JEE Chemistry Chemical Kinetics MCQs Set 08

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

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Multiple Choice Questions

Question 1: For the reaction N2(g) + 3H2(g) → 2NH3(g), identify the correct relationship among the rate expressions.
(a) \( \text{Rate} = -\dfrac{d[N_2]}{dt} = -\dfrac13\dfrac{d[H_2]}{dt} = \dfrac12\dfrac{d[NH_3]}{dt} \)
(b) \( \text{Rate} = -\dfrac{d[N_2]}{dt} = -\dfrac{3d[H_2]}{dt} = \dfrac{2d[NH_3]}{dt} \)
(c) \( \text{Rate} = \dfrac{d[N_2]}{dt} = \dfrac13\dfrac{d[H_2]}{dt} = \dfrac12\dfrac{d[NH_3]}{dt} \)
(d) \( \text{Rate} = -\dfrac{d[N_2]}{dt} = -\dfrac{d[H_2]}{dt} = \dfrac{d[NH_3]}{dt} \)
Show Answer & Explanation

Answer: (a) \( \text{Rate} = -\dfrac{d[N_2]}{dt} = -\dfrac13\dfrac{d[H_2]}{dt} = \dfrac12\dfrac{d[NH_3]}{dt} \)

Explanation:
1. Divide each rate of change by the stoichiometric coefficient.
2. Reactants have a negative sign (decreasing) and products a positive sign.
3. Rate = −d[N2]/dt = −(1/3)d[H2]/dt = (1/2)d[NH3]/dt.

Teacher's Note:
1. General: for aA + bB → cC + dD, rate = −(1/a)d[A]/dt = (1/c)d[C]/dt.
2. Do not multiply by the coefficient.

Question 2: In the contact process 2SO2 + O2 → 2SO3, the rate is given by −d[O2]/dt = 2.5 × 10−4 mol L−1 s−1. The rate of disappearance of SO2 is
(a) 5 × 10−4 mol L−1 s−1
(b) −2.25 × 10−4 mol L−1 s−1
(c) 3.75 × 10−4 mol L−1 s−1
(d) 50.0 × 10−4 mol L−1 s−1
Show Answer & Explanation

Answer: (a) 5 × 10−4 mol L−1 s−1

Explanation:
1. −(1/2)d[SO2]/dt = −d[O2]/dt.
2. −d[SO2]/dt = 2 × 2.5 × 10−4 = 5 × 10−4 mol L−1 s−1.

Teacher's Note:
1. Two moles of SO2 are used for every mole of O2.
2. The rate of disappearance is quoted as a positive number.

Question 3: In Haber's process the rate of appearance of NH3 is 2.5 × 10−4 mol L−1 s−1. The rate of disappearance of H2 will be
(a) 2.5 × 10−4 mol L−1 s−1
(b) 1.25 × 10−4 mol L−1 s−1
(c) 3.75 × 10−4 mol L−1 s−1
(d) 5 × 10−4 mol L−1 s−1
Show Answer & Explanation

Answer: (c) 3.75 × 10−4 mol L−1 s−1

Explanation:
1. N2 + 3H2 → 2NH3: (1/3)(−d[H2]/dt) = (1/2)(d[NH3]/dt).
2. −d[H2]/dt = (3/2) × 2.5 × 10−4 = 3.75 × 10−4 mol L−1 s−1.

Teacher's Note:
1. Ratio of rates equals ratio of coefficients: H2 : NH3 = 3 : 2.
2. The same data also gives rate of reaction = 1.25 × 10−4.

Question 4: Which statement is correct for the reaction X + 2Y → Product?
(a) The rate of disappearance of X is twice the rate of disappearance of Y
(b) The rate of disappearance of X is 1/2 the rate of appearance of products
(c) The rate of appearance of products is 1/2 the rate of disappearance of Y
(d) The rate of appearance of products is 1/2 the rate of disappearance of X
Show Answer & Explanation

Answer: (c) The rate of appearance of products is 1/2 the rate of disappearance of Y

Explanation:
1. Rate = −d[X]/dt = −(1/2)d[Y]/dt = d[P]/dt.
2. So d[P]/dt = (1/2)(−d[Y]/dt) and −d[X]/dt = d[P]/dt.

Teacher's Note:
1. Y is used twice as fast as X, and the product forms at the same rate as X is used.
2. Write the rate expression first, then compare.

Question 5: For N2O5 → 2NO2 + 1/2 O2, with −d[N2O5]/dt = K1[N2O5], d[NO2]/dt = K2[N2O5] and d[O2]/dt = K3[N2O5], the relation between K1, K2 and K3 is
(a) 2K1 = K2 = 4K3
(b) K1 = K2 = K3
(c) 2K1 = 4K2 = K3
(d) none of these
Show Answer & Explanation

Answer: (a) 2K1 = K2 = 4K3

Explanation:
1. For the equation as written, d[NO2]/dt = 2(−d[N2O5]/dt), so K2 = 2K1.
2. d[O2]/dt = (1/2)(−d[N2O5]/dt), so K3 = K1/2.
3. Hence K2 = 2K1 = 4K3.

Teacher's Note:
1. Convert each constant using the coefficients of the balanced equation.
2. K1 : K2 : K3 = 1 : 2 : 1/2.

Question 6: The rate of formation of SO3 in 2SO2 + O2 → 2SO3 is 1.6 × 10−3 kg min−1. At what rate does SO2 react? (S = 32, O = 16)
(a) 1.6 × 10−3 kg min−1
(b) 8.0 × 10−4 kg min−1
(c) 3.2 × 10−3 kg min−1
(d) 1.28 × 10−3 kg min−1
Show Answer & Explanation

Answer: (d) 1.28 × 10−3 kg min−1

Explanation:
1. SO2 and SO3 react and form in equal mole ratio (2 : 2).
2. Moles of SO3 per min are proportional to 1.6/80; the same number of moles of SO2 (M = 64) is used.
3. Rate = 1.6 × 10−3 × 64/80 = 1.28 × 10−3 kg min−1.

Teacher's Note:
1. Rates in mass units must be converted through molar masses.
2. Equal moles do not mean equal mass.

Question 7: For a general change 2A + 3B → products, the rates of disappearance of A and B are r1 and r2. The relation between r1 and r2 is
(a) 3r1 = 2r2
(b) r1 = r2
(c) 2r1 = 3r2
(d) r12 = 2r2
Show Answer & Explanation

Answer: (a) 3r1 = 2r2

Explanation:
1. Rate = r1/2 = r2/3.
2. Cross-multiplying: 3r1 = 2r2.

Teacher's Note:
1. Divide each disappearance rate by its coefficient and equate.
2. B is consumed faster, so r2 > r1.

Question 8: For N2 + 3H2 → 2NH3, the rate of appearance of NH3 is 2.5 × 10−4 mol L−1 s−1. The rate of reaction and the rate of disappearance of H2 are respectively (in mol L−1 s−1)
(a) 3.75 × 10−4, 1.25 × 10−4
(b) 1.25 × 10−4, 2.5 × 10−4
(c) 1.25 × 10−4, 3.75 × 10−4
(d) 5.0 × 10−4, 3.75 × 10−4
Show Answer & Explanation

Answer: (c) 1.25 × 10−4, 3.75 × 10−4

Explanation:
1. Rate of reaction = (1/2) × 2.5 × 10−4 = 1.25 × 10−4.
2. −d[H2]/dt = 3 × rate = 3.75 × 10−4.

Teacher's Note:
1. Rate of reaction is a single number: the rate of change divided by the coefficient.
2. Disappearance rate of H2 is 3 times the reaction rate.

Question 9: Which statement is not correct for 4A + B → 2C + 2D?
(a) The rate of disappearance of B is twice the rate of appearance of C
(b) The rate of disappearance of B is one fourth the rate of disappearance of A
(c) The rate of formation of D is half the rate of consumption of A
(d) The rates of formation of C and D are equal
Show Answer & Explanation

Answer: (a) The rate of disappearance of B is twice the rate of appearance of C

Explanation:
1. Rate = −(1/4)d[A]/dt = −d[B]/dt = (1/2)d[C]/dt = (1/2)d[D]/dt.
2. So −d[B]/dt is half (not twice) of d[C]/dt, and statement (a) is wrong.

Teacher's Note:
1. Check each statement against the rate expression.
2. C and D have the same coefficient, so they form at the same rate.

Question 10: For a gaseous reaction, rate = k[A][B]. If the volume of the container is reduced to 1/4 of the initial volume, the rate will be ____ times the initial rate.
(a) \( \dfrac18 \)
(b) 8
(c) \( \dfrac1{16} \)
(d) 16
Show Answer & Explanation

Answer: (d) 16

Explanation:
1. Reducing the volume to 1/4 raises each concentration 4 times.
2. Rate = k(4[A])(4[B]) = 16k[A][B].

Teacher's Note:
1. For gases, volume reduction multiplies every concentration by the same factor.
2. Second-order overall, so the rate changes by the square.

Question 11: For the elementary reaction 2A(g) + B(g) → product, if the pressure is increased three times, the rate of the forward reaction will become ____ the previous value.
(a) 9 times
(b) 27 times
(c) 1/9 times
(d) 1/27 times
Show Answer & Explanation

Answer: (b) 27 times

Explanation:
1. Elementary, so rate = k[A]2[B].
2. Pressure ×3 means each concentration ×3: 32 × 3 = 27.

Teacher's Note:
1. For an elementary step, the exponents equal the coefficients.
2. Total order 3 gives a 33 increase.

Question 12: For the reaction 2A + B → 3C + D, which expression does not describe the rate of the reaction?
(a) \( -\dfrac13\dfrac{d[C]}{dt} \)
(b) \( -\dfrac{d[B]}{dt} \)
(c) \( \dfrac{d[D]}{dt} \)
(d) \( -\dfrac12\dfrac{d[A]}{dt} \)
Show Answer & Explanation

Answer: (a) \( -\dfrac13\dfrac{d[C]}{dt} \)

Explanation:
1. Rate = −(1/2)d[A]/dt = −d[B]/dt = (1/3)d[C]/dt = d[D]/dt.
2. Option (a) has a wrong sign: C is a product, so its rate is +(1/3)d[C]/dt.

Teacher's Note:
1. Products increase, so d[C]/dt is positive.
2. A negative sign belongs only before reactant terms.

Question 13: For N2 + 3H2 → 2NH3, the correct relation between d[NH3]/dt and −d[H2]/dt is
(a) \( \dfrac{d[NH_3]}{dt} = -\dfrac13\dfrac{d[H_2]}{dt} \)
(b) \( \dfrac{d[NH_3]}{dt} = -\dfrac23\dfrac{d[H_2]}{dt} \)
(c) \( \dfrac{d[NH_3]}{dt} = -\dfrac32\dfrac{d[H_2]}{dt} \)
(d) \( \dfrac{d[NH_3]}{dt} = -\dfrac{d[H_2]}{dt} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{d[NH_3]}{dt} = -\dfrac23\dfrac{d[H_2]}{dt} \)

Explanation:
1. (1/2)d[NH3]/dt = −(1/3)d[H2]/dt.
2. So d[NH3]/dt = −(2/3)d[H2]/dt.

Teacher's Note:
1. NH3 forms at 2/3 the rate at which H2 disappears.
2. Cross-check with coefficients 2 : 3.

Question 14: In BrO3− + 5Br− + 6H+ → 3Br2 + 3H2O, the rate of appearance of bromine is related to the rate of disappearance of bromide ions as
(a) \( \dfrac{d[Br_2]}{dt} = \dfrac35\dfrac{d[Br^-]}{dt} \)
(b) \( \dfrac{d[Br_2]}{dt} = -\dfrac35\dfrac{d[Br^-]}{dt} \)
(c) \( \dfrac{d[Br_2]}{dt} = -\dfrac53\dfrac{d[Br^-]}{dt} \)
(d) \( \dfrac{d[Br_2]}{dt} = \dfrac53\dfrac{d[Br^-]}{dt} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{d[Br_2]}{dt} = -\dfrac35\dfrac{d[Br^-]}{dt} \)

Explanation:
1. (1/3)d[Br2]/dt = −(1/5)d[Br−]/dt.
2. d[Br2]/dt = −(3/5)d[Br−]/dt; the negative sign appears because Br− decreases.

Teacher's Note:
1. Br− decreases, so its derivative is negative; the minus sign makes the equality true.
2. Ratio of magnitudes 3 : 5 follows the coefficients.

Question 15: For N2O5(g) → 2NO2(g) + 1/2 O2(g), the rate of disappearance of N2O5 is 6.25 × 10−3 mol L−1 s−1. The rates of formation of NO2 and O2 are respectively
(a) 1.25 × 10−2 and 6.25 × 10−3
(b) 6.25 × 10−3 and 6.25 × 10−3
(c) 1.25 × 10−2 and 3.125 × 10−3
(d) 6.25 × 10−3 and 3.125 × 10−3
Show Answer & Explanation

Answer: (c) 1.25 × 10−2 and 3.125 × 10−3

Explanation:
1. NO2 forms at twice the disappearance rate: 2 × 6.25 × 10−3 = 1.25 × 10−2.
2. O2 forms at half the rate: 3.125 × 10−3.

Teacher's Note:
1. Use the coefficients 1 : 2 : 1/2 from the equation as written.
2. Rewriting the equation with whole numbers changes the rate of reaction but not these rates.

Question 16: The rate of the reaction 2N2O5 → 4NO2 + O2 can be written as −d[N2O5]/dt = k[N2O5], d[NO2]/dt = k′[N2O5], d[O2]/dt = k″[N2O5]. The relation between k and k′, and k and k″ is
(a) k′ = k; k″ = k
(b) k′ = 2k; k″ = k
(c) k′ = 2k; k″ = k/2
(d) k′ = 2k; k″ = 2k
Show Answer & Explanation

Answer: (c) k′ = 2k; k″ = k/2

Explanation:
1. Per N2O5 consumed, 2 NO2 and 1/2 O2 are produced.
2. So d[NO2]/dt = 2(−d[N2O5]/dt) gives k′ = 2k, and d[O2]/dt = (1/2)(−d[N2O5]/dt) gives k″ = k/2.

Teacher's Note:
1. Rate constants defined by different species differ by stoichiometric factors.
2. The rate of reaction itself is unique.

Question 17: The rate of a hypothetical reaction is r = −d[A]/dt = K[A]1/2[B]1/3[C]1/4. The order of the reaction is
(a) 1
(b) 1/2
(c) 2
(d) 13/12
Show Answer & Explanation

Answer: (d) 13/12

Explanation:
1. Order = sum of exponents = 1/2 + 1/3 + 1/4 = (6 + 4 + 3)/12 = 13/12.

Teacher's Note:
1. Order can be fractional and is found only from the rate law.
2. Add the powers of all concentration terms.

Question 18: Which of the following rate laws has an overall order of 0.5 for a reaction involving substances x, y and z?
(a) Rate = K(Cx)(Cy)(Cz)
(b) Rate = K(Cx)0.5(Cy)0.5(Cz)0.5
(c) Rate = K(Cx)1.5(Cy)−1(Cz)0
(d) Rate = K(Cx)(Cy)0/(Cz)2
Show Answer & Explanation

Answer: (c) Rate = K(Cx)1.5(Cy)−1(Cz)0

Explanation:
1. (a) 3; (b) 1.5; (c) 1.5 − 1 + 0 = 0.5; (d) 1 + 0 − 2 = −1.
2. Only (c) has order 0.5.

Teacher's Note:
1. Order = sum of exponents; a denominator term counts as a negative exponent.

Question 19: The rate of a chemical reaction is directly proportional to the concentration of one reactant and inversely proportional to the concentration of the other. The order of the reaction is
(a) 1
(b) 2
(c) 0
(d) unpredictable
Show Answer & Explanation

Answer: (c) 0

Explanation:
1. Rate = k[A]1[B]−1.
2. Order = 1 − 1 = 0.

Teacher's Note:
1. Negative exponents are allowed in empirical rate laws.
2. They often signal an inhibitor or a pre-equilibrium.

Question 20: For H2(g) + Br2(g) → 2HBr(g), the experimental rate is Rate = k[H2][Br2]1/2. The order of the reaction is
(a) 2
(b) 3/2
(c) 1
(d) 5/2
Show Answer & Explanation

Answer: (b) 3/2

Explanation:
1. Order = 1 + 1/2 = 3/2.

Teacher's Note:
1. Stoichiometry would suggest order 2, but experiment gives 1.5.
2. This is a chain mechanism.

Question 21: Time required to complete half of a reaction varies inversely with the concentration of the reactant. The order of the reaction is
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. For order n, t1/2 ∝ a1−n.
2. Inverse proportionality means 1 − n = −1, so n = 2.

Teacher's Note:
1. Zero order: t1/2 ∝ a; first order: independent of a; second order: ∝ 1/a.

Question 22: The rate law for the single-step reaction 2A → 2C is given by
(a) Rate = K[A][B]
(b) Rate = K[A]2
(c) Rate = K[2A][B]
(d) Rate = K[A]2[B]0
Show Answer & Explanation

Answer: (b) Rate = K[A]2

Explanation:
1. For a single-step (elementary) reaction, the rate law follows the stoichiometry: rate = k[A]2.
2. (d) is equivalent but written unnecessarily; the simplest correct form is (b).

Teacher's Note:
1. B does not appear in the reaction.
2. Elementary: powers equal the coefficients.

Question 23: For the elementary process 2X + Y → Z + W, the molecularity is
(a) 2
(b) 1
(c) 3
(d) unpredictable
Show Answer & Explanation

Answer: (c) 3

Explanation:
1. Molecularity is the number of species colliding in the elementary step: 2 + 1 = 3.

Teacher's Note:
1. Molecularity is always a whole number (1, 2 or rarely 3).
2. It is defined only for elementary steps.

Question 24: The specific rate constant of a first order reaction depends on the
(a) concentration of the reactant
(b) concentration of the product
(c) time
(d) temperature
Show Answer & Explanation

Answer: (d) temperature

Explanation:
1. k is independent of concentration and time.
2. It depends on temperature (Arrhenius equation) and on a catalyst.

Teacher's Note:
1. Rate depends on concentration; rate constant does not.
2. Only temperature and catalyst change k.

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