JEE Chemistry Chemical Kinetics MCQs Set 33

Practice 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.

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Multiple Choice Questions

Question 1: Which statement is correct for the reaction X + 2Y → Product?
(a) Rate of disappearance of X = twice the rate of disappearance of Y
(b) Rate of disappearance of X = ½ × rate of appearance of products
(c) Rate of appearance of products = ½ × rate of disappearance of Y
(d) Rate of appearance of products = ½ × rate of disappearance of X
Show Answer & Explanation

Answer: (c) Rate of appearance of products = ½ × rate of disappearance of Y

Explanation:
1. Rate = −d[X]/dt = −(1/2)d[Y]/dt = d[P]/dt.
2. So d[P]/dt = ½ × (−d[Y]/dt), and X disappears at half the rate of Y and at the same rate as P appears.
3. Answer: (c)

Teacher's Note:
1. The coefficient of the species fixes its relative rate.
2. Y is consumed twice as fast as X.

Question 2: For N2O5 → 2NO2 + ½O2, given −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. d[NO2]/dt = 2 × (−d[N2O5]/dt), so K2 = 2K1.
2. d[O2]/dt = ½ × (−d[N2O5]/dt), so K3 = K1/2.
3. Hence K2 = 2K1 = 4K3.
4. Answer: (a)

Teacher's Note:
1. Relate each rate to the loss of N2O5 using the coefficients.
2. Compare the constants only after dividing by the same concentration.

Question 3: The rate of formation of SO3 in 2SO2 + O2 → 2SO3 is 1.6 × 10−3 kg min−1. The rate at which SO2 reacts is:
(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 are used and formed in the ratio 1 : 1 by moles.
2. Mass rate of SO2 = 1.6 × 10−3 × 64/80 = 1.28 × 10−3 kg min−1 (M of SO3 is 80, SO2 is 64).
3. Answer: (d)

Teacher's Note:
1. Convert mass rate to mole rate before using coefficients.
2. Equal moles do not mean equal masses.

Question 4: In N2(g) + 3H2(g) → 2NH3(g) 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 (in mol L−1 s−1) are:
(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. Rate of disappearance of H2 = 3 × rate of reaction = 3.75 × 10−4.
3. Answer: (c)

Teacher's Note:
1. Rate of reaction is the rate divided by the coefficient.
2. It is a single number for the whole reaction.

Question 5: For a gaseous reaction, rate = k[A][B]. If the volume of the container is reduced to 1/4 of its initial value, the rate of the reaction becomes ____ times the initial rate:
(a) 1/8
(b) 8
(c) 1/16
(d) 16
Show Answer & Explanation

Answer: (d) 16

Explanation:
1. Volume reduced to 1/4 makes each concentration 4 times larger.
2. New rate = k(4[A])(4[B]) = 16 × initial rate.
3. Answer: (d)

Teacher's Note:
1. Concentration of a gas is inversely proportional to volume at constant moles.
2. Multiply the factors for each reactant.

Question 6: 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 CxCyCz
(b) Rate = k Cx0.5Cy0.5Cz0.5
(c) Rate = k Cx1.5Cy−1Cz0
(d) Rate = k CxCz0 / Cy2
Show Answer & Explanation

Answer: (c) Rate = k Cx1.5Cy−1Cz0

Explanation:
1. Add the exponents: (a) 3, (b) 1.5, (c) 1.5 − 1 + 0 = 0.5, (d) 1 + 0 − 2 = −1.
2. Only (c) has overall order 0.5.
3. Answer: (c)

Teacher's Note:
1. Overall order is the algebraic sum of exponents in the rate law.
2. Negative exponents are allowed.

Question 7: A chemical reaction involves two reacting species. The rate of reaction is directly proportional to the concentration of one of them and inversely proportional to the concentration of the other. The order of the reaction is:
(a) 1
(b) 2
(c) Zero
(d) Unpredictable
Show Answer & Explanation

Answer: (c) Zero

Explanation:
1. Rate = k[A]/[B] = k[A]1[B]−1.
2. Order = 1 + (−1) = 0.
3. Answer: (c)

Teacher's Note:
1. Inverse proportionality gives a −1 exponent.
2. Order is the sum of exponents.

Question 8: Calculate the order of the reaction in A and B from the data: ([A], [B], rate) = (0.05, 0.05, 1.2 × 10−3), (0.10, 0.05, 2.4 × 10−3), (0.05, 0.10, 1.2 × 10−3).
(a) 1 and 0
(b) 1 and 1
(c) 0 and 1
(d) None of these
Show Answer & Explanation

Answer: (a) 1 and 0

Explanation:
1. Doubling [A] with [B] fixed doubles the rate, so order in A is 1.
2. Doubling [B] with [A] fixed leaves the rate unchanged, so order in B is 0.
3. Answer: (a)

Teacher's Note:
1. Compare runs that differ in one concentration.
2. Zero order means the rate is independent of that species.

Question 9: For A + B → product, the order is one with respect to each of A and B. From the data (rate, [A], [B]) = (0.10, 0.20 M, 0.05 M), (0.40, x, 0.05 M), (0.80, 0.40 M, y), the values of x and y are:
(a) 0.20, 0.80
(b) 0.80, 0.40
(c) 0.80, 0.20
(d) 0.40, 0.20
Show Answer & Explanation

Answer: (c) 0.80, 0.20

Explanation:
1. Rate = k[A][B]: k = 0.10/(0.20 × 0.05) = 10 L mol−1 s−1.
2. x = 0.40/(10 × 0.05) = 0.80 M.
3. y = 0.80/(10 × 0.40) = 0.20 M.
4. Answer: (c)

Teacher's Note:
1. Find k from the first row.
2. Then solve each row for the unknown.

Question 10: If the time required to complete half of a reaction varies inversely with the concentration of the reactant, the order of the reaction is:
(a) Zero
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. t1/2 ∝ a1−n.
2. t1/2 ∝ 1/a means 1 − n = −1, so n = 2.
3. Answer: (c)

Teacher's Note:
1. Zero order: t½ ∝ a; first order: independent; second order: ∝ 1/a.
2. Use t1/2 ∝ a1−n.

Question 11: The rate law for the single-step reaction 2A + B → 2C is:
(a) Rate = k[A][B]
(b) Rate = k[A]2[B]
(c) Rate = k[2A][B]
(d) Rate = k[A]2[B]0
Show Answer & Explanation

Answer: (b) Rate = k[A]2[B]

Explanation:
1. For an elementary (single-step) reaction exponents equal the stoichiometric coefficients.
2. Rate = k[A]2[B]1.
3. Answer: (b)

Teacher's Note:
1. Only elementary reactions follow the molecularity-order match.
2. Do not write the coefficient inside the brackets.

Question 12: For A + B → products, doubling the concentration of A makes the rate four times as great, while doubling B does not affect the rate. The rate equation is:
(a) Rate = k[A][B]
(b) Rate = k[A]2
(c) Rate = k[A]2[B]
(d) Rate = k[A]2[B]2
Show Answer & Explanation

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

Explanation:
1. 2x = 4 gives order 2 in A.
2. B has no effect, order 0.
3. Rate = k[A]2.
4. Answer: (b)

Teacher's Note:
1. Match the factor to 2order.
2. Order 0 species drop out of the rate law.

Question 13: Point out the incorrect statement.
(a) Rate law is an experimental result
(b) Law of mass action is a theoretical proposal
(c) Rate law is more informative than the law of mass action for developing a mechanism
(d) Rate law is always different from the expression of the law of mass action
Show Answer & Explanation

Answer: (d) Rate law is always different from the expression of the law of mass action

Explanation:
1. For an elementary reaction the rate law and the mass-action expression are the same.
2. So 'always different' is incorrect.
3. Answer: (d)

Teacher's Note:
1. They coincide only for single-step reactions.
2. Rate laws must be found by experiment.

Question 14: For an 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 reacting species colliding in the elementary step: 2 + 1 = 3.
2. Answer: (c)

Teacher's Note:
1. Molecularity applies only to elementary steps.
2. It is always a whole number.

Question 15: The rate law for A + B → product is rate = k[A]1[B]2. Which statement is false?
(a) If [B] is held constant while [A] is doubled, the reaction proceeds twice as fast
(b) If [A] is held constant while [B] is reduced to one quarter, the rate will be halved
(c) If [A] and [B] are both doubled, the reaction proceeds 8 times as fast
(d) This is a third order reaction
Show Answer & Explanation

Answer: (b) If [A] is held constant while [B] is reduced to one quarter, the rate will be halved

Explanation:
1. [B] reduced to 1/4: rate falls by (1/4)2 = 1/16, not 1/2, so (b) is false.
2. (a) 2×, (c) 2 × 4 = 8×, (d) order 1 + 2 = 3 are true.
3. Answer: (b)

Teacher's Note:
1. Apply the factor raised to the order.
2. One quarter squared is one sixteenth.

Question 16: For A → B, the rate of reaction doubles when the concentration of A is increased 8 times. The order with respect to A is:
(a) 3
(b) 1/2
(c) 1/3
(d) Zero
Show Answer & Explanation

Answer: (c) 1/3

Explanation:
1. 8n = 2 gives 23n = 21, so n = 1/3.
2. Answer: (c)

Teacher's Note:
1. Orders can be fractional.
2. Write the factor as a power of 2.

Question 17: The hypothetical reaction A2 + B2 → 2AB follows the mechanism: A2 ⇌ A + A (fast); A + B2 → AB + B (slow); A + B → AB (fast). The order of the overall reaction is:
(a) 2
(b) 1
(c) 1½
(d) Zero
Show Answer & Explanation

Answer: (c) 1½

Explanation:
1. Slow step: rate = k[A][B2].
2. From the fast equilibrium [A] = (K[A2])1/2.
3. Rate = k′[A2]1/2[B2], order = 1.5.
4. Answer: (c)

Teacher's Note:
1. Intermediates must not appear in the final rate law.
2. Half-order comes from the dissociation equilibrium.

Question 18: The rate of the reaction RCl + NaOH(aq) → ROH + NaCl is given by rate = k1[RCl]. The rate of the reaction is:
(a) Doubled on doubling the concentration of NaOH
(b) Halved on reducing the concentration of RCl to half
(c) Decreased on increasing the temperature of the reaction
(d) Unaffected by increasing the temperature of the reaction
Show Answer & Explanation

Answer: (b) Halved on reducing the concentration of RCl to half

Explanation:
1. Rate depends only on [RCl] (first order), so halving [RCl] halves the rate.
2. NaOH does not appear, and a higher temperature increases k.
3. Answer: (b)

Teacher's Note:
1. This is SN1-type kinetics.
2. Order in NaOH is zero.

Question 19: For NO2 + CO → CO2 + NO, the rate expression is rate = k[NO2]2. The number of molecules of CO involved in the slowest step is:
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. CO is absent from the rate law, so it takes part only after the slow step.
2. Hence zero CO molecules are involved in the slowest step (2NO2 → NO3 + NO).
3. Answer: (a)

Teacher's Note:
1. Species in the rate law are those in the slow step plus anything before it.
2. Zero order in CO.

Question 20: The rate law of the reaction A + 2B → product is d[P]/dt = k[A]2[B]. If A is taken in large excess, the order of the reaction will be:
(a) Zero
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. [A] stays practically constant, so rate = k′[B] with k′ = k[A]2.
2. This is a pseudo first order reaction.
3. Answer: (b)

Teacher's Note:
1. Excess reactant concentration is absorbed into the constant.
2. Only [B] changes appreciably.

Question 21: In the reaction sequence A → B → C → D with rate constants K1, K2, K3 respectively, K3 > K2 > K1. The rate determining step is:
(a) A → B
(b) C → D
(c) B → C
(d) A → D
Show Answer & Explanation

Answer: (a) A → B

Explanation:
1. The slowest step controls the overall rate; it has the smallest rate constant K1.
2. That is A → B.
3. Answer: (a)

Teacher's Note:
1. Smallest k means slowest step.
2. Later steps are faster and have no influence on the rate.

Question 22: The rate constant is numerically the same for three reactions of first, second and third order respectively. Which is true for the rates r1, r2, r3 if the concentration of reactants is the same and greater than 1 M?
(a) r1 = r2 = r3
(b) r1 > r2 > r3
(c) r1 < r2 < r3
(d) All of the above
Show Answer & Explanation

Answer: (c) r1 < r2 < r3

Explanation:
1. r1 = kc, r2 = kc2, r3 = kc3.
2. For c > 1, c < c2 < c3, so r1 < r2 < r3.
3. Answer: (c)

Teacher's Note:
1. For c < 1 the order reverses.
2. Compare powers of the concentration.

Question 23: K for a zero order reaction is 2 × 10−2 mol L−1 s−1. If the concentration of the reactant after 25 s is 0.5 M, the initial concentration must have been:
(a) 0.5 M
(b) 1.25 M
(c) 12.5 M
(d) 1.0 M
Show Answer & Explanation

Answer: (d) 1.0 M

Explanation:
1. Zero order: [A] = [A]0 − kt.
2. [A]0 = 0.5 + 2 × 10−2 × 25 = 1.0 M.
3. Answer: (d)

Teacher's Note:
1. Zero order concentration falls linearly with time.
2. kt = 0.5 M here.

Question 24: For the reaction X → Y, the concentrations of X (falling) and Y (rising) are plotted against time. The point where the two curves intersect (that is, [X] = [Y]) represents:
(a) t1/2
(b) t3/4
(c) t2/3
(d) Data are insufficient to predict
Show Answer & Explanation

Answer: (a) t1/2

Explanation:
1. At the intersection [X] = [Y] and [X] + [Y] = initial concentration, so half of X has reacted.
2. This time is the half life, t1/2.
3. Answer: (a)

Teacher's Note:
1. Equal concentrations of reactant and product mean 50% completion.
2. True for X → Y with no side reactions.

Question 25: A plot of log(a − x) versus time t is a straight line. This indicates that the reaction is of:
(a) Second order
(b) First order
(c) Zero order
(d) Third order
Show Answer & Explanation

Answer: (b) First order

Explanation:
1. First order: log(a − x) = log a − kt/2.303 is linear in t with slope −k/2.303.
2. Answer: (b)

Teacher's Note:
1. Linear log plot means first order.
2. The slope gives the rate constant.

Question 26: The rate constant of a first order reaction is 4 × 10−3 s−1. At a reactant concentration of 0.02 M, the rate of reaction would be:
(a) 8 × 10−5 M s−1
(b) 4 × 10−3 M s−1
(c) 2 × 10−1 M s−1
(d) 4 × 10−1 M s−1
Show Answer & Explanation

Answer: (a) 8 × 10−5 M s−1

Explanation:
1. Rate = k[A] = 4 × 10−3 × 0.02 = 8 × 10−5 M s−1.
2. Answer: (a)

Teacher's Note:
1. Rate = k × concentration for first order.
2. Check the unit of the answer.

Question 27: A first order reaction has a half life of 69.3 s. At 0.10 mol L−1 reactant concentration, the rate will be:
(a) 10−4 M s−1
(b) 10−3 M s−1
(c) 10−1 M s−1
(d) 6.93 × 10−1 M s−1
Show Answer & Explanation

Answer: (b) 10−3 M s−1

Explanation:
1. k = 0.693/69.3 = 0.01 s−1.
2. Rate = 0.01 × 0.10 = 10−3 M s−1.
3. Answer: (b)

Teacher's Note:
1. Find k from the half life, then the rate.
2. k = 0.693/t1/2.

Question 28: The half life for the first order reaction N2O5 → 2NO2 + ½O2 is 24 h at 30°C. Starting with 10 g of N2O5, how many grams will remain after 96 h?
(a) 1.25 g
(b) 0.63 g
(c) 1.77 g
(d) 0.5 g
Show Answer & Explanation

Answer: (b) 0.63 g

Explanation:
1. 96 h = 4 half lives.
2. Mass left = 10/24 = 0.625 g ≈ 0.63 g.
3. Answer: (b)

Teacher's Note:
1. Divide time by t1/2.
2. Halve the mass that many times.

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