JEE Chemistry Chemical Kinetics MCQs Set 31

Multiple Choice Questions (MCQs) for JEE Chemistry: Chemical Kinetics

Access targeted multiple-choice questions for Chemical Kinetics designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Chemical Kinetics MCQs for JEE Chemistry

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: A reactant A undergoes two parallel first order reactions A → 2B (k1) and A → 3C (k2), starting with pure A. At time t the correct material balance is
(a) [A]0 = [A]t + [B]t/2 + [C]t/3
(b) [A]0 = (2/3)[A]t + [B]t + [C]t
(c) [A]0 = [A]t + 2[B]t + 3[C]t
(d) [A]0 = [A]t + [B]t + [C]t
Show Answer & Explanation

Answer: (a) [A]0 = [A]t + [B]t/2 + [C]t/3

Explanation:
1. Each mole of A that reacts gives 2 mol B or 3 mol C.
2. Moles of A reacted = [B]/2 + [C]/3.
3. So [A]0 = [A]t + [B]t/2 + [C]t/3.

Teacher's Note:
1. Divide the product amount by its stoichiometric coefficient.
2. Atoms of A are conserved.

Question 2: The rate of 2SO2(g) + O2(g) → 2SO3(g) over platinum is rate = k[SO2]/[SO3]1/2. Which statements are correct? (A) The overall order is -1/2. (B) The overall order is +1/2. (C) The reaction slows down as SO3 builds up. (D) The rate does not depend on the concentration of SO3.
(a) C only
(b) A, B and C only
(c) B only
(d) B and C only
Show Answer & Explanation

Answer: (d) B and C only

Explanation:
1. Order = 1 - 1/2 = +1/2, so B is correct and A is wrong.
2. The rate varies as [SO3]-1/2, so it falls as SO3 accumulates (C true, D false).

Teacher's Note:
1. Add the exponents with their signs.
2. A negative exponent is an inhibiting effect.

Question 3: For aG + bH → products, doubling the concentrations of both G and H increases the rate eight times, while doubling G alone doubles the rate. The overall order is
(a) 0
(b) 3
(c) 2
(d) 1
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. Doubling G alone doubles the rate: first order in G.
2. Doubling both multiplies by 8 = 23, so order in H = 3 - 1 = 2.
3. Overall order = 1 + 2 = 3.

Teacher's Note:
1. Separate the contributions of each reactant.
2. Rate multiplies by 2^n for order n.

Question 4: The concentration of R in R → P is 1.0, 0.75, 0.40 and 0.10 molar at times 0.0, 0.05, 0.12 and 0.18 min. The order of the reaction is
(a) 1
(b) 2
(c) 0
(d) 3
Show Answer & Explanation

Answer: (c) 0

Explanation:
1. Fall in concentration per unit time: 0.25/0.05 = 5, 0.35/0.07 = 5, 0.30/0.06 = 5 M per min.
2. The rate is constant, so the reaction is zero order.

Teacher's Note:
1. Compare the rate over successive intervals.
2. Constant rate means zero order.

Question 5: For a reaction obeying the Arrhenius equation, a plot of ln k against 1/(RT) is a straight line with gradient -y (in energy units). The energy required to activate the reactant is
(a) -y
(b) y
(c) yR
(d) y/R
Show Answer & Explanation

Answer: (b) y

Explanation:
1. ln k = ln A - Ea x (1/RT), so the gradient is -Ea.
2. -Ea = -y, hence Ea = y.

Teacher's Note:
1. Treat 1/RT as the x variable.
2. The factor R is already in the axis.

Question 6: Which of the following are true about the rate of a reaction? (A) The rate of a reaction cannot be negative. (B) The rate of a reaction in terms of a reactant is the change in its concentration per unit time. (C) Average rate and instantaneous rate can never be equal. (D) The rate depends on the surface area of the reactants.
(a) A, B, C and D only
(b) A, B and D only
(c) A and D only
(d) B and D only
Show Answer & Explanation

Answer: (b) A, B and D only

Explanation:
1. Rate of reaction is always positive (a negative sign is only used for reactants in the derivative).
2. It is a change in concentration per unit time.
3. Average and instantaneous rate are equal when the rate is constant (zero order), so C is false.
4. For heterogeneous reactions rate depends on surface area, so D is true.

Teacher's Note:
1. Check each statement separately.
2. A zero order reaction has equal average and instantaneous rates.

Question 7: A reaction A → B is studied without a catalyst: at temperature T1 the rate constant is k1 and activation energy Ea1; at T2 they are k2 and Ea2 (activation energy independent of temperature). Which statement is correct?
(a) if T1 > T2, then k1 < k2 for an exothermic reaction
(b) if T1 > T2, then k1 > k2 only for an endothermic reaction
(c) Ea1 is greater than Ea2
(d) if T1 > T2, then k1 > k2 always
Show Answer & Explanation

Answer: (d) if T1 > T2, then k1 > k2 always

Explanation:
1. With Ea > 0, k = A e-Ea/RT increases with temperature for every reaction, whether endothermic or exothermic.
2. So k1 > k2 whenever T1 > T2, and Ea1 = Ea2.

Teacher's Note:
1. Thermicity affects the equilibrium constant, not the sign of the temperature dependence of k.
2. Activation energy is temperature independent in this model.

Question 8: The rate constant for 2N2O5 → 4NO2 + O2 is 3 × 10-5 s-1. If the rate is 2.4 × 10-5 mol L-1 s-1, the concentration of N2O5 in mol L-1 is
(a) 1.4
(b) 0.8
(c) 0.04
(d) 1.2
Show Answer & Explanation

Answer: (b) 0.8

Explanation:
1. The unit s-1 shows a first order reaction: rate = k[N2O5].
2. [N2O5] = 2.4 × 10-5 / 3 × 10-5 = 0.8 mol L-1.

Teacher's Note:
1. Use the unit of k to find the order.
2. Divide rate by k.

Question 9: A substance of initial concentration 1.386 mol dm-3 becomes half in 40 s by first order kinetics and in 20 s by zero order kinetics under the same conditions. The ratio k1/k0 of the first order to the zero order rate constant is
(a) 2.0 mol-1 dm3
(b) 0.5 mol-1 dm3
(c) 1.0 mol-1 dm3
(d) 1.5 mol-1 dm3
Show Answer & Explanation

Answer: (b) 0.5 mol-1 dm3

Explanation:
1. k1 = 0.693/40 = 0.01733 s-1.
2. k0 = (1.386/2)/20 = 0.03465 mol dm-3 s-1.
3. k1/k0 = 0.5 mol-1 dm3.

Teacher's Note:
1. Zero order: t1/2 = a/2k0.
2. Keep track of units.

Question 10: 80% of a radioactive element is left after a certain time t starting from the pure element. The whole number of such periods that must elapse from the start for the amount left to be just over 50% is
(a) 2
(b) 3
(c) 5
(d) 4
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. After n periods the fraction left is (0.8)^n.
2. (0.8)3 = 0.512 (just over 50%) while (0.8)4 = 0.410.
3. So n = 3.

Teacher's Note:
1. Compute successive powers of 0.8 until the value drops below 0.5.
2. Exactly 50% would need n = 3.1.

Question 11: The rate constant of a reaction varies as log k (min-1) = 5 - 2000/T. Which conclusions are correct (take R = 2 cal K-1 mol-1)? (A) The pre-exponential factor A is 5. (B) Ea is 2000 kcal. (C) The pre-exponential factor A is 105 min-1. (D) Ea is 9.212 kcal.
(a) C and D only
(b) A, C and D only
(c) B, C and D only
(d) D only
Show Answer & Explanation

Answer: (a) C and D only

Explanation:
1. log A = 5, so A = 105 min-1.
2. Ea = 2.303 x R x 2000 = 2.303 × 2 × 2000 = 9212 cal = 9.212 kcal.

Teacher's Note:
1. The intercept is log A, not A.
2. Slope = -Ea/(2.303 R).

Question 12: If 50% of a reaction occurs in 100 s and 75% in 200 s, the order of the reaction is
(a) 1
(b) 2
(c) 0
(d) 3
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. The second 100 s again halves the remaining reactant: constant half-life of 100 s.
2. Constant half-life means first order.

Teacher's Note:
1. 75% is two half-lives.
2. For zero order 100% would be done in 200 s.

Question 13: 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. At t = t1 the pressure of gas B is
(a) 2.5 bar
(b) data is insufficient
(c) 1.25 bar
(d) 5.0 bar
Show Answer & Explanation

Answer: (a) 2.5 bar

Explanation:
1. Let initial moles be A = 2u and C = u (mole fraction of C = 1/3); initial pressure 3u.
2. At infinity A is gone: B = 4u, C = 3u, total 7u = 35 bar, so u = 5 bar.
3. At t1 total = 17.5 bar = 3.5u. If y is the decrease in A (in units u): 3 + 2y = 3.5, y = 0.25, so p(B) = 2y u = 0.5 × 5 = 2.5 bar.

Teacher's Note:
1. Express the pressures in a unit u fixed by the final pressure.
2. Each mole of A gives 2 B and 1 °C, a net gain of 2.

Question 14: At high temperature acetic acid decomposes simultaneously by CH3COOH → CH4 + CO2 (k1 = 3 s-1) and CH3COOH → CH2CO + H2O (k2 = 4 s-1). The fraction of acetic acid that reacts by the first path is
(a) 4/7
(b) none of these
(c) 3/4
(d) 3/7
Show Answer & Explanation

Answer: (d) 3/7

Explanation:
1. For parallel first order paths the fractions are in the ratio of the rate constants.
2. Fraction by path 1 = k1/(k1 + k2) = 3/7.

Teacher's Note:
1. Both paths consume the same acid, so their rates add.
2. The ratio of products equals k1 : k2.

Question 15: Select the correct statements about reaction mechanisms. (A) The molecularity of an elementary reaction indicates how many reactant molecules take part in the step. (B) The rate law of an elementary reaction can be written from its stoichiometry. (C) The slowest elementary step governs the overall rate of formation of the product. (D) A rate law is often derived from a proposed mechanism using the steady state approximation or a pre-equilibrium.
(a) A, B and D only
(b) A, C and D only
(c) A, B, C and D only
(d) A, B and C only
Show Answer & Explanation

Answer: (c) A, B, C and D only

Explanation:
1. All four are standard results of mechanism analysis.
2. For an elementary step the exponents equal the stoichiometric coefficients.

Teacher's Note:
1. Think of the rate determining step as a bottleneck.
2. Steady state and pre-equilibrium are the two usual tools.

Question 16: The rate of radioactive disintegration (-dN/dt) is also known as (A) activity (B) radioactivity (C) half-life period (D) average life period
(a) A, B and C only
(b) A and B only
(c) A only
(d) B only
Show Answer & Explanation

Answer: (b) A and B only

Explanation:
1. -dN/dt gives the number of disintegrations per unit time, called the activity or radioactivity of the sample.
2. Half-life and average life are times, not rates.

Teacher's Note:
1. Units of activity are becquerel or curie.
2. Rate means per unit time.

Question 17: The rate of a first order reaction is 0.04 mol L-1 s-1 at 10 minutes and 0.03 mol L-1 s-1 at 20 minutes after initiation. The half-life of the reaction is about
(a) 6 min
(b) 12 min
(c) 48 min
(d) 24 min
Show Answer & Explanation

Answer: (d) 24 min

Explanation:
1. Rate is proportional to concentration: r1/r2 = ek(10) = 4/3.
2. k = ln(4/3)/10 = 0.02877 min-1.
3. t1/2 = 0.693/0.02877 = 24.1 min.

Teacher's Note:
1. For first order the rate falls in the same ratio as concentration.
2. The interval is 10 min.

Question 18: The reaction 2N2O5(g) → 4NO2(g) + O2(g) follows first order kinetics. The pressure in a vessel containing only N2O5 increased from 50 mm Hg to 87.5 mm Hg in 30 min. The total pressure after 60 min will be
(a) 106.25 mm Hg
(b) 125 mm Hg
(c) 150 mm Hg
(d) 116.25 mm Hg
Show Answer & Explanation

Answer: (a) 106.25 mm Hg

Explanation:
1. If p of N2O5 reacts, total = 50 + p(4/2 + 1/2 - 1) = 50 + 1.5p. 87.5 = 50 + 1.5p gives p = 25 mm.
2. N2O5 left = 25 mm = half of 50 in 30 min, so t1/2 = 30 min.
3. At 60 min N2O5 left = 12.5 mm, reacted 37.5 mm: total = 50 + 1.5 × 37.5 = 106.25 mm Hg.

Teacher's Note:
1. Each 2 mol N2O5 gives 5 mol of gas, a net gain of 1.5 per mole reacted.
2. Find the half-life first.

Question 19: For N2(g) + 3H2(g) → 2NH3(g), the correct relation between rate expressions is
(a) Rate = -d[N2]/dt = -(1/3) d[H2]/dt = -(1/2) d[NH3]/dt
(b) Rate = d[N2]/dt = d[H2]/dt = d[NH3]/dt
(c) Rate = -d[N2]/dt = -(1/3) d[H2]/dt = (1/2) d[NH3]/dt
(d) Rate = -d[N2]/dt = -3 d[H2]/dt = 2 d[NH3]/dt
Show Answer & Explanation

Answer: (c) Rate = -d[N2]/dt = -(1/3) d[H2]/dt = (1/2) d[NH3]/dt

Explanation:
1. Rate of reaction = (-1/a) d[reactant]/dt = (1/b) d[product]/dt, with a and b the coefficients.
2. Hence the first option; reactants have a minus sign and the product a plus sign.

Teacher's Note:
1. Divide by the stoichiometric coefficient.
2. Products form (positive), reactants are consumed (negative).

Question 20: Order of reaction (Column-1): (I) first, (II) second, (III) third, (IV) zero. Rate constant expression (Column-2): (i) k = (1/2t)[1/(a-x)2 - 1/a2], (ii) k = (1/t)[1/(a-x) - 1/a], (iii) k = x/t, (iv) k = (1/t) ln[a/(a-x)]. Half-life (Column-3): (P) proportional to 1/a2, (Q) proportional to 1/a, (R) doubled on doubling the initial concentration, (S) the 50% reaction takes the same time even if the concentration is halved or doubled. The correct combination for the reaction A → B with rate constant 2 × 10-3 min-1 is
(a) II - ii - Q
(b) IV - iv - S
(c) I - iii - R
(d) I - iv - S
Show Answer & Explanation

Answer: (d) I - iv - S

Explanation:
1. The unit min-1 shows the reaction is first order (I).
2. First order: k = (1/t) ln[a/(a-x)] (iv), and the half-life is independent of concentration (S).
3. So I - iv - S.

Teacher's Note:
1. Check the unit of k to find the order.
2. Match formula and half-life with the same order.

Question 21: In a plot of ln k against 1/T for a reaction the slope is -4606 K. If the rate constant at 400 K is 10-5 s-1, the rate constant at 500 K is
(a) 4 × 10-4 s-1
(b) 10-6 s-1
(c) 2 × 10-4 s-1
(d) 10-4 s-1
Show Answer & Explanation

Answer: (d) 10-4 s-1

Explanation:
1. ln(k2/k1) = 4606 (1/400 - 1/500) = 4606 × 5 × 10-4 = 2.303.
2. k2/k1 = 10, so k2 = 10-4 s-1.

Teacher's Note:
1. Slope = -Ea/R.
2. ln 10 = 2.303.

Question 22: In the first order gaseous reaction A(g) → 2B(g) + C(g) the initial pressure was 400 mm and it became 1000 mm (total) after 20 min. Which are correct? (A) The half-life of A is 10 min. (B) The rate constant is 0.0693 min-1. (C) The partial pressure of C at 30 min is 350 mm. (D) The total pressure after 30 min is 1100 mm.
(a) A, C and D only
(b) A, B, C and D only
(c) B, C and D only
(d) A, B and D only
Show Answer & Explanation

Answer: (b) A, B, C and D only

Explanation:
1. Total = 400 + 2x = 1000, x = 300 mm, so A left = 100 mm = 1/4 in 20 min, i.e. 2 half-lives; t1/2 = 10 min and k = 0.693/10 = 0.0693 min-1.
2. At 30 min A left = 400/8 = 50 mm, x = 350 mm, so p(C) = 350 mm.
3. Total = 400 + 2(350) = 1100 mm. All four are correct.

Teacher's Note:
1. Pressure rise = 2x for this stoichiometry.
2. 30 min is 3 half-lives.

Question 23: For the same reaction as above (2X(g) → 3Y(g) + 2Z(g); p(X) = 800, 400, 200 mm Hg at 0, 100, 200 min), the rate constant and the time for 75% completion are
(a) 3.47 × 10-3 min-1 and 200 min
(b) 6.93 × 10-2 min-1 and 20 min
(c) 6.93 × 10-3 min-1 and 300 min
(d) 6.93 × 10-3 min-1 and 200 min
Show Answer & Explanation

Answer: (d) 6.93 × 10-3 min-1 and 200 min

Explanation:
1. First order with t1/2 = 100 min: k = 0.693/100 = 6.93 × 10-3 min-1.
2. 75% completion = two half-lives = 200 min.

Teacher's Note:
1. 75% always needs two half-lives for first order.
2. k = 0.693/t1/2.

Question 24: In dilute aqueous H2SO4 the complex diaquodioxalatoferrate(II) is oxidised by MnO4-. For this reaction, the ratio of the rate of change of [H+] to the rate of change of [MnO4-] is
(a) 8
(b) 24
(c) 3
(d) 5
Show Answer & Explanation

Answer: (a) 8

Explanation:
1. Each complex ion [Fe(C2O4)2(H2O)2]2- loses 5 electrons (1 for Fe(II) to Fe(III) and 4 for two oxalates to 4 CO2).
2. MnO4- + 8H+ + 5e → Mn2+ + 4H2O, so one MnO4- oxidises one complex ion and uses 8 H+.
3. Ratio of rates = 8.

Teacher's Note:
1. Balance the electrons first.
2. Count only the H+ taken by permanganate.

Question 25: The initial rate of hydrolysis of methyl acetate (1 M) by a weak acid HA (1 M) is 1/100 of that by a strong acid HX (1 M) at 25 °C. The Ka of HA is about
(a) 1 × 10-3
(b) 1 × 10-4
(c) 1 × 10-6
(d) 1 × 10-5
Show Answer & Explanation

Answer: (b) 1 × 10-4

Explanation:
1. The rate is proportional to [H+].
2. For HX, [H+] = 1 M; for HA the rate is 1/100, so [H+] = 0.01 M.
3. Ka = [H+]2/[HA] = (0.01)2/(1 - 0.01) = 1.0 × 10-4.

Teacher's Note:
1. Strong acid is fully ionised.
2. Use Ka = x2/(c - x).

Question 26: In a bimolecular reaction the steric factor P was found experimentally to be 4.5. Which options are correct? (A) The activation energy of the reaction is unaffected by the value of the steric factor. (B) The experimentally determined value of the frequency factor is higher than that predicted by the Arrhenius equation. (C) The value of the frequency factor predicted by the Arrhenius equation is higher than that determined experimentally. (D) Since P = 4.5, the reaction will not proceed unless an effective catalyst is used.
(a) A, B and D only
(b) B only
(c) A and B only
(d) A, B and C only
Show Answer & Explanation

Answer: (c) A and B only

Explanation:
1. P = A(experimental)/A(theory). P = 4.5 means the experimental frequency factor is 4.5 times that predicted by simple collision theory.
2. Ea is not affected by P (A true, B true).
3. C is the opposite and D is incorrect, since the reaction proceeds without a catalyst.

Teacher's Note:
1. P greater than 1 is possible when the theory underestimates A.
2. Ea comes from the temperature dependence of k.

Question 27: In the Haber process N2 + 3H2 → 2NH3 the rate of formation of NH3 is 2.0 × 10-4 mol L-1 s-1. The rate of disappearance of N2 is p x 10-4 mol L-1 s-1. The value of p is
(a) 0.5
(b) 2
(c) 1
(d) 4
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. Rate of reaction = (1/2) d[NH3]/dt = -d[N2]/dt.
2. So -d[N2]/dt = (1/2)(2.0 × 10-4) = 1.0 × 10-4, hence p = 1.

Teacher's Note:
1. Divide each rate by its stoichiometric coefficient.
2. Two moles of NH3 form for every mole of N2 used.

Question 28: A undergoes two parallel first order reactions A → 3B (k1) and A → 4C (k2), starting from pure A. The ratio of the rate of production of B to the rate of production of C is
(a) dependent on the initial concentration of A
(b) independent of temperature
(c) independent of time
(d) independent of the mechanism of the reaction
Show Answer & Explanation

Answer: (c) independent of time

Explanation:
1. d[B]/dt = 3k1[A] and d[C]/dt = 4k2[A].
2. The ratio 3k1/(4k2) contains no [A] or t, so it is independent of time (and of the initial concentration).
3. It does depend on temperature through k1 and k2.

Teacher's Note:
1. [A] cancels in the ratio.
2. k1 and k2 usually have different activation energies.

Question 29: In the biologically catalysed oxidation of ethanol the concentration of ethanol falls (first order) from 800 mol dm-3 to 50 mol dm-3 in 2 × 104 s. The rate constant in s-1 is
(a) 5.00 × 10-5
(b) 1.00 × 10-4
(c) 1.38 × 10-4
(d) 3.45 × 10-5
Show Answer & Explanation

Answer: (c) 1.38 × 10-4

Explanation:
1. k = (2.303/t) log(800/50) = (2.303/(2 × 104)) log 16.
2. log 16 = 1.204, so k = 2.303 × 1.204/(2 × 104) = 1.386 × 10-4 s-1.

Teacher's Note:
1. 800/50 = 16 = 24, which is four half-lives.
2. k = 4 × 0.693/t.

Question 30: The half-life of ammonia decomposition on a tungsten wire is 3.52, 1.82 and 0.93 min for initial pressures 50, 100 and 200 (same units). The order of the reaction with respect to the data given is
(a) 1
(b) 0
(c) 3
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. t1/2 is proportional to p1-n.
2. Doubling p roughly halves t1/2 (3.52 → 1.82 → 0.93), so p1-n = p-1.
3. 1 - n = -1, n = 2.

Teacher's Note:
1. For zero order the half-life would increase with pressure; for first order it stays constant.
2. Use two successive readings to find the exponent.

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