Practice Reaction Kinetics MCQs for Air University Entry Test Chemistry — topic-wise sets with solved answers.
Q1. For a first-order reaction, the rate constant is 0.01 s^-1. What is the half-life?
Answer: 69.3 s
Explanation: For a first-order reaction, t1/2 = 0.693 / k = 0.693 / 0.01 = 69.3 s, using the formula t1/2 = ln(2) / k.
Q2. The rate of reaction A + B -> products is given by rate = k[A]²[B]. The order of reaction is
Answer: 3
Explanation: Order is the sum of exponents in the rate law, here 2 + 1 = 3, so the order is 3.
Q3. For the reaction 2NO + O2 -> 2NO2, the rate law is rate = k[NO]²[O2]. If [NO] is doubled and [O2] is halved, the rate
Answer: becomes 2 times
Explanation: New rate = k(2[NO])²([O2]/2) = k(4[NO]²)([O2]/2) = 2k[NO]²[O2], so rate becomes 2 times the original.
Q4. The unit of rate constant for a zero-order reaction is
Answer: mol L^-1 s^-1
Explanation: For zero-order, rate = k, so k has units of rate, which is concentration/time, i.e., mol L^-1 s^-1.
Q5. The half-life of a radioactive substance is 20 years. What fraction remains after 1 / 2 40 years?
Answer: 1/16
Explanation: After 20 years, 1/2 remains; after 40 years, 1/4; after 60 years, 1/8; after 80 years (2 * 40 years), 1/16 remains.
Q6. For a reaction with rate law rate = k[A][B]², if [A] is doubled and [B] is tripled, the rate increases by a factor of
Answer: 18
Explanation: New rate = k(2[A])(3[B])² = k(2[A])(9[B]²) = 18k[A][B]², so rate increases by 18 times.
Q7. The rate constant of a reaction is 0.05 s^-1 at 25°C and 0.15 s^-1 at 35°C. The activation energy is
Answer: 60.3 kJ/mol
Explanation: Using Arrhenius equation, ln(k2/k1) = (Ea/R)(1/T1 - 1/T2), we get Ea = 60.3 kJ/mol, with R = 8.314 J/(mol K).
Q8. The reaction A -> B is first-order with k = 0.02 min^-1. If [A]0 = 1 M, [A] at t = 50 min is
Answer: 0.368 M
Explanation: For first-order, [A] = [A]0 * e^(-kt) = 1 * e^(-0.02*50) = e^(-1) = 0.368 M.
Q9. For a second-order reaction, the half-life is
Answer: inversely proportional to initial concentration
Explanation: For second-order, t1/2 = 1/(k[A]0), so t1/2 is inversely proportional to [A]0.
Q10. The rate of reaction 2A + B -> products is given by rate = k[A][B]. If [A] = [B] = 0.1 M, the rate is 0.01 M/s. The rate constant k is
Answer: 1 M^-1 s^-1
Explanation: k = rate / ([A][B]) = 0.01 / (0.1 * 0.1) = 1 M^-1 s^-1.
Q11. The activation energy for a reaction is 80 kJ/mol. At what temperature will the rate be double at 300 K?
Answer: 310 K
Explanation: Using Arrhenius equation and given that k2 = 2k1, we can find T2 = 310 K, with Ea = 80 kJ/mol and T1 = 300 K.
Q12. For the reaction A + 2B -> C, the rate law is rate = k[A]². The order with respect to B is
Answer: 0
Explanation: The rate law does not contain [B], so the order with respect to B is 0.
Q13. The decomposition of N2O5 is a first-order reaction with k = 0.001 s^-1 at 25°C. The half-life is
Answer: 693 s
Explanation: For first-order, t1/2 = 0.693 / k = 0.693 / 0.001 = 693 s.
Q14. For a reaction with rate law rate = k[A]²[B], if [A] and [B] are both doubled, the rate increases by a factor of
Answer: 8
Explanation: New rate = k(2[A])²(2[B]) = k(4[A]²)(2[B]) = 8k[A]²[B], so rate increases by 8 times.
Q15. The unit of rate constant for a third-order reaction is
Answer: L² mol^-2 s^-1
Explanation: For third-order, rate = k[A]³, so k has units of L² mol^-2 s^-1, derived from rate / [A]³.
Q16. For the reaction 2NO2 -> 2NO + O2, the rate law is rate = k[NO2]². The order is
Answer: 2
Explanation: The order is the exponent in the rate law, here 2, so the order is 2.
Q17. The rate constant of a reaction at 25°C is 0.1 s^-1. The activation energy is 50 kJ/mol. The rate constant at 35°C is
Answer: 0.2 s^-1
Explanation: Using Arrhenius equation, we can find k2 = 0.2 s^-1, with Ea = 50 kJ/mol, T1 = 298 K, and T2 = 308 K.
Q18. The half-life of a first-order reaction is 100 s. The rate constant k is
Answer: 0.00693 s^-1
Explanation: For first-order, k = 0.693 / t1/2 = 0.693 / 100 = 0.00693 s^-1.
Q19. For a reaction A -> B, the rate law is rate = k[A]. If [A]0 = 1 M and k = 0.1 min^-1, [A] after 10 min is
Answer: 0.368 M
Explanation: For first-order, [A] = [A]0 * e^(-kt) = 1 * e^(-0.1*10) = e^(-1) = 0.368 M.
Q20. The rate of reaction A -> B is given by rate = k[A]^n. If the rate is doubled when [A] is doubled, the order n is
Answer: 1
Explanation: For rate = k[A]^n, if doubling [A] doubles the rate, then n = 1, since 2 = 2^n implies n = 1.
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