CHAPTER CHEMISTRY

1 HYDROCARBONS
Name and draw the structural formulas for four
smallest Alkanes.

Y
ANSWER:

EM
Methane

D
6 A
48 C
97 E A
40 IN
Ethane
31 L
23 ON
+9 R
A

Propane
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 1

CHAPTER CHEMISTRY
1 HYDROCARBONS
Butane

Y
EM
D
6 A
48 C
97 E A
Question 2
What is the main component of natural gas?
40 IN
Answer:
31 L
23 ON

The largest component of natural gas is methane, a compound

with one carbon atom and four hydrogen atoms
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 2

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 3
What is meant by the term saturated, when applied to

Y
hydrocarbons?

EM
Answer:

D
Saturated Hydrocarbons are hydrocarbons that contain

6 A
only single bonds between carbon-carbon atoms. They are

48 C
simplest class of hydrocarbons. They are called saturated

97 E A
because each carbon atom is boned to as many hydrogen atom
as possible like Methane, Ethane, Propane etc.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 3

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 4
What is meant by fractional distillation?

Y
EM
ANSWER:
Fractional distillation is the separation of a

D
mixture into its component parts, or

6 A
fractions. Chemical compounds are separated by heating them

48 C
to a temperature at which one or more fractions of the mixture

97 E A
will vaporize.
40 IN
WITH RESPECT TO HYDROCARBONS
31 L
Fractional distillation is the process by which oil refineries
23 ON

separate crude oil into different, more useful hydrocarbon

products based on their relative molecular weights in
a distillation tower
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 4

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 5
Where in the fractioning column will the longest

Y
hydrocarbon condense?

EM
Answer:

D
The fractionating column is hot at the bottom and cool at

6 A
the top. Substances with high boiling points condense at

48 C
the bottom and substances with lower boiling

97 E A
points condense on the way to the top. The crude oil is
40 IN
evaporated and it vapours condense at different
temperatures in the fractionating column.
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 5

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 6
Name and draw the structural formula for the four

Y
smallest alkenes?

EM
Answer:

D
Ethene

6 A
48 C
97 E A
Propene
40 IN
31 L
23 ON
+9 R

Butene
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 6

CHAPTER CHEMISTRY
1 HYDROCARBONS
Pentene

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 7

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 7
can represent an alkane or an alkene.

Y
Explain.

EM
Answer:

D
It represents Alkene because for general formula for

6 A
Alkanes is .

48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 8

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 8
What does the term aromatic imply about an organic

Y
molecule?

EM
Answer:

D
The term aromatic compounds simply refers to a substance that

6 A
has a ring structure and bonding characteristics and properties

48 C
related to those of benzene, which means that all aromatic

97 E A
compounds have some delocalized bonding.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 9

 CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 9
Which of the following could be the formula of a

Y
cyclic alkane?

EM
(a)

D
(b)

6 A
(c)

48 C
(d)

97 E A
Answer:
40 IN
31 L

Examples of Simple Cycloalkanes

23 ON

Name Cyclopropane Cyclobutane Cyclopentane Cyclohexane Cycloheptane Cycloalkane

+9 R

Molecular
C3H6 C4H8 C5H10 C6H12 C7H14 CnH2n
Formula
A
ST

Structural
(CH2)n
Formula
G
N

Line
SI

Formula
RI

So, only

RISING STAR ONLINE ACADEMY +92331-4097486 Page 10

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 10 & 11 (COMBINED)
How many carbons are in the longest chain of each

Y
of the following compounds? Write the name of the

EM
longest alkane chain for each.

D
Answer:

6 A
a.

48 C
97 E A
40 IN
31 L
23 ON
+9 R
A

b.
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 11

CHAPTER CHEMISTRY
1 HYDROCARBONS
c.

Y
EM
D
6 A
48 C
97 E A
d.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 12

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 12
What does the term normal imply when used in

Y
Alkanes?

EM
ANSWER:

D
Normal (n) in alkanes means a straight chain, i.e

6 A
no branches in the structure.

48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 13

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 13
Identify the following as an alkane, alkene, or

Y
alkyne. Assume that the compounds are not cyclic.

EM
(a)

D
(b)

6 A
(c)

48 C
(d)

97 E A
(e)
(f) 40 IN
(g)
31 L
ANSWERS:
23 ON

(a) Alkene
(b) Alkane
+9 R

(c) Alkene
A

(d) Alkyne
ST

(e) Alkane
(f) Alkene
G

(g) Alkane
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 14

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 14
Explain why the name 1-propene is incorrect. What

Y
is the proper name of this molecule?

EM
Answer:

D
1-propene is incorrect as there are no position

6 A
isomers possible hence its not necessary to write 1 as

48 C
97 E A
it is obvious here. It's correct name is Propene.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 15

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 15
Explain why the name 3-butene is incorrect. What is

Y
the proper name of this molecule?

EM
Answer:

D
The name for this compound should be 1 – butene , not 3 –

6 A
butene because double bond should be given the last possible

48 C
number.

97 E A
ANSWER:
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 16

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 16
Name and draw the structural formula for each

Y
isomer of pentene.

EM
Answer:

D
Changing the position of the double bond in an alkene makes a

6 A
different isomer. There are five possible isomers of Pentene.

48 C
STRUCTURES

97 E A
1
pent−1−ene
40 IN
31 L
23 ON

2
pent−2−ene
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 17

CHAPTER CHEMISTRY
1 HYDROCARBONS
3
2−methylbut−1−ene

Y
EM
D
6 A
48 C
97 E A
4
3−methylbut−1−ene
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 18

CHAPTER CHEMISTRY
1 HYDROCARBONS
5
2−methylbut−2−ene

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 19

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 17
Name and draw the structural formula for each

Y
isomer of hexyne.

EM
Answer:

D
6 A
1

48 C
97 E A
40 IN
31 L
23 ON

2
+9 R
A
ST
G

3
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 20

CHAPTER CHEMISTRY
1 HYDROCARBONS
4

Y
EM
D
6 A
5

48 C
97 E A
40 IN
31 L
23 ON

6
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 21

CHAPTER CHEMISTRY
1 HYDROCARBONS

Question 18

Y
EM
Write a chemical equation for the reaction between
methane and bromine?

D
Answer:

6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 22

CHAPTER CHEMISTRY
1 HYDROCARBONS

Question 19

Y
EM
Write a chemical equation for the reaction between
ethane and chlorine?

D
Answer:

6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 23

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 20
Draw the structure of the product of the reaction of

Y
EM
a) Bromine with propene,
Solution:

D
Propene, CH3CH=CH2, reacts with bromine. Predict

6 A
the structure of the product formed.

48 C
97 E A
The structure will be: CH3CHBrCH2Br.
40 IN
b) Chlorine with 2-butene,
31 L
23 ON

Solution:
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 24

CHAPTER CHEMISTRY
1 HYDROCARBONS
c) Hydrogen with 1-butene,

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON

d) Hydrogen with 2-pentene.

+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 25

CHAPTER CHEMISTRY
1 HYDROCARBONS

Question 21

Y
EM
Write the balanced chemical equation for the
combustion of,

D
a) Heptane

6 A
Answer:

48 C
97 E A
b) Benzene 40 IN
Answer:
31 L
23 ON

c) Ethyne
+9 R

Answer:
A
ST

d) Hexene
Answer:
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 26

CHAPTER CHEMISTRY
1 HYDROCARBONS

Question 22

Y
EM
Draw the structure for each of the following
compound.

D
a) 3,4-diethyloctane

6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 27

CHAPTER CHEMISTRY
1 HYDROCARBONS

b) 3-methylpentane

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON

c) 4-ethyl-2-methylheptane
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 28

CHAPTER CHEMISTRY
1 HYDROCARBONS
d) n-hexane

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L

e) 3-isopropylhexane
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 29

CHAPTER CHEMISTRY
1 HYDROCARBONS

f) 2,4,5-trimethyloctane

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 30

CHAPTER CHEMISTRY
1 HYDROCARBONS
g) dimethyl-4-t-butylnonane

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 31

CHAPTER CHEMISTRY
1 HYDROCARBONS
h) 2,2-dimethyl-4-propylnonane

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 32

CHAPTER CHEMISTRY
1 HYDROCARBONS

i) 3-ethyl-4-methyl-3-heptene

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 33

CHAPTER CHEMISTRY
1 HYDROCARBONS
j) 3,3-diethyl-1-pentyne

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 34

CHAPTER CHEMISTRY
1 HYDROCARBONS
k) 4-ethyl-4-propyl-2-octyne

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 35

CHAPTER CHEMISTRY
1 HYDROCARBONS

l)5-butyl-2,2-dimethyldecane

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 36

CHAPTER CHEMISTRY
1 HYDROCARBONS

m) 3,4-diethyl-1-hexyne

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 37

CHAPTER CHEMISTRY
1 HYDROCARBONS

n) 4-propyl-3-ethyl-2-methyloctane

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 38

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 23
The name 2-ethylhexane is incorrect. Draw the

Y
structure and write the correct name for this

EM
molecule?

D
Answer:

6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 39

CHAPTER CHEMISTRY
1 HYDROCARBONS

Question 24

Y
EM
The name 3-butyl-7-methyloctane is incorrect. Draw
the structure and write the correct name for this

D
molecule.

6 A
Answer:

48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 40

CHAPTER CHEMISTRY
1 HYDROCARBONS

Question 25

Y
EM
The following skeletal structures represent alkenes
or alkynes. Fill in the proper number of hydrogen on

D
each carbon.

6 A
Answer:

48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 41

CHAPTER CHEMISTRY
1 HYDROCARBONS

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 42

CHAPTER CHEMISTRY
1 HYDROCARBONS

Question 26

Y
EM
Write the condensed structures of a compound with
seven carbon atoms -in each of the four hydrocarbon

D
classes.

6 A
Answer:

48 C
97 E A
Alkane

Alkene
40 IN
31 L
23 ON

Alkynes ( )

Arenes
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 43

CHAPTER CHEMISTRY
1 HYDROCARBONS

Question 27

Y
EM
Explain the process of polymerization.

D
Answer:

6 A
Polymerisation is a process of reacting monomer

48 C
97 E A
molecules together in a chemical reaction to form
polymer chains or three-dimensional networks.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 44

CHAPTER CHEMISTRY
1 HYDROCARBONS
CHEMISTRY EXERCISE

Y
CHAPTER 1

EM
Question 1

D
Briefly describe the important differences between an

6 A
alkene and an alkyne. How are they similar?

48 C
97 E A
Answer:

40 IN Differences
Alkenes Alkynes
31 L
Alkenes are hydrocarbons with Alkynes are hydrocarbons with
carbon-carbon double bond. carbon-carbon Triple bond.
23 ON

Less soluble in water More soluble in water

No acidic hydrogen Acidic hydrogen present
+9 R

Bond angle is 120 Bond angle is 180

Similarities
A
ST

1. Both undergoes addition reaction.

2. Both are unsaturated hydrocarbons.
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 45

CHAPTER CHEMISTRY
1 HYDROCARBONS

Question 2

Y
EM
Do alkynes show cis – trnas isomerisim?
Explain .

D
6 A
Answer:

48 C
No; A Triple bonded carbon atom can form only one other bond. It

97 E A
would have to have two groups attached to show cis-trans isomerism.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 46

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 3
Draw the structure for each compound

Y
EM
Answer:

D
(a) Acetylene

6 A
(b) 3-methyl-1-hexyne

48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 47

CHAPTER CHEMISTRY
1 HYDROCARBONS
(c) 4-methyl-2-hexyne

Y
EM
D
6 A
48 C
97 E A
40 IN
(d) 3-octyne
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 48

CHAPTER CHEMISTRY
1 HYDROCARBONS
Question 4

Y
Name each Alkynes.

EM
Answer:

D
(a) 1 – Pentyne or pent-1-yne

6 A
(b) 2 – hexyne or hex-2-yne

48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ONLINE ACADEMY +92331-4097486 Page 49

CHAPTER CHEMISTRY
1 HYDROCARBONS

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

NOTICE : If ANYONE find any error or mistake in notes …

Please let me Know at WhatsApp.
JAZAK ALLAH KHAIR
RISING STAR ONLINE ACADEMY +92331-4097486 Page 50
CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 1

Y
What are the characteristics of alcohols and ethers?

EM
Answer:
1. Ethers are similar in structure to alcohols, and

D
both ethers and alcohols are similar in structure to water.

6 A
2. An ether is more volatile than an alcohol having the same

48 C
molecular formula.

97 E A
3. They have same molecular formula.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 1

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 2

Y
How are alcohols classified?

EM
Answer:
Alcohols are classified into FOUR categories,

D
according to the number of hydroxyl group attach to the

6 A
48 C
chain.

97 E A
1. Mono-hydroxy Alcohols
2. Di-hydroxy Alcohols
40 IN
3. Tri-hydroxy Alcohols
31 L
4. Poly-hydroxy Alcohols
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 2

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 3

Y
What are the properties of alcohols and ethers?

EM
Answer:
Properties of Alcohols:

D
1) Solubility in Water

6 A
Alcohol are soluble in water.

48 C
97 E A
2) Boiling Point.
The size of alcohol determines its boiling point. The larger the
size of the alcohol, the higher boiling point.
40 IN
3) Flammability
31 L

The flammability of alcohols decreases as the size and mass of

23 ON

the molecules increases.

+9 R

Physical Properties of Ethers:

A

1) Flammability
ST

Ethers are generally nonreactive substance but are extremely

flammable.
G

2) Boiling Point.
N

Ethers have lower boiling point than alcohols.

SI

3) Solubility in Water
RI

Ethers are soluble in water.

Ethers are appreciably soluble in organic solvents like alcohol,
benzene, acetone etc.

RISING STAR ACADEMY +92-331-4097486 Page 3

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 4

Y
State the IUPAC name for diethyl ether.

EM
Answer:
Ethoxy Ethane

D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 4

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 5

Y
What products are formed when an alcohol undergoes

EM
dehydration?
Answer:

D
Alcohols can be dehydrated to form either alkenes or ethers.

6 A
Primary alcohols are oxidized to form aldehydes.

48 C
Secondary alcohols are oxidized to form ketones.

97 E A
Tertiary alcohols are not readily oxidized.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 5

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 6

Y
What products are formed when an alcohol is

EM
oxidized?
Answer:

D
Primary alcohols can be oxidized to form aldehydes and

6 A
carboxylic acids;

48 C
Secondary alcohols can be oxidized to give ketones.

97 E A
Tertiary alcohols, in contrast, cannot be oxidized without
breaking the molecule's C–C bonds.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 6

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 7

Y
Draw structures for the four constitutional isomers of

EM
molecular formula C4H10O that contain an OH group.
Answer:

D
1

6 A
48 C
97 E A
40 IN
31 L
23 ON

2
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 7

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Y
3

EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON

4
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 8

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 8

Y
Classify each alcohol as 1o, 2o, or 3o.

EM
(a) CH3CH2CH2OH
(b) (CH3CH2)3COH (3o)

D
(c) CH3CH2CHCHCH3 (2o)

6 A
Answer:

48 C
(a) Primary 1o

97 E A
(b) Tertiary 3o
(c) Secondary 2o
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 9

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 9

Y
Give the IUPAC name for each alcohol.

EM
(a)

D
6 A
48 C
97 E A
40 IN
31 L
(b)
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 10

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS
(c)

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 11

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 10

Y
Give the structure corresponding to each IUPAC

EM
name.
a) 3-methy1-3-pentanol

D
6 A
48 C
97 E A
40 IN
31 L
23 ON

b) 4-methy 1-2-pentanol
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 12

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS
c) 2,4-dimethyl-2-hexanol

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
d) 1,3-propanediol
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 13

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS
e) 3,5-dimethylcyclohexanol

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R

f) 6,6-diethy 1-4-nonanol
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 14

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 11

Y
Give an acceptable name IUPAC or common for each below.

EM
(a) CH3CH2OCH2CH2CH2CH3
(b) CH3CH2CH2CHCH3

D
|

6 A
OCH2CH3

48 C
Answer:

97 E A
(a) 1 – ethoxy butane
(b) 2 – ethoxypentane
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 15

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 12

Y
Give the structure corresponding to each IUPAC

EM
name.
Answers:

D
a) 2-methoxypropane

6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R

b) cyclobutyl ethyl ether

A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 16

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS
c) 1-ethoxy-2-ethylcyclohexane

Y
EM
D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 17

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 13

Y
Rank the compounds in each group in order of

EM
increasing boiling point.
a) CH3CH2OCH2CH3 ; CH3 (CH2)3 CH3 ;

D
CH3CH2CH2CH2OH

6 A
b) CH3CH2CH2CH2OH ; CH3CH2CH2OH ; CH3CH2OH

48 C
c) CH3CH2CH2CH2OH ; CH3CH2CH2CH3

97 E A
Answers:
(a) Alcohols have higher boiling points than both ethers and
40 IN
hydrocarbons.
31 L
CH3(CH2)3CH3 CH3CH2OCH2CH3 CH3CH2CH2CH2OH
23 ON

(b) The given compounds all are alcohols. The boiling point
of alcohols increases with increasing molar mass.
CH3CH2OH CH3CH2CH2OH < CH3 CH2CH2CH2OH
+9 R
A

(c) Alcohols have higher boiling points than both ethers and
ST

hydrocarbons,
G
N

CH3CH2CH2CH3 CH3CH2CH2CH2OH
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 18

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Question 14

Y
Explain the following observation. Dimethyl ether

EM
[(CH3)2O] and ethanol (CH3CH2OH) are both water
soluble compounds. The boiling point of ethanol is

D
78oC, however, is much higher than the boiling point of

6 A
dimethyl ether (-24oC).

48 C
Answer:

97 E A
In case of ethanol hydrogen bonding is possible as below which
is not possible in dimethyl ether. So ethanol boils at higher
40 IN
temperature compared to dimethyl ether.
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 19

CHAPTER CHEMISTRY
EXERCISE ALCOHOLS, PHENOLS AND
2 ETHERS

Y
Question 15

EM
Methanol (CH3OH) is used as a fuel for some types of
race cars, because it readily combusts in air to form

D
CO2 and H2O.

6 A
Write a balanced equation for the complete combustion of

48 C
methanol.

97 E A
Answer: 40 IN
Alkanes undergoes combustion to form Carbon dioxide and
water.
31 L

Balanced chemical equation for complete combustion of

23 ON

methanol is shown below.

+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 20

CHEMISTRY ALDEHYDES, KETONES &
CHAPTER 3 CARBOXYLIC ACIDS
Question 1
Define each of the following:

Y
EM
a) Aldehydes
Aldehydes are Organic compounds which contains aldehydic

D
6 A
Carbonyl ( ) as a functional group. General Formula is

48 C
97 E A
b) Ketones
40 IN
Ketones are Organic compounds which contains
31 L
Carbonyl ( ) as a functional group.General Formula is
23 ON

c) Carboxylic acids
+9 R
A

Carboxylic acids are Organic acids which contains a

ST

Carboxyl group. General Formula is

G
N
SI
RI

RISING STAR ACADEMY -92-331-4097486 Page 1

CHEMISTRY ALDEHYDES, KETONES &
CHAPTER 3 CARBOXYLIC ACIDS

Question 2

Y
EM
How are aldehydes and ketones named?

D
Answer:

6 A
We can name Aldehyde as Alkanal and Ketones as

48 C
Alkanone.

97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY -92-331-4097486 Page 2

CHEMISTRY ALDEHYDES, KETONES &
CHAPTER 3 CARBOXYLIC ACIDS

Question 3

Y
EM
What are the characteristics of aldehydes and ketones?

D
Answer:

6 A
1. Boiling Point:

48 C
97 E A
Aldehydes and ketones have lower boiling points than
alcohols of similar mass because not having Hydrogen
bond between them.
40 IN
31 L

2. Solubility:
23 ON

Aldehydes and ketones having less than 6 carbons are

soluble in both organic solvents and water.
+9 R

BUT
A

Aldehydes and ketones having 6 or more than 6 carbons are

ST

soluble in organic solvents and insoluble in water.

G
N
SI
RI

RISING STAR ACADEMY -92-331-4097486 Page 3

CHEMISTRY ALDEHYDES, KETONES &
CHAPTER 3 CARBOXYLIC ACIDS

Question 4

Y
EM
Give examples of useful aldehydes and ketones.

D
Answer:

6 A
Aldehydes:

48 C
97 E A
1. Formalin is used a disinfectant and preservative.
2. Acetaldehyde is used for production of acetic acid.
40 IN
3. Formal Aldehyde is starting material for synthesis of many
resins like plastics.
31 L
23 ON

Ketones:
1. Used as industrial solvent and starting material in the
+9 R

synthesis of some organic polymer.

A
ST
G
N
SI
RI

RISING STAR ACADEMY -92-331-4097486 Page 4

CHEMISTRY ALDEHYDES, KETONES &
CHAPTER 3 CARBOXYLIC ACIDS

Question 5

Y
EM
What products are formed when aldehydes and ketones are
reduced?

D
6 A
Answer:

48 C
1. Aldehydes (RCHO) are reduced to Primary Alcohols or 1o

97 E A
alcohols (RCH2OH) by using the process of Catalytic
Hydrogenation using catalyst like Pd (Palladium), Pt (Platinum) or
40 IN
Ni(Nickel).
2. Ketones (RCOR) are reduced to Secondary Alcohol or 2o
31 L

alcohols (R2CHOH) by using the process of Catalytic

23 ON

Hydrogenation using catalyst like Pd (Palladium), Pt (Platinum) or

Ni(Nickel).
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY -92-331-4097486 Page 5

CHEMISTRY ALDEHYDES, KETONES &
CHAPTER 3 CARBOXYLIC ACIDS
Question 6
Sketch the hydrogen bonding between formaldehyde and water.

Y
EM
Answer:
In formaldehyde, carbon atom is bonded to one Oxygen atom and two

D
hydrogen atoms.

6 A
48 C
97 E A
40 IN
31 L
23 ON

Due to presence of lone pair of electrons on oxygen atom, it is a H –

+9 R

bond acceptor. Hydrogen atoms of water form H – bond with carbonyl

A

Oxygen.
ST
G
N
SI
RI

RISING STAR ACADEMY -92-331-4097486 Page 6

CHEMISTRY ALDEHYDES, KETONES &
CHAPTER 3 CARBOXYLIC ACIDS

Question 7

Y
EM
Classify each of the following as an aldehyde or ketone.

D
Answer:

6 A
1. Structure (B) and (C) are classified as aldehyde

48 C
2. Structure (A) and (D) are classified as ketone

97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY -92-331-4097486 Page 7

CHAPTER CHEMISTRY
4 BIOCHEMISTRY
Question # 1

Y
What is the true structure of DNA?

EM
Answer: Double Helix

D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 1

CHAPTER CHEMISTRY
4 BIOCHEMISTRY
Question # 2

Y
What nucleotide base does RNA contain

EM
but DNA does not?

D
Answer:

6 A
Uracil base

48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 2

CHAPTER CHEMISTRY
4 BIOCHEMISTRY
Question # 3

Y
What molecules do both DNA and RNA

EM
contain?

D
Answer:

6 A
The purines adenine (A) and guanine (G) and the

48 C
97 E A
pyrimidine cytosine (C) are present in both DNA and
RNA. 40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 3

CHAPTER CHEMISTRY
4 BIOCHEMISTRY

Y
Question # 4

EM
What is the function of nucleic acids?

D
Answer:

6 A
48 C
Nucleic acids (both DNA and RNA) are responsible

97 E A
for protein synthesis in a cell.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 4

CHAPTER CHEMISTRY
4 BIOCHEMISTRY

Y
Question # 5

EM
Define

D
a) Carbohydrates

6 A
Carbohydrates (also called saccharides) are

48 C
97 E A
molecular compounds made from just three
elements: carbon, hydrogen and oxygen.
40 IN
b) Nucleic acid
31 L

a complex organic substance present in living

23 ON

cells, especially DNA or RNA.

c) Proteins
+9 R
A

Large molecules composed of one or more

ST

chains of amino acids

d) Lipids
G

A lipid is any of various organic compounds

N
SI

that are insoluble in water. They include fats,

RI

waxes, oils.

RISING STAR ACADEMY +92-331-4097486 Page 5

CHAPTER CHEMISTRY
4 BIOCHEMISTRY

Y
Question # 6

EM
What are the structural difference

D
between ribose and deoxyribose?

6 A
Answer:

48 C
97 E A
Ribose sugar has one −𝑂𝐻 on each carbon
40 IN
and deoxyribose, the second carbon atom
lacks the −𝑂𝐻group.
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 6

CHAPTER CHEMISTRY
4 BIOCHEMISTRY

Y
Question # 7

EM
What are the structural difference

D
between DNA and RNA?

6 A
Answer:

48 C
97 E A
DNA is Double stranded while RNA is Single
Stranded.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 7

CHAPTER CHEMISTRY
4 BIOCHEMISTRY

Y
EM
Question # 8

D
Why some amino acids are essential?

6 A
48 C
Answer:

97 E A
Some amino acids are essential because human body
40 IN
cannot make.
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 8

CHAPTER CHEMISTRY
4 BIOCHEMISTRY

Y
EM
Question # 9

D
State at least two biologically important

6 A
proteins and their functions.

48 C
97 E A
Answer:
40 IN
1. Enzymes are catalysts in biochemical reactions
like digestion.
31 L
23 ON

2. Hormones: coordinate the activity of different

body system.
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 9

CHAPTER CHEMISTRY
4 BIOCHEMISTRY

Y
EM
Question # 10

D
What is protein denaturation, and give

6 A
example of denaturing agents?

48 C
97 E A
Answer:
40 IN
The process of breaking down the 3D structure of
the proteins in called denaturation. Heat, UV-rays,
31 L
23 ON

X-rays are the denaturing agents.

+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 10

CHAPTER CHEMISTRY
4 BIOCHEMISTRY

Y
EM
Question # 11

D
Identify the individual amino acids used

6 A
to form the following tri-peptide.

48 C
97 E A
Answer: Structures are NOT given
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 11

CHAPTER CHEMISTRY
4 BIOCHEMISTRY

Y
EM
Question # 12

D
What is the name of the tri-peptide?

6 A
48 C
Answer: Structures of Tri-peptide is NOT given

97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 12

CHAPTER CHEMISTRY
4 BIOCHEMISTRY

Y
Question # 13

EM
Draw a triacylglycerol formed from two

D
molecules of myristic acid and one

6 A
molecule of palmitic acid?

48 C
97 E A
Answer:
Formula of Palmtitic Acid : CH3(CH2)14COOH
40 IN
Formula of Myristic Acid : CH3(CH2)12COOH
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 13

CHAPTER CHEMISTRY
4 BIOCHEMISTRY

Y
EM
Question # 14

D
Give an example of each type of lipid.

6 A
48 C
a) a monounsaturated fatty acid : Oleic Acid

97 E A
b) saturated triacylglycerol : Palmitic Acid
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 14

CHAPTER CHEMISTRY
4 BIOCHEMISTRY

Y
EM
D
Question # 15

6 A
48 C
97 E A
40 IN
31 L
23 ON

Answer:
+9 R

(A) Glycerophospholipid
A

(B) Sphingolipid
ST

(C) Phospholipid
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 15

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 1

Y
EM
Define boron family.
Answer:

D
6 A
In Boron family, we have five elements Boron, Aluminum,

48 C
Gallium, Indium, and Thallium. All five have three electrons in

97 E A
their outermost or valence shell. Only one member of
this family is a metalloid i.e. Boron.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 1

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 2

Y
EM
Identify the position of boron family in the
periodic table.

D
Answer:

6 A
48 C
Boron is located in Group 13.

97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 2

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 3

Y
EM
Write at least three physical properties of boron
family?

D
Answer:

6 A
48 C
1. Boron has a high melting point.

97 E A
2. Aluminium is amphoteric.
3. Indium has a lesser nuclear radius than Thallium
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 3

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 4

Y
EM
State at least three common uses of boron
family.

D
Answer:

6 A
48 C
1. Aluminium is used for making cooking utensils.

97 E A
2. Thalium is extremely toxic and has no commercial use.
3. Indium is also used in making semi-conductor devices.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 4

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 5

Y
EM
Define noble gases?
Answer:

D
6 A
Noble gases are inert gases. They do not react to make

48 C
compounds because they have no free electrons in its

97 E A
outmost shell. These gases are, Helium, Neon, Argon,
Krypton, Xenon, Radon are these 6 gases.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 5

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 6

Y
EM
What is the position of noble gases in the
periodic table?

D
Answer:

6 A
48 C
Noble gases are lying in 18th Group.

97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 6

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 7

Y
EM
Write at least three the physical properties of
noble gases?

D
Answer:

6 A
48 C
Noble gases are colorless, odorless, tasteless, and

97 E A
nonflammable gases under standard conditions.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 7

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 8

Y
EM
State three common uses of noble gases.
Answer:

D
6 A
1. Helium is used in filling balloons.

48 C
2. Oxygen -Helium mixture is used in the treatment of

97 E A
asthma.
3. Neon is used for filling sodium vapour lamps.
4. Argon is used as a carrier gas in gas chromatography.
40 IN
5. Krypton is used in high efficiency miner's cap lamps.
31 L
6. Radon is used in radioactive research.
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 8

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 9

Y
EM
Choose two elements from the group and write
notes about them.

D
Answer:

6 A
48 C
Aluminium is the third most abundant element in the earth's crust. It is

97 E A
commonly used in the household as aluminum foil, in crafts such as
dyeing and pottery, and also in construction to make alloys. In its purest
40 IN
form the metal is bluish-white and very ductile. It is an excellent
conductor of heat and electricity and finds use in some wiring.
31 L
23 ON

Indium has atomic number 49. It has the +1 or +3 oxidation state but
the +3 state is more common. It is a soft, malleable metal that is similar
to gallium. Indium found in photoconductors in optical instruments.
+9 R

Indium is soluble in acids, but does not react with oxygen at room
A

temperature.
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 9

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 10

Y
EM
Define p-block elements?
The p-block elements are found on the right side of the periodic

D
table. There are 35 p-block elements, all of which are in p orbital

6 A
with valence electrons. They include the boron, carbon,

48 C
nitrogen, oxygen and flourine families in addition to the noble

97 E A
gases.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 10

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 11

Y
EM
What are the types of elements in p-block
elements?

D
Answer:

6 A
48 C
The p block consists of non-metals at the right, metalloids in the

97 E A
middle, and metals at the left.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 11

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 12

Y
EM
What is the valence shell electronic
configuration

D
a) Boron family: ns2np1

6 A
48 C
b) Carbon family: ns2np2

97 E A
c) Nitrogen family: ns2np3
40 IN
e) Halogens: ns2np5
31 L

d) Oxygen family: ns2np4

23 ON

f) Noble gases: ns2np6

+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 12

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 13

Y
EM
What elements of boron family do not react
with nitrogen?

D
Answer:

6 A
48 C
Gallium (Ga), Indium (In), Thallium (Tl)

97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 13

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 14

Y
EM
List the element of
a) Carbon family

D
Carbon, Silicon, Germanium, Tin, Lead,

6 A
48 C
and Flerovium.

97 E A
b)Oxygen family40 IN
Oxygen, Sulfur, Selenium, Tellurium and Polonium.
31 L
23 ON

c) Halogens
Fluorine, Chlorine, Bromine, Iodine, Astatine
+9 R
A

d)Noble gases
ST

Helium, Neon, Argon, Krypton, Xenon, Radon

G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 14

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 15

Y
EM
Why are noble gases chemically stable?
Answer:

D
6 A
They are the most stable due to having the maximum number of

48 C
valence electrons OR having no free electron in its outermost or

97 E A
valence shell to make bond.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 15

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS

Question # 16

Y
EM
What happens to melting points, boiling points
and density of noble gases?

D
Answer:

6 A
48 C
The melting point of noble gases increases down the group.

97 E A
The boiling point of noble gases increases down the group.
The density of noble gases increases down the group.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 16

CHEMISTRY
CHAPTER 5 P BLOCK ELEMENTS
Choose the circle the correct answer:
1. Boron Group

Y
2. 3

EM
3. ns2np1

D
4. In

6 A
5. Al

48 C
6. Noble gases

97 E A
7. Noble gases
8. Halogens 40 IN
9. Alkali metals
31 L
10. VIIIA
23 ON

11. The electronic configuration of their outermost

electrons is ns2np6.
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-331-4097486 Page 17

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS
LESSON EXERCISE:
Question # 1

Y
Do chemical reactions usually occur in a single stop?

EM
Answer:

D
No, Chemical reactions rarely occur in one simple step. The overall

6 A
balanced equation for a chemical reaction does not always tell us how
a reaction actually proceeds. In many cases, the overall reaction takes

48 C
place in a series of small steps called mechanism.

97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 1

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS
Question # 2
What does the overall balanced equation not tell us?

Y
Answer:

EM
It does not tell us about the reaction kinetic order.

D
6 A
48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 2

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 3

Y
How do overall reactions all proceeds?

EM
Answer:

D
Every chemical reaction proceeds according to a reaction mechanism,

6 A
which is a step-by-step description of what occurs during a reaction on
the molecular level. Each step of the mechanism is known as an

48 C
elementary process, which describes a single moment during a reaction

97 E A
in which molecules break and/or form new bonds.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 3

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 4

Y
Explain an intermediate species?

EM
Answer:

D
An intermediate is a species which appears in the mechanism of a

6 A
reaction, but not in the overall balanced equation. An intermediate is
always formed in an early step in the mechanism and consumed in a

48 C
later step.

97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 4

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 5

Y
Describe the steps that a mechanism of a reaction consists of?

EM
Answer:

D
A reaction mechanism is the sequence of elementary steps by which a

6 A
chemical reaction occurs. A reaction that occurs is two or more
elementary steps is called a multistep or complex reaction.

48 C
97 E A
In this reaction, the reactant in first step converted to an intermediate
which then gives the product. The slowest step is the rate determining
40 IN
step.
𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡 → 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖𝑎𝑡𝑒(𝑈𝑛𝑠𝑡𝑎𝑏𝑙𝑒) → 𝑃𝑟𝑜𝑑𝑐𝑢𝑡(𝑆𝑡𝑎𝑏𝑙𝑒)
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 5

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS
CHAPTER REVIEW QUESTIONS
Question # 1

Y
Explain the following

EM
a) Reaction Rate
It is the speed at which a chemical reaction proceeds.

D
It is often expressed in terms of either the concentration

6 A
(amount per unit volume) of a product that is formed in a unit of

48 C
time or the concentration of a reactant that is consumed in a unit

97 E A
of time.
b) Collision theory
40 IN
It is used to predict the rates of chemical reactions, particularly
for gases. The collision theory is based on the assumption that
31 L
for a reaction to occur it is necessary for the reacting species
23 ON

(atoms or molecules) to come together or collide with one

another.
+9 R

c) Activation energy
A

It is the minimum amount of energy provided to the system to

system for reaction to occur.
ST

d) Catalyst
Catalyst is any substance that increases the rate of a reaction
G

without itself being consumed.

N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 6

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 2

Y
In the following reaction:

EM
2𝐶𝑎 + 𝑂2 − −−> 2𝐶𝑎𝑂

D
1g of calcium is oxidized. the reaction takes 10 minutes to reach

6 A
completion. What is the rate of the formation of CaO.

48 C
Solution:

97 E A
Molar mass of Ca= 40g
1
No of moles = 40 IN
40
We are assuming volume of container to be 1Litre
∆𝑡 = 10 𝑚𝑖𝑛𝑠
31 L

𝑑[𝐶𝑎] 1 1
23 ON

𝑇ℎ𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑓𝑜𝑟𝑚𝑎𝑖𝑡𝑜𝑛 𝑜𝑓 𝐶𝑎 = = =

𝑑𝑡 40 × 60 2400

1 𝑑[𝐶𝑎] 𝑑[𝑂2 ] 1 𝑑[𝐶𝑎𝑂]

+9 R

𝑅𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = = =
2 𝑑𝑡 𝑑𝑡 2 𝑑𝑡
A

1 𝑑[𝐶𝑎] 1 𝑑[𝐶𝑎𝑂]
ST

=> =
2 𝑑𝑡 2 𝑑𝑡
𝑑[𝐶𝑎𝑂] 1 1
=> = ×2= 𝑚𝑜𝑙/𝐿𝑠
G

𝑑𝑡 2 × 40 × 60 × 10 24000
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 7

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 3

Y
10g of Mg reacts with 0.15mol.dm-3 solution of HCl at a

EM
temperature of 250 𝐶
a) Write a balanced chemical equation for the reaction.
b) State two ways of increasing the average rate of production of H2(g)

D
solution:

6 A
(a) 𝐌𝐠(𝐬) + 𝟐𝐇𝐂𝐥(𝐚𝐪) → 𝐌𝐠𝐂𝐥𝟐 (𝐚𝐪) + 𝐇𝟐 (𝒈)

48 C
97 E A
(b) Two ways of increasing average rate of reaction
40 IN
(i) By increasing the concentration of HCl.
31 L
23 ON

(ii) By increasing the temperature. We know that, according to

the temperature coefficient for every 10C raise in
temperature, the rate of reaction doubles.
+9 R

STOICHIOMETRY
A
ST

0.15 mol/dm3 of HCl = 0.15 × 36.5 = 5.475 g

Mg + 2HCl → MgCl2 + H2
G

Relative mass of Mg = 24.3 g

N
SI

Molecular Mass of HCl = 73g

10𝑔×73𝑔
RI

HCl required = = 30.04𝑔

24.3𝑔

So, HCl is the limiting Reagent.

Increasing the concentration of HCl, increases the Production of H2.

RISING STAR ACADEMY +92-3314097486 Page 8

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 4

Y
A table of results is given below:

EM
Time Elapsed Volume of H2(g) cm3
0 0

D
0.5 17

6 A
1.0 25

48 C
1.5 30
2.0 33

97 E A
2.5 35
3.0
40 IN 35
31 L
Answer:
23 ON

INCOMPLETE DISCRIPTION …
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 9

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 5

Y
Construct the following hypothetical reaction

EM
𝐴(𝑔) + 𝐵(𝑔) → 𝐴𝐵(𝑔)

D
Increasing the average rate of reaction, which of the following you

6 A
would do?

48 C
(a) Decrease the concentration of B

97 E A
(b) Grind A into fine powder
(c) Decrease the temperature
40 IN
Answer:
31 L
(a) By decreasing the concentration of B, the possible effective
23 ON

collision decreases and rate decreases.

(b) Due to grinding, the surface area increases and thus more
+9 R

effective collision are possible and thus the average rate of

A

reaction increases.
(c) By decreasing the temperature, the rate will decreases as the
ST

kinetic energy decreases, the number of effective collision

decreases.
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 10

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 7

Y
The reaction between the peroxydisulfate ion (𝑆2 𝑂8 2− ) and

EM
the iodide ion (𝐼3 − ) is:
(a) From the following data (Table below), which were

D
collected at a set temperature, determine the rate law and

6 A
calculate the specific rate constant.

48 C
Experiment [𝑆2 𝑂8 2− ](𝑀) [𝐼− ](𝑀) Initial Rater

97 E A
(M/s)
1 0.080 0.034 2.2× 10−4
2 0.080 0.017 1.1× 10−4
40 IN
3 0.16 0.017 2.2× 10−4
31 L
23 ON

(b) What is the initial rate of reaction if the concentrations of both

𝑆2 𝑂8 2− and 𝐼− are 0.5M?
+9 R

Answer:
A

(a) 𝑆2 𝑂8 2− (𝑎𝑞) + 𝐼3 − (𝑎𝑞) → 2𝑆𝑂4 2− (𝑎𝑞) + 𝐼3 − (𝑎𝑞)

ST

Determination of the rate law of the equation,

G

𝑅𝑎𝑡𝑒 = 𝑘 × [𝑆2 𝑂8 2− ] × [𝐼− ]3

N
SI

Given from the chart,

RI

𝑀
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑅𝑎𝑡𝑒 = 𝑟 = 2.2 × 10−4
𝑠
[𝑆2 𝑂8 2− ] = 0.080 𝑀

RISING STAR ACADEMY +92-3314097486 Page 11

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS
[𝐼− 3 ] = 0.034 𝑀

Y
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑅𝑎𝑡𝑒 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑘)
𝑟

EM
=
𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑛𝑎𝑡

D
𝑀
2.2 × 10−4
𝑠

6 A
𝑘 = 3
= 7 × 10−11 𝑀−3 𝑠 −1
0.080𝑀 × (0.034𝑀)

48 C
97 E A
b) 𝐼𝑓 𝑡ℎ𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑜𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑏𝑒𝑐𝑜𝑚𝑒
40 IN
[𝑆2 𝑂8 2− ] 𝑎𝑛𝑑 [𝐼− 3 ] = 0.5 𝑀
31 L

The rate of the reaction becomes,

23 ON

𝑟 = 7 × 10−11 × 0.5 × 0.53 = 4.37 × 10−12 𝑀/𝑠

Thus, the initial rate of the reaction will be 4.37 × 10−12 𝑀/𝑠
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 12

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 8

Y
Explain the following:

EM
(a) Hydrogen peroxide 𝐻2 𝑂2 decomposes to water and
oxygen very slowly. When solid manganese dioxide

D
is added, however, oxygen evolves rapidly.

6 A
Answer:

48 C
Hydrogen peroxide is a very unique substance due to its

97 E A
molecular structure. It consists atoms of oxygen in oxidation
state of -1 unlike many substances, where oxygen occurs in
40 IN
oxidation state of 0 or -2. This means that this substance can be
used as both an oxidizing and a reducing agent, depending of
31 L
pH of its solution.
23 ON

Due to those properties, particles of hydrogen peroxide can

decompose via reaction of disproportionation Manganese oxide
+9 R

(MnO2) is widely used in heterogeneous reaction of

A

decomposition of hydrogen peroxide. Its excellent catalytic

efficiency in this reaction. It is usually used in the form of a
ST

powder.
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 13

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

(b) People who are lactose intolerant cannot break down

milk sugar (lactose).

Y
Answer:

EM
Lactose intolerance is when your body can't break
down or digest lactose. Lactose is a sugar found in

D
milk and milk products. Lactose intolerance happens when your

6 A
small intestine does not make enough of a digestive enzyme

48 C
called lactase. Lactase breaks down the lactose in food so your

97 E A
body can absorb it. People who are lactose intolerant have
unpleasant symptoms after eating or drinking milk or milk
products. These symptoms include bloating, diarrhea, and gas.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 14

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 9

Y
Indicate the following changes will increase, decrease or

EM
have no effect on rate of reaction:
(a) Increasing the temperature

D
(b) Increasing the number of reacting molecules

6 A
(c) Adding a catalyst

48 C
Answer:

97 E A
(a) By increasing the temperature, the rate of reaction will be
40 IN
increase.
(b) By increasing the number of reacting molecules, the rate of
31 L
reaction will be increase
23 ON

(c) By adding the catalyst, the rate of reaction will be increase.

+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 15

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 10

Y
Why does bread grow mold more quickly at room temperature

EM
than in the refrigerator?
Answer:

D
6 A
The reason that bread kept at room temperature molds faster than bread
kept in the fridge is due to the fact that warmer temperatures promote

48 C
more to grow. Mold flourishes in room temperature and has a harder

97 E A
time in colder climates.
OR
40 IN
The rate of a reaction is the speed at which reactants are used to form a
31 L
23 ON

product or products.
The rate of reaction is affected by the change in the temperature.
+9 R

At higher temperatures, like room, the rate of a reaction is high, and at

A

lower temperatures, like refrigerator, the rate of reaction is small.

ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 16

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 11

Y
How does a catalyst affect the activation energy?

EM
Answer:

D
A catalyst is used to speed up the reaction by lowering its activation

6 A
energy.

48 C
Activation energy is the energy required to break apart the bonds of the

97 E A
reacting molecules.
As the activation energy is lowered, more molecules collide with each
40 IN
other, so they form product faster. So the catalyst lowers the activation
energy and it can help to precede the reaction at a faster rate.
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 17

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 12

Y
For the decomposition reaction 𝑋 → 𝑌 + 𝑍, the reaction is

EM
first-order with respect to X, and the value of the specific rate
constant is 0.0296/s.

D
Calculate the initial reaction rate when starting with the following

6 A
concentration of X.

48 C
(a) [𝑋] = 0.410 𝑀

97 E A
Solution:
According to Rate Law, for given reaction
40 IN
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = 𝑘[𝑋]
31 L
For [X] = 0.410 M
23 ON

Rate of reaction = 0.0296 × 0.410 = 0.0121 𝑀/𝑠

(b) [𝑋] = 0.0223𝑀

+9 R
A

Solution:
ST

According to Rate Law, for given reaction

𝑅𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = 𝑘[𝑋]
G

For [X] = 0.0223 M

N

Rate of reaction = 0.0296 × 0.0223 = 6.60 × 10−4 𝑀/𝑠

SI
RI

RISING STAR ACADEMY +92-3314097486 Page 18

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Y
Question # 13

EM
In the following reaction, what happens to the number of
collisions when more 𝐵𝑟2(𝑔) molecules are added?

D
6 A
𝐻2(𝑔) + 𝐵𝑟2(𝑔) → 2𝐻𝐵𝑟(𝑔)

48 C
Answer:

97 E A
In the above reaction, both reactants H2 and Br2 are the reactants.
40 IN
The addition of more Br2 molecules increases the concentration of Br2.
By this reason the number of collisions between H2 and Br2 molecules
31 L

also increased, therefore the reaction will occur more rapidly.

23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 19

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 14

Y
In the following reaction, what happens to the number of

EM
collision when the temperature of the reaction is decreased?
𝐶(𝑠) + 𝑂2(𝑔) → 𝐶𝑂2(𝑔)

D
6 A
Answer:

48 C
In the above reaction, both C and O2, are reactants.

97 E A
When the temperature of the reaction decreases, the kinetic energy of the
reactant molecules decreases. Due to this reason, they are moving
40 IN
slowly, and the number of collisions between C and O2 molecules get
31 L
decreased. Therefore, the rate of the formation of CO2 also decreased.
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 20

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 15

Y
How would each of the following change the rate of reaction

EM
shown here?
2𝑆𝑂2(𝑔) + 𝑂2(𝑔) → 2𝑆𝑂3(𝑔)

D
6 A
Answer:

48 C
In this reaction, SO2 and O2 are reactants and SO3 is the product

97 E A
(a) Adding some 𝑺𝑶𝟐 (𝒈)
When more SO2 molecules are added in the reaction, the
40 IN
concentration of reactants increases (that is SO2 increases).
31 L
As the number of SO2 molecules increases, the collision
23 ON

between SO2 and O2 will increase. Due to more collisions, the

rate of the formation of SO3 will be more. This means that the
reaction goes faster
+9 R

Hence, adding more SO2 will increase the rate of the reaction.
A

(b) Increasing the temperature

ST

When the temperature of the reaction increases, the kinetic

energy of the molecule increases. Due to this, the SO2 and O2,
G

molecules move faster, so there will be more collisions between

N

them. This gives the reacting molecules of SO2 and O2

SI

sufficient energy to form SO3. As a result, the rate of formation

RI

of SO3 increases. Hence, the raising temperature increases the

rate of the formation of SO3.

RISING STAR ACADEMY +92-3314097486 Page 21

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Y
EM
(c) Adding a catalyst
A catalyst is a substance that increases the rate of the

D
reaction by lowering the activation energy. When a catalyst is

6 A
added in the reaction, its activation energy lowers the energy

48 C
provided by the collisions between SO2 and O2. Hence, they

97 E A
more easily form SO3. Hence, adding a catalyst to the reaction
increases the rate of the formation of SO3.
(d) Removing some 𝑶𝟐(𝒈)
40 IN
In the above reaction, the SO2 and H2 molecules are the
31 L
reactants. The removal of SO2 decreases the concentration of the
23 ON

reactant (SO2), so the chances of collisions between SO2 and O2

will decrease. A decrease in collisions between SO2 and O2
decreases the rate of the formation of products (SO3). Hence, the
+9 R

removal of O2 decreases the rate of the reaction.

A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 22

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 16

Y
How would each of the following change the rate of reaction

EM
shown here?
2𝑁𝑂(𝑔) + 2𝐻2(𝑔) → 𝑁2(𝑔) + 2𝐻2 𝑂(𝑔)

D
6 A
Answer:

48 C
In this reaction, NO and H2 are reactants and N2 and H2O are the product

97 E A
(a) Adding some 𝑵𝑶(𝒈)
When more NO molecules are added in the reaction, the
40 IN
concentration of the reactants increases (that is NO increases).
31 L
As the number of NO molecules increases the collisions
23 ON

between NO and H2 will increase. Due to more collisions, the

rate of the formation N2 and H2O will be more. This means that
the reaction goes faster. Hence, adding more NO will increases
+9 R

the rate of the reaction.

A
ST

(b) Decreasing the temperature

When the temperature of the reaction decreases, the kinetic
G

energy of the molecule decreases. Due to this, the NO and H2

N

molecules move slowly, so there will be less collisions between

SI

them. This gives the reacting molecules of NO and H2

insufficient energy to form N2 and H2O. Thus, the rate of the
RI

formation of N2 and H2O is decreases. Hence, lowering the

temperature decreases the rate of the reaction.

RISING STAR ACADEMY +92-3314097486 Page 23

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Y
(c) Removing some 𝑯𝟐(𝒈)

EM
In the above reaction, NO and H2 molecules are the
reactants. The removal of H2 decreases the

D
concentration of the reactant H2, thus the chances of collisions

6 A
between N2 and H2O will decrease. A decrease in collisions

48 C
between NO and H2 decreases the rate of the formation of the

97 E A
products. Hence, removing some H2 decreases the rate of the
reaction.
40 IN
(d) Adding a catalyst
31 L
When a catalyst is added in the reaction, its activation energy
23 ON

lowers the energy provided by the collisions between NO and

H2. Hence, they more easily form N2 and H2O.
Hence, adding a catalyst increases the rate of reaction.
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 24

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 17

Y
Classify the following processes as spontaneous or

EM
nonspontaneous. Explain your answers in terms of whether
energy must be continually supplied to keep the process

D
going:

6 A
(a) Water is decomposed into hydrogen and oxygen gas by passing

48 C
electricity through the liquid.

97 E A
Non-spontaneous
(b) An explosive detonates after being struck by a falling rock
40 IN
Spontaneous
(c) A coating of magnesium oxide forms on a clean piece of
31 L
magnesium exposed to air
23 ON

Spontaneous
(d) A light bulb emits light when an electric current is passed
+9 R

through it
A

Non-spontaneous
(e) A cube of sugar dissolves in a cup of hot coffee.
ST

Spontaneous
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 25

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 18

Y
Classify the following processes according to their rates as

EM
fast, slow or very slow:
(a) The souring of milk stored in a refrigerator

D
Very Slow

6 A
(b) The cooking of an egg in boiling water

48 C
Slow

97 E A
(c) The ripening of a banana stored at room temperature
Very slow 40 IN
(d) The rising of bread dough in a warm room
Slow
31 L
(e) The melting of butter put into a hot pan
23 ON

Fast
(f) The change of apple juice to cider
+9 R

Very slow
A

(g) The movement of sound from a slammed door to your ear

ST

Fast
(h) The ringing of your phone after you step into the shower
G

Fast
N

(i) The combustion of gasoline in the engine of your car

SI

Fast
RI

(j) The perceived passage of time when you are doing something
enjoyable
Fast

RISING STAR ACADEMY +92-3314097486 Page 26

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 19

Y
Describe the observations or measurements that could be made

EM
to allow you to follow the rate of the following processes:
(a) The melting of a block of ice

D
Answer:

6 A
To monitor the rate of ice melting, you could measure the

48 C
volume of water formed with respect to time or the decreasing

97 E A
mass of the ice with respect to time. Also you could be
measured the physical dimensions of the melting block.
40 IN
(b) The setting (hardening) of concrete
Answer:
31 L
To monitor the setting of concrete, you could measure the
23 ON

ability of a tool to penetrate the concrete with respect to time.

Also you could be measured the force required to break a
+9 R

specific sized piece.

A

(c) The burning of candle

Answer:
ST

To monitor the burning of a candle with respect to time, you

could measure the height of the candle before its lit and then
G

again at certain intervals.

N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 27

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 20

Y
Consider the following hypothetical reaction:

EM
𝐴+𝐵 →𝐶

D
Calculate the average rate of the reaction on the basis of the following

6 A
information:

48 C
(a) Pure A and B are mixed, and after 12.0 minutes the measured

97 E A
concentration of C is 0.396 mol/L.
Solution: 40 IN
Use the expression to calculate the rate of reaction
∆𝐶
𝑅𝑎𝑡𝑒 =
31 L
∆𝑡
23 ON

𝐶𝑡 − 𝐶𝑜
𝑅𝑎𝑡𝑒 =
∆𝑡
0.396 − 0.000
+9 R

𝑅𝑎𝑡𝑒 = = 0.033 𝑀/𝑚𝑖𝑛

12 𝑚𝑖𝑛
A

(b) Pure A, B and C are mixed together at equal concentration of 0.300M.

ST

After 8.00 minutes, the concentration of C is found to be 0.455M.

Solution:
G

Use the expression to calculate the rate of reaction

N

∆𝐶
𝑅𝑎𝑡𝑒 =
SI

∆𝑡
𝐶𝑡 − 𝐶𝑜
RI

𝑅𝑎𝑡𝑒 =
∆𝑡
0.455 𝑀 − 0.300 𝑀
𝑅𝑎𝑡𝑒 = = 0.0194 𝑀/𝑚𝑖𝑛
8 𝑚𝑖𝑛

RISING STAR ACADEMY +92-3314097486 Page 28

CHAPTER CHEMISTRY
6 CHEMICAL KINETICS

Question # 21

Y
Use energy diagram to compare catalyzed and uncatalyzed

EM
reactions.
Answer:

D
6 A
Catalyst: Catalyst is substance that increases the rate of reaction without
being used up in the reaction.

48 C
97 E A
Catalyst increases the rete of reaction by providing an alternative
reaction path way. Catalyst decreases the activation energy of the
40 IN
reaction; hence the rate of reaction increases. The activation energy of
reaction without catalyst is having more value and the activation energy
31 L
of reaction with the catalyst is less value. The energy profile diagram of
23 ON

catalyzed and uncatalyzed reaction is shown below:

+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92-3314097486 Page 29

CHAPTER CHEMICAL
7 EQUILIBRIUM
LESSON EXERCISE # 1

Y
Question # 1

EM
Define chemical equilibrium. Give an example.
Answer:

D
6 A
A reaction is in chemical equilibrium when the rate of the forward
reaction equals the rate of the reverse reaction. There are

48 C
many examples of chemical equilibrium all around you. One example is

97 E A
a bottle of fizzy cold drink. In the bottle there is carbon dioxide (CO2)
dissolved in the liquid.
40 IN
Question # 2
31 L
23 ON

What are the conditions of chemical equilibrium?

Answer:
+9 R

1. Equilibrium is a dynamic state.

A

2. The forward and reverse reactions are happening at equal rates.

ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 1

CHAPTER CHEMICAL
7 EQUILIBRIUM
Question # 3
Explain what is meant when it is said that chemical

Y
equilibrium is dynamic.

EM
Answer:

D
Dynamic equilibrium only occurs in reversible reactions, and it’s when

6 A
the rate of the forward reaction is equal to the rate of the reverse

48 C
reaction. These equations are dynamic because the forward and reverse

97 E A
reactions are still occurring, but the two rates are equal and unchanging,
so they’re also at equilibrium.
40 IN
Dynamic equilibrium is an example of a system in a steady state. This
means the variables in the equation are unchanging over time (since the
31 L

rates of reaction are equal).

23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 2

CHAPTER CHEMICAL
7 EQUILIBRIUM
LESSON EXERCISE # 2

Y
Question # 1

EM
What are the differences between homogenous and
heterogeneous equilibrium?

D
Answer:

6 A
Homogenous equilibrium:

48 C
97 E A
If reactants and products both are in the same phase in an equilibrium
reaction then that equilibrium is called as homogeneous equilibrium.
40 IN
Example: 2𝑆𝑂2(𝑔) + 𝑂2(𝑔) ⇌ 2𝑆𝑂3(𝑔)
31 L
In this equilibrium, reactants and products are in GASOEOUS STATE,
23 ON

so it is an homogenous equilibrium.
Heterogeneous Equilibrium:
+9 R
A

If all the species in an equilibrium are not in same phase, then it is

referred as heterogeneous equilibrium.
ST

𝐶𝑎𝐶𝑂3(𝑠) ⇌ 𝐶𝑎2+ (𝑎𝑞) + 𝐶𝑂32− (𝑎𝑞)

G

In this equilibrium, reactant is in solid phase and products are in liquid

N

phase, so it is a heterogeneous equilibrium.

SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 3

CHAPTER CHEMICAL
7 EQUILIBRIUM
Question # 2
Classify each of the following reactions as homogeneous

Y
and heterogeneous.

EM
(a) 𝑃4(𝑠) + 6𝐶𝑙2(𝑔) ⇋ 4𝑃𝐶𝑙3(𝑙)
(b) 𝐻𝐶𝑙(𝑔) + 𝑁𝑎𝑂𝐻(𝑎𝑞) ⇋ 𝑁𝑎𝐶𝑙(𝑎𝑞) + 𝐻2 𝑂(𝑙)

D
𝑂3(𝑔) + 𝑁𝑂(𝑔) ⇋ 𝑂2(𝑔) + 𝑁𝑂2(𝑔)

6 A
(c)

48 C
Answer:

97 E A
(a) Heterogeneous Equilibrium
(b) Heterogeneous Equilibrium
40 IN
(c) Homogeneous Equilibrium
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 4

CHAPTER CHEMICAL
7 EQUILIBRIUM
LESSON EXERCISE # 3

Y
Question # 1

EM
Why are solids and liquids not included in chemical
equilibrium?

D
Answer:

6 A
Pure solids or liquids are excluded from the equilibrium expression

48 C
because their effective concentrations stay constant throughout

97 E A
the reaction.
Question # 2
40 IN
Write the equilibrium expressions for the following reactions in the
31 L
23 ON

space below:
(a) 𝑂3(𝑔) + 𝑁𝑂(𝑔) ⇋ 𝑂2(𝑔) + 𝑁𝑂2(𝑔)
+9 R

(b) 2𝐶𝑂(𝑔) + 𝑂2(𝑔) ⇋ 2𝐶𝑂2(𝑔)

A

(c) 𝑁𝐻4 𝑁𝑂3(𝑠) ⇋ 𝑁2 𝑂(𝑔) + 2𝐻2 𝑂(𝑙)

ST

Answer:
𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠
G

𝐾=
𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
N

[𝑂2 ][𝑁𝑂2 ]
(a) 𝐾=
SI

[𝑂3 ][𝑁𝑂]
[𝐶𝑂2 ]2
RI

(b) 𝐾 = [𝑂 2
2 ][𝐶𝑂]
(c) 𝐾 = [𝑁𝑂2 ]

RISING STAR ACADEMY +92- 331- 4097486 Page 5

CHAPTER CHEMICAL
7 EQUILIBRIUM
Question # 3
Indicate whether reactants or products are likely to be

Y
favored. Justify your decisions.

EM
(a) 𝐾𝑒𝑞 = 1 × 1012
(b) 𝐾𝑒𝑞 = 1.5

D
6 A
48 C
Answer:

97 E A
(a) When we have 𝐾𝑒𝑞 > 1, then reaction has a high yield and is
said to be right shifted or favored to products.
40 IN
(b) When we have 𝐾𝑒𝑞 > 1, then reaction has a high yield and is
31 L
said to be right shifted or favored to products.
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 6

CHAPTER CHEMICAL
7 EQUILIBRIUM
LESSON EXERCISE # 4

Y
Question # 1

EM
What is the equilibrium constant, 𝐾𝑒𝑞, if the equilibrium
concentrations are as follow:

D
𝑃𝐶𝑙5(𝑔) ⇋ 𝑃𝐶𝑙3(𝑔) + 𝐶𝑙2(𝑔)

6 A
48 C
Answer:

97 E A
[𝑃𝐶𝑙3 ][𝐶𝑙2 ]
𝐾𝑒𝑞 =
40 IN [𝑃𝐶𝑙5 ]
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 7

CHAPTER CHEMICAL
7 EQUILIBRIUM
Question # 2

Acetic acid dissociates in water. If 𝐾𝑒𝑞, = 1.80 × 10−5 and

Y
the equilibrium concentration of acetic acid is 0.09986M,

EM
what is the concentration of 𝐻 + (𝑎𝑞) and 𝐶𝐻2 𝐶𝑂𝑂 − (𝑎𝑞) ?

D
𝐶𝐻3 𝐶𝑂𝑂𝐻(𝑎𝑞) ⇋ 𝐻 + (𝑎𝑞) + 𝐶𝐻3 𝐶𝑂𝑂 − (𝑎𝑞)

6 A
Answer:

48 C
97 E A
𝐶𝐻3 𝐶𝑂𝑂𝐻 + 𝐻2 𝑂 ⇋ 𝐻3 𝑂+ + 𝐶𝐻3 𝐶𝑂𝑂 −

𝐶𝐻3 𝐶𝑂𝑂𝐻
40 IN 𝐻2 𝑂 𝐻3 𝑂+ 𝐶𝐻3 𝐶𝑂𝑂 −
I 0.09986 M - - -
C −𝑥 - +𝑥 +𝑥
31 L
E 0.09986 − 𝑥 𝑥 𝑥
23 ON

[𝐻3 𝑂+ ][𝐶𝐻3 𝐶𝑂𝑂− ]

𝐾𝑎 =
+9 R

[𝐶𝐻3 𝐶𝑂𝑂𝐻 ]
A

[𝑥 ][𝑥 ]
1.80 × 10−5 =
ST

[0.09986 − 𝑥 ]
𝑥 2 = 1.80 × 10−5 × (0.09986 − 𝑥)
G

𝑥 2 + 1.80 × 10−5 𝑥 − 0.1797 × 10−5 = 0

N

𝐴𝑝𝑝𝑙𝑦 𝑄𝑢𝑑𝑟𝑎𝑡𝑖𝑐 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑡𝑜 𝑠𝑜𝑙𝑣𝑒

SI

𝑥 = 0.00133 = [ 𝐻3 𝑂+ ]
[𝐻 + ] = 1.33 × 10−3 𝑀
RI

[𝐶𝐻3 𝐶𝑂𝑂 − ] = 1.33 × 10−3 𝑀

RISING STAR ACADEMY +92- 331- 4097486 Page 8

CHAPTER CHEMICAL
7 EQUILIBRIUM

Question # 3

Y
EM
Hydrogen sulfide gas decomposes into its elements and
establishes equilibrium at 1400C. One liter of this gas
mixtures contains 0.18 mol of hydrogen sulfide, 0.014 mol of hydrogen

D
gas and 0.035 mol of sulfur. Calculate the equilibrium constant 𝐾𝑒𝑞 for

6 A
this reaction. Are products or reactants favored? How do you know?

48 C
97 E A
2𝐻2 𝑆(𝑔) ⇋ 2𝐻2(𝑔) + 𝑆2(𝑔)

Solution: 40 IN
Given
31 L
Number of moles of 𝐻2(𝑔) = 0.014 mol
23 ON

Number of moles of 𝑆2(𝑔) = 0.035 mol

Number of moles of 𝐻2 𝑆(𝑔) = 0.18 mol
+9 R

𝐾𝑐 = 𝐾𝑛 × 𝑉 −∆𝑛𝑔
A

(𝑛𝐻2 )2 × (𝑛𝑆2 )
𝐾𝑐 = × 𝑉 −(3−2)
ST

(𝑛𝐻2 𝑆 )2

(0.014)2 × (0.035) 1
𝐾𝑐 = ×
G

(0.18)2 𝑉
N

1
𝐾𝑐 = 2.11 × 10−4 × (𝑎𝑠 𝑉𝑜𝑙𝑢𝑚𝑒 𝑖𝑠 1 𝑙𝑖𝑡𝑒𝑟)
SI

1
𝐾𝑐 = 2.11 × 10−4
RI

So, 𝐾𝑒𝑞 for the reaction will be 2.11 × 10−4

This value is too small, so reaction is favored to reactants.

RISING STAR ACADEMY +92- 331- 4097486 Page 9

CHAPTER CHEMICAL
7 EQUILIBRIUM

Question # 4

Y
EM
Consider the reaction:
2𝑁𝑂𝐶𝑙(𝑔) ⇋ 2𝑁𝑂(𝑔) + 𝐶𝑙2(𝑔)

D
3.00 mol NOCl, 1.00 mol NO, and 2.00 mol Cl2 are mixed in a 10.0 L

6 A
flask. After the system has reached equilibrium, the concentrations at

48 C
35C are observed to be:

97 E A
[Cl2] = 1.52 × 10−1 M;
[NO] = 4.00 × 10−3 M;
40 IN
31 L
[NOCl] = 3.96 × 10−1 M;
23 ON

Calculate the value of K for this reaction at 35C.

Solution:
+9 R
A

[𝐶𝑙2 ][𝑁𝑂]2
𝐾=
ST

[𝑁𝑂𝐶𝑙 ]2
(1.52 × 10−1 )(4.0 × 10−3 )2
= 1.55 × 10−5
G

𝐾= −1 2
(3.96 × 10 )
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 10

CHAPTER CHEMICAL
7 EQUILIBRIUM
LESSON EXERCISE # 5

Y
Question # 1

EM
For the synthesis of ammonia the value of K is 6 × 10−2 at
500C. In an experiment, 0.50 mol of Nitrogen, 1.0 × 10−2

D
mol of Hydrogen, and 1.0 × 10−4 mol of Ammonia are mixed at 500C.

6 A
In which direction will the system proceed to reach equilibrium?

48 C
𝑁2(𝑔) + 3𝐻2(𝑔) ⇋ 2𝑁𝐻3(𝑔)

97 E A
Solution:
40 IN
𝑁2(𝑔) + 3𝐻2(𝑔) ⇋ 2𝑁𝐻3(𝑔)
31 L
Given Kc = 6 × 10−2 at 500oC
23 ON

[𝑁𝐻3 ]2
𝐾𝑐 =
[𝑁2 ][𝐻2 ]3
+9 R

Here, [NH3], [H2] and [N2] are in equilibrium concentration.

A
ST

Reaction Quotient
[𝑁𝐻3 ]2 [1.0 × 10−4 ]2 1
G

𝑄𝑐 = 3
= −2 3
= = 0.2 = 2 × 10−1
[𝑁2 ][𝐻2 ] [0.5][1 × 10 ] 5
N

So, 𝐾𝑐 < 𝑄𝑐 , it means the reaction goes toward reactant side or

SI

leftwards.
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 11

CHAPTER CHEMICAL
7 EQUILIBRIUM

Question # 2

Y
EM
Find the value of reaction quotient and determine which
direction of the reaction will shift in order to reach the
equilibrium.

D
6 A
Given:

48 C
𝑁2(𝑔) + 3𝐻2(𝑔) ⇋ 2𝑁𝐻3(𝑔)

97 E A
[N2] = 0.04M ; [H2] = 0.09M ; K = 0.040

Solution:
40 IN
Initial concentration of N2 = 0.04M
31 L
23 ON

Initial concentration of H2 = 0.09

𝑁2(𝑔) 𝐻2(𝑔) 𝑁𝐻3(𝑔)

I 0.04 0.09 0
+9 R

C −𝑥 −𝑥 +𝑥
A

E 0.04 − 𝑥 0.09 − 𝑥 𝑥
ST

From the reaction,

[𝑁𝐻3 ]2
𝐾𝑐 =
G

[𝑁2 ][𝐻2 ]3
N

Putting the values,

SI

𝑥2
0.040 =
(0.04 − 𝑥)(0.09 − 𝑥)3
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 12

CHAPTER CHEMICAL
7 EQUILIBRIUM

‘x’ is very small change, we can neglect in denominator

Y
𝑥2

EM
0.040 =
(0.04)(0.09)3

𝑥 2 = 116 × 10−8

D
6 A
Taking Square root on both sides

48 C
𝑥 = 10.7 × 10−4 = 0.00107

97 E A
[𝑥 ]2
Reaction Quotient =
[0.04−𝑥][0.09−𝑥]3
40 IN
[0.00107]2
Reaction Quotient =
[0.04−0.00107][0.09−0.00107]3
= 0.042
31 L
23 ON

From these results, we can see that

Kc < Qc
+9 R
A

So, 𝐾𝑐 < 𝑄𝑐 , it means the reaction goes toward reactant side or

ST

leftwards.
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 13

CHAPTER CHEMICAL
7 EQUILIBRIUM
LESSON EXERCISE # 6

Y
Question # 1

EM
Given this reaction at equilibrium:
𝐶𝑂(𝑔) + 𝐵𝑟2(𝑔) ⇋ 𝐶𝑂𝐵𝑟2(𝑔)

D
6 A
In which direction, toward reactants or toward products – does the

48 C
reaction shift if the equilibrium is stressed by each change?

97 E A
(a) 𝐵𝑟2 is removed
(b) 𝐶𝑂𝐵𝑟2 is removed
40 IN
Answer:
31 L

(a) If we remove the Br2, then reaction would shift towards the left
23 ON

side or reactant side.

(b) If we remove the Br2, then reaction would shift towards the left
+9 R

side or reactant side.

A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 14

CHAPTER CHEMICAL
7 EQUILIBRIUM
LESSON EXERCISE # 7

Y
Question # 1

EM
Define the Le Chatelier’s principle.
Answer:

D
Le Chatelier’s principle states that when equilibrium is stressed, the

6 A
equilibrium shifts to minimize that stress.

48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 15

CHAPTER CHEMICAL
7 EQUILIBRIUM
Question # 2
What is meant by a stress? State how equilibrium can be

Y
stressed?

EM
Answer:

D
When we change something in a chemical reaction at equilibrium, it

6 A
means we put stress on the equilibrium. When this occurs, the

48 C
reaction will no longer be in equilibrium and the reaction

97 E A
itself will begin changing the concentrations of reactants and products
until the reaction comes to a new position of equilibrium.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 16

CHAPTER CHEMICAL
7 EQUILIBRIUM
Question # 3
Given this equilibrium, predict the direction of shift for each

Y
stress.

EM
𝐻2(𝑔) + 𝐼2(𝑔) ⇄ 2𝐻𝐼(𝑔) + 53 𝑘𝐽

D
a) decreased temperature

6 A
b) increased pressure

48 C
97 E A
c) removal of HI
Answer:
(a)
40 IN
Towards reactant
(b) Towards reactant
31 L

(c) Towards product

23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 17

CHAPTER CHEMICAL
7 EQUILIBRIUM
Question # 4
What is the effect on this equilibrium if pressure is

Y
decreased?

EM
3𝑂2(𝑔) ⇄ 2𝑂3(𝑔)

D
Answer:

6 A
Reaction shift toward reactants.

48 C
97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 18

CHAPTER CHEMICAL
7 EQUILIBRIUM

Question # 5

Y
EM
Given this equilibrium, predict the direction of shift for
each stress.

D
𝐻2(𝑔) + 𝐹2(𝑔) ⇄ 2𝐻𝐹(𝑔) + 546 𝑘𝐽

6 A
a. Increased temperature

48 C
b. Adding of H2

97 E A
c. Decreased pressure
Answer:
40 IN
𝐻2(𝑔) + 𝐹2(𝑔) ⇄ 2𝐻𝐹(𝑔) + 546 𝑘𝐽
31 L
23 ON

It is an exothermic reaction, so ∆𝐻 = −𝑣𝑒

(a) Equilibrium will shift in backwards direction
+9 R

(b) Equilibrium will shift in forward direction to minimize the

A

effect of increased H2
ST

(c) No effect because ∆𝑛𝑔 = 0

G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 19

CHAPTER CHEMICAL
7 EQUILIBRIUM

Question # 6

Y
EM
Given this equilibrium, predict the direction of shift for each
stress.

D
2𝑆𝑂2(𝑔) + 𝑂2(𝑔) ⇄ 2𝑆𝑂3(𝑔) + 196 𝑘𝐽

6 A
a. Removal of SO3

48 C
b. Addition of O2

97 E A
c. Decreased temperature
40 IN
Answer:
31 L

It is an exothermic reaction. So,

23 ON

(a) Equilibrium would shift towards product or rightwards

(b) Equilibrium would shift towards product or rightwards
+9 R

(c) Equilibrium would shift towards product or rightwards

A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 20

CHAPTER CHEMICAL
7 EQUILIBRIUM

Question # 7

Y
EM
Given this equilibrium, predict the direction of shift for each
stress.

D
𝐶𝑂2(𝑔) + 𝐶(𝑠) + 171 𝑘𝐽 ⇄ 2𝐶𝑂(𝑔)

6 A
a. Addition of CO

48 C
b. Increased Pressure

97 E A
c. Addition of a catalyst
Answer:
40 IN
Reaction is Endothermic.
31 L
23 ON

(a) Equilibrium will shift in backward direction

(b) Equilibrium will shift in backward direction
(c) No change in equilibrium
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 21

CHAPTER CHEMICAL
7 EQUILIBRIUM

Question # 8

Y
EM
𝑆𝑂2(𝑔) ⇋ 2𝑁𝑂 is endothermic. What occurs with
increasing temperature?

D
Answer:

6 A
When we increase the temperature, equilibrium will shift in forward

48 C
direction.

97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 22

CHAPTER CHEMICAL
7 EQUILIBRIUM

Question # 9

Y
EM
Predict the effect of decreasing the temperature on this
equilibrium.

D
𝑁2 𝑂4 + 𝑒𝑛𝑒𝑟𝑔𝑦 ⇋ 2𝑁𝑂2

6 A
Answer:

48 C
97 E A
It is an endothermic reaction
By decreasing the temperature, equilibrium will shift backward
40 IN
direction.
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 23

CHAPTER CHEMICAL
7 EQUILIBRIUM

Question # 10

Y
EM
If the temperature is decreased for the reaction:
2𝐶𝑂2 ⇋ 2𝐶𝑂 + 𝑂2 ∆𝐻 = 566 𝑘𝐽

D
a) Will the equilibrium shift left, or right?

6 A
b) Does 𝐾𝑐 become larger or smaller?

48 C
97 E A
Answer:
(a) Decreasing the temperature , will shift the equilibrium towards
40 IN
rightwards.
[𝐶𝑂]2 [𝑂2 ]
(b) 𝐾𝑐 = , so Kc becomes larger.
31 L
[𝐶𝑂2 ]2
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 24

CHAPTER CHEMICAL
7 EQUILIBRIUM
CHAPTER REVIEW QUESTIONS

Y
Question # 1

EM
Indicate whether reactants or products are likely to be
favored. Justify your decision

D
(a) 𝐾𝑒𝑞 = 1

6 A
(b) 𝐾𝑒𝑞 = 5.6 × 10−4

48 C
97 E A
Answer:
(a) When we have 𝐾𝑒𝑞 = 1, then reaction will show no change
40 IN
(b) When we have 𝐾𝑒𝑞 < 1, then reaction has a high yield and is
said to be left shifted or favored to reactants.
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 25

CHAPTER CHEMICAL
7 EQUILIBRIUM
Question # 2
Calculate the number of moles of chlorine produced at

Y
equilibrium in a 10.0L flask when 2.0 mol of phosphorus

EM
pentachloride is heated to 150𝑜 𝐶.
(𝐾𝑐 = 4.10 × 10−6 )

D
6 A
𝑃𝐶𝑙5(𝑔) ⇋ 𝑃𝐶𝑙3(𝑔) + 𝐶𝑙2(𝑔)

48 C
Answer:

97 E A
𝑃𝐶𝑙5(𝑔) ⇋ 𝑃𝐶𝑙3(𝑔) + 𝐶𝑙2(𝑔)
40 IN
𝑃𝐶𝑙5 𝑃𝐶𝑙3 𝐶𝑙2
I 2 0 0
31 L

C −𝑥 𝑥 𝑥
23 ON

E 2−𝑥 𝑥 𝑥
+9 R

[𝑃𝐶𝑙3 ][𝐶𝑙2 ]
A

𝐾𝑐 =
[𝑃𝐶𝑙5 ]
ST

𝑥 𝑥
( )( )
4.10 × 10−6 = 10 10
G

2−𝑥
( )
10
N

𝑥2
SI

−6
4.10 × 10 × 10 =
2−𝑥
RI

−6
𝑥2
41 × 10 =
2−𝑥
‘x’ are small, hence neglected
RISING STAR ACADEMY +92- 331- 4097486 Page 26
CHAPTER CHEMICAL
7 EQUILIBRIUM

𝑥 2 = 41 × 2 × 10−6

Y
𝑥 2 = 82 × 10−6

EM
Taking square root on both sides

D
𝑥 = 9.05 × 10−3 mol

6 A
48 C
Cl2 produced = x = 9.05 × 10−3 mol = 0.009 mol

97 E A
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 27

CHAPTER CHEMICAL
7 EQUILIBRIUM

Question # 3

Y
EM
What is the effect on this equilibrium if pressure is
increased?

D
𝑁2(𝑔) + 3𝐻2(𝑔) ⇋ 2𝑁𝐻3(𝑔)

6 A
Answer:

48 C
97 E A
When pressure is increase, equilibrium shifts towards the side with less
number of gaseous moles, so equilibrium shifts toward the right side.
40 IN
31 L
23 ON
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 28

CHAPTER CHEMICAL
7 EQUILIBRIUM

Question # 4

Y
EM
At 700K, carbon monoxide reacts with water to form Carbon
Dioxide and hydrogen:

D
𝐶𝑂(𝑔) + 𝐻2 𝑂(𝑔) ⇋ 𝐶𝑂2 (𝑔) + 𝐻2(𝑔)

6 A
The equilibrium constant for this reaction is 5.10.

48 C
97 E A
Consider an experiment in which 0.500 mol of carbon monoxide and
0.500 mol of water vapor are mixed together in a 0.500 L flask.
Calculate the concentrations of all species at equilibrium.
40 IN
Answer:
31 L
23 ON

CO H2O CO2 H2
I 0.50 0.50 - -
C −𝑥 −𝑥 𝑥 𝑥
+9 R

E 0.50 − 𝑥 0.50 − 𝑥 𝑥 𝑥
A
ST

[𝐶𝑂2 ][𝐻2 ]
𝐾𝑐 =
[𝐶𝑂][𝐻2 𝑂]
G

𝑥2
N

𝑥×𝑥
5.10 = =
SI

(0.50 − 𝑥)(0.50 − 𝑥) (0.50 − 𝑥)2

RI

𝑥2
5.10 =
(0.50 − 𝑥)2
Taking Square root on both sides

RISING STAR ACADEMY +92- 331- 4097486 Page 29

CHAPTER CHEMICAL
7 EQUILIBRIUM
𝑥2
√ = √5.10
(0.50 − 𝑥)2

Y
EM
𝑥
= 2.26
0.50 − 𝑥

D
𝑥 = 2.26(0.50 − 𝑥 )

6 A
𝑥 = 1.13 − 2.26𝑥

48 C
97 E A
3.36𝑥 = 1.13
1.13
40 IN𝑥= = 0.336𝑀
3.36
Concentration of CO = 0.50 – x = 0.50 – 0.336 = 0.164
31 L
23 ON

Concentration of H2O = 0.50 – x = 0.50 – 0.336 = 0.164

Concentration of CO2 = x = 0.336
+9 R

Concentration of H2 = x = 0.336
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 30

CHAPTER CHEMICAL
7 EQUILIBRIUM

Question # 5

Y
EM
For the synthesis of hydrogen fluoride from hydrogen and
fluorine, 3.000 mol of hydrogen and 6.000 mol of fluorine
are mixed in a 3.000L flask. The equilibrium constant for the reaction at

D
this temperature is 1.15 × 102 . Calculate the equilibrium concentration

6 A
of each component.

48 C
97 E A
Answer:

40 IN 𝐻2 + 𝐹2 → 2𝐻𝐹
3.00
Initial Concentration of H2 = = 𝑚𝑜𝑙/𝐿
3.00
31 L

6.00
23 ON

Initial Concentration of F2 = = 2 𝑚𝑜𝑙/𝐿

3.00

[𝐻𝐹] 1
𝑄= = = 0.5
+9 R

[𝐻2 ][𝐹2 ] 1 × 2
A

As, Q < Kc, so equilibrium shifts to right side

ST

H2 F2 HF
G

I 1.00 2.00 0
C −𝑥 −𝑥 2𝑥
N

E 1−𝑥 2−𝑥 2𝑥
SI
RI

[𝐻𝐹]2 (2𝑥)2
𝐾𝑐 = =
[𝐻2 ][𝐹2 ] (1 − 𝑥)(2 − 𝑥)

RISING STAR ACADEMY +92- 331- 4097486 Page 31

CHAPTER CHEMICAL
7 EQUILIBRIUM
(2𝑥)2
1.15 × 102 =
(1−𝑥)(2−𝑥)

Y
4𝑥 2
115 =

EM
2 − 3𝑥 + 𝑥 2
4𝑥 2 = 115(2 − 3𝑥 + 𝑥 2 )

D
115𝑥 2 − 4𝑥 2 − 345𝑥 + 230 = 0

6 A
48 C
111𝑥 2 − 345𝑥 + 230 = 0

97 E A
𝑥 = 0.9683
Concentration of each component
40 IN
Concentration of H2 = 1 – x = 1 – 0.9683 = 0.0317
31 L
23 ON

Concentration of F2 = 2 – x = 1 – 0.9683 = 1.0317

Concentration of HF = 2x = 2(0.9683) = 1.9366
+9 R
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 32

CHAPTER CHEMICAL
7 EQUILIBRIUM
Choose the Correct Answer:
1–B

Y
EM
2–A

D
3–A

6 A
4–D

48 C
97 E A
5–A
40 IN
6–D
31 L

7–A
23 ON

8–B
+9 R

9–B
A
ST
G
N
SI
RI

RISING STAR ACADEMY +92- 331- 4097486 Page 33