NCERT Solutions for Class 12 Chemistry Chapter 3 Chemical Kinetics

NCERT
SOLUTIONS
CLASS - 12th

aglase .co

Class : 12th
Subject : Chemistry
Chapter : 4
Chapter Name : Chemical Kinetics

Q4.1 For the reaction R → P, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using
units of time both in minutes and seconds.

Δ[R]
Average rate of reaction = − Δt
[ R ]2 − [ R ]
= − t2 − t1
0.02 − 0.03
= − 25
Mmin − 1
− 0.01
= − 25
M min − 1

= 4 × 10 − 4Mmin − 1
4 × 10 − 4
= 60
Ms − 1

= 6.67 × 10 − 6Ms − 1

Page : 100 , Block Name : Intext Questions

Q4.2 In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L − 1 to 0.4 mol L − 1 in 10 minutes. Calculate the rate during this
interval?

1 Δ[A]
Answer. Average rate = − 2 Δt
1 [A] 2 − [A] 1
= −
2 t2 − t1
1 0.4 − 0.5
= −
2 10
1 − 0.1
= −
2 10
= 0.005molL − 1min − 1
= 5 × 10 − 3Mmin − 1

Page : 100 , Block Name : Intext Questions

Q4.3 For a reaction, A + B → Product; the rate law is given by,r = k[A] 1 / 2[B] 2. What is the order of the reaction?

1
The order of the reaction = 2 + 2
1
= 22

= 2.5

Page : 105 , Block Name : Intext Questions

Q4.4 The conversion Of molecules X to Y follows second order kinetics. If concentration Of X is increased to three times how will it affect the rate of
formation of Y?

Answer. The reaction X → Y follows second order kinetics. Therefore. the rate equation for this reaction will be:
Rate = k[X] 2(1)
then equation (i) can be written as:
Let [X] = a mol L − 1
2
Rate 1 = k ⋅ (a)

= ka 2
If the concentration Of X is increased to three times, then [X] = 3amolL − 1

( )
NOW, the rate equation Will be: Rate = k(3a) 2 = 9 ka 2 Hence, the rate of formation will increase by 9 times.

Page : 105 , Block Name : Intext Questions

Q4.5 A rst order reaction has a rate constant 1.1510 − 3s − 1. How long will 5 g of this reactant take to reduce to 3 g?

From the question, we can write down the following information:
Initial amount = 5g
Final concentration = 39
Rate constant = 1.1510 − 3s − 1
We know that for a 1 s ∗ order reaction,
2.303 [R] 0
t= log
k [R]
2.303 5
= log
1.15 × 10 − 3 3
2.303
= × 0.2219
1.15 × 10 − 3
= 444.38s
= 444s( approx )

Page : 113 , Block Name : Intext Questions

Q4.6 Time required to decompose SO 2Cl 2 to half Of its initial amount is 60 minutes. If the decomposition is a rst order reaction, calculate the rate
constant of the reaction.

We know that for a 1 st order reaction,
0.693
t1 / 2 = k

It is given that t 1 / 2 = 60min
0.693
∴k = t
1/2

0.693
= 60

= 0.01155min − 1
= 1.155min − 1
Ork k = 1.925 × 10 − 4s − 1

Page : 113 , Block Name : Intext Questions

Q4.7 What will be the effect of temperature on rate constant?

Answer. The rate Constant of a reaction is nearly doubled with a 10 ∘ rise in temperature. However, the exact dependence of the rate of a chemical
reaction on temperature is guen by Arrhenius equation,
k = Ae − Ea / RT
Where,
A is the Arrhenius factor or the frequency factor
T is the temperature
R is the gas constant
E a is the activation energy

Page : 118 , Block Name : Intext Questions

Q4.8 The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E a

It is given that T 1 = 298K
∴ T 2 = (298 + 10)K
= 308K
We also know that the rate Of the reaction doubles when temperature is increased by
10 ∘ .
Therefore, let us take the value of k 1 = k and that of k 2 = 2k

Also, R = 8.3143K − 1mol − 1

Now, substituting these values in the equation:

k2 Ea
log k = 2.303R
1 [ ] T2 − T1
T 1T 2

we get:

[ ]
2k Ea 10
log k = 2.303 × 8.314 298 × 308
2.303 × 8.314 × 298 × 308 × log 2
⇒ En = 10

= 52897.78Jmol − 1
= 52.9k3mol − 1

Page : 118 , Block Name : Intext Questions

Q4.9 The activation energy for the reaction 2HI ( g ) → H 2 + I 2 ( g ) is 209.5k3mol − 1 at 581KCalculate the fraction of molecules of reactants having
energy equal to or greater than activation energy?

In the given case:
E 8 = 209.5kJmol − 1 = 209500Jmol − 1
T = 581K
R = 8.314JK − 1mol − 1
Now, the fraction of molecules of reactants having energy equal to or greater than
activation energy is given as:
x = e − Ea / RT
⇒ lnx = − E a / RT
Ea
⇒ logx = − 2.303RT
209500Jmol − 1
⇒ logx = = 18.8323
2.303 × 8.314JK − 1mol − 1 × 581
Now, x = Anti log(18.8323)
¯
= Anti log19.1677
= 1.471 × 10 − 19

Page : 118 , Block Name : Intext Questions

Q4.1 From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 3NO(g) → N 2O(g) Rate = k[NO] 2

[ ] [I − ]
(ii) H 2O 2(aq) + 3I − (aq) + 2H + → 2H 2O(l) + I 3 Rate = k H 2O 2

(iii) CH 3CHO(g) → CH 4(g) + CO(g) Rate = k [CH 3CHO ] 3 / 2

(iv) C 2H 5Cl(g) → C 2H 4(g) + HCl(g) Rate = k [C 2H 5Cl ]

(i) Given rate = k[NO] 2
Therefore, order of the reaction = 2
Rate
Dimension of k=
[ NO ] 2
molL − 1s − 1
=
(molL )−1 2

molL − 1s − 1
=
mol 2L − 2
= Lmol − 1s − 1
(ii) Given rate = k H 2O 2 [ ] [I − ]
Therefore, order of the reaction = 2
Rate
Dimension of k =
[ H 2O 2 ] [ I − ]

molL − 1s − 1
=
(molL )(molL )
−1 −1

= Lmol − 1s − 1
(iii) Given rate = k CH 3CHO 3 / 2 [ ]
3
Therefore, order of reaction = 2
Rate
Dimension of k = 3

[ CH3CHO ] 2

mol L − 1s − 1
= 3

( mol L ) −1 2

mol L − 1s − 1
= 3 3
mol 2 L 2
1 1
= L 2 mol − 2 s − 1
(iv) Given rate = k C 2H 5Cl [ ]
Therefore, order of the reaction = 1
Rate
Dimension of k =
[ C 2H , Cl ]
mol L − 1s − 1
=
mol L − 1

= s −1

Page : 119 , Block Name : Exercise

Q4.2 For the reaction :
2A + B → A 2B
The rate = k[A][B] 2 with k = 2.0 × 10 − 6mol − 2L 2s − 1. Calculate the initial rate of reaction when [A] = 0.1molL − 1, [B] = 0.2molL − 1
After [A] is reduce to 0.06 molL − 1.

The initial rate of the reaction is
Rate = k[A][B] 2

(
= 2.0 × 10 − 6mol − 2L 2s − 1 )(0.1molL )(0.2molL )
−1 −1 2

= 8.0 × 10 − 9mol − 2L 2s − 1
When [A] is reduced from 0.1 mol L − 1to 0.06 mol L − 1, the concentration of A reacted = (0.1 − 0.06)molL − 1 = 0.04molL − 1
1
Therefore, concentration of B reacted = 2 × 0.04molL − 1 = 0.02 mol L − 1
Then concentration of B available [B]= (0.2 - 0.02) mol L − 1= 0.18 mol L − 1
After [A] is reduced to 0.06 mol L − 1, the rate of the reaction is given by,
Rate = k[A][B] 2

(
= 2.0 × 10 − 6mol − 2L 2s − 1 )(0.06molL )(0.18molL )
−1 −1 2

= 3.89molL − 1s − 1

Page : 119 , Block Name : Exercise

Q4.3 The decomposition of dimethvl ether leads to the formation of CH., and CO and the reaction rate is given by

[
= k CH 3OCH 3 3 / 2 ]
The rate of reaction is followed bv increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of
dimethyl ether, i.e.,

( )
3
Rate = k p ch , 0CH 2
3

If the pressure is measured in bar and time in minutes, then What are the units Of rate and rate constants?

Answer. If pressure is measured in bar and time in minutes, then
Unit of rate = bar min − 1

(
Rate = k p CM , 0cH 3 / 2 )
bar min − 1
Therefore, unit of rate constants =
bar 3

= bar − 1 / 2 min − 1

Page : 119 , Block Name : Exercise

Q4.5 Mention the factors that affect the rate of a chemical reaction.

Answer. The factors that affect the rate Of reaction are as follows.
(i) Concentration Of reactants (pressure in case Of gases)
(ii) Temperature
(iii) Presence Of a catalyst

Page : 119 , Block Name : Exercise

Q4.6 A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
(i) Doubled (ii) reduced to half?

Let the concentration of the reactant be [A] = a
Rate of reaction, R = k[A] 2
= ka 2
(i)lf the concentration Of the reactant is doubled, i.e. [A] = 2a, then the rate of the
reaction would be
R ′ = k(2a) 2
= 4ka 2
= 4R
Therefore, the rate of the reaction would increase by 4 times.
1
(ii) If the concentration of the reactant is reduced to half, i.e. [A] = 2 a, then the rate of the reaction would be

R ′′ = k
( )1 2
2
a

1
= ka
4
1
= R
4
Therefore, the rate of the reaction would be reduced to 4 th

Page : 120 , Block Name : Exercise

Q4.7 What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented
quantitatively?

Answer. The rate constant is nearly doubled with a rise in temperature by 10 ∘ for a chemical reaction. The temperature effect on the rate constant
can be represented quantitatively by Arrhenius equation,
k = Ae − Ea / RT
Where, k is the rate constant
A is the Arrhenius factor or the frequency factor,
R is the gas constant,
T is the temperature, and
E a is the energy of activation for the reaction

Page : 120 , Block Name : Exercise

Q4.8 In a pseudo rst order hydrolysis Of ester in water, the following results were obtained:

(i) Calculate the average rate Of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo rst Order rate constant for the hydrolysis Of ester.

Answer. (i) Average rate Of reaction between the time interval, 30 to 60 seconds,
d [ Ester ]
= dt
0.31 − 0.17
= 60 − 30
0.14
= 30

= 4.67 × 10 − 3molL − 1s − 1

(ii) For a pseudo rst order reaction,
2.303 [ R ]0
k= t
log [ R ]
2.303 0.55
t = 30s, k 1 = 30 log 0.31
= 1.911 × 10 − 2s − 1
2.303 0.55
k2 = log
60 0.17
= 2.075 × 10 − 2s − 1
k1 + k2 + k3
Then, average rate constant, = 3

( 1.91 × 10 − 2 )+ ( 1.957 × 10 − 2 ) + ( 2.075 × 10 )
−2

= 3
−2 −1
= 1.98 × 10 s

Page : 120 , Block Name : Exercise

Q4.9 A reaction is rst order in A and second order in B,
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) HOW is the rate affected When the concentrations Of both A and a are doubled?

(i) The differential rate equation will be
d[R]
− dt = k[A][B] 2
(ii) If the concentration of B is increased three times, then
d[R]
− = k[A][3B] 2
dt
= 9 ⋅ k[A][B] 2
Therefore, the rate of reaction will increase 9 times.
d[R]
(iii) When the concentrations of both A and B are doubled, − dt
= k[A][B] 2 = k[2A][B] 2 Therefore, the rate of reaction will increase 8 times.

Page : 120 , Block Name : Exercise

Q4.10 In a reaction between A and B, the initial rate of reaction (r0 ) was measured for different initial concentrations of A and B as given below:

What is the order of the reaction with respect to A and B?

Answer. Let the order of the reaction With respect to A be x and with respect to 3 be y.
Therefore.
r 0 = k[A] x[B] y
5.07 × 10 − 5 = k[0.20] x[0.30] y.........(1)
5.07 × 10 − 5 = k[0.20] x[0.10] y..........(2)
1.43 × 10 − 4 = k[0.40] x[0.05] y...........(3)
Dividing equation (1) by (2), we obtain
5.07 × 10 − 5 k [ 0.20 ] x [ 0.30 ] y
=
5.07 × 10 − 5 k [ 0.20 ] x [ 0.10 ] y
[ 0.30 ] y
⇒1=
[ 0.10 ] y

( ) ( )
0.30 0 0.30 y
⇒ 0.10 = 0.10

⇒y=0
Dividing equation (3) by (2), we obtain
1.43 × 10 − 4 k [ 0.40 ] x [ 0.05 ] y
=
5.07 × 10 − 5 k [ 0.20 ] x [ 0.30 ] y

⇒
1.43 × 10 − 4
5.07 × 10 − 5
=
[ 0.40 ] x
[ 0.20 ] x [ since y = 0
[0.05] y = [0.30] y = 1 ]
⇒ 2.821 = 2 x

⇒ log2.821 = xlog2 (Taking log on both sides )
log 2.821
⇒x= log 2
= 1.496
= 1.5 (approximately)
Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.

Page : 120 , Block Name : Exercise

Q4.11 The following results have been obtained during the kinetic studies of the reaction:
2A + B → C + D

Determine the rate law and the rate constant for the reaction.

Answer. Let the order of the reaction with respect to A be x and with respect to be y,
Therefore, rate of the reaction is given by,
Rate = = k[A] x[B] y
6.0 × 10 − 3 = k[0.1] x[0.1] y...... (i)
7.2 × 10 − 2 = k[0.3] x[0.2] y......(ii)
2.88 × 10 − 1 = k[0.3] x[0.4] y......(iii)
2.40 × 10 − 2 = k[0.4] x[0.1] y......(iv)
Dividing equation (iv) by (i), we obtain
2.40 × 10 − 2 k [ 0.4 ] x [ 0.1 ] v
=
6.0 × 10 − 3 k [ 0.1 ] x [ 0.1 ] v
[ 0.4 ] x
⇒4=
[ 0.1 ] x

⇒4=
( ) 0.4 x
0.1

⇒ (4) 1 = 4 x
⇒x=1
Dividing equation (iii) by (ii), we obtain
2.88 × 10 − 1 k [ 0.3 ] x [ 0.4 ] y
=
7.2 × 10 − 2 k [ 0.3 ] x [ 0.2 ] y

⇒4=
( ) 0.4 y
0.2

⇒ 4 = 2y
⇒ 22 = 2y
⇒y=2
Therefore, the rate law is
Rate = k[A][B] 2
Rate
k=
[A][B] 2
⇒ From experiment I, we obtain
6.0 × 10 − 3molL − 1min − 1
k=
( 0.1molL ) ( 0.1molL )
−1 −1 2

= 6.0L 2mol − 2min − 1
From experiment II, we obtain
7.2 × 10 − 1molL − 1min − 1
k=
( 0.3molL ) ( 0.2molL )
−1 −1 2

2.88 × 10 − 1molL − 1min − 1
k=
( 0.3molL ) ( 0.4molL )
−1 −1 2

= 6.0L 2mol − 2min − 1
From experiment IV, we obtain
2.40 × 10 − 2molL − 1min − 1
k=
( 0.4molL ) ( 0.1molL )
−1 −1 2

= 6.0L 2mol − 2min − 1
Therefore, rate constant, k = 6.0L 2mol 2min − 1

Page : 120 , Block Name : Exercise

Q4.12 The reaction between A and a is rst order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

Answer. The given reaction is of the rst order with respect to A and of zero order with respect to B
Therefore, the rate of the reaction is given by,
Rate = k[A] 1[B] 0
⇒ Rate = k[A]
From experiment I, we obtain

2.0 × 10 − 2molL − 1min − 1 = k 0.1molL − 1 ( )
⇒ k = 0.2min − 1
From experiment II, we obtain
4.0 × 10 − 2molL − 1min − 1 = 0.2min − 1[A]
⇒ [A] = 0.2molL − 1
From experiment III, we obtain
Rate = 0.2min − 1 × 0.4molL − 1
= 0.08molL − 1min − 1 = 0.2min − 1[A]
⇒ [A] = 0.1molL − 1

Page : 120 , Block Name : Exercise

Q4.13 Calculate the half-life of a rst order reaction from their rate constants given below:
(i)200s − 1(ii)2min − 1(iii)4years − 1

0.693
Answer. (i) Half life, t 1 / 2 = k
0.693
=
200s − 1
= 3.47s (approximately)
0.693
(ii) Half life, t 1 / 2 = k
0.693
=
2min − 1

= 0.35min (approximately)
0.693
(iii) Half life, t 1 / 2 = k
0.693
=
4 years − 1
= 0.173 years (approximately)

Page : 121 , Block Name : Exercise

Q4.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% Of the 14C found in a living
tree. Estimate the age Of the sample.

0.693 0.693
k= t Here, 1 / 2 = 5730 years − 1 It is known that,
1/2

2.303 [ R ]0 2.303 100
t= k
log [ R ] = 0.303 log 80 = 1845 years (approximately) Hence, the age of the sample is 1845 years.
5730

Page : 121 , Block Name : Exercise

Q4.15 The experimental data for decomposition of N 2O 5,

[2N2O5 ⟶ 4NO2 + O2 ]
in gas phase at 318K are given below:

(i) Plot N 2O 5 against t.
(ii) Find the half-life period for the reaction.

[
(iii) Draw a graph between log N 2O 5 and t. ]
(iv) What is the rate law?
(v) Calculate the rate constant.
(Vi) Calculate the half-life genod from k and compare it with (ii).

1.630 × 10 2
(ii) Time corresponding to the concentration, 2
molL − 1 = 81.5molL − 1,
life. From the graph, the half life is obtained as 1450s .

(iv) The given reaction is of the first order as the plot, log N 2O 5 [ ] v / st is a straight line.
Therefore, the rate law of the reaction is

[
Rate = k N 2O 5 ]
(v) From the plot, log N 2O 5[ ] v / st, we obtain
− 2.46 − ( − 1.79)
Slope =
3200 − 0
− 0.67
=
3200
Again, slope of the line of the plot log N 2O 5[ ]v / st is given by
k
− 2.303

Therefore, we obtain,
k 0.67
− 2.303 = − 3200

⇒ k = 4.82 × 10 − 4s − 1
(vi) Half-life is given by,
0.639
t1 / 2 =
k
0.693
= s
4.82 × 10 − 4
= 1.438 × 10 3s
= 1438s
This value, 1438 s, is very close to the value that was obtained from the graph.

Page : 121 , Block Name : Exercise

Q4.16 The rate constant for a rst order reaction is 60 s − 1. How much time will it take to reduce the initial concentration of the reactant to its 1 / 16 th
value?

It is known that
2.303 [R] 0
t= log
k [R]
2.303 1
= log
60s − 1 1/6
2.303
= log16
60s − 1
= 4.6 × 10 2s (approximately)
−

Hence, the required time is 4.6 × 10 − 2s

Page : 121 , Block Name : Exercise

Q4.17 During nuclear explosion, one of the products is 90 Sr with half-life of 28.1 years. If 1µg of 90 Sr was absorbed in the bones of a newly born baby
instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

0.693 0.693
k= t = 28.1 y − 1
1/2

Here,
It is known that,
2.303 [ R ]0
t= k
log [ R ]
2.303 1
⇒ 10 = 0.303 log [ R ]
28.1
2.303
⇒ 10 = ( − log[R])
0.693
10 × 0.693
⇒
2.303 × 28.1
⇒ [R] = antilog ( − 0.1071)
¯
= antilog(1.8929)
= 0.7814μg
Therefore, 0.7814μg of 90Sr will remain after 10 years.
Again,
2.303 [ R ]0
t= k
log [ R ]
2.303 1
⇒ 60 = 0.303 log [ R ]
28.1
60 × 0.693 ¯
⇒ log[R] = − 2.303 × 28.1 ⇒ [R] = antilog ( − 0.6425) = antilog (1.3575) = 0.2278μg Therefore, 0.2278mug of 90Sr will remain after 60 years.

Page : 121 , Block Name : Exercise

Q4.18 For a rst order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

2.303 100 2.303 2.303
For a first order reaction, the time required for 99% completion is t1 = k
log 100 − 99 = k
log100 = 2 × k
2.303 100 2.303 2.303
For a first order reaction, the time required for 90% completion is t 2 =
k
log 100 − 90 = k log10 = k Therefore, t 1 = 2t 2
Hence, the time required for 99% completion Of a rst order reaction is twice the time required for the completion Of 90% Of the reaction.

Page : 121 , Block Name : Exercise

Q4.19 A rst order reaction takes 40 min for decomposition. Calculate t 1 / 2

Answer. For rst order reaction
2.303 [R] 0
t= log
k [R]
2.303 100
k= log
40min 100 − 30
2.303 10
= log
40min 7
= 8.918 × 10 − 3min − 1

0.693 0.693
Therefore, t 1 / 2 of the decomposition reaction is t 1 / 2 = k
= min = 77.7min (approximately)
8.918 × 10 − 3

Page : 121 , Block Name : Exercise

Q4.20 For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

Calculate the rate constant.

Page : 121 , Block Name : Exercise

Q4.21 The following data were obtained during the rst order thermal decomposition of SO 2Cl 2 at a constant volume.
sO 2Cl 2(g) ⟶ SO 2(g) + Cl 2(g)

Calculate the rate Of the reaction when total pressure is 0.65 atm.

Page : 121 , Block Name : Exercise

Q4.22 The rate constant for the decomposition Of N 2O 5, at various temperatures is given below:

Draw a graph between in K and 1/ T and calculate the values of A and E s.
Predict the rate constant at 30 ∘ and 50 ∘ C.

Answer. For the given data, we obtain

Slope of the line,
y2 − y1
x2 − x1
= − 12.301K

According to Arrhenius equation,
Ea
Slope = − R

⇒ E a = − Slope R

(
= − ( − 12.301K) × 8.314JK − 1mol − 1 )
= 102.27kJmol − 1
Again
Ea
lnk = lnA − RT
Eo
lnA = lnk + RT
When T = 273K
In k = − 7.147
102.27 × 10 3
Then, lnA = − 7.147 +
8.314 × 273
= 37.911
Therefore, A = 2.91 × 10 6
When T = 30 + 273K = 303K
1
= 0.0033K = 3.3 × 10 − 3K
T
1
at T = 3.3 × 10 − 3K

lnk = − 2.8
Therefore, k = 6.08 × 10 − 2s − 1
Again, when T = 50 + 273K = 323K
1
T
= 0.0031K = 3.1 × 10 − 3K
1
Then, at T = 3.1 × 10 − 3K

In k = − 0.5
Therefore, k = 0.607s − 1

Page : 122 , Block Name : Exercise

Q4.23 The rate constant for the decomposition of hydrocarbons is 2.418 × 10 − 5s − 1 at 546K. If the energy of activation is 179.9 kJ/mol, what will be
the value of pre-exponential factor.

k = 2.418 × 10 − 5s − 1
T = 546K
E s = 179.9kJmol − 1 = 179.9 × 10 3Jmol − 1
According to the Arrhenius equation,
k = Ae − E / RT
Ee
⇒ lnk = lnA − RT
Ea
⇒ logk = logA − 2.303RT
Ea
⇒ logA = logk + 2.303RT
179.9 × 10 3Jmol − 1
(
= log 2.418 × 10 − 4s − 1 +)
2.303 × 8.314Jk − 1mol − 1 × 546K
= (0.3835 − 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 × 10 12s − 1 (approximately)

Page : 122 , Block Name : Exercise

Q4.24
Consider a certain reaction A → products with k = 2.0 × 10 − 2s − 1 . Calculate the concentration of A remaining after 100s if the initial concentration of A is 1.0molL

k = 2.0 × 10 − 2s − 1
T = 100s
[A] 0 = 1.0moL − 1

since the unit of k is s − 1, the given reaction is a first order reaction.
2.303 [ A ]0
k= t
log [ A ]
2.303 1.0
⇒ 2.0 × 10 − 2s − 1 = 100s log [ A ]
2.303
⇒ 2.0 × 10 − 2s − 1 = 100s ( − log[A])
2.0 × 10 − 2 × 100
⇒ − log[A] = 2.303

⇒ [A] = anti log −
( 2.0 × 10 − 2 × 100
2.303 )
= 0.135 mol L − 1 (approximately)
Hence, the remaining concentration of A is 0.135 mol L − 1

Page : 122 , Block Name : Exercise

Q4.25 Sucrose decomposes in acid solution into glucose and fructose according to the rst order rate law, with t 1 / 2 = 3.00 hours. What fraction of
sample of sucrose remains after 8 hours?

For a first order reaction,
2.303 [ R ]0
k= t
log [ R ]

It is given that, t 1 / 2 = 3.00 hours
0.693
k= t
1/2

Therefore
0.693
= 3
h −1

= 0.231h − 1
2.303 [R]
Then 0.231h − 1 = 8h log [ R ] k
[ R ]0 0.231h − 1 × 8h
⇒ log [ R ] = 2.303
[ R ]0
⇒ [ R ] = antilog(0.8024)
[R] 0
⇒ = 6.3445
[R]
[R]
⇒ = 0.1576( approx )
[R] 0
= 0.158
Hence, the fraction of sample of sucrose that remains after 8 hours is O. 158.

Page : 122 , Block Name : Exercise

( )
Q4.26 The decomposition of hydrocarbon follows the equation k = 4.5 × 10 11s − 1 e − 28000k / T Calculate E a

The given equation is

( )
k = 4.5 × 10 11s − 1 e − 28000k / T(i)
Arrhenius equation is given by,
k = Ae − E , kτ(ii)
From equation (i) and (ii), we obtain
Ea 28000K
RT
= T
⇒ E 0 = R × 28000K

= 8.314JK − 1mol − 1 × 28000K
= 232792Jmol − 1
= 232.792kJmol − 1

Page : 122 , Block Name : Exercise

Q4.27
The rate constant for the first order decomposition of H 2O 2 is given by the following equation: logk = 14.34 − 1.25 × 10 4K / T Calculate E a for this reaction and at w

Arrhenius equation is given by,
k = Ae − E / kT
Ee
⇒ lnk = lnA − RT
Ea
⇒ lnk = logA − RT
Ea
⇒ logk = logA − 2.303RT ........(i)
The given equation is
.......... (ii)
logk = 14.34 − 1.25 × 10 4K / T
From equation (i) and (ii), we obtain
Ea 1.25 × 10 4K
2.303RT = T

⇒ E y = 1.25 × 10 4K × 2.303 × R
= 1.25 × 10 4K × 2.303 × 8.314JK − 1mol − 1
= 239339.3Jmol − 1 (approximately)
= 239.34kJmol − 1
Also, when t 1 / 2 = 256 minutes,
0.693
k= t
1/2

0.693
= 256

= 2.707 × 10 − 3min − 1
= 4.51 × 10 − 5s − 1
It is also g ven that, logk = 14.34 − 1.25 × 10 4K / T
1.25 × 10 4K
(
⇒ log 4.51 × 10 − 5 = 14.34 − ) T
1.25 × 10 4K
⇒ T = 18.686
1.25 × 10 4K
⇒T= 18.686

= 668.95K
= 669K (approximately)

Page : 122 , Block Name : Exercise

Q4.28 The decomposition of A into product has value of k as 4.5 × 10 3s − 1 at 10 ∘ C and energy of activation 60kmol − 1 At what temperature would k be
1.5 × 10 4s − 1 ?

From Arrhenius equation, we obta

k2
log k = 2.303R
1
Ea

( ) T2 − T1
T 1T 2

Also, k 1 = 4.5 × 10 3s − 1
T 1 = 273 + 10 = 283K

k 2 = 1.5 × 10 4s − 1

E a = 60kJmol − 1 = 6.0 × 10 4Jmol − 1
Then,

log
1.5 × 10 4
4.5 × 10 3
=
6.0 × 10 4Jmol − 1
2.303 × 8.314JK − 1mol − 1 ( ) T 2 − 283
283T 2

⇒ 0.5229 = 3133.627
( ) T 2 − 283
283T 2

0.5229 × 283T 2
⇒ 3133.627
= T 2 − 283
⇒ 0.0472T 2 = T 2 − 283
⇒ 0.9528T 2 = 283
⇒ T 2 = 297.019K (approximately)
= 297k
= 24 ∘ C
Hence, k would be 1.5 × 10 4s − 1 at 24 ∘ C .

Page : 122 , Block Name : Exercise

Q4.29
The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K . If the value of A is 4 × 10 10s − 1 . Ca

For a first order reaction,
2.303 a
t= k
log a − x
2.303 100
298K, t= k
log 90
0.1054
= k
2.303 100
At 308K, t′ = log 75
k′
2.2877
=
k′

According to the question,
t = t′
0.1054 0.2877
⇒ k
=
k′
k′
⇒ k = 2.7296

k′
log k = 2.303R
Ea

( ) T′ − T
TT ′

( )
Ea 308 − 298
log(2.7296) = 2.303 × 8.314 298 × 308

2.303 × 8.314 × 298 × 308 × log(2.7296)
Eo =
308 − 298
= 76640.096Jmol − 1
= 76.64kJmol − 1
To calculate k at 318K
It is given that, A = 4 × 10 10s − 1, T = 318K
Again, from Arrhenius equation, we obtain
Ea
logk = logA − 2.303RT
76.64 × 10 3
(
= log 4 × 10 10 − ) 2.303 × 8.314 × 318
= (0.6021 + 10) − 12.5876
= − 1.9855
Therefore, k = Antilog ( − 1.9855)
= 1.034 × 10 − 2s − 1

Page : 122 , Block Name : Exercise

Q4.30 The rate Of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy Of activation Of the reaction
assuming that it does not change With temperature.

From Arrhenius equation, we obtain

k2 Ea
log k = 2.303R
1 ( )
T2 − T1
T 1T 2

It is given that, k 2 = 4k 1
T 1 = 293K
T 2 = 313K

( )
4k 1 Ea 313 − 293
Therefore, log k = 2.303 × 8.314 293 × 313
2

20 × E o
⇒ 0.6021 = 2.303 × 8.314 × 293 × 313
0.6021 × 2.303 × 8.314 × 293 × 313
⇒ Ea =
20
= 52863.33Jmol − 1
= 52.86kJmol − 1
Hence, the required energy of activation is 52.86kJmol − 1

Page : 122 , Block Name : Exercise