JEE Chemistry Chemical Kinetics MCQs Set 28

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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 half-life of a certain reaction A → products is 1 hour when the initial concentration is 2.0 mol L-1. If it is a zero order reaction, how long does it take for the concentration to fall from 0.50 to 0.25 mol L-1?
(a) 1 h
(b) 0.5 h
(c) 4 h
(d) 0.25 h
Show Answer & Explanation

Answer: (d) 0.25 h

Explanation:
1. For zero order t1/2 = a/2k, so k = 2.0/(2 × 1) = 1.0 mol L-1 h-1.
2. Time = (0.50 - 0.25)/1.0 = 0.25 h.

Teacher's Note:
1. Zero order concentration falls linearly.
2. Do not apply the constant half-life of a first order reaction.

Question 2: The reaction A(s) → 2B(g) + C(g) is first order. The total pressure after 20 min and after a very long time are 150 mm Hg and 225 mm Hg. The rate constant and the pressure after 40 minutes are
(a) 0.05 ln 3 min-1, 200 mm Hg
(b) 0.05 ln 3 min-1, 300 mm Hg
(c) 0.5 ln 2 min-1, 300 mm Hg
(d) 0.05 ln 1.5 min-1, 200 mm Hg
Show Answer & Explanation

Answer: (a) 0.05 ln 3 min-1, 200 mm Hg

Explanation:
1. The solid does not contribute to the pressure; P(infinity) = 225 corresponds to complete reaction.
2. k = (1/20) ln[225/(225 - 150)] = (1/20) ln 3 = 0.05 ln 3 min-1.
3. After 40 min, 225 - P = 225 (1/3)2 = 25, so P = 200 mm Hg.

Teacher's Note:
1. The amount of solid left is proportional to P(infinity) - P(t).
2. After 40 min, two intervals of 20 min.

Question 3: In BrO3-(aq) + 5Br-(aq) + 6H+(aq) → 3Br2 + 3H2O, the rate of appearance of Br2 is related to the rate of disappearance of Br- by
(a) d[Br2]/dt = -(5/3) d[Br-]/dt
(b) d[Br2]/dt = d[Br-]/dt
(c) d[Br2]/dt = -d[Br-]/dt
(d) d[Br2]/dt = -(3/5) d[Br-]/dt
Show Answer & Explanation

Answer: (d) d[Br2]/dt = -(3/5) d[Br-]/dt

Explanation:
1. Rate of reaction = (1/3) d[Br2]/dt = -(1/5) d[Br-]/dt.
2. So d[Br2]/dt = -(3/5) d[Br-]/dt.

Teacher's Note:
1. Divide each rate by its coefficient.
2. Br2 is formed more slowly than Br- is used up.

Question 4: At 518 °C, the rate of decomposition of gaseous acetaldehyde, initially at 363 Torr, was 1.00 Torr s-1 when 5% had reacted and 0.5 Torr s-1 when 33% had reacted. The order of the reaction is
(a) 3
(b) 1
(c) 0
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. Pressures of reactant: 363 × 0.95 = 345 Torr and 363 × 0.67 = 243 Torr.
2. Rate ratio 1.00/0.5 = 2 = (345/243)^n = (1.42)^n.
3. n = ln 2/ln 1.42 = 1.98, so n = 2.

Teacher's Note:
1. Compare two rates at known concentrations.
2. Take logarithms to find the exponent.

Question 5: For A2 + B2 → 2AB the activation energies of the forward and reverse reactions are 180 and 200 kJ mol-1. A catalyst lowers both by 100 kJ mol-1. The enthalpy change of the reaction in the presence of the catalyst is
(a) +120 kJ mol-1
(b) +280 kJ mol-1
(c) -20 kJ mol-1
(d) +300 kJ mol-1
Show Answer & Explanation

Answer: (c) -20 kJ mol-1

Explanation:
1. ΔH = Ef - Eb = 180 - 200 = -20 kJ mol-1.
2. A catalyst lowers both activation energies equally and does not change ΔH.

Teacher's Note:
1. ΔH = Eforward - Ereverse.
2. Catalysts do not alter thermodynamic quantities.

Question 6: Two reactions R1 and R2 have identical pre-exponential factors, and the activation energy of R1 exceeds that of R2 by 10 kJ mol-1. If k1 and k2 are the rate constants at 300 K, ln(k2/k1) is (R = 8.314 J mol-1 K-1)
(a) 8
(b) 12
(c) 4
(d) 6
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. ln(k2/k1) = (Ea1 - Ea2)/RT = 10 000/(8.314 × 300) = 4.0.

Teacher's Note:
1. Pre-exponential factors cancel.
2. The reaction with lower Ea is faster.

Question 7: A reactant A forms two products, A → B (activation energy Ea1) and A → C (activation energy Ea2), and Ea2 = 2Ea1. Assuming equal pre-exponential factors, k1 and k2 are related as
(a) k2 = k1 e^Ea2/RT
(b) k1 = A k2 e^-Ea1/RT
(c) k2 = k1 e^-Ea1/RT
(d) k1 = 2 k2 e^-Ea2/RT
Show Answer & Explanation

Answer: (c) k2 = k1 e^-Ea1/RT

Explanation:
1. k1 = A e^-Ea1/RT and k2 = A e^-Ea2/RT = A e-2Ea1/RT.
2. Dividing, k2 = k1 e^-Ea1/RT.

Teacher's Note:
1. Write both Arrhenius expressions.
2. The common factor A cancels.

Question 8: In the decomposition of N2O5 in CCl4, 2N2O5 → 2N2O4 + O2, the maximum volume of O2 obtained was 100 cm3. In 500 minutes, 90 cm3 of O2 had been evolved. The first order rate constant for the disappearance of N2O5 (in min-1) is
(a) (100/90)/500
(b) (2.303/500) log(100/10)
(c) (2.303/500) log(100/90)
(d) 2.303/500
Show Answer & Explanation

Answer: (b) (2.303/500) log(100/10)

Explanation:
1. Remaining N2O5 is proportional to 100 - 90 = 10 cm3.
2. k = (2.303/500) log[100/(100 - 90)] = (2.303/500) log 10 = 4.6 × 10-3 min-1.

Teacher's Note:
1. Take V∞ as 100 and Vt as 90.
2. Do not use log(100/90): that would be the amount reacted.

Question 9: The rate constant, activation energy and frequency factor of a reaction at 25 °C are 3.0 × 10-4 s-1, 104.4 kJ mol-1 and 6.0 × 1014 s-1. The value of k as T tends to infinity is
(a) 6.0 × 1014 s-1
(b) 2.0 × 1018 s-1
(c) infinite
(d) 3.6 × 1030 s-1
Show Answer & Explanation

Answer: (a) 6.0 × 1014 s-1

Explanation:
1. k = A e-Ea/RT.
2. As T → infinity the exponent tends to zero, so k → A = 6.0 × 1014 s-1.

Teacher's Note:
1. The frequency factor is the limiting value of k.
2. The exponential factor cannot exceed 1.

Question 10: In the Arrhenius equation k = A e-Ea/RT, which statement is correct?
(a) Ea is the activation energy
(b) A is the adsorption factor
(c) k is the equilibrium constant
(d) R is the Rydberg constant
Show Answer & Explanation

Answer: (a) Ea is the activation energy

Explanation:
1. k is the rate constant, A the frequency (pre-exponential) factor, Ea the activation energy and R the gas constant.

Teacher's Note:
1. Know each symbol in the Arrhenius equation.
2. R = 8.314 J mol-1 K-1.

Question 11: The reaction 2NO2 + O3 → N2O5 + O2 is first order in each of NO2 and O3. Mechanism I: NO2 + O3 → NO3 + O2 (slow), NO3 + NO2 → N2O5 (fast). Mechanism II: O3 ⇌ O2 + O (fast equilibrium), NO2 + O → NO3 (slow), NO3 + NO2 → N2O5 (fast). Which mechanism is consistent with the observed rate law?
(a) I only
(b) II only
(c) both I and II
(d) neither I nor II
Show Answer & Explanation

Answer: (a) I only

Explanation:
1. Mechanism I: the slow step gives rate = k[NO2][O3], matching the observed law.
2. Mechanism II: rate = k[NO2][O] with [O] = K[O3]/[O2], so rate = kK[NO2][O3]/[O2], which has an inverse dependence on [O2] and is not the observed law.

Teacher's Note:
1. The rate-determining step gives the rate law; replace intermediates using equilibrium steps.
2. Observed law has no [O2] term.

Question 12: For a first order reaction, the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is
(a) 1.73 × 10-5 M min-1
(b) 3.47 × 10-4 M min-1
(c) 3.47 × 10-5 M min-1
(d) 1.73 × 10-4 M min-1
Show Answer & Explanation

Answer: (b) 3.47 × 10-4 M min-1

Explanation:
1. A fall to one-quarter in 40 min means t1/2 = 20 min, so k = 0.693/20 = 0.03466 min-1.
2. Rate = k[A] = 0.03466 × 0.01 = 3.47 × 10-4 M min-1.

Teacher's Note:
1. Find k from the half-life.
2. Rate = k times the instantaneous concentration.

Question 13: In the reaction 2SO2 + O2 → 2SO3 the rate of formation of SO3 is 100 g min-1. The rate of disappearance of O2 is (molar masses: SO3 = 80, O2 = 32)
(a) 40 g min-1
(b) 200 g min-1
(c) 50 g min-1
(d) 20 g min-1
Show Answer & Explanation

Answer: (d) 20 g min-1

Explanation:
1. Rate of formation of SO3 = 100/80 = 1.25 mol min-1.
2. From the equation, O2 disappears at half this rate: 0.625 mol min-1.
3. In grams: 0.625 × 32 = 20 g min-1.

Teacher's Note:
1. Convert to moles before using stoichiometric ratios.
2. Rate of reaction = (1/2)(d[SO3]/dt) = -d[O2]/dt.

Question 14: The rate of a reaction doubles when the temperature changes from 300 K to 310 K. The activation energy is (R = 8.314 J K-1 mol-1, log 2 = 0.301)
(a) 53.6 kJ mol-1
(b) 60.5 kJ mol-1
(c) 58.5 kJ mol-1
(d) 48.6 kJ mol-1
Show Answer & Explanation

Answer: (a) 53.6 kJ mol-1

Explanation:
1. log(k2/k1) = (Ea/2.303R)(1/T1 - 1/T2) = (Ea/2.303R)(10/(300 × 310)).
2. Ea = 2.303 × 8.314 × 0.301 × 9300 = 53 600 J mol-1 = 53.6 kJ mol-1.

Teacher's Note:
1. Use the two-temperature Arrhenius equation.
2. 1/300 - 1/310 = 10/93 000.

Question 15: For a first order reaction with rate = k[A], the concentration of the reactant at time t = 1/k, in terms of the initial concentration C0, is
(a) C0/e2
(b) 1/C0
(c) C0 e
(d) C0/e
Show Answer & Explanation

Answer: (d) C0/e

Explanation:
1. C = C0 e-kt.
2. At t = 1/k, kt = 1 and C = C0 e-1 = C0/e.

Teacher's Note:
1. 1/k is the mean life of a first order reaction.
2. The concentration falls to 36.8% in one mean life.

Question 16: For an endothermic reaction X → Y with activation energies Eb (backward) and Ef (forward), in general
(a) ΔH = ΔU
(b) ΔH < ΔU
(c) Eb < Ef
(d) ΔH > ΔU
Show Answer & Explanation

Answer: (c) Eb < Ef

Explanation:
1. ΔH = Ef - Eb, and it is positive for an endothermic reaction.
2. So Ef > Eb, i.e. Eb < Ef.

Teacher's Note:
1. Draw the energy profile: products lie above reactants.
2. ΔH = Ef - Eb.

Question 17: For a reaction (1/2)A → 2B, the rate of disappearance of A is related to the rate of appearance of B by
(a) -d[A]/dt = (1/4) d[B]/dt
(b) -d[A]/dt = (1/2) d[B]/dt
(c) -d[A]/dt = d[B]/dt
(d) -d[A]/dt = 4 d[B]/dt
Show Answer & Explanation

Answer: (a) -d[A]/dt = (1/4) d[B]/dt

Explanation:
1. Rate = -(1/(1/2)) d[A]/dt = -2 d[A]/dt = (1/2) d[B]/dt.
2. So -d[A]/dt = (1/4) d[B]/dt.

Teacher's Note:
1. Divide each rate by the coefficient, which is 1/2 for A.
2. Be careful with fractional coefficients.

Question 18: Statements: (1) the unit of the rate of disappearance is M s-1; (2) the unit of the rate of reaction is M s-1; (3) the unit of k depends on the order; (4) the unit of k for a first order reaction is M s-1. How many are correct?
(a) 3
(b) 4
(c) 2
(d) 1
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. Statements 1, 2 and 3 are correct.
2. For first order k has unit s-1, not M s-1, so statement 4 is false.
3. Three statements are correct.

Teacher's Note:
1. Unit of k = (mol L-1)1-n s-1.
2. M s-1 is the unit of k for zero order.

Question 19: The mechanism for NO + Br2 → NOBr is NO + Br2 ⇌ NOBr2 (fast equilibrium) and NOBr2 + NO → 2NOBr (slow). The order with respect to NO is
(a) 0
(b) 1
(c) 3
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. Rate = k[NOBr2][NO].
2. From the fast equilibrium [NOBr2] = K[NO][Br2].
3. Rate = kK[NO]2[Br2], so the order in NO is 2.

Teacher's Note:
1. Replace the intermediate using the equilibrium.
2. Add the NO powers from both steps.

Question 20: For the elementary reaction 2A + B → A2B, the volume of the vessel is suddenly reduced to half of its original volume. The rate of reaction will
(a) increase four times
(b) remain unchanged
(c) decrease eight times
(d) increase eight times
Show Answer & Explanation

Answer: (d) increase eight times

Explanation:
1. Halving the volume doubles every concentration.
2. Rate = k[A]2[B] increases by 22 × 2 = 8 times.

Teacher's Note:
1. Concentration is inversely proportional to volume.
2. Order is 3.

Question 21: The following data are for A + B → C. Experiment 1: [A] = 0.012, [B] = 0.035, rate = 0.10. Experiment 2: [A] = 0.024, [B] = 0.035, rate = 0.80. Experiment 3: [A] = 0.012, [B] = 0.070, rate = 0.10. Experiment 4: [A] = 0.024, [B] = 0.070, rate = 0.80. The rate law is
(a) rate = k[A][B]4
(b) rate = k[A]2[B]2
(c) rate = k[A]3
(d) rate = k[B]3
Show Answer & Explanation

Answer: (c) rate = k[A]3

Explanation:
1. Experiments 1 and 2: [A] is doubled and the rate rises eight times, so the order in A is 3.
2. Experiments 1 and 3: doubling [B] leaves the rate unchanged, so the order in B is zero.
3. Rate = k[A]3.

Teacher's Note:
1. Change one concentration at a time.
2. 23 = 8.

Question 22: For 2NO(g) + O2(g) → 2NO2(g), the volume is suddenly reduced to half by increasing the pressure. The reaction is first order in O2 and second order in NO. The rate of reaction will
(a) increase to eight times
(b) diminish to one-fourth
(c) increase to four times
(d) diminish to one-eighth
Show Answer & Explanation

Answer: (a) increase to eight times

Explanation:
1. Halving the volume doubles all concentrations.
2. Rate = k[NO]2[O2] increases by 22 × 2 = 8 times.

Teacher's Note:
1. Total order is 3.
2. Rate increases by 23.

Question 23: A reaction involving two different reactants can never be
(a) first order
(b) bimolecular
(c) second order
(d) unimolecular
Show Answer & Explanation

Answer: (d) unimolecular

Explanation:
1. A unimolecular elementary step involves a single molecule.
2. A reaction with two different reactants can be of first order (as a pseudo-first-order or experimental order) but cannot be a unimolecular step.

Teacher's Note:
1. Molecularity is the number of molecules in one elementary step.
2. It cannot be 1 for two different reactants.

Question 24: How many times faster would a reaction with activation energy 65 kJ mol-1 proceed at 25 °C than at 0 °C?
(a) 5 times
(b) 11 times
(c) 2 times
(d) 16 times
Show Answer & Explanation

Answer: (b) 11 times

Explanation:
1. ln(k2/k1) = (Ea/R)(1/T1 - 1/T2) = (65000/8.314)(1/273 - 1/298) = 7818 × 3.073 × 10-4 = 2.40.
2. k2/k1 = e2.40 = 11.

Teacher's Note:
1. Use the Arrhenius equation between two temperatures.
2. Temperatures must be in kelvin.

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