Chapter Test: Chemical Kinetics

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CBSE Class 12 Chapter Test: CHEMICAL KINETICS

Time Allowed: 1.5 Hours Maximum Marks: 35

General Instructions:

1. This question paper consists of 16 questions divided into 5 sections: A, B, C, D, and E.
2. Section A comprises 6 MCQs and 2 Assertion-Reasoning questions of 1 mark each.
3. Section B comprises 2 Very Short Answer questions of 2 marks each.
4. Section C comprises 3 Short Answer questions of 3 marks each.
5. Section D comprises 1 Case-Based question of 4 marks.
6. Section E comprises 2 Long Answer questions of 5 marks each.
7. Use of log tables is permitted.

Section A (1 Mark Each)

[1] 1. Which of the following represents the unit of the rate constant for a zero-order reaction?

• s^-1
• mol L^-1 s^-1
• L mol^-1 s^-1
• L² mol^-2 s^-1

[1] 2. The half-life period of a first-order reaction is:

• Directly proportional to the initial concentration of reactants
• Inversely proportional to the initial concentration of reactants
• Independent of the initial concentration of reactants
• Directly proportional to the square of initial concentration

[1] 3. According to the Arrhenius equation, the slope of the plot of ln k versus 1/T is equal to:

• -E_a / R
• E_a / R
• -E_a / 2.303R
• E_a / 2.303R

[1] 4. For a general reaction A + B → Products, the rate of reaction doubles when the concentration of B is doubled. The rate increases by a factor of 8 when the concentrations of both A and B are doubled. The overall order of the reaction is:

• 1
• 2
• 3
• 0

[1] 5. For the reaction 2N₂O₅(g) → 4NO₂(g) + O₂(g), the rate of disappearance of N₂O₅ is related to the rate of appearance of NO₂ by the expression:

• -d[N₂O₅]/dt = d[NO₂]/dt
• -1/2 d[N₂O₅]/dt = 1/4 d[NO₂]/dt
• -2 d[N₂O₅]/dt = 4 d[NO₂]/dt
• -d[N₂O₅]/dt = 1/2 d[NO₂]/dt

[1] 6. The role of a catalyst in a reversible reaction is to:

• Increase the equilibrium constant
• Decrease the activation energy of both forward and backward reactions equally
• Alter the enthalpy change (ΔH) of the reaction
• Make a non-spontaneous reaction spontaneous

Directions for Q7 & Q8: These questions consist of two statements, each printed as Assertion (A) and Reason (R). Select the correct answer from the codes (a), (b), (c), and (d) as given below:

• Both A and R are true and R is the correct explanation of A.
• Both A and R are true but R is not the correct explanation of A.
• A is true but R is false.
• A is false but R is true.

[1] 7. Assertion (A): The acid-catalyzed hydrolysis of ethyl acetate in excess water is a pseudo first-order reaction.
Reason (R): Water is present in such a large excess that its concentration remains virtually constant during the course of the reaction.

[1] 8. Assertion (A): The molecularity of a reaction can never be fractional or zero.
Reason (R): Molecularity represents the number of reacting species colliding simultaneously to bring about a chemical change in an elementary reaction.

Section B (2 Marks Each)

[2] 9. Write any two differences between the order of a reaction and the molecularity of a reaction.

[2] 10. The rate constant for a first-order reaction is 60 s^-1. How much time will it take to reduce the initial concentration of the reactant to 1/16^th of its initial value?

Section C (3 Marks Each)

[3] 11. Show that in a first-order reaction, the time required for completion of 99% of the reaction is twice the time required for completion of 90% of the reaction.

[3] 12. A reaction is first order with respect to reactant A and second order with respect to reactant B.
(i) Write the differential rate equation.
(ii) How is the rate of reaction affected if the concentration of B is increased three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?

[3] 13. The rate of a chemical reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the activation energy of the reaction assuming that it does not change with temperature.
(R = 8.314 J K^-1 mol^-1)

Section D: Case-Based (4 Marks)

[4] 14. Read the passage carefully and answer the questions that follow:

According to the Collision Theory of chemical kinetics, for a reaction to occur, the reactant molecules must collide with each other. However, not all collisions lead to the formation of products. The collisions must be "effective". For a collision to be effective, the colliding molecules must possess a certain minimum amount of kinetic energy and they must collide with the proper orientation. The Arrhenius equation mathematically relates the rate constant of a reaction to the absolute temperature and the activation energy. The addition of a catalyst significantly alters the kinetics of the reaction without being permanently consumed.

(a) Define the term "Threshold Energy". [1 Mark]
(b) What is meant by the "steric factor" (or probability factor) in collision theory? [1 Mark]
(c) Explain graphically or theoretically how the addition of a positive catalyst increases the rate of a chemical reaction. Does it alter the Gibbs free energy (ΔG) of the reaction? [2 Marks]

Section E: Long Answer (5 Marks Each)

[5] 15. (a) Derive the integrated rate equation for a zero-order reaction: R → Products. [2 Marks]

(b) A first-order reaction takes 40 minutes for 30% decomposition. Calculate its half-life (t_1/2).
(Given: log 10 = 1, log 7 = 0.845) [3 Marks]

[5] 16. (a) Define a pseudo first-order reaction. Give one chemical equation as an example. [2 Marks]

(b) The decomposition of NH₃ on a platinum surface is a zero-order reaction:
2NH₃(g) → N₂(g) + 3H₂(g)
If the rate constant (k) is 2.5 × 10^-4 mol L^-1 s^-1, what are the rates of production of N₂ and H₂? [3 Marks]

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SOLUTIONS & MARKING SCHEME

1. Answer: (b) mol L^-1 s^-1

Reasoning: The unit of rate constant (k) is given by (mol L^-1)^1-n s^-1, where n is the order of the reaction. For a zero-order reaction (n=0), the unit is (mol L^-1)^1-0 s^-1 = mol L^-1 s^-1.

2. Answer: (c) Independent of the initial concentration of reactants

Reasoning: For a first-order reaction, the formula for half-life is t_1/2 = 0.693 / k. As seen in the formula, there is no concentration term [R]₀, proving it is independent of the initial concentration.

3. Answer: (a) -E_a / R

Reasoning: Taking the natural log of the Arrhenius equation gives: ln k = -E_a/(RT) + ln A. Comparing this to the equation of a straight line y = mx + c (where y = ln k and x = 1/T), the slope (m) is -E_a/R. (Note: If the plot was log k vs 1/T, the slope would be -E_a/2.303R).

4. Answer: (c) 3

Reasoning: Let Rate = k[A]^x[B]^y.
If [B] is doubled, rate doubles → 2^y = 2 → y = 1.
If both [A] and [B] are doubled, rate increases by 8 → 2^x × 2¹ = 8 → 2^x = 4 → x = 2.
Overall order = x + y = 2 + 1 = 3.

5. Answer: (b) -1/2 d[N₂O₅]/dt = 1/4 d[NO₂]/dt

Reasoning: To express the unique rate of reaction, the rate of change of concentration of any species is divided by its stoichiometric coefficient. For reactants, a negative sign is added. Hence, Rate = -1/2 d[N₂O₅]/dt = +1/4 d[NO₂]/dt.

6. Answer: (b) Decrease the activation energy of both forward and backward reactions equally

Reasoning: A catalyst provides an alternate reaction pathway with a lower activation energy barrier. It lowers the activation energy for both the forward and backward reactions by the exact same amount, which is why it doesn't change the equilibrium constant or ΔH.

7. Answer: (a) Both A and R are true and R is the correct explanation of A.

Reasoning: The molecularity of hydrolysis of an ester is 2, but the order is 1 because the concentration of water is so large it remains effectively constant. This makes it a pseudo first-order reaction, perfectly explained by the reason.

8. Answer: (a) Both A and R are true and R is the correct explanation of A.

Reasoning: Because molecularity is the physical count of molecules colliding simultaneously in a single step, you cannot have half a molecule colliding, nor can you have zero molecules colliding to produce a reaction.

9. Solution:

• Order: It is the sum of powers of the concentration terms in the experimentally determined rate law. It can be zero, fractional, or an integer.
• Molecularity: It is the number of reacting species colliding simultaneously in an elementary reaction. It is a theoretical concept and is always a whole number (never zero or fractional).

10. Solution:

Given: k = 60 s^-1, [R] = [R]₀ / 16.
Formula: t = (2.303 / k) log([R]₀ / [R])
t = (2.303 / 60) log([R]₀ / ([R]₀ / 16))
t = (2.303 / 60) log(16)
Since log(16) = log(2⁴) = 4 × 0.301 = 1.204
t = (2.303 / 60) × 1.204 = 2.77 / 60 = 0.046 s.
(Alternative method: 1/16 is 4 half-lives. t_1/2 = 0.693/60 = 0.01155 s. Total time = 4 × 0.01155 = 0.046 s).

11. Solution:

Case 1: 99% completion
Let [R]₀ = 100. Amount remaining [R] = 100 - 99 = 1.
t_99% = (2.303 / k) log (100 / 1) = (2.303 / k) log(10²) = (2.303 / k) × 2  --- (Eq 1)

Case 2: 90% completion
Let [R]₀ = 100. Amount remaining [R] = 100 - 90 = 10.
t_90% = (2.303 / k) log (100 / 10) = (2.303 / k) log(10¹) = (2.303 / k) × 1  --- (Eq 2)

Dividing Eq 1 by Eq 2:
t_99% / t_90% = 2 / 1 ⇒ t_99% = 2 × t_90%. (Hence Proved).

12. Solution:

(i) Differential rate equation: Rate = k [A]¹ [B]².

(ii) If [B] is increased three times ([B]' = 3[B]):
New Rate = k [A] (3[B])² = 9 k[A][B]² = 9 times the original rate.

(iii) If both [A] and [B] are doubled:
New Rate = k (2[A]) (2[B])² = k(2[A])(4[B]²) = 8 k[A][B]² = 8 times the original rate.

13. Solution:

Given: T₁ = 293 K, T₂ = 313 K, k₂ / k₁ = 4.
Using Arrhenius Equation: log(k₂/k₁) = (E_a / 2.303R) × [ (T₂ - T₁) / (T₁T₂) ]
log(4) = [E_a / (2.303 × 8.314)] × [ (313 - 293) / (293 × 313) ]
0.602 = (E_a / 19.147) × [ 20 / 91709 ]
0.602 = E_a × (20 / 1755951)
E_a = (0.602 × 1755951) / 20 = 52854 J mol^-1 = 52.85 kJ mol^-1.

14. Solution (Case-Based):

(a) Threshold Energy: It is the total minimum amount of energy that the colliding molecules must possess in order for the collision to be effective and result in a chemical reaction. (Threshold Energy = Initial Kinetic Energy + Activation Energy).

(b) The steric factor (P), or probability factor, accounts for the fact that colliding molecules must have proper spatial orientation during the collision to break old bonds and form new ones.

(c) A positive catalyst provides an alternate pathway or reaction mechanism by lowering the activation energy barrier. As the barrier is lowered, a much larger fraction of molecules possess enough energy to cross it, thereby drastically increasing the rate of reaction.
No, a catalyst does not alter the Gibbs free energy (ΔG) of the reaction. It only affects the kinetics, not the thermodynamics.

15. Solution:

(a) Derivation for Zero Order:
For reaction R → P, Rate = - d[R]/dt = k[R]⁰.
Since [R]⁰ = 1, we get: - d[R]/dt = k ⇒ d[R] = -k dt.
Integrating both sides: ∫ d[R] = -k ∫ dt ⇒ [R] = -kt + I (where I is integration constant).
At t=0, [R] = [R]₀. Therefore, I = [R]₀.
Substituting I back: [R] = -kt + [R]₀ OR k = ([R]₀ - [R]) / t.

(b) Numerical:
Given: 30% decomposition means 70% remains. Let [R]₀ = 100, [R] = 70, t = 40 min.
Step 1: Find k
k = (2.303 / t) log([R]₀ / [R]) = (2.303 / 40) log(100 / 70)
k = (2.303 / 40) log(1.428) OR (2.303 / 40) × [log(10) - log(7)]
k = (2.303 / 40) × [1 - 0.845] = (2.303 / 40) × 0.155 = 0.00892 min^-1.

Step 2: Find t_1/2
t_1/2 = 0.693 / k = 0.693 / 0.00892 = 77.7 minutes.

16. Solution:

(a) Pseudo first-order reaction: It is a reaction whose molecularity is 2 or more, but its order is 1. This occurs when one of the reactants is present in such a large excess that its concentration remains virtually unchanged, acting as a constant.
Example: Acid-catalyzed hydrolysis of ethyl acetate.
CH₃COOC₂H₅ + H₂O (excess) →(H⁺) CH₃COOH + C₂H₅OH.
Rate = k'[CH₃COOC₂H₅][H₂O] ⇒ Rate = k[CH₃COOC₂H₅].

(b) Numerical:
Reaction: 2NH₃ → N₂ + 3H₂.
For a zero-order reaction, Rate of reaction = k = 2.5 × 10^-4 M s^-1.
From stoichiometry, the overall rate is given by:
Rate = -1/2 d[NH₃]/dt = d[N₂]/dt = 1/3 d[H₂]/dt.

Rate of production of N₂:
d[N₂]/dt = Rate = k = 2.5 × 10^-4 mol L^-1 s^-1.

Rate of production of H₂:
1/3 d[H₂]/dt = Rate ⇒ d[H₂]/dt = 3 × Rate = 3 × (2.5 × 10^-4)
d[H₂]/dt = 7.5 × 10^-4 mol L^-1 s^-1.

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