can be paid for by standing orders
  • Rent payments
  • Subscriptions
  • Insurance’s
  • Hire purchase payment
  • Loan payments
  • Mortgage repayments

Any other relevant point                    1 x 4 =  any four pts 1mks each

 

  1. Why the Kenyan government is involved in licencing.
  • Regulate the number of businesses in any one line of activity to control unhealthy competition.
  • Ensure that traders engage in only those business activities for which they have licences.
  • Control illegal businesses in the country
  • Ensure that those who engage in professional activities meet the requirements
  • Control the type of goods that should enter or leave the country.

any four pts 1mks each 

 

  1. Principles of investment.
  • Should be secure
  • Liquidity ie ability to be converted into cash
  • Ability to grow over a period of time
  • Should generate income / returns
  • Ability to spread risks over several types of securities. 1 x 4
  1. Par value –  This is the face value or nominal value of a share
  2. Blue chips –  Refers to high grade equity shares ie the ordinary shares of a company which have had a good dividend record extending over the years.
  • Cum Right or Ex – Right –  Used in reference to a new rights issue attaching to the shares.  Ex-right refers to that right retained by the seller while cum – Right is the right transferred to the buyer.
  1. Gilt Edged Securities –  Refers to securities sold by the government considered absolutely safe.                                                                             1 x 4

 

  1.        Ryan Traders

Cash book for the month

of April 2002.

Date

2/4/ 02

3/4/ 02

5/4/ 02

19/4/ 02

23/4/ 02

 

 

25/4/ 02

 

 

 

25/4/ 02

 

Particulars

Bal b/f

Sales

Debtors

Sales

Cash ( c)

 

 

Bal c/d

 

 

 

Bal b/d

Cash

12,000

6,000

16,000

 

 

 

 

23,000

 

57,000

Bank

150,000

 

 

40,000

20,000

 

 

 

 

210,000

 

146,000

Date

12/4/ 02

13/4/ 02

15/4/ 02

17/4/ 02

18/4/ 02

23/4/ 02

25/4/ 02

25/4/ 02

 

 

 

25/4/02

 

Particulars

Motor expenses

Office

Salaries

Furniture

Expenses

Bank ( c)

Creditor

Bal c/d

 

 

 

Bal b/d

 

Cash

7,000

 

15,000

 

15,000

20,000

 

 

 

57,000

 

23,000

Bank

 

30,000

 

18,000

 

16,000

146,000

 

 

210,000

8mks

 

  1. Why the government trains businessmen.
  • Expose businessmen to modern developments in management
  • Educate businessmen of efficient methods of operating business eg effective methods of sales promotion
  • Identify and discuss current problems facing businesses and look for possible solution
  • Import proper business ethics eg good customer relations
  • Educate businessmen on how to use available resources to minimize costs and maximize profits
  • Inform businessmen on various available profitable business opportunities in their surroundings
  • Expose them to government policies regarding business activities in the country

1 x 4  =   any four pts 1mks each

 

  1. Calculation of capital

Initial capital                    200,000

Less Drawings                    50,000

150,000

Add net profit                     80,000

Net capital                                    230,000                                   1 x 4 mks

 

 

  1. Other methods of sales promotion
  • Free samples
  • Gifts
  • After – sales service
  • Credit facilities
  • Price reductions Any other relevant point                    1 x 4 mks

 

  1. Problems of canals.
  • Canals lack speed hence cant be used to transport perishable or urgently required goods
  • Canals can only serve a limited area where water exists
  • Canals may dry in drought seasons and freeze in very cold seasons
  • Maintenance charges are high

Any other relevant point                    1 x 4mks

 

  1. Calculation of total income.

Shs.                             Shs.

Basic Salary                                              10,000                         10,000

On the first sales of 100,000                    2/100 x 100,000                          2,000

On the second sales of 200,000                5/100 x 150,000                        10,000

On the balance of sales 150,000              10/100 x 150,000           15,000

Total income                                                                          37,000             1 x 4mks

 

  1. Methods used by government to influence supply.
  • Taxes
  • Subsidies
  • Price control
  • Production quotas
  • Creation of monopolies through licencing             1 x 4mks

 

  1. Item                                       Type of Resources

Forest                                      Natural

Road                                        Man-made

Doctor                                     Human

Wildlife                                  Natural                        1 x 4 = any four pts 1mks each

 

  1. Methods of price determination.
  • Haggling ie bargaining over a price range
  • Auction
  • Price mechanism
  • Government policy

 

  1. Short notes.
  2. Active partner –  One who takes an active role in the running of the partnership business
  3. Minor partner –  One who is under 18 years of age, shares profits but cant take part in management.
  4. Quasi partner –  One who does not contribute capital but allows the firms to use his name as a partner.  Shares profits but not losses.
  5. Outgoing partner – One who is retiring from a partnership business with the consent of all other partners or in accordance with any previous agreement.

any four pts 1mks each   =   1 x 4

  1. Functions of an office
  • Collecting and recording information
  • Distributing information
  • Mailing
  • Reproduction of documents
  • Storing information
  • Safeguarding an organisations property

 

  1. Factors to consider when setting up a small scale retail shop in town.
  • Extent of demand / market
  • Level of competition
  • Security level
  • Accessibility
  • Type of business
  • Availability of raw materials
  • Availability of water and power Any other relevant point                    1 x 4

 

 

TEST PAPER  VIII

 

  1. a) Explain any five methods that the Kenyan government may adopt in order to control

the volume of imports.                                                                                               10mks

  1. b) Discuss the main sources of capital to small scale enterprises.                                10mks

 

  1. a) What do you understand by economies to scale.  Discuss any four internal economies

of scale                                                                                                                       10mks

  1. Clearly define the term sales promotion. Discuss the importance of Trade Fairs as

part of sales promotion.                                                                                      10mks

 

  1. Opportunities exist for purchasing of necessary durable goods by installment.
  2. Describe two methods commonly used for this purpose.             4mks
  3. Mention three advantages of each to:
  4. The manufacturers             6mks
  5. The consumer             6mks
  6. Give any two differences of the two methods.             4mks

 

 

  1. Prepare a trading profit and loss account for 2000 and a balance sheet as at 31st Dec 2000 from the following trial balance.
  2. Kabas Co.

Trial Balance

As at 31st Dec 2000

 

Capital

Land and buildings

Motor vehicle

Drawings

Furniture and fittings

Debtors

Creditors

Stock on 1st Jan 2000

Purchases

Sales

Returns inward

Transport

Return outwards

Telephone and Electricity

Rates and Insurance

Licence and Legal expenses

Rent income

Salaries

Advertising

Carriage inwards

Stationery and office expenses

General expenses

DR (Kshs)

 

50,000

38,000

32,485

11,200

38,459

 

37,129

273,538

 

7,832

35,223

 

9,835

4,942

1,956

 

37,280

6,342

12,345

950

  2,484

  600,000

CR (Kshs)

150,000

 

 

 

 

 

46,253

 

 

384,803

 

 

10,544

 

 

 

8,400

 

 

 

 

 

     600,000

 

Cost of unsold stock on 31st Dec 2000 was Kshs. 43,210.                                        20mks

 

  1. a) Explain any five functions of commercial banks.                                                10mks
  2. Assume that you are a policy maker in Kenya today, explain the reasons which would

make you recommend rural industrialization policy.                                                     10mks

 

  1. a) Account for the low bargaining power of trade unions in Kenya today.             10mks
  2. Outline any five benefits that a person may derive from being a member of savings

and credit cooperative society.                                                                            10mks

 

 

 

TEST PAPER VIII

MARKING SCHEME

 

  1. a) Methods of controlling volume of imports
  • Tariffs –  Involves levying import duty on goods that are entering a country.
  • Quotas –   Is a quantitative restriction permitting only a given number of units of a

commodity to be imported during a specific period.

  • Exchange control – Refers to control by the government on the exchange value of the currency of a country. Importers can only import goods when they seek permission for acquiring foreign exchange.
  • Trade agreements –  is an agreement between different countries regarding the foreign trade under this system, goods can only be imported from some specific countries only.
  • Total ban/ Embargo –  Involves imposing a complete ban on import of some specific commodities.                                    2 x 5

 

  1. Sources of capital to small scale enterprise
  • Personal funds / owners capital
  • Borrowing from friends and relatives
  • Members contribution
  • Loan from banks and other non bank financial institutions
  • Going public ie borrowing from the public by the use of stocks
  • Venture capital ie a group of rich people with a pool of financial resources which they led to viable small scale business. Any other relevant point

 

  1. a) Economies of scale refers to the benefits that a firm enjoys due to expanding its scale of

production.

Internal economies

  • Administration: When a firm expands, its administrative expenses are distributed over greater output and the per unit cost decreases.
  • Financial: Large firms have a good reputation hence can get loans from financial institutions.
  • Marketing: Large firms sell their products easily since they are able to engage in sales promotion methods
  • Technical: Due to their financial capacity, large firms can afford capital goods that enables them to produce superior products.                          2 x 4

 

  1. b) Sales promotion: refers to the process whereby the demand for the products of a

business enterprise is created in an attempt to increase sales.  It

has to do with informing the consumers about the existence of a

particular product, its quality and features, places of availability as

well as the price of the product.                2mks

 

Importance of Trade fairs.

  • It’s a means of advertising products to the wholesale and retail trade
  • Brings new products on view to create a demand
  • Builds up an advance order book within the trade and hence ensure continuity of production over a period of time
  • Enables new business contacts to be made Any other relevant point   2 x 4

 

 

  1. a) Methods of buying goods in instalment.

–  Hire purchase –   A system where the buyer agrees to pay for the commodity in

instalments.  The buyer obtains the possession of the article on the down payment (deposit) but the title of ownership remains wit the seller until the last instalment is made.

  • Credit sales / Deferred payment –  under this system the article becomes the property of the

buyer with the payment of the first instalment.  The buyer is supposed to pay

the remaining amount in instalments.                                                           2 x 2

 

  1. Advantages of Hire Purchase to:
  2. The manufacturer.
  • He is able to increase his volume of sales.
  • Profit earned on hire purchase is higher than profit earned on cash basis.
  • Goods belong to the seller until the last instalment is paid.

 

  1. The buyer.
  • He takes possession of goods and their use immediately after down payment.
  • Instalments to be made are predetermined enabling him to plan and budget for payment of the amounts.
  • He is able to acquire goods he could not have afforded on cash terms
  • The buyer can possess more goods including expensive items. 1 x 6

 

Advantages of credit sale to:

  1. The manufacturer
  • He is able to increase his volume of sales
  • Profits realized are higher since customers must pay higher prices
  • He can reclaim the amount due from the buyer under a court order.

 

  1. The buyer
  • He is able to buy expensive goods on credit
  • Ownership of the article posses to the buyer after the first instalment
  • He is able to plan on acquisition of certain goods.

Any other relevant point                                                        1 x 6.

 

  1. Differences:

                Hire purchase

–     Goods can be repossessed if the buyer breeches the contract

–     Ownership remains with the seller till the last instalment is made

                    Credit sale

–       Goods cant be repossessed but the buyer can be sued for damages

–       Ownership posses to the buyer when the first instalment is affected.

Any other relevant point        1 x 4

 

 

 

  1.                                                      Kabs Co

Trading, Profit and loss A/c

For the year ended 31st Dec 2000

  Shs. Shs.   Shs. Shs
Opening stock

Add:  purchase

–   Returns outwards

+  Carriage inwards

Goods available for sale

Less closing stock

Cost of goods sold

Gross profit c/d

 

Expenses.

Transport

Telephone & Electricity

Rates and Insurance

Licence and legal expenses

Salaries

Advertisement

Stationery & office expenses

General expenses

Net profit c/d

 

273538

 

12345

 

 

 

 

 

 

35223

9835

4942

1956

37280

6342

 

950

2484

 

 

37129

 

 

312468

43210

269258

107713

376971

 

 

 

 

 

 

 

 

 

99012

17101

116113

Sales

Less returns inwards

 

Net sales

 

 

 

 

Gross profit b/d

Add Rent income

 

 

 

 

 

 

 

 

Net profit b/d

 

 

 

384803

7832

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

376971

 

 

 

376971

107713

8400

 

 

 

 

 

 

 

 

116113

17101

 

 

  1. Kabs Co

Balance sheet

At at 31st Dec 2000

  Shs Shs   Shs Shs
Fixed Assets

Land & Building

Motor vehicle

Furniture

 

Current assets

Debtors

Stock

 

50000

38000

11200

 

 

38459

43210

 

 

 

99200

 

 

 

81669

180869

Capital

Less Drawings

Add Net profit

Net capital

Add creditors

 

150,000

32485

17101

 

 

 

134616

46253

 

 

 

180869

 

  1. a) Function of commercial Banks
  • Receiving deposits
  • Advancing loans
  • Transferring money
  • Agents of the stock exchange market
  • Custodian of valuable items.

Any other relevant point                    2 x 5

 

 

 

  1. Reasons for rural industrialization policy.
  • Creation of job opportunities in rural areas
  • Provision of market for raw materials produced in rural areas.
  • Balanced growth in the economy
  • Reduced rural – urban migration / influx
  • Increase incomes of rural population
  • Rural population will get modern facilities like electricity, piped water, etc.

Any other relevant point.

 

  1. a) Problems of trade unions.
  • Poor management
  • Political interference
  • Insufficient funds
  • Lack of strike benefits
  • High levels of unemployment
  • Lack of patience during the negotiation period.

Any other relevant point                    2 x 5

 

  1. Benefits of savings and credit co-operative society (SACCO)
  • Education
  • Sharing of additional profits
  • Loans can be advanced
  • Savings
  • Improved living standards
  • Credit facilities eg Bank overdraft.

Any other relevant point                    2 x 5

 

 

TEST PAPER IX

 

  1. Explain the following terms as used in commerce       4mks
  2. Registered capital
  3. Subscribed capital
  • Issued capital
  1. Paid up capital
  2. Outline any four circumstances under which a credit note may be used.       4mks
  3. The main objective of commerce is profit maximization. Highlight any four ways in

which businesses can achieve this objective.                                                                  4mks

  1. Give any four functions of commercial attaches.       4mks
  2. Mention four investment services offered by the Kenya Postal Corporation.       4mks
  3. Give any four reasons that may lead to dishonour of a cheque.       4mks
  4. Highlight any four benefits that accrue to a seller who uses personal selling as a method

of sales promotion.                                                                                                           4mks

  1. Give any four categories of goods that are suitable for direct distribution from producers

to retailers.                                                                                                                      4mks

  1. Highlight four ways in which Insurance companies make profits.       4mks
  2. State any four sources of capital to co-operative societies.       4mks

 

 

 

  1. The following balances relates to a business for the year ended 31st Dec 2000.

Capital as at 31st Dec 2000           170,000

Drawings for the year                 5,000

Net profit for the year                      5,000

Extra capital introduced                 52,000

Purchases                                        20,000

 

  1. Determine the amount of capital as at 1st Jan 2001.       3mks
  2. Calculate the Net profit ratio       3mks

 

  1. Outline any four features of a fixed deposit account.       4mks
  2. Give any four reasons why most business people prefer to transport their finished

products via road transport.                                                                                             4mks

  1. Give any four benefits that you may reap as a consumer from an advertising campaign. 4mks
  2. Mention any four remittance services offered by the post office.       4mks
  3. Give four reasons why an office may prefer storing information in micro-films instead

of files                                                                                                                              4mks

  1. Assume that you want to start photocopying business in your nearest town. Cite any

four factors you would consider before buying your copies.                                    4mks

 

  1. The following machines are found in a mailroom. State the function of each.       5mks
Machine Functions
Addressing machine  
Shredding machine  
Guillotine  
Franking machine  
Composite machine  

 

  1. Give any four reasons why the use of Electronic mail (E-mail has become very popular

in Kenya today.                                                                                                                4mks

  1. The following balances were extracted from the books of MM traders on 31st January 2002.
Item                                   Amount

Sales                                 10,000

Purchases                          6,000

Bank loan                          2,000

Buildings                          30,000

Machinery                        40,000

Premises                           60,000

Debtors                              5,000

Creditors                            4,000

Opening Stock                  7,000

Capital                          132,000

 

You are required to prepare a trial balance for MM traders as at 31st January 2002 from

the above information.                                                                                               6mks

  1. Outline any four functions of money as a medium of exchange.       4mks
  2. Give any four reasons why the government would need to train businessmen. 4mks
  3. Identify three reasons that can lead to dissolution of partnership.       3mks

 

TEST PAPER IX

MARKING SCHEME

 

  1. i) Registered Capital –  Amount of capital that a company can raise from its registered

members

  1. ii) Subscribed Capital –  The portion of issued share capital that has been accepted by the

shareholders

iii)   Issued Capital    –   The portion of the authorised share capital that has been offered to

the shareholders.

  1. iv) Paid up capital –   The amount of subscribed share capital paid up by the shareholders of a

company.

 

  1. When to use a credit note:
  • When there are damaged goods
  • When packing cases and containers are returned
  • When there is overcharge resulting from an arithmetic error.
  • When goods delivered are not in accordance with the order.

1 x 4

  1. How to achieve profit maximisation objective.
  • Reduce prices to increase sales
  • Reduce the cost of production
  • Diversify and expand production
  • Expand market
  • Increase prices of products
  • Increase sales promotion campaigns. 1 x 4

 

  1. Functions of commercial attaches.
  • Search for Kenyan market abroad
  • Inform business abroad of Kenyan goods
  • Negotiate to expand Kenyan market abroad.
  • Liase with the ministry of commerce to educate Kenyan producers/ manufacturers.

1 x 4

 

  1. Investment services by posta corporation.
  • Premium bonds scheme
  • Operate S.A.Y.E. (save as you earn)
  • Have fixed deposit facilities
  • Have saving facilities

 

  1. Why cheques are dishonoured
  • Insufficient funds
  • Post – dated cheque
  • Stale cheque.
  • Amount in words differ from amount in figures.
  • Signature of the account holder may differ from the specimen in the bank
  • Alterations that have not been signed against
  • The drawer has stopped payment of the cheque.

 

 

  • If any one of the requirements for a properly drawn cheque is missing or wrongly written.
  • A crossed cheque will be dishonoured if the depositor does not abide by the conditions of the crossings. 1 x 4

 

  1. Benefits of personal selling.
  • Creates a personal touch between the seller and the buyer
  • The seller explains to the buyer since he/ she can ask questions regarding the products
  • The seller gets to know how to improve the quality of his products depending on lamentations made by the buyer
  • The buyer knows exactly whom to turn to incase of any problem.
  • The seller can sell on credit to the credit worth customers

Any other relevant point        1 x 4

 

  1. Goods suitable for direct distribution.
  • Goods which require sufficient after –sales service eg electronics
  • Goods which the manufacturer wants to exercise some control during distribution
  • Goods with high profit margin and the retailers purchase in bulk
  • Goods which are very expensive eg motor vehicles
  • Goods which are highly perishable eg milk.
  • Branded goods and goods which are widely advertised eg coca-cola.

Any other relevant point        1 x 4

 

  1. How insurance companies make profits
  • Investing their funds
  • Differences in claims
  • From forfeited funds by policy holders
  • By charging those who breach the contract of the policy.

Any other relevant point        1 x 4

 

  1. Sources of capital to co-operative societies.
  • Members contributions
  • Retained profits
  • Entrance fee deposits
  • Credit facilities
  • Surplus and reserves. 1 x 4

 

  1. Capital = ( opening capital + Extra capital introduced + net profit)

=    ( 170,000 + 52,000 + 5000)  –   5000

=   227,000 – 5000

=  222,000

 

  1. Net profit ratio.

NPR   =  Net Profit           x  100

Sales

Sales         =  Net profit + purchases

=   5,000 + 20,000  =  25,000

\   N.P.R   =  5000  x  100

25,000                        =  20%            1 x 3

 

  1. Features of a fixed deposit account.
  • Opened for a specific period of time
  • Opened with a specific amount for a specific period
  • Higher interest rates are paid
  • Used by commercial bank to advance loan to others
  • Can be used as a security to apply for an overdraft against the current account of the same person. 1 x 4

 

  1. Why road transport is preferred.
  • Its cheaper and faster for short distances
  • All areas including remote areas can be accessed by road.
  • Suitable for delivering perishable goods like vegetables
  • Handling is minimal hence cuts down breakage and pilferage
  • Its flexible hence goods may be sold on route.
  • Insurance charges are lower compared to sea transport
  • Convenient since vehicles need not run on scheduled time like with rail transport.

Any relevant point.     1 x 4

 

  1. Advantages of advertising to the consumer.
  • Getting information about what is available in the market
  • Information about price, quality, size and other features of the product.
  • Explanations on the use and value of the product.
  • The consumer gets to know quickly what he/she wants to buy
  • Enables the consumer to compare prices and other features of different products
  • Gets entertained by the advertising compaign.

Any relevant point.     1 x 4

 

  1. Remittance services offered by the post office.
  • Postal order
  • Money order
  • Telegraphic money order
  • Registered mail
  • 1 x 4

 

  1. Advantages of micro-films.
  • Saves space and weight since bulky files are replaced by compact carbons of films.
  • Documents can be sent abroad more cheaply
  • More durable compared to files
  • Little risk of misplacing
  • Saves filing equipment and also floor space.
  • Can be enlarged on paper to provide quick and accurate duplicate copies of the original documents
  • Provides a more permanent record since its more.

Any relevant point.     1 x 4

 

  1. Factors to consider when buying a copier.
  • the cost ie both initial and operating cost
  • availability for the intended purpose
  • likelihood of obsolescence
  • Room for expansion or to accommodate higher work load

 

  • Availability of credit or hire purchase terms offered by the supplier.

Any relevant point.     1 x 4

 

Machine

Function.

Addressing machine

Print names and addresses of customers

Shredding machine

Destroy unwanted materials

Guillotine

Cutting and trimming papers

Franking machine

Print postal stamp impressions on the envelop.

Composite machine

Fold document, insert in envelops and seal the envelops.

            1 x 5

 

  1.         MM Traders

Trial balance

As at 31st January 2002

 

Details Debit (Kshs.) Credit (Kshs)
Sale   10,000
Purchases 6,000  
Bank Loan   2,000
Buildings 30,000  
Machinery 40,000  
Premises 60,000  
Debtors 5,000  
Creditors   4,000
Opening stock 7,000  
Capital   132,000
  148,000 148,000

½ x 12

 

  1. Importance of a balance sheet.
  • Helps the shareholders to know whether the business is in a position to meet its claims
  • Helps the investor to determine whether their funds are invested wisely
  • Helps to compare the performance of the business with that of previous years or similar business’s.
  • Helps the government to determine the tax to impose on the business
  • Help potential investors in making informed decisions on the share to buy in the stock exchange market.
  • Helps the management of the business to identify where urgent actions needs to be taken in order to promote the business. 1 x 4
  1. Functions of money.
  • Medium of exchange
  • Standard measure of value
  • Store of value
  • Unit of account
  • Standard of deferred payment. 1 x 4

 

  1. Why businessmen need to be trained.
  • Expose them to modern development and management
  • Import efficient business operating methods
  • Identify problems facing businesses and look for solutions
  • Import proper business ethics and good customer relations
  • Educate them on how to utilize the scarce resources
  • Instil in them the need to minimize costs inorder to maximise profits

Any other relevant point                    1 x 4

 

  1. Reasons for dissolution.
  • Completion of the intended purpose or period agreed upon
  • Loss of one partner
  • Request of dissolution by one partner
  • A change in the country’s law disqualifying the business.
  • A court order.
  • Inability of the business to meet its claims / liabilities.

Any other relevant point                    1 x 3

 

 

TEST PAPER X

 

  1. a) Discuss  the factors that should be considered in choosing a channel of distribution

for any  particular product.                                                                                        10mks

  1. b) Explain any five types of wholesalers who may be found in our country today.            10mks

 

  1. a) Assume that you are the sales manager of Kenya Breweries limited.  Explain the

qualities you would look for every time you are recruiting your sales executives.            12mks

  1. What do you understand by the term production. Explain any four factors of

production.                                                                                                           8mks

  1. a) The following balances were extracted from the books of spikes Ltd on 31st Dec 2001.

  Shs.

Fixed Assets                           42,000

Stock                                       48,000

Debtors                                   50,000

Bank overdraft                        25,000

Creditors                                 40,000

Cash in hand                           2,000

Capital                                                77,000

 

During the month of January, the following transactions took place.

 

 

  1. Bought motor vehicle on credit valued at Kshs. 80,000 from General motors.
  2. Withdrew sh. 28,000 for personal use.

Prepare a balance sheet as at 31st Jan 2002 for spikes Ltd.

  1. c) Give any five reasons to justify that the central Bank is not a Commercial Bank.

10mks

 

  1. a) Discuss circumstances under which a wholesaler may be eliminated from the chain

of distribution.                                                                                                             10mks

  1. b) Explain the essentials of a good warehouse.             10mks

 

  1. a) Advertising media refers to the means used to convey the sales message to the customers /

users of any given product.  Explain any five.                                                          10mks

  1. b) What do you understand by ‘containerization’.  Briefly explain the benefits associated

with  containerisation.                                                                                                            10mks

 

  1. a) Explain the role played by consumer co-operative societies in Kenya today.      10mks
  2. b) To a certain extent, public corporations are to considered desirable. Explain      10mks

 

 

TEST PAPER X

MARKING SCHEME

 

  1. a) Factors considered in choosing a channel of distribution for any particular product.

     –      Nature of goods:   Technical, Industrial, Perishable or goods that require servicing and

installation should be sold directly to ultimate consumers or through retailers.

  • Marketing costs: Goods that require special marketing facilities like storage, refrigeration, etc should be sold through intermediaries in order to cost share the marketing costs.
  • Size of the market: where the size of market is large enough, direct selling is preferred.
  • Nature of the market: where the market is local and the producer can reach his customers with ease and make large sales then direct selling is the best.
  • Scale of production: Small producers like small scale farmers should sell their products directly
  • Marketing risks: Where these risks are high such as spoilage and fall in value, its advisable to sell through intermediaries in order to distribute these risks.

Any five well explained points  ( 2 x 5)

 

  1. b) Types of wholesalers who may be found in our country today.
  • National wholesalers – set up by government to distribute a specified product in the country. Have branches all over the country e.g K.M.T.C.
  • Regional wholesalers: Cover a given region such as province, districts, etc. and distribute a given product in that region.  No other party is allowed to distribute similar goods in the region.
  • Specialised wholesalers: Sell a particular type of a product eg books but offer great variety within their particular field.
  • General wholesalers: Stock a variety of related goods eg a hardware shop stocks nails, iron sheets, cement, etc.
  • Truck wholesalers: Carry only a limited range of stock and combine selling, delivery and collection function in one operation.

 

 

  • Cash and carry: Operate as self service stores.  Require retailers to come to the wholesale shop, pick their requirements, pay and carry away their own purchases.
  • Rack jobbers sell particular products to certain types of retail stores.

2 x 5

 

  1. a) Qualities of Sales Executives.
  • Attractive personality
  • Ability to assess the customers
  • Politeness and courtesy
  • Thorough knowledge of the product
  • Aggressiveness
  • Any other relevant point                    2 x 5

 

  1. b) Production refers to the process of creating finished goods from raw materials to satisfy

human wants.

 

Factors of production.

Land:   Includes all natural resources that are given to man free of charge eg soil, water, forest

etc paid rent.

Labour:   Physical and mental human effort employed in the process of production eg. Teacher,

manager, etc paid wages

Capital:     Man made goods that assist in the process of production increases productivity of

land and labour.  Eg machinery paid interest.

Enterpreneurship:  Ability to organise other factors of production in appropriate proportions for

effective production.   Paid profit.

4 well explained points          2 x 4

 

  1. a) Spikes ltd

Balance sheet

At at 31st Jan 20

 

 

Kshs.                                                    Kshs.

Fixed Assets   42,000             Capital                        77,000

Motor vehicle 80,000             Less Drawings                        28,000

   Net capital                     49,000

Current Assets                                    Current Liabilities

Stock               48,000             Bank overdraft                        25,000

Debtors           50,000             creditors                                  40,000

Cash in hand   2,000              loan (withdrawal)                   28,000

222,000                                                           222,000

1 x 10

 

  1. b) Why central Bank is not a Commercial Bank.
  • Does not advance loans to individuals
  • Does not receive deposits from individuals nor run accounts of individuals
  • Does not discount bills of exchange
  • Does not offer money transfer services
  • Its not a custodian of valuables. 2 x 5

 

 

 

  1. a) When wholesaler may be eliminated.
  • When prices of goods are high like those of motor vehicles, the wholesaler may be eliminated since the manufacture can bear both selling and distribution costs.
  • Where goods are highly perishable eg milk
  • Where goods are branded and widely advertised
  • Where goods have high profit margins.

Any other relevant point                                                        2 x 5

 

  1. b) Essentials of a good warehouse.
  • Ideal location where goods are produced in greater quantity to be used later.
  • Suitable building which is large enough and appropriate to store goods of different kinds more safely
  • Should be well equipped with necessary facilities to store goods for a longer period
  • Should have well trained and efficient staff to ensure that goods are protected against losses and damages
  • Good transport system to enhance movement to and from the warehouse
  • Should have appropriate protection measures to elleviate losses, damages and other calamities.                              2 x 5

 

  1. a) Advertising media
  • The press
  • Radio
  • Television
  • Outdoor publicity
  • Window display
  • Fair and exhibitions Any five well explained points                       2 x 5

 

  1. b) Containerization refers to the packing of the cargo in standardised containers which are

sealed by the  exporter or his agent.  The sealed containers are loaded on to the

ship and delivered to the importer sealed.                                                   2mks

 

Benefits:

  • Takes little space since containers are of standardised sizes.
  • Enhances simple and cheep movement of goods
  • Enables the use of modern cargo – handling machinery
  • Safe since containers are sealed
  • Insurance premium is low due to lower risks involved
  • Assists road transport
  • Convenient since different containers are for specific goods. 2 x 4.

 

  1. a) Role of consumer co-operatives.
  • Provide members with goods and services at fair prices and of good quality
  • Protect members against exploitation, inefficient and dishonest trade practices by individual retailers
  • Provide channel through which ideas regarding quality and use of products can be propagated
  • Bring services to remote areas where independent traders are afraid.
  • Accept members deposits and assists them to buy or acquire land by purchase or lease.
  • Remove excess profits enjoyed by independent traders and pass the same to the members.

            2 x 5

 

  1. b) Problems associated with public corporations
  • Poor services and low quality goods due to absence of competition
  • Wastes are encouraged since losses are financed by tax payers.
  • Poor bad business decisions are likely to be made due to political motives
  • Where rival private firms exist, they are subjected to unfair terms and conditions
  • Fear of political repercussions may lead to avoidance of risky yet important projects
  • Though inefficient, they may not be criticized openly due to fear of political influences.

Any other relevant point.                   2 x 5

 

 

TEST PAPER XI

 

  1. State any four advantages of a wholesaler to a consumer       4mks
  2. List four qualities of good filing system       4mks
  3. Study the diagram below and answer the questions that follow

 

 

 

 

 

 

 

 

 

  1. What type of demand is potrayed by curve DD                                                             2 mks
  2. Outline the two causes of this curve       2mks
  3. Why would an exporter prefer to use container services as a method of transport       4mks
  4. State any four after sales services business firm selling electronic goods sets can offer

to the customers                                                                                                               4mks

  1. The following information was extracted from the books of a trader :

Creditors       sh           60,000

Cash at bank                20,000

Debtors                        40,000

Stock                           70,000

Calculate :

  • Current ratio                                                                                                                        3mks
  • Quick ratio 2mks
  1. Give four reasons why the budget is an important government tool of control to the economy                   4mks
  2. Differentiate between Balance of trade and Balance of payments                        4mks
  3. The following information was extracted from the books of Roadside kiosk as at

Sept 30th  1999:

Cash at hand  Shs. 30,000

Capital                 170,000

Creditors               40,000

Stock                     70,000

Purchases             150,000

Sales                    205,000

Debtors                  65,000

Motor vehicles    100,000

Prepare a trial balance as at 30th Sept 1999.                                                                       5mks

  1. What are the four methods of personal selling carried out by salesmen              4mks
  2. In the recent past many countries have adopted trade liberalization as the way forward to development. Highlight any four adverse effects that such nations may experience due

to liberalization                                                                                                                                 4mks

12.Outline any four functions of the Nairobi stock exchange                                            4mks

  1. List four methods used by the central bank of Kenya in credit control            4mks
  2. Consumer protection is safeguarding consumers from exploitation by producers and  businessmen . Give four reasons why this is necessary                                          4mks
  3. The following is a demand/ supply schedule for kerosene. Draw a diagram to illustrate

using your own scale                                                                                                       5mks

Also find the equilibrium price / quantities

 

Price                Quantity Demanded               Quantity Supplied

Per day                         Per  day

 

5                                60                                   10

7                                50                                   15

10                               40                                   20

15                                35                                   25

20                                30                                   30

25                                25                                   35

30                                20                                   40

 

  1. Name four characteristics that should be posed by goods in order for them to be described

as wealth.                                                                                                                                     4 mks

  1. Limited companies are Large Organizations which attract large profits. However these undertakings are still not popular in Kenya and sole proprietorship seems  to be more

popular. Give four reasons to explain this                                                                    4 mks

  1. Give four reasons why most organization prefer oral interviews whenever they are

recruiting new employees/staff                                                                                     4 mks

  1. The law of demand states that ‘Other things being equal, a fall in price will lead to an

increase in quantity demanded .’Highlight any Four circumstances under which this

law may not hold.                                                                                                           4 mks

 

  1. In the spaces provided below indicate the level of production that relates to each of the following activities

Activity                                                           Level of production

 Selling goods in a shop                                 __________________________________

 Fishing  from a river                                     __________________________________

       Growing maize                                                    __________________________________

       Making leather belts                                           __________________________________

4 mks

  1. Enter the following transaction into a cash book and balance it as at 20th March 1995.

March 2nd started business with Kshs. 28,000 of which Kshs. 8000 was banked .

March 4. Bought stock from Nekesa worth Kshs 2000 and paid by cheque.

March 6. Paid  wages in cash Ksh  4000.

March 10.Received a cheque for Kshs. 3000 from a debtor

March 20. Deposited cash in bank from  cash till Kshs. 3000.                                                  5 mks

  1. State any four benefits accruing as a result of SmithKline Beecham merging with Glaxowellcome                                                                                                                 4 mks

 

  1. The sales from Mambo stores for the year 1999was Ksh 99,000. Given that goods are

usually sold at a margin of 20%, what was the gross  profit for the year.                     3 mks

  1. Differentiate between
  • W.O and C.O.D 2 mks
  • O.B and F.A.S 2 mks

 

 

TEST PAPER XI

MARKING SCHEME

 

  • Breaking of bulk
  • Offers a variety of  goods
  • Price stabilisation
  • Advertising creates awareness
  • Steady supply – (storage)
  • After sales services ( 1×4 = 4mks)

 

  1. – Simplicity
  • Compactness
  • Accessibility
  • Safety
  • Elasticity/ flexibility/ Expansion
  • Cross reference
  • Economy   (1 x 4 = 4marks)

 

  1. a) An abnormal demand curve
  2. b) (i) Geffen goods
  • (ii)  Conspicuous consumption /luxurious
  • iii) Speculative demand – Future expectations of a rise or fall in prices

 (1 x 2  = 2mks)

 

  1. – Easy handling of goods – containers are designed to enable use of machinery
  • Less damage to goods
  • Safety – They are sealed until they reach the importers
  • Low premium insurance costs, this results due to less risks
  • Use of specially designed containers for special goods (.e.g tankers)
  • Reduced inspection hence less disturbance to the goods

(1 x 4 = 4marks )

 

  1. – Availability of spare parts
  • Guarantee- repair of the articles for a specific period of time after purchase
  • Technical advise to the consumer on the use of items
  • Helping to install machinery or equipment
  • Servicing of goods from time to time
  • Delivery / transporting the goods to customers premises.

( 1 x 4 = 4 marks)

 

 

 

 

  1. (i) Current ratio = Current assets / Current liabilities

=  130000 / 60000

=  2.2 / 1

» 2 :1

 

  • Quick ratio =  Quick assets

Current liabilities

130000- 70000

60000

60000

60000

=  1:1                       ( ½  x  4  =  2marks)

 

  1. – A necessary aspect of planning
  • A tool for correcting development plans that are not feasible
  • A standard measure of government performances
  • Forces the government to prioritise its projects and plans
  • Reflects government monetary policies used to control the supply of money
  • Evaluates the success and failures of the economy
  • Motivates the facilitators to meet the set objective by working hard

( 1 x 4 = 4 marks)

 

  1. Balance of trade : The difference between the values of visible exports and visible imports of a country over a given period    ( 2mks)

Balance of payments : The difference between the record  of a country’s visible and invisible exports against the value of her invisible imports over a period .

 

  1. ROADSIDE KIOSK TRIAL BALANCE AS AT 30TH SEPT .
  2. CR.

Purchases                                150,000

Capital                                                            170,000

Sales                                                               205,000

Debtors                                   65,000

Creditors                                                         40,000

Motor vehicles                       100,000

Stock                                        70,000

Cash at bank                           30,000

415000            415000

(½  x 10 tics  = 5marks)

10 –      Show room sales

  • Display of goods in a room attendants
  • Provision of detailed information about the products
  • Trade fairs and exhibitions
  • Manufacturers / Agriculturist stalls ( outlets)
  • Visiting customers in homes or towns (e.g insurance)
  • Posting sales girls / boys in supermarkets shops .

  ( 1 x 4 = 4 marks )

 

 

 

 

  1. – Encourages single sided development since industries with comparative

advantage  develops to the neglect of others

  • Substandard and harmful products may be produced and traded leading to dilution of social welfare
  • Dumping is encouraged which discourages development of local industries
  • May lead to exploitation and colonisation
  • Promotes unproductive competition which leads to closure of infant industries
  • Presupposes the existence of leissez faire and working of price mechanism under perfect market which are unrealistic           ( 1x 4 = 4  marks)

 

  1. – Marketing of securities which facilitate flow of new capital into the industry.
  • Protect investors by maintaining rules and discipline of the brokers.
  • Advise and guide companies seeking capital
  • Publishes information regarding prices of securities
  • Transfer the title to any quoted security ( 1 x 4 = 4 marks)

 

  1. – Open market operation
  • Bank rate method
  • Margin requirement
  • Selective credit
  • Reserve requirements

 

  1. – Ensure goods are of good quality
  • Ensure the right quantity and size
  • Ensure health standards are met
  • Ensure safety standards are met
  • Ensure fair pricing
  • Protect them against breach of contract
  • Protect from false advertising

 

15 –

 

 

 

 

 

 

 

 

 

 

 

 

  • Equilibrium quantity  30 units
  • Equilibrium price 20 shillings

 

  1. – Must have utility
  • Should be scarce
  • Must have money value
  • Should be transferable

 

  1. – Profits in sole proprietorship aren’t shared
  • In sole proprietorship there is personal contact with customers unlike in joint stock companies
  • Formation procedure is easier in single owned business than joint stock companies
  • Decision making in single owned business are easier than in joint sock companies                                  ( 1 x 4 = 4marks)

 

  1. – Immediate feedback
  • Leaves no room for alteration /misinterpretation of the message
  • Creates a personal touch or healthy relationship between the interviewer and the interviewee.
  • Saves time since information is passed on and a response to it received instantly enabling a decision  to be made
  • Enables the interviewer to access extra information about the interviewee by asking probing questions.
  • The interviewer is able to assess the confidence levels of the interviewee

( 1 x 4 = 4marks)

  1. – Incase of the essential goods that must be consumed
  • Ostentatious goods that must be consumed
  • Luxurious goods like cars
  • Inferior goods i.e goods for the poor
  • Geffen goods e.g bread              ( 1 x 4 = 4marks)

 

  1. – Tertiary level
  • Primary level
  • Secondary level

 

  1. CASH BOOK

 

 

Date         Particulars          Cash      Bank             Date     Particulars        Cash     Bank

 

 

March  2nd    B/f                   20,000  8,000              4th        Purchases                          2000

Wages                4000

10th               Debtors                          3,000                        Bank                   3000

20th               Cash                               3,000                        Balance c/d      13000     12000

 

 

 

20th              Bal c/d            20,000  14,000                                                  20,000   14,000

13000     12000

  • – Higher production leading to low cost of goods per unit thus higher profits
  • Large market due to lower prices and no competition
  • Reduced competition increases sales
  • Advantages of specialisation
  • Better management as a result of combined pool of specialist / specialised labour
  • Cheap purchases of raw materials as a result of bulk buying at a discount
  • Will be able to attract external finance easily for expansion
  • Control of prices due to information of cartel .

 

 

 

 

 

  1. Sales =  Shs. 100,000

Margin =  Gross Profit  x 100

Sales

  • = GP  x 100

100,000

GP =  Ksh  20,000.

 

  1. a) W.O  –  Cash with order – The buyer  must  pay for the goods when he makes an order.

C.O.D –     Cash on Delivery –  Payment is effected when goods are delivered to the buyer.

  1. O.B – Free on board . Prices  quoted includes expenses  from the sellers warehouse up to the

docks .

 

TEST PAPER XII

 

  1. A company intends to promote its products by arranging a prize winning competition through the media. Prizes can only be won through buying the products.
  • Explain five benefits that the company would get by using this method of sales promotion                                                                                                                          10mks

(b) Explain four factors determining the market prices of commodities.                      10mks

 

  1. (a) The government of Kenya has been adopting the policy of privatisation of firms.

Discuss some  of the benefits that the Kenyan people reap from this policy            10mks

  • The following balance were extracted from the books of Charles & Sons

as at May  31st 2002 .

Capital                        600,000

Net profit                    130,000

Plant  fixtures                   160,000

Equipment’s                     100,000

Buildings                                      240,000

Office furniture                  10,000

Stock                                 192,000

Cash                                  118,000

Debtors                             158,000

Pre-paid  expenses              22,000

Creditors                           130,000

Outstanding expenses         20,000

Bank overdraft                  110,000

 

Required :

(i)  Prepare a balance sheet as at  31st may 2002                                                                 7mks

  • Calculate
  • Working capital
  • Current ratio
  • Net profit as percentage to capital     3mks

 

  1. (a) Explain five circumstances under which an insurance company would not compensate

the insured in the event of loss                                                                                  10mks

(b) Odhiambo’s publishing  company recently joined the Nairobi stock exchange.

Describe five benefits                                                                                                            10mks

 

  1. Opportunities exist for purchasing of necessary durable goods by instalment
  • Describe two methods commonly used for this purpose                            4mks
  • Mention three advantages of each to:
  • Manufacturers
  • Consumers      12mks

(c)  Give any two differences of the two methods                                                    4mks

 

  1. (a) Explain any five methods that the Kenyan government may adopt in order to control the

volume of imports                                                                                                  10mks

(b)  Discuss the main sources of capital to small scale enterprises                          10mks

 

  1. (a) Explain five reasons why it is necessary for a country to control international  trade

10mks

(b) Explain five diseconomies that may be suffered by firms due to large scale production

10mks

 

TEST PAPER XII

MARKING SCHEME

 

  1. – Increased awareness of the consumers of the product advertised

– Increased sales volume of the product

– More customers / wider market to the  goods

– Retention /loyalty of the already existing customers

– Increased market due to large market and consumers

– Better quality due to competition created by advertising

 

  • – Price mechanism- The interaction of demand and supply in the market .

–  Government policy – Through taxation subsidies and price controls

– Auction – Prices suggested by prospective buyers

(Mentioning 1 mk)         (Explanation 1 mk)

 

  1. (a) Benefits of privatization
  • Kenyans enjoy improved products and services due to research carried out by privatised firms
  • The spirit of competition enhances good management hence efficiency
  • Consumers enjoy fair prices of products and services due  to minimal wastes if any.
  • Privatisation calls upon many firms entering the market leading to product diversity    hence Consumers have a wide variety to choose from.
  • Due to stiff competition, privatised firms engage in aggressive sales promotion methods which keep consumers informed of the products in the market.
  • The need for greater profits makes them engage in risky projects which may be  of great importance to the consumers.

                      Any other relevant point 2×5 = 10

 

 

 

 

 

 

 

  1. b) Billy and Sons Co.

Balance sheet

As at 31st May 2002

 

Shs              shs                                                            Shs.             Shs.

Fixed Assets                                                              Capital                                      600,000

Plant& Fixtures          160,000                                    Add net profit                          140,000

Equipment’s                100,000                                  Net capital                                       740,000

Buildings                    240,000                                   Current Liabilities

Office furniture            10,000      510,000                Creditors                        130,000                                                                                                        Outstanding Exp.             20,000

Current  Assets

Cash                           118,000                                   Bank overdraft             110,000    260,000

Debtors                       158,000

Pre-paid expenses         22,000

Stock                           192,000              490,000

1,000,000                                            1,000,000

 

  1. Working capital = Current  Assets – Current liabilities

=  490,000 – 260,000

=  230,000

 

  1. Current Ratio =  Current Assets

Current liabilities

400,000

260,000

=  49.26                                                                                               (1mk)

 

  1. Net profit as a percentage to capital

Net Profit  x 100     =  140,000  x 100

Capital                       600,000

=  23.3%

 

3 a) –     Losses caused by the insurers carelessness i.e not accidental

  • Incase the insured has no insurable interest in the property
  • Incase of accident / loss is not directly related to the risk insured
  • Incase the insured has not paid premiums as expected
  • If on investigation the insurance company realises that false information was given by the insured
  • A car accident which is caused by a third person who is not insured.

 

  1. – The company knows value of its shares in the market
  • The company may sell more shares since its is known to the public.
  • The company raises more capital by floating more shares in the stock exchange
  • The company uses more information gathered from stock exchange to improve its performances
  • Helps to attract competent personnel in the company
  • Promotes company image .

 

  1. HIRE PURCHASE :- A  system where the buyer agrees to pay for the commodity in installment. The buyer obtains the possession of the article on the down payment (deposit) but the title of ownership remains with the seller until the last installment is made .

 

CREDIT SALES/DEFFERED PAYMENT- Under this system the article becomes

the property of the buyer with the payment of the first installment. The

buyer is supposed to pay the remaining amount in installments.                                                                                                                                                           (2 x 2 )

  • Advantages

Manufacturer.

  • He is able to increase his volume of sales .
  • Profit earned on hire purchase is higher than profit earned on cash basis
  • Goods belong to the seller until the last installment is paid .

 

The buyer.

  • He takes possession of goods immediately after down payment
  • Installments to be made are predetermined enabling him to plan and budget for payment  of the amounts
  • He is able to acquire goods he would not have afforded on cash terms.
  • The buyer can possess more goods including expensive items.

               (1 x 6 = 6marks)

 

Advantages of credit sale to:

  1. The manufacturer:
  • He is able to increase his volume of sales
  • Profits realised are higher since customers must pay higher prices
  • He can reclaim the amount due from the buyer under a court order.

 

  1. The buyer
  • He is able to expensive goods on credit
  • Ownership of the article passes to the buyer after the first installment .
  • He is able to plan on acquisition of certain goods ( 1 x 6 = 6 marks)

 

Differences

Hire purchase                                                   Credit sale    

–  Goods can be repossessed if the              –  Goods cant but the buyer can be  sued for

buyer breaches the contract.                          damages

–  Ownership remains with the seller          –   Ownership passes to the buyer installment is

when  the first  till the last installment         is effected.

Is made.                                        Any other relevant 1 x 4 = 4 mks.

 

  1. a) Methods of controlling volume of imports
  • Tariffs – Involves levying import duty on goods that are entering a country
  • Quotas – Is a quantitative restriction permitting only a given number of units of a commodity to be imported during a specific period
  • Exchange control – Refers to the control by the government on the exchange value of the currency of a country. Importers can only import goods when they seek permission for acquiring foreign exchange.
  • Trade agreements – Is an agreement between different countries regarding the foreign trade.
  • Total ban / Embargo – Involves imposing a complete ban on import of some specific commodities . 2 x 5 = 10marks.
  1. Sources of capital to small scale enterprises.
  • Personal funds /owners capital
  • Borrowing from friends and relatives
  • Members contribution
  • Loans from bank and other non bank financial institutions
  • Going public i.e borrowing from public by use of stocks.
  • Venture capital i.e  a group of rich people with a pool of financial resources which  they lead to viable small scale businesses.

(any other relevant point)

 

  1. To enable infant industries to grow without  interference from foreign industries.
  • To avoid entry into the country of commodities which might be harmful to the moral culture and health standard of its people.
  • To eliminate dumping of inferior or substandard goods in country.
  • To correct balance of payment deficit .
  • To enable a country to fully exploit its resources which otherwise might be idle from undue competition .
  • To raise revenue for government projects in the country
  • To reduce the dependency of a country on another which might mistreat her incase of misunderstanding between them .
  • To protect key industries which promotes faster development and lead to establishment of other industries
  • To diversify industrial growth by allowing production of different commodities to prevent dependence on other countries .

 

  1. Decision making take too long
  • Size of market may be too small
  • Complexity of the organisation may create difficulty in management
  • Lack of enthusiasm due to size which kills commitment- no personal contact
  • High risks due to change in fashion, taste or falling prices .
  • Formation of monopolies leading to exploitation of consumers
  • Poor staff and employee welfare are due to large number.

 

 

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