Chapter 3 Chemical Kinetics Exercises NCERT Solutions Class 12 Chemistry PDF Download
Page No. 85
3.1. From the rate expression for the following reaction, determine their order of reaction and the dimensions of the rate constants.
(i) 3 NO(g) → N_{2}O(g) Rate = k[NO]^{2}
(ii) H_{2}O_{2 }(aq) + 3 I^{− }(aq) + 2 H^{+} → 2 H_{2}O (l) + I3 Rate = k[H_{2}O_{2}][I^{−}]
(iii) CH_{3}CHO(g) → CH_{4}(g) + CO(g) Rate = k [CH_{3}CHO]^{3/2}
(iv) C_{2}H_{5}Cl(g) → C_{2}H_{4}(g) + HCl(g) Rate = k [C_{2}H_{5}Cl]
Solution
(i) Given rate = k [NO]^{2}
Therefore, order of the reaction = 2
= L mol^{1}s^{1}
(ii) Given rate = k [H_{2}O_{2}] [I^{−}]
Therefore, order of the reaction = 2
= L mol^{1}s^{1}
(iii) Given rate = k [CH_{3}CHO]^{3/2}
Therefore, order of reaction = 3/2
= L^{1/2}mol^{1/2}s^{1}
(iv) Given rate = k [C_{2}H_{5}Cl]
Therefore, order of the reaction = 1
= s^{1}
3.2. For the reaction: 2A + B ⟶ A2B the rate = k[A][B]^{2} with k = 2.0 × 10^{−6} mol^{−2} L^{2} s^{−1}. Calculate the initial rate of the reaction when [A] = 0.1 mol L^{−1}, [B] = 0.2 mol L^{−1}. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^{−1}.
Solution
The initial rate of the reaction is
Rate = k [A][B]^{2}
= (2.0 × 10^{−6} mol^{−2} L^{2} s^{−1}) (0.1 mol L^{−1}) (0.2 mol L^{−1})^{2}
= 8.0 × 10^{−9} mol^{−2} L^{2} s^{−1}
When [A] is reduced from 0.1 mol L^{−1} to 0.06 mol^{−1}, the concentration of A reacted
= (0.1 − 0.06) mol L^{−1}
= 0.04 mol L^{−1}
Therefore, concentration of B reacted = ½ × 0.04 mol L^{}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^{−6} mol^{−2} L^{2} s^{−1}) (0.06 mol L^{−1}) (0.18 mol L^{−1})^{2}
= 3.89 × 10^{−9} mol L^{−1} s^{−1}
3.3. The decomposition of NH_{3} on platinum surface is zero order reaction. What are the rates of production of N_{2} and H_{2} if k = 2.5 × 10^{−4} mol^{−1} L s^{−1}?
Solution
The decomposition of NH_{3} on platinum surface is represented by the following equation.
Therefore
However, it is given that the reaction is of zero order.
Therefore,
= 2.5 × 10^{4} mol L^{1} s^{1}
Therefore, the rate of production of N_{2} is
And, the rate of production of H_{2} is
= 7.5 × 10^{−4} mol L^{−1} s^{−1}
3.4. The decomposition of dimethyl ether leads to the formation of CH_{4}, H_{2} and CO and the reaction rate is given by
Rate = k [CH_{3}OCH_{3}]^{3/2}.
The rate of reaction is followed by 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., Rate = k(pCH_{3}OCH_{3})^{3/2}
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
Solution
If pressure is measured in bar and time in minutes, then
Unit of rate = bar min^{−1}
3.5. Mention the factors that affect the rate of a chemical reaction.
Solution
The factors that affect the rate of a reaction are as follows.
(i) Concentration of reactants (pressure in case of gases)
(ii) Temperature
(iii) Presence of a catalyst
3.6. 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?
Solution
Let the concentration of the reactant be [A] = a
Rate of reaction, R = k [A]^{2}
= ka^{2}
(i) If 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}
= 4 R
Therefore, the rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half, i.e. [A] = 1/2 a, then the rate of the reaction would be
Therefore, the rate of the reaction would be reduced to 1/4th.
3.7. What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
Solution
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
3.8. In a pseudo first order hydrolysis of ester in water, the following results were obtained:
t/s 
0 
30 
60 
90 
[Ester]mol L^{−1} 
0.55 
0.31 
0.17 
0.085 
(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Solution
(i) Average rate of reaction between the time interval,
= 4.67×10^{3} mol L^{1} s^{1}
(ii) For a pseudo first order reaction,
For t = 30 s,
= 1.911 × 10^{−2} s^{−1}
For t = 60 s,
= 1.957 × 10^{−2} s^{−1}
For t = 90 s,
= 2.075 × 10^{−2} s^{−1}
Then,
=1.98 × 10^{2} s^{1}
3.9. A reaction is first 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 B are doubled?
Solution
(i) The differential rate equation will be
(ii) If the concentration of B is increased three times, then
= 9.k[A][B]^{2}
Therefore, the rate of reaction will increase 9 times.
(iii) When the concentrations of both A and B are doubled,
= k[2A][2B]^{2}
= 8.k[A][B]^{2}
Therefore, the rate of reaction will increase 8 times.
3.10. In a reaction between A and B, the initial rate of reaction (r_{0}) was measured for different initial concentrations of A and B as given below:
A/mol L^{−1} 
0.20 
0.20 
0.40 
B/mol L^{−1} 
0.30 
0.10 
0.05 
r_{0}/mol L^{−1} s^{−1} 
5.07 × 10^{−5} 
5.07 × 10^{−5} 
1.43 × 10^{−4} 
What is the order of the reaction with respect to A and B?
Solution
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore,
r_{0} = k[A]^{x}[B]^{y}
5.07×10^{}5 = k[0.20]^{x}[0.30]^{y} ...(i)
5.07×10^{}5 = k[0.20]^{x}[0.10]^{y} ...(ii)
1.43×10^{4} = k[0.40]^{x}[0.05]^{y} ...(iii)
Dividing equation (i) by (ii), we obtain
⇒ y = 0
Dividing equation (iii) by (ii), we obtain
[Since y = 0 [0.05]y = [0.30]y = 1]
⇒ 2.821 = 2^{x}
⇒ log 2.821 = x log 2 (Taking log on both sides)
⇒ x = (log2.821)/xlog 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.
3.11. The following results have been obtained during the kinetic studies of the reaction:
2A + B → C + D
Experiment 
A/mol L^{−1} 
B/mol L^{−1} 
Initial rate of formation of D/mol L^{−1} min^{−1} 
I 
0.1 
0.1 
6.0 × 10^{−3} 
II 
0.3 
0.2 
7.2 × 10^{−2} 
III 
0.3 
0.4 
2.40 × 10^{−2} 
IV 
0.4 
0.1 

Determine the rate law and the rate constant for the reaction.
Solution
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, rate of the reaction is given by,
Rate = k[A]^{x}[B]^{y}
According to the question,
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
⇒ (4)^{1} = 4^{x}
⇒ x = 1
Dividing equation (iii) by (ii), we obtain
⇒ 4 = 2^{y}
⇒ 2^{2} = 2^{y}
⇒ y = 2
Therefore, the rate law is
Rate = k [A] [B]^{2}
From experiment I, we obtain
= 6.0 L^{2} mol^{−2} min^{−1}
From experiment II, we obtain
= 6.0 L^{2} mol^{−2} min^{−1}
From experiment III, we obtain
= 6.0 L^{2} mol^{−2} min^{−1}
From experiment IV, we obtain
= 6.0 L^{2} mol^{−2} min^{−1}
Therefore, rate constant, k = 6.0 L^{2} mol^{−2} min^{−1}
3.12. The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
Experiment 
A/ mol L^{−1} 
B/ mol L^{−1} 
Initial rate/mol L^{−1} min^{−1} 
I 
0.1 
0.1 
2.0 × 10^{−2} 
II 
 
0.2 
4.0 × 10^{−2} 
III 
0.4 
0.4 
 
IV 
 
0.2 
2.0 × 10^{−2} 
Solution
The given reaction is of the first 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^{−2} mol L^{−1} min^{−1} = k (0.1 mol L^{−1})
⇒ k = 0.2 min^{−1}
From experiment II, we obtain
4.0 × 10^{−2} mol L^{−1} min^{−1} = 0.2 min^{−1} [A]
⇒ [A] = 0.2 mol L^{−1}
From experiment III, we obtain
Rate = 0.2 min^{−1} × 0.4 mol L^{−1}
= 0.08 mol L^{−1} min^{−1}
From experiment IV, we obtain
2.0 × 10^{−2} mol L^{−1} min^{−1} = 0.2 min^{−1} [A]
⇒ [A] = 0.1 mol L^{−1}
3.13. Calculate the halflife of a first order reaction from their rate constants given below:
(i) 200 s^{−1}
(ii) 2 min^{−1}
(iii) 4 years^{−1}
Solution
(i) Half life, t_{1/2} = 0.693/k
= 0.693/200s^{1}
= 3.47 ×10^{3} s (approximately)
(ii) Half life, t_{1/2} = 0.693/k
=0.693/2min^{1}
= 0.35 min (approximately)
(iii) Half life, t_{1/2} = 0.693k
= 0.693/4year^{1}
= 0.173 years (approximately)