JEE Chemistry Chemical Kinetics MCQs Set 15

Download JEE MCQs for JEE Chemistry: Chemical Kinetics

Explore reliable objective questions for Chemical Kinetics tailored for JEE learners. Utilizing these Chemistry multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Chapter-wise Objective Questions: Chemical Kinetics

Access the complete set of multiple-choice questions for Chemical Kinetics below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: The rate of \( \ce{RCl + NaOH(aq) -> ROH + NaCl} \) is \( k_1[RCl] \). The rate is :
(a) doubled on doubling the concentration of NaOH
(b) halved on reducing the concentration of RCl to half
(c) decreased on increasing the temperature of the reaction
(d) unaffected by increasing the temperature of the reaction
Show Answer & Explanation

Answer: (b) halved on reducing the concentration of RCl to half

Explanation:
1. Rate depends only on [RCl].
2. Halving [RCl] halves the rate.
3. Rate does not depend on NaOH and rises with temperature.

Teacher's Note:
1. Read the rate law and ignore species that do not appear.

Question 2: For \( \ce{NO2 + CO -> CO2 + NO} \), rate = k[NO₂]². The number of molecules of CO involved in the slowest step is :
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. CO does not appear in the rate law, so it is not in the slow step.
2. Zero molecules of CO take part in the slowest step.

Teacher's Note:
1. Only species in the slow step appear in the rate law.

Question 3: The rate law of \( \ce{A + 2B -> product} \) is \( d[P]/dt = k[A]^2[B] \). If A is taken in large excess, the order of the reaction becomes :
(a) zero
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. A is in large excess, so [A] stays constant and merges into the rate constant.
2. Rate = k'[B], a pseudo first order reaction.

Teacher's Note:
1. Excess reagent gives a pseudo order.
2. Count only the species that change.

Question 4: The acid hydrolysis of an ester is :
(a) pseudo first order
(b) bimolecular
(c) pseudo unimolecular
(d) all of these
Show Answer & Explanation

Answer: (d) all of these

Explanation:
1. Water is in large excess, so its concentration is constant.
2. The reaction is bimolecular but behaves as pseudo first order (pseudo unimolecular).

Teacher's Note:
1. Excess solvent gives pseudo orders.
2. All statements apply.

Question 5: In the sequence \( \ce{A ->[K1] B ->[K2] C ->[K3] D} \) with \( K_3 > K_2 > K_1 \), the rate-determining step is :
(a) A → B
(b) B → C
(c) C → D
(d) A → D
Show Answer & Explanation

Answer: (a) A → B

Explanation:
1. The slowest step determines the rate.
2. The smallest rate constant is \( K_1 \), so A → B is rate determining.

Teacher's Note:
1. Smallest k means slowest step.

Question 6: For \( \ce{P -> R} \): \( \ce{P <=>[K1] 2Q} \) (fast) and \( \ce{2Q + P ->[K2] R} \) (slow). The rate law is :
(a) k₁[P][Q]
(b) k₁k₂[P]
(c) k₁k₂[P]²
(d) k₁k₂[a]
Show Answer & Explanation

Answer: (c) k₁k₂[P]²

Explanation:
1. Slow step: rate = k₂[Q]²[P].
2. From the fast step \( [Q]^2 = K_1[P] \).
3. Rate = k₁k₂[P]².

Teacher's Note:
1. Replace [Q]² by K₁[P].
2. Combine the constants.

Question 7: The reaction \( \ce{2A + B -> P} \) follows the mechanism \( \ce{2A <=> A2} \) (fast), \( \ce{A2 + B -> P} \) (slow). The order of the reaction is :
(a) 1.5
(b) 3
(c) 1
(d) 2
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. Slow step: rate = k[A₂][B].
2. From the fast equilibrium \( [A_2] = K[A]^2 \).
3. Rate = kK[A]²[B], so the order is 3.

Teacher's Note:
1. Substitute the intermediate.
2. Add the powers.

Question 8: For \( \ce{2NO + Cl2 -> 2NOCl} \) the mechanism is \( \ce{NO + Cl2 <=> NOCl2} \) (fast) and \( \ce{NOCl2 + NO -> 2NOCl} \) (slow). The rate law is :
(a) Rate = k[NO]²[Cl₂]
(b) Rate = k[NO][Cl₂]²
(c) Rate = k[NOCl₂]
(d) Rate = k[NOCl]²
Show Answer & Explanation

Answer: (a) Rate = k[NO]²[Cl₂]

Explanation:
1. Slow step: rate = k[NOCl₂][NO].
2. \( [NOCl_2] = K[NO][Cl_2] \) from the fast step.
3. Rate = kK[NO]²[Cl₂].

Teacher's Note:
1. Express the intermediate in terms of the reactants.

Question 9: The rate constant is numerically the same for three reactions of first, second and third order respectively. At a moment when the concentration of the reactants is the same and greater than 1 M for all three, which is true?
(a) r₁ = r₂ = r₃
(b) r₁ > r₂ > r₃
(c) r₁ < r₂ < r₃
(d) all of these
Show Answer & Explanation

Answer: (c) r₁ < r₂ < r₃

Explanation:
1. r₁ = kC, r₂ = kC², r₃ = kC³.
2. For C greater than 1, C < C² < C³, so r₁ < r₂ < r₃.

Teacher's Note:
1. Compare powers of a number greater than 1.

Question 10: K for a zero order reaction is \( 2 \times 10^{-2} \) mol L⁻¹ s⁻¹. If the concentration of the reactant after 25 s is 0.5 M, the initial concentration must have been :
(a) 0.5 M
(b) 1.25 M
(c) 12.5 M
(d) 1.0 M
Show Answer & Explanation

Answer: (d) 1.0 M

Explanation:
1. Zero order: \( [A] = [A]_0 - kt \).
2. \( [A]_0 = 0.5 + 0.02 \times 25 = 1.0 \) M.

Teacher's Note:
1. Concentration falls linearly with time.
2. Add the amount used.

Question 11: The decomposition \( \ce{2N2O5 -> 4NO2 + O2} \) follows first order kinetics. Hence :
(a) the reaction is bimolecular
(b) the reaction is unimolecular
(c) \( t_{1/2} \propto a^0 \)
(d) \( t_{1/2} \propto a^2 \)
Show Answer & Explanation

Answer: (c) \( t_{1/2} \propto a^0 \)

Explanation:
1. For a first order reaction \( t_{1/2} = 0.693/k \), which is independent of the initial concentration a.
2. So \( t_{1/2} \propto a^0 \).

Teacher's Note:
1. First order half-life is constant.

Question 12: The rate constant of a first order reaction is \( 4 \times 10^{-3} \) s⁻¹. At a reactant concentration of 0.02 M the rate would be :
(a) \( 8 \times 10^{-5} \) M s⁻¹
(b) \( 4 \times 10^{-3} \) M s⁻¹
(c) \( 2 \times 10^{-1} \) M s⁻¹
(d) \( 4 \times 10^{-1} \) M s⁻¹
Show Answer & Explanation

Answer: (a) \( 8 \times 10^{-5} \) M s⁻¹

Explanation:
1. Rate = k[A] = 4 × 10⁻³ × 0.02 = 8 × 10⁻⁵ M s⁻¹.

Teacher's Note:
1. First order rate = k × concentration.

Question 13: A first order reaction has a half life of 69.3 s. At 0.10 mol L⁻¹ reactant concentration, the rate is :
(a) \( 10^{-4} \) M s⁻¹
(b) \( 10^{-3} \) M s⁻¹
(c) \( 10^{-1} \) M s⁻¹
(d) \( 6.93 \times 10^{-1} \) M s⁻¹
Show Answer & Explanation

Answer: (b) \( 10^{-3} \) M s⁻¹

Explanation:
1. k = 0.693/69.3 = 0.01 s⁻¹.
2. Rate = k[A] = 0.01 × 0.10 = 10⁻³ M s⁻¹.

Teacher's Note:
1. Find k from the half-life first.

Question 14: For a first order reaction \( \ce{A -> products} \), the rate at [A] = 0.2 M is \( 1 \times 10^{-2} \) mol L⁻¹ min⁻¹. The half-life of the reaction is :
(a) 832 min
(b) 440 min
(c) 416 min
(d) 14 min
Show Answer & Explanation

Answer: (d) 14 min

Explanation:
1. k = rate/[A] = 0.01/0.2 = 0.05 min⁻¹.
2. \( t_{1/2} = 0.693/0.05 = 13.9 \approx 14 \) min.

Teacher's Note:
1. Rate divided by concentration gives k.

Question 15: The half-life of the first order reaction \( \ce{N2O5 -> 2NO2 + 1/2 O2} \) is 24 h at 30°C. Starting with 10 g of \( \ce{N2O5} \), how many grams remain after 96 hours?
(a) 1.25 g
(b) 0.63 g
(c) 1.77 g
(d) 0.5 g
Show Answer & Explanation

Answer: (b) 0.63 g

Explanation:
1. 96 h = 4 half-lives.
2. Amount left = 10/2⁴ = 10/16 = 0.625 ≈ 0.63 g.

Teacher's Note:
1. Divide the time by the half-life.

Question 16: For a given first order reaction it takes 20 minutes for the concentration to drop from 1 M to 0.6 M. The time needed for the concentration to drop from 0.6 M to 0.36 M will be :
(a) more than 20 min
(b) less than 20 min
(c) equal to 20 min
(d) infinity
Show Answer & Explanation

Answer: (c) equal to 20 min

Explanation:
1. The ratio 1/0.6 equals 0.6/0.36.
2. For a first order reaction, equal ratios take equal time, so 20 min.

Teacher's Note:
1. First order time depends on the ratio, not on the absolute drop.

Question 17: A first order reaction is carried out with an initial concentration of 10 mol L⁻¹ and 80% of the reactant changes into product. If the same reaction is carried out with an initial concentration of 5 mol L⁻¹ for the same period, the percentage of reactant changing to product is :
(a) 40
(b) 80
(c) 160
(d) cannot be calculated
Show Answer & Explanation

Answer: (b) 80

Explanation:
1. For a first order reaction the fraction converted in a given time does not depend on the initial concentration.
2. The percentage remains 80.

Teacher's Note:
1. k fixes the fraction converted in time t.

Question 18: The reaction L → M is started with 10 g/L. After 30 minutes and 90 minutes, 5 g/L and 1.25 g/L are left respectively. The order of the reaction is :
(a) 0
(b) 2
(c) 1
(d) 3
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. 10 → 5 takes 30 min, so t₁/₂ = 30 min.
2. 5 → 1.25 g/L is two more half-lives, which takes 60 min (30 → 90 min).
3. Constant half-life shows first order.

Teacher's Note:
1. Constant half-life means first order.

Question 19: The half-life for the catalytic decomposition of \( \ce{AB3} \) at 50 mm is 4 hours and at 100 mm it is 2 hours. The order of the reaction is :
(a) 3
(b) 1
(c) 2
(d) 0
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. Doubling the pressure halves the half-life, so t₁/₂ ∝ 1/p.
2. That is a second order reaction.

Teacher's Note:
1. t₁/₂ ∝ 1/a means second order.

Question 20: The rate constant for a reaction is \( 10.8 \times 10^{-5} \) mol L⁻¹ s⁻¹. The reaction obeys :
(a) first order
(b) zero order
(c) second order
(d) all are wrong
Show Answer & Explanation

Answer: (b) zero order

Explanation:
1. The unit of k is concentration per time for a zero order reaction.
2. So the reaction is zero order.

Teacher's Note:
1. Check the unit of k.

Question 21: A substance A decomposes in solution by first order kinetics. Flask I contains 1 L of 1 M A and flask II contains 100 mL of 0.6 M A. After 8 hours the concentration of A in flask I becomes 0.25 M. The time for the concentration of A in flask II to become 0.3 M is :
(a) 0.4 hr
(b) 2.4 hr
(c) 4.0 hr
(d) unpredictable as the rate constant is not given
Show Answer & Explanation

Answer: (c) 4.0 hr

Explanation:
1. In flask I, 1 → 0.25 M is two half-lives in 8 h, so t₁/₂ = 4 h.
2. In flask II, 0.6 → 0.3 M is one half-life, so 4.0 h.

Teacher's Note:
1. For first order, half-life is independent of the volume and the concentration.

Question 22: The rate constant for the reaction \( \ce{2A + B -> product} \) was found to be \( 2.5 \times 10^{-5} \) L mol⁻¹ s⁻¹ after 15 s, \( 2.60 \times 10^{-5} \) after 30 s and \( 2.55 \times 10^{-5} \) after 50 s. The order of the reaction is :
(a) 2
(b) 3
(c) zero
(d) 1
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. The value of k is nearly constant, which confirms the integrated rate law used.
2. The unit L mol⁻¹ s⁻¹ is that of a second order constant.

Teacher's Note:
1. Check the unit of the constant.

Question 23: The rate constant of a second order reaction is \( 8 \times 10^{-5} \) M⁻¹ min⁻¹. The time to reduce a 1 M solution to 0.5 M is :
(a) \( 8.665 \times 10^{3} \) min
(b) \( 8 \times 10^{-3} \) min
(c) \( 1.25 \times 10^{4} \) min
(d) \( 4 \times 10^{-5} \) min
Show Answer & Explanation

Answer: (c) \( 1.25 \times 10^{4} \) min

Explanation:
1. \( t = \dfrac{1}{k}\left(\dfrac{1}{[A]} - \dfrac{1}{[A]_0}\right) = \dfrac{1}{8 \times 10^{-5}}(2 - 1) = 1.25 \times 10^{4} \) min.

Teacher's Note:
1. Use the second order integrated equation.

Question 24: A graph between t₁/₂ and concentration for an nth order reaction is a horizontal straight line (t₁/₂ independent of concentration). The reaction is 50% complete in 10 minutes when the concentration is 2 mol L⁻¹. It is decomposed 50% in t minutes at 4 mol L⁻¹. The values of n and t are respectively :
(a) 0, 20 min
(b) 1, 10 min
(c) 1, 20 min
(d) 0, 5 min
Show Answer & Explanation

Answer: (b) 1, 10 min

Explanation:
1. A constant half-life means a first order reaction (n = 1).
2. The half-life is then 10 min whatever the concentration, so t = 10 min.

Teacher's Note:
1. Horizontal line means first order.

Question 25: In a first order reaction, 75% of the reactant disappeared in 1.388 s. The rate constant is :
(a) 1 s⁻¹
(b) \( 2.8 \times 10^{-4} \) s⁻¹
(c) \( 17.2 \times 10^{-3} \) s⁻¹
(d) \( 1.8 \times 10^{-3} \) s⁻¹
Show Answer & Explanation

Answer: (a) 1 s⁻¹

Explanation:
1. \( k = \dfrac{2.303}{1.388}\log 4 = \dfrac{2.303 \times 0.602}{1.388} \approx 1.0 \) s⁻¹.

Teacher's Note:
1. 75% completion means the concentration becomes 1/4.

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Multiple Choice Questions (MCQs) for JEE Chemistry Chemical Kinetics

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