JEE Chemistry Chemical Kinetics MCQs Set 32

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Chapter-wise Objective Questions: Chemical Kinetics

Navigate directly to the 50 objective questions for Chemical Kinetics using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: For a first order reaction, the time required for 99.0% completion is x times the time required for 90% completion. The value of x is
(a) 1.5
(b) 10
(c) 2
(d) 3
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. t = (2.303/k) log(100/(100 - %)).
2. t(99%) = (2.303/k) log 100 = 2 (2.303/k), t(90%) = (2.303/k) log 10 = 1 (2.303/k).
3. x = 2.

Teacher's Note:
1. Each extra '9' of completion adds one unit of 2.303/k.
2. The ratio does not depend on k.

Question 2: According to the Arrhenius equation, which statements are correct? (A) A high activation energy usually implies a fast reaction. (B) The rate constant increases with temperature because a greater number of collisions have energy exceeding the activation energy. (C) A higher activation energy means a stronger temperature dependence of the rate constant. (D) The pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy.
(a) B, C and D only
(b) B and C only
(c) B and D only
(d) C and D only
Show Answer & Explanation

Answer: (a) B, C and D only

Explanation:
1. High Ea means few molecules can react, so usually slow (A false).
2. B, C and D are the standard interpretation of k = A e-Ea/RT.

Teacher's Note:
1. Read each statement against k = A exp(-Ea/RT).
2. d ln k/dT = Ea/RT2.

Question 3: Uranium-238 (half-life 4.5 × 109 years) decays finally to lead-206. An ore contains U-238 and Pb-206 in the weight ratio 1 : 0.9, all the lead coming from uranium. The age of the ore is about
(a) 9.0 × 109 years
(b) 1.9 × 109 years
(c) 2.3 × 109 years
(d) 4.62 × 109 years
Show Answer & Explanation

Answer: (d) 4.62 × 109 years

Explanation:
1. Moles of U = 1/238, moles of Pb = 0.9/206, so original U = 1/238 + 0.9/206 and N0/N = 1 + (0.9/206)(238) = 2.04.
2. λ = 0.693/(4.5 × 109) y-1.
3. t = (1/λ) ln(2.04) = 6.49 × 109 × 0.7129 = 4.63 × 109 years, about 4.62 × 109 years.

Teacher's Note:
1. The lead present equals the uranium that has decayed.
2. Convert weights to moles before taking the ratio.

Question 4: The rate of reaction A doubles when the temperature is raised from 300 K to 310 K. If the activation energy of reaction B is twice that of A, by how much should the temperature of B be raised from 300 K for its rate to double?
(a) 4.92 K
(b) 9.84 K
(c) 2.45 K
(d) 19.67 K
Show Answer & Explanation

Answer: (a) 4.92 K

Explanation:
1. For A: ln 2 = (Ea/R)(1/300 - 1/310).
2. For B: ln 2 = (2Ea/R)(1/300 - 1/T2), so 1/300 - 1/T2 = (1/2)(1/300 - 1/310) = 5.376 × 10-5.
3. 1/T2 = 3.2796 × 10-3, T2 = 304.92 K: rise = 4.92 K.

Teacher's Note:
1. Double Ea means half the required change in 1/T.
2. Use 1/300 - 1/310 = 1.0753 × 10-4.

Question 5: For a first order reaction X → product, the concentration falls from 0.01 mol L-1 to 0.0025 mol L-1 in 40 min. The initial rate in mol L-1 min-1 is
(a) 1.73 × 10-5
(b) 3.47 × 10-5
(c) 1.73 × 10-4
(d) 3.47 × 10-4
Show Answer & Explanation

Answer: (d) 3.47 × 10-4

Explanation:
1. The concentration falls to 1/4 in 40 min, i.e. two half-lives, t1/2 = 20 min.
2. k = 0.693/20 = 0.03466 min-1.
3. Initial rate = k[X]0 = 0.03466 × 0.01 = 3.47 × 10-4 mol L-1 min-1.

Teacher's Note:
1. Rate = k x concentration.
2. Two half-lives bring 0.01 to 0.0025.

Question 6: For A(g) + B(g) ⇌ AB(g), the activation energy of the backward reaction exceeds that of the forward reaction by 2RT (J mol-1), and the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction. Given ln 2 = 0.7 and RT = 2500 J mol-1 at 300 K, the absolute value of the standard Gibbs energy change in J mol-1 is
(a) 5000
(b) 1700
(c) 10000
(d) 8500
Show Answer & Explanation

Answer: (d) 8500

Explanation:
1. K = kf/kb = (Af/Ab) e(Eb - Ef)/RT = 4 e2.
2. ln K = ln 4 + 2 = 1.4 + 2 = 3.4.
3. |ΔG| = RT ln K = 2500 × 3.4 = 8500 J mol-1.

Teacher's Note:
1. Equilibrium constant is the ratio of forward to reverse rate constants.
2. ΔG = -RT ln K.

Question 7: For a first order reaction A → P the half-life is 10 days. The time in days required for 1/4th conversion of A is (ln 2 = 0.693, ln 3 = 1.1)
(a) 5
(b) 3.2
(c) 2.5
(d) 4.1
Show Answer & Explanation

Answer: (d) 4.1

Explanation:
1. k = 0.693/10 = 0.0693 day-1.
2. After 1/4 conversion, 3/4 is left: t = (1/k) ln(4/3) = (2 ln 2 - ln 3)/0.0693 = (1.386 - 1.1)/0.0693 = 4.1 days.

Teacher's Note:
1. ln(4/3) = ln 4 - ln 3.
2. t = (1/k) ln(a/(a - x)).

Question 8: Taking A + 2B → products to be of second order overall, which rate expressions are possible? (A) dx/dt = k[A]2 (B) dx/dt = k[A][B]2 (C) dx/dt = k[A][B] (D) dx/dt = k1[A] + k2[B]2
(a) A, B and C only
(b) A only
(c) C only
(d) A and C only
Show Answer & Explanation

Answer: (d) A and C only

Explanation:
1. Overall order is the sum of the exponents.
2. k[A]2 and k[A][B] have order 2.
3. k[A][B]2 is third order and k1[A] + k2[B]2 has no single order.

Teacher's Note:
1. The exponents need not equal the stoichiometric coefficients.
2. Add the powers.

Question 9: Initial rates for A + B → P: ([A]0, [B]0, r0) = (0.1, 0.1, 0.05), (0.2, 0.1, 0.10), (0.1, 0.2, 0.05), concentrations in mol L-1 and rates in mol L-1 s-1. The rate equation and the rate constant are
(a) rate = k[A]2, k = 5 L mol-1 s-1
(b) rate = k[A][B], k = 5 L mol-1 s-1
(c) rate = k[B], k = 0.5 s-1
(d) rate = k[A], k = 0.5 s-1
Show Answer & Explanation

Answer: (d) rate = k[A], k = 0.5 s-1

Explanation:
1. Doubling [A] doubles the rate: first order in A.
2. Doubling [B] leaves the rate unchanged: zero order in B.
3. rate = k[A]; k = 0.05/0.1 = 0.5 s-1.

Teacher's Note:
1. Compare runs in which only one concentration changes.
2. Unit of k for first order is s-1.

Question 10: For the reaction 3A + 2B → C + D, the differential rate law is correctly written as
(a) -3 d[A]/dt = d[C]/dt = k[A]^n[B]^m
(b) -d[A]/dt = d[C]/dt = k[A]^n[B]^m
(c) -(1/3) d[A]/dt = d[C]/dt = k[A]^n[B]^m
(d) (1/3) d[A]/dt = d[C]/dt = k[A]^n[B]^m
Show Answer & Explanation

Answer: (c) -(1/3) d[A]/dt = d[C]/dt = k[A]^n[B]^m

Explanation:
1. Rate of reaction = -(1/3) d[A]/dt = -(1/2) d[B]/dt = d[C]/dt = d[D]/dt.
2. It equals k[A]^n[B]^m with n and m found experimentally.

Teacher's Note:
1. Divide each rate by its coefficient.
2. Reactant terms carry a minus sign.

Question 11: The decomposition of X shows a rate constant of 0.05 g per year. The number of years needed for 5 g of X to decompose to 2.5 g is
(a) 20
(b) 40
(c) 25
(d) 50
Show Answer & Explanation

Answer: (d) 50

Explanation:
1. The unit g per year shows a zero order reaction.
2. Amount decomposed = 2.5 g, so t = 2.5/0.05 = 50 years.

Teacher's Note:
1. The unit of k reveals the order.
2. Zero order: x = kt.

Question 12: The reaction 2X → B is zero order. For initial [X] = 0.2 M the half-life is 6 h. When the initial concentration of X is 0.5 M, the time to reach 0.2 M is
(a) 18.0 h
(b) 12.0 h
(c) 7.2 h
(d) 9.0 h
Show Answer & Explanation

Answer: (a) 18.0 h

Explanation:
1. Zero order: t1/2 = a/(2k) gives k = 0.2/(2 × 6) = 1/60 M h-1.
2. Time = (0.5 - 0.2)/k = 0.3 × 60 = 18 h.

Teacher's Note:
1. Rate is constant in zero order.
2. Time = concentration change/k.

Question 13: The rate of a reaction quadruples when the temperature changes from 300 K to 310 K. The activation energy of this reaction is (ln 2 = 0.693, R = 8.314 J mol-1 K-1)
(a) 214.4 kJ mol-1
(b) 107.2 kJ mol-1
(c) 53.6 kJ mol-1
(d) 26.8 kJ mol-1
Show Answer & Explanation

Answer: (b) 107.2 kJ mol-1

Explanation:
1. ln 4 = (Ea/R)(1/300 - 1/310) = (Ea/R)(1.0753 × 10-4).
2. Ea = 8.314 × 1.386/1.0753 × 10-4 = 1.072 × 105 J mol-1 = 107.2 kJ mol-1.

Teacher's Note:
1. ln 4 = 2 ln 2 = 1.386.
2. Convert J to kJ.

Question 14: Which of the following can be used to determine the activation energy of a reaction? (A) Evaluating the rate constant at the standard temperature. (B) Evaluating the rates (at equal concentrations) at two different temperatures. (C) Evaluating the rate constant at two different temperatures. (D) Changing the concentration of the reactants.
(a) A, B and C only
(b) B and C only
(c) B, C and D only
(d) C only
Show Answer & Explanation

Answer: (b) B and C only

Explanation:
1. ln(k2/k1) = (Ea/R)(1/T1 - 1/T2) needs k at two temperatures.
2. At equal concentrations the ratio of rates equals the ratio of rate constants, so B also works.
3. A single temperature or a concentration change does not give Ea.

Teacher's Note:
1. Two temperatures are needed.
2. Rates at the same composition are proportional to k.

Question 15: Gas A decomposes by a first order reaction A(g) → 2B(g) + C(g) at constant volume and temperature. At t = 0 gas A is present along with some gas C, and the mole fraction of C is 1/3. At t = t1 the total pressure is half of the final total pressure at t = infinity, which is 35 bar. The rate constant is k = (log 64 - log 49) s-1 (logarithms to base 10, used with the 2.303 form of the first order law). The value of t1 in seconds is
(a) 2.3 s
(b) 1.15 s
(c) 1.5 s
(d) 2.15 s
Show Answer & Explanation

Answer: (b) 1.15 s

Explanation:
1. At t1 the pressure of A = 2u - 0.25u = 1.75u, against 2u initially.
2. 2.303 log(2/1.75) = k t, with 2.303 log(8/7) = 0.1335.
3. k = log(64/49) = 2 log(8/7) = 2 × 0.0580 = 0.1160 s-1, so t1 = 0.1335/0.1160 = 1.15 s.

Teacher's Note:
1. Use the same log base on both sides.
2. A falls from 2u to 1.75u.

Question 16: For 2N2O5 → 4NO2 + O2 the rate equation is written as -d[N2O5]/dt = k[N2O5] and also d[NO2]/dt = k'[N2O5]. The relation between k and k' is
(a) k = k'
(b) k = 2k'
(c) 2k = k'
(d) k = 4k'
Show Answer & Explanation

Answer: (c) 2k = k'

Explanation:
1. d[NO2]/dt = 2 × (-d[N2O5]/dt) because 4 NO2 form for 2 N2O5 used.
2. So k'[N2O5] = 2k[N2O5], k' = 2k.

Teacher's Note:
1. NO2 forms twice as fast as N2O5 disappears.
2. Use the stoichiometric ratio 4 : 2.

Question 17: A reaction A → B has rate constants k1 (T1) and k2 (T2) without a catalyst (activation energy Ea1) and k3 (T1) and k4 (T2) with a positive catalyst (Ea lowered). Which of the following is correct?
(a) k4 < k3 and k2 < k1, if T2 > T1
(b) k4 > k3 and k2 > k1, if T2 < T1
(c) k4 < k3 and k2 > k1, if T2 < T1
(d) k4 > k3 and k2 > k1, if T2 > T1
Show Answer & Explanation

Answer: (d) k4 > k3 and k2 > k1, if T2 > T1

Explanation:
1. If T2 > T1, raising the temperature raises k with or without catalyst: k2 > k1 and k4 > k3.
2. If T2 < T1 both pairs fall, so k2 < k1 and k4 < k3. Only the first option is consistent.

Teacher's Note:
1. The direction of the temperature change decides the sign of the change in k.
2. A catalyst does not reverse the temperature trend.

Question 18: For 2A + B → products the initial rates (rate of reaction defined as -(1/2) d[A]/dt) are: (A 0.10, B 0.20, 6.93 × 10-3); (A 0.10, B 0.25, 6.93 × 10-3); (A 0.20, B 0.30, 1.386 × 10-2), concentrations in mol L-1 and rates in mol L-1 min-1. The time in minutes to consume half of A is
(a) 1
(b) 5
(c) 10
(d) 100
Show Answer & Explanation

Answer: (b) 5

Explanation:
1. Runs 1 and 2: rate independent of [B]. Runs 1 and 3: doubling [A] doubles the rate. So rate = k[A], k = 6.93 × 10-3/0.10 = 0.0693 min-1.
2. Since rate = -(1/2) d[A]/dt, -d[A]/dt = 2k[A], so the first order constant for A is 0.1386 min-1.
3. t1/2 = 0.693/0.1386 = 5 min.

Teacher's Note:
1. Rate of reaction differs from rate of consumption of A by the factor 2.
2. Half-life of A uses the constant for A.

Question 19: The inversion of cane sugar in excess water with an acid catalyst is a reaction that is
(a) first order with respect to cane sugar
(b) second order
(c) third order
(d) zero order with respect to cane sugar
Show Answer & Explanation

Answer: (a) first order with respect to cane sugar

Explanation:
1. Water is present in large excess, so its concentration is effectively constant.
2. Rate = k[sucrose][H2O] reduces to k'[sucrose], a pseudo first order law.

Teacher's Note:
1. Such reactions are called pseudo first order.
2. Molecularity is two but the order is one.

Question 20: For 2A + B → products, doubling the concentrations of both A and B raises the rate from 0.3 to 2.4 mol L-1 s-1, and doubling A alone raises it from 0.3 to 0.6 mol L-1 s-1. Which statement is correct?
(a) The order with respect to B is 2
(b) The order with respect to B is 1
(c) The total order of the reaction is 4
(d) The order with respect to A is 2
Show Answer & Explanation

Answer: (a) The order with respect to B is 2

Explanation:
1. Doubling A alone doubles the rate: order in A = 1.
2. Both doubled gives 8 times: 2 × 2^m = 8, so m = 2.
3. Order in B is 2 and total order 3.

Teacher's Note:
1. Separate the contributions of A and B.
2. 2.4/0.3 = 8.

Question 21: Acid catalysed hydrolysis of a compound A at 30 °C has a half-change time of 100 min in a buffer of pH 5 and 10 min at pH 4. Both half-times are independent of the initial concentration of A. If rate = k[A]^a[H+]^b, the value of (a + b) is
(a) 2
(b) 1
(c) 3
(d) 0
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. Half-time independent of [A]: the reaction is first order in A, a = 1.
2. Between pH 5 and 4 [H+] increases 10 times and the half-time falls 10 times, so rate is proportional to [H+]: b = 1.
3. a + b = 2.

Teacher's Note:
1. In a buffer [H+] is constant, so the reaction behaves as pseudo first order.
2. Half-time is inversely proportional to the effective rate constant.

Question 22: For the equilibrium A(g) ⇌ B(g), ΔH = -40 kJ mol-1. If the ratio of the activation energies of the forward and reverse reactions Ef : Eb is 2 : 3, then
(a) Ef = 60 kJ mol-1; Eb = 100 kJ mol-1
(b) Ef = 30 kJ mol-1; Eb = 70 kJ mol-1
(c) Ef = 80 kJ mol-1; Eb = 120 kJ mol-1
(d) Ef = 70 kJ mol-1; Eb = 30 kJ mol-1
Show Answer & Explanation

Answer: (c) Ef = 80 kJ mol-1; Eb = 120 kJ mol-1

Explanation:
1. ΔH = Ef - Eb = -40 kJ.
2. With Ef = 2x, Eb = 3x: 2x - 3x = -40, x = 40.
3. Ef = 80 kJ mol-1, Eb = 120 kJ mol-1.

Teacher's Note:
1. ΔH equals Ef minus Eb.
2. Exothermic means Eb is larger.

Question 23: N2O5 decomposes to NO2 and O2 by first order kinetics. After 50 min the pressure inside the vessel rises from 50 mm Hg to 87.5 mm Hg. The pressure of the gaseous mixture after 100 min at constant temperature is
(a) 106.25 mm Hg
(b) 116.25 mm Hg
(c) 175.0 mm Hg
(d) 136.25 mm Hg
Show Answer & Explanation

Answer: (a) 106.25 mm Hg

Explanation:
1. Total = 50 + 1.5p (p = N2O5 reacted); 87.5 gives p = 25 mm, half of 50: t1/2 = 50 min.
2. At 100 min (two half-lives) N2O5 left = 12.5 mm, p = 37.5 mm: total = 50 + 56.25 = 106.25 mm Hg.

Teacher's Note:
1. Two half-lives leave one quarter.
2. Use the pressure gain 1.5 per unit of N2O5 reacted.

Question 24: An organic compound decomposes by first order kinetics. The times to fall to 1/8 and to 1/10 of the initial concentration are t1/8 and t1/10. The value of 10 x t1/8 / t1/10 is (log10 2 = 0.3)
(a) 30
(b) 8
(c) 3
(d) 9
Show Answer & Explanation

Answer: (d) 9

Explanation:
1. t = (2.303/k) log(c0/c).
2. t1/8 = (2.303/k) log 8 = (2.303/k)(0.9); t1/10 = (2.303/k)(1).
3. 10 × 0.9 = 9.

Teacher's Note:
1. log 8 = 3 log 2 = 0.9.
2. The ratio is independent of k.

Question 25: The rate law for A + B → product is rate = k[A][B]. If the concentration of B is increased from 0.1 to 0.3 mol while that of A is kept at 0.1 mol, the rate constant will be
(a) 9k
(b) 3k
(c) k
(d) k/3
Show Answer & Explanation

Answer: (c) k

Explanation:
1. The rate constant depends only on temperature (and catalyst).
2. Changing a concentration changes the rate, not k.

Teacher's Note:
1. Rate triples but k stays the same.
2. Do not confuse rate and rate constant.

Question 26: Copper-64 (half-life 12.8 h) decays by β-minus emission (38%), β-plus emission (19%) and electron capture (43%). The partial half-life for β-minus emission, to two decimals, is
(a) 33.68 h
(b) 29.77 h
(c) 12.8 h
(d) 67.37 h
Show Answer & Explanation

Answer: (a) 33.68 h

Explanation:
1. Partial decay constant = fraction x total decay constant, so partial half-life = t1/2/fraction.
2. For β-minus: 12.8/0.38 = 33.68 h. (For β-plus 67.37 h, electron capture 29.77 h.)

Teacher's Note:
1. The three paths are parallel first order processes.
2. Smaller fraction means longer partial half-life.

Question 27: 3A(g) → 2B(g) + 2C(g) follows first order kinetics. Initially only A is present. The total pressure after 20 min is 3.5 atm and after infinite time is 4 atm. Which are true? (A) t50% = 20 min (B) t75% = 40 min (C) t99% = 64/3 min (D) t87.5% = 60 min
(a) A and B only
(b) A and D only
(c) A, B, C and D only
(d) A, B and D only
Show Answer & Explanation

Answer: (d) A, B and D only

Explanation:
1. Let p0 be the initial pressure; each 3 atm of A gives 4 atm of products, so total at infinity = (4/3)p0 = 4 and p0 = 3 atm.
2. After 20 min, if 3y atm of A has reacted, total = 3 + y = 3.5, y = 0.5, so A left = 3 - 1.5 = 1.5 atm = half.
3. So t50% = 20 min, t75% = 40 min, t87.5% = 60 min. t99% = 20 log2(100) = 133 min, not 64/3 min.

Teacher's Note:
1. Find p0 from the final pressure first.
2. Each half-life is 20 min.

Question 28: For 2SO2 + O2 → 2SO3 the rate of the reaction is measured as d[O2]/dt = -2.5 × 10-4 mol L-1 s-1. The rate of change of [SO2] in mol L-1 s-1 is
(a) -1.25 × 10-4
(b) -5.00 × 10-4
(c) -2.50 × 10-4
(d) -3.75 × 10-4
Show Answer & Explanation

Answer: (b) -5.00 × 10-4

Explanation:
1. -d[SO2]/dt = 2 (-d[O2]/dt) from the stoichiometry 2 : 1.
2. d[SO2]/dt = -5.00 × 10-4 mol L-1 s-1.

Teacher's Note:
1. SO2 is used up twice as fast as O2.
2. Keep the minus sign for a reactant.

Question 29: For A + 2B → C with rate = k[A][B], the concentration of A is kept the same and that of B is doubled. The rate
(a) is halved
(b) is doubled
(c) is quadrupled
(d) remains the same
Show Answer & Explanation

Answer: (b) is doubled

Explanation:
1. Rate is first order in B.
2. Doubling [B] doubles the rate.

Teacher's Note:
1. The order, not the stoichiometric coefficient, matters.
2. The coefficient 2 does not enter the rate law here.

Question 30: Gas A decomposes by a first order reaction A(g) → 2B(g) + C(g) at constant volume and temperature. At t = 0 gas A is present along with some gas C, and the mole fraction of C is 1/3. At t = t1 the total pressure is half of the final total pressure at t = infinity, which is 35 bar. The ratio of the rate constants at t = 0, t = t1 and t = infinity is
(a) 2 : 3 : 4
(b) 1 : 2 : 3
(c) 1 : 3 : 5
(d) 1 : 1 : 1
Show Answer & Explanation

Answer: (d) 1 : 1 : 1

Explanation:
1. The rate constant of a reaction depends only on temperature (and catalyst).
2. The temperature is constant, so k is the same at all three times: 1 : 1 : 1.

Teacher's Note:
1. Do not confuse the rate constant with the rate.
2. The rate falls as A is used up, k does not.

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