JEE Chemistry Chemical Kinetics MCQs Set 14

Chemistry Objective Questions and Answers: 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: For \( \ce{N2(g) + 3H2(g) -> 2NH3(g)} \), which expression of the rate in terms of the concentrations is correct?
(a) \( \text{Rate} = -\dfrac{d[N_2]}{dt} = -\dfrac{1}{3}\dfrac{d[H_2]}{dt} = \dfrac{1}{2}\dfrac{d[NH_3]}{dt} \)
(b) \( \text{Rate} = -\dfrac{d[N_2]}{dt} = -\dfrac{3\,d[H_2]}{dt} = \dfrac{2\,d[NH_3]}{dt} \)
(c) \( \text{Rate} = \dfrac{d[N_2]}{dt} = \dfrac{1}{3}\dfrac{d[H_2]}{dt} = \dfrac{1}{2}\dfrac{d[NH_3]}{dt} \)
(d) \( \text{Rate} = -\dfrac{d[N_2]}{dt} = -\dfrac{d[H_2]}{dt} = \dfrac{d[NH_3]}{dt} \)
Show Answer & Explanation

Answer: (a) \( \text{Rate} = -\dfrac{d[N_2]}{dt} = -\dfrac{1}{3}\dfrac{d[H_2]}{dt} = \dfrac{1}{2}\dfrac{d[NH_3]}{dt} \)

Explanation:
1. Divide each rate of change by its stoichiometric coefficient.
2. Reactants carry a negative sign because their concentrations fall, products a positive sign.
3. This gives the first expression.

Teacher's Note:
1. Rate of reaction = (change of concentration / coefficient) / time.
2. Reactants take a minus sign.

Question 2: In the contact process \( \ce{2SO2(g) + O2(g) -> 2SO3(g)} \), the rate is \( -\dfrac{d[O_2]}{dt} = 2.5 \times 10^{-4} \) mol L⁻¹ s⁻¹. The rate of disappearance of \( \ce{SO2} \) is :
(a) \( 5 \times 10^{-4} \) mol L⁻¹ s⁻¹
(b) \( -2.25 \times 10^{-4} \) mol L⁻¹ s⁻¹
(c) \( 3.75 \times 10^{-4} \) mol L⁻¹ s⁻¹
(d) \( 50.0 \times 10^{-4} \) mol L⁻¹ s⁻¹
Show Answer & Explanation

Answer: (a) \( 5 \times 10^{-4} \) mol L⁻¹ s⁻¹

Explanation:
1. \( \ce{SO2} \) is consumed twice as fast as \( \ce{O2} \).
2. Rate for \( \ce{SO2} \) = 2 × 2.5 × 10⁻⁴ = 5 × 10⁻⁴ mol L⁻¹ s⁻¹.

Teacher's Note:
1. Use the ratio of coefficients to compare rates.
2. A rate of disappearance is quoted as a positive number.

Question 3: For a reaction \( \ce{X + 2Y -> product} \), which statement is correct?
(a) The rate of disappearance of X is twice the rate of disappearance of Y
(b) The rate of disappearance of X is half the rate of appearance of products
(c) The rate of appearance of products is half the rate of disappearance of Y
(d) The rate of appearance of products is half the rate of disappearance of X
Show Answer & Explanation

Answer: (c) The rate of appearance of products is half the rate of disappearance of Y

Explanation:
1. Rate of reaction = −d[X]/dt = −½ d[Y]/dt = d[P]/dt (one mole of product).
2. So Y disappears twice as fast as X and the product appears at the same rate as X disappears.
3. The product forms at half the rate of Y's disappearance.

Teacher's Note:
1. Compare coefficients to compare rates.
2. Y has twice the coefficient of X.

Question 4: For \( \ce{N2O5 -> 2NO2 + 1/2 O2} \) it is given that \( -\dfrac{d[N_2O_5]}{dt} = K_1[N_2O_5] \), \( \dfrac{d[NO_2]}{dt} = K_2[N_2O_5] \) and \( \dfrac{d[O_2]}{dt} = K_3[N_2O_5] \). The relation between \( K_1, K_2, K_3 \) is :
(a) \( 2K_1 = K_2 = 4K_3 \)
(b) \( K_1 = K_2 = K_3 \)
(c) \( 2K_1 = 4K_2 = K_3 \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( 2K_1 = K_2 = 4K_3 \)

Explanation:
1. \( \ce{NO2} \) forms twice as fast as \( \ce{N2O5} \) is used, so \( K_2 = 2K_1 \).
2. \( \ce{O2} \) forms at half that rate, so \( K_3 = K_1/2 \).
3. Hence \( 2K_1 = K_2 = 4K_3 \).

Teacher's Note:
1. Link each rate to the coefficient.
2. Compare with \( K_1 \).

Question 5: The rate of formation of \( \ce{SO3} \) in \( \ce{2SO2 + O2 -> 2SO3} \) is \( 1.6 \times 10^{-3} \) kg min⁻¹. The rate at which \( \ce{SO2} \) reacts is :
(a) \( 1.6 \times 10^{-3} \) kg min⁻¹
(b) \( 8.0 \times 10^{-4} \) kg min⁻¹
(c) \( 3.2 \times 10^{-3} \) kg min⁻¹
(d) \( 1.28 \times 10^{-3} \) kg min⁻¹
Show Answer & Explanation

Answer: (d) \( 1.28 \times 10^{-3} \) kg min⁻¹

Explanation:
1. \( \ce{SO2} \) and \( \ce{SO3} \) react in equal moles (2 : 2).
2. Molar masses are 64 and 80 g mol⁻¹.
3. Mass rate of \( \ce{SO2} \) = \( 1.6 \times 10^{-3} \times 64/80 = 1.28 \times 10^{-3} \) kg min⁻¹.

Teacher's Note:
1. The mole ratio, not the mass ratio, follows the coefficients.
2. Convert using molar masses.

Question 6: For a general reaction \( \ce{2A + 3B -> products} \), the rates of disappearance of A and B are \( r_1 \) and \( r_2 \). They are related by :
(a) \( 3r_1 = 2r_2 \)
(b) \( r_1 = r_2 \)
(c) \( 2r_1 = 3r_2 \)
(d) \( r_1^2 = 2r_2^2 \)
Show Answer & Explanation

Answer: (a) \( 3r_1 = 2r_2 \)

Explanation:
1. \( \dfrac{r_1}{2} = \dfrac{r_2}{3} \) because both equal the rate of reaction.
2. Cross-multiplying gives \( 3r_1 = 2r_2 \).

Teacher's Note:
1. Divide each rate by its coefficient.
2. Equate them.

Question 7: For \( \ce{N2(g) + 3H2(g) -> 2NH3(g)} \) the rate of appearance of \( \ce{NH3} \) is \( 2.5 \times 10^{-4} \) mol L⁻¹ s⁻¹. The rate of reaction and the rate of disappearance of \( \ce{H2} \) (in mol L⁻¹ s⁻¹) are respectively :
(a) \( 3.75 \times 10^{-4},\ 1.25 \times 10^{-4} \)
(b) \( 1.25 \times 10^{-4},\ 2.5 \times 10^{-4} \)
(c) \( 1.25 \times 10^{-4},\ 3.75 \times 10^{-4} \)
(d) \( 5.0 \times 10^{-4},\ 3.75 \times 10^{-4} \)
Show Answer & Explanation

Answer: (c) \( 1.25 \times 10^{-4},\ 3.75 \times 10^{-4} \)

Explanation:
1. Rate of reaction = (rate of appearance of \( \ce{NH3} \))/2 = \( 1.25 \times 10^{-4} \).
2. Rate of disappearance of \( \ce{H2} \) = 3 × rate of reaction = \( 3.75 \times 10^{-4} \).

Teacher's Note:
1. Rate of reaction is the common value after dividing by coefficients.
2. Keep the order asked in the question.

Question 8: Which statement is not correct for the reaction \( \ce{4A + B -> 2C + 2D} \)?
(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 = −¼ d[A]/dt = −d[B]/dt = ½ d[C]/dt = ½ d[D]/dt.
2. So B disappears at half the rate at which C appears, not twice.
3. The other three statements follow from the same relation.

Teacher's Note:
1. Convert every statement to rates divided by coefficients.
2. Spot the one that does not match.

Question 9: The rate of a hypothetical reaction \( \ce{A + B + C -> products} \) is \( r = k[A]^{1/2}[B]^{1/3}[C]^{1/4} \). 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 is the sum of the exponents.
2. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12.

Teacher's Note:
1. Add the powers over a common denominator.
2. Order need not be an integer.

Question 10: 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 \( C_x C_y C_z \)
(b) Rate = k \( C_x^{0.5} C_y^{0.5} C_z^{0.5} \)
(c) Rate = k \( C_x^{1.5} C_y^{-1} C_z^{0} \)
(d) Rate = k \( C_x C_z^{0}/C_z^{2} \)
Show Answer & Explanation

Answer: (c) Rate = k \( C_x^{1.5} C_y^{-1} C_z^{0} \)

Explanation:
1. Add the exponents in each option: 3, 1.5, 1.5 − 1 + 0 = 0.5 and 1 − 2 = −1 in the last (with x and z).
2. Only the third option gives 0.5.

Teacher's Note:
1. Order is the sum of exponents.
2. Negative exponents are allowed.

Question 11: The following data were obtained for \( \ce{A + B -> C} \): (1) [A] = 0.012, [B] = 0.035, rate = 0.10; (2) [A] = 0.024, [B] = 0.070, rate = 1.6; (3) [A] = 0.024, [B] = 0.035, rate = 0.20; (4) [A] = 0.012, [B] = 0.070, rate = 0.80. The rate law is :
(a) Rate = k[B]³
(b) Rate = k[B]⁴
(c) Rate = k[A][B]³
(d) Rate = k[A]²[B]²
Show Answer & Explanation

Answer: (c) Rate = k[A][B]³

Explanation:
1. Experiments 1 and 3: A doubles with B fixed and the rate doubles, so order in A = 1.
2. Experiments 1 and 4: B doubles with A fixed and the rate rises 8 times, so order in B = 3.
3. Rate = k[A][B]³ (check: experiment 2 gives 2 × 8 × 0.10 = 1.6).

Teacher's Note:
1. Change one variable at a time to find each order.
2. Verify with the remaining experiment.

Question 12: For \( \ce{X + 3Y -> XY3} \) the initial rates are: [X] = 0.1 M, [Y] = 0.1 M, rate = 0.002 M s⁻¹; [X] = 0.2, [Y] = 0.1, rate = 0.002; [X] = 0.3, [Y] = 0.2, rate = 0.008; [X] = 0.4, [Y] = 0.3, rate = 0.018. The rate law is :
(a) r = k[X][Y]³
(b) r = k[X]⁰[Y]²
(c) r = k[X][Y]
(d) r = [X]⁴[Y]³
Show Answer & Explanation

Answer: (b) r = k[X]⁰[Y]²

Explanation:
1. Rows 1 and 2: doubling [X] does not change the rate, so order in X = 0.
2. Rows 1 and 3: [Y] doubles and rate goes from 0.002 to 0.008, four times, so order in Y = 2.
3. Row 4 check: (3)² × 0.002 = 0.018.

Teacher's Note:
1. An unchanged rate means zero order in that species.
2. Use another row to confirm.

Question 13: For \( \ce{2A + B -> C + D} \): (1) [A] = 0.1, [B] = 0.1, rate = 7.5 × 10⁻³; (2) [A] = 0.3, [B] = 0.2, rate = 9.0 × 10⁻²; (3) [A] = 0.3, [B] = 0.4, rate = 3.6 × 10⁻¹; (4) [A] = 0.4, [B] = 0.1, rate = 3.0 × 10⁻² (mol L⁻¹ min⁻¹). The rate law is :
(a) Rate = k[A]²[B]
(b) Rate = k[A][B]²
(c) Rate = k[A][B]³
(d) Rate = k[A][B]
Show Answer & Explanation

Answer: (b) Rate = k[A][B]²

Explanation:
1. Experiments 2 and 3: [B] doubles at fixed [A] and the rate becomes 4 times, so order in B = 2.
2. Experiments 1 and 4: [A] becomes 4 times and the rate becomes 4 times, so order in A = 1.
3. Check experiment 2: 3 × 4 × 7.5 × 10⁻³ = 9.0 × 10⁻².

Teacher's Note:
1. Pick pairs where only one concentration changes.
2. Always check the unused row.

Question 14: For \( \ce{A + B -> C} \): [A] = 0.01, [B] = 0.01, rate = 1.0 × 10⁻⁴; [A] = 0.01, [B] = 0.03, rate = 9.0 × 10⁻⁴; [A] = 0.03, [B] = 0.03, rate = 2.70 × 10⁻³ (mol L⁻¹ min⁻¹). The rate law is :
(a) r = k[A]²[B]
(b) r = k[A][B]²
(c) r = k[A][B]
(d) none of these
Show Answer & Explanation

Answer: (b) r = k[A][B]²

Explanation:
1. Rows 1 and 2: [B] triples and the rate becomes 9 times, so order in B = 2.
2. Rows 2 and 3: [A] triples and the rate becomes 3 times, so order in A = 1.
3. Rate = k[A][B]².

Teacher's Note:
1. Triple the concentration and see the factor: 3² = 9, 3¹ = 3.
2. Take rows with one change.

Question 15: The rate data for a reaction are: [A] = 0.05, [B] = 0.05, rate = 1.2 × 10⁻³; [A] = 0.10, [B] = 0.05, rate = 2.4 × 10⁻³; [A] = 0.05, [B] = 0.10, rate = 1.2 × 10⁻³. The orders with respect to A and B are :
(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] doubles the rate, so order in A = 1.
2. Doubling [B] leaves the rate unchanged, so order in B = 0.

Teacher's Note:
1. No change in rate means zero order.
2. A doubling gives order 1.

Question 16: The time required to complete a half fraction 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. For order n, \( t_{1/2} \propto a^{1-n} \).
2. Inverse proportionality \( t_{1/2} \propto a^{-1} \) gives \( 1 - n = -1 \), so n = 2.

Teacher's Note:
1. Use the half-life dependence on initial concentration.
2. Second order gives 1/a.

Question 17: For a reaction \( \ce{A + B -> product} \), doubling the concentration of A makes the rate four times as great, but doubling the amount of B has no effect. 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. Doubling [A] gives a factor of 4 = 2², so the order in A is 2.
2. No effect from [B] means zero order in B.
3. Rate = k[A]².

Teacher's Note:
1. Find each order separately.
2. Omit the species with zero order.

Question 18: For an elementary process \( \ce{2X + Y -> Z + W} \), 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 in the elementary step.
2. 2 + 1 = 3.

Teacher's Note:
1. Count the reacting molecules in an elementary step.
2. This is a termolecular step.

Question 19: For \( \ce{A + B -> products} \), the rate doubled when the concentration of A was doubled. The rate was again doubled when the concentrations of both A and B were doubled. The orders with respect to A and B are :
(a) 1, 1
(b) 2, 0
(c) 1, 0
(d) 0, 1
Show Answer & Explanation

Answer: (c) 1, 0

Explanation:
1. Doubling A alone doubles the rate, so order in A = 1.
2. Doubling A and B together gives the same factor 2 overall (no extra factor from B), so B has order 0.

Teacher's Note:
1. Separate the effects of A and B.
2. Compare with the factor from A alone.

Question 20: For \( \ce{A + B -> products} \), the order of A is 1 and the order of B is 1/2. When the concentrations of both A and B are increased four times, the rate increases by a factor of :
(a) 16
(b) 8
(c) 6
(d) 4
Show Answer & Explanation

Answer: (b) 8

Explanation:
1. Rate = k[A][B]^{1/2}.
2. Factor = 4¹ × 4^{1/2} = 4 × 2 = 8.

Teacher's Note:
1. Multiply the factors raised to each order.
2. 4^{1/2} = 2.

Question 21: The rate law for \( \ce{A + B -> product} \) is rate = k[A][B]². Which statement is false?
(a) If [B] is constant and [A] is doubled the reaction proceeds twice as fast
(b) If [A] is constant and [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 constant and [B] is reduced to one quarter, the rate will be halved

Explanation:
1. Reducing [B] to 1/4 gives (1/4)² = 1/16, so the rate becomes 1/16, not 1/2.
2. The other statements are correct: doubling A doubles the rate, doubling both gives 2 × 4 = 8 and the order is 1 + 2 = 3.

Teacher's Note:
1. Square the factor for B.
2. 1/2 would need [B] to be 1/√2.

Question 22: For \( \ce{A -> B} \), the rate doubles when the concentration of A is increased 8 times. The order of the reaction with respect to A is :
(a) 3
(b) 1/2
(c) 1/3
(d) 0
Show Answer & Explanation

Answer: (c) 1/3

Explanation:
1. \( 8^n = 2 \), so \( 2^{3n} = 2^1 \) and n = 1/3.

Teacher's Note:
1. Write the factor as a power of 2.
2. Equate the exponents.

Question 23: The specific rate constant of a first order reaction depends on :
(a) concentration of the reactant
(b) concentration of the product
(c) time
(d) temperature
Show Answer & Explanation

Answer: (d) temperature

Explanation:
1. The rate constant is independent of concentration and time.
2. It depends only on temperature (and on the catalyst).

Teacher's Note:
1. k changes with temperature as in the Arrhenius equation.
2. Concentration changes the rate, not k.

Question 24: A mechanism is proposed for \( \ce{2A + B -> D + E} \): \( \ce{A + B -> C + D} \) (slow) and \( \ce{A + C -> E} \) (fast). The rate law is :
(a) r = k[A]²[B]
(b) r = k[A][B]
(c) r = k[A]²
(d) r = k[A][C]
Show Answer & Explanation

Answer: (b) r = k[A][B]

Explanation:
1. The slowest step controls the rate.
2. Its molecularity gives r = k[A][B].

Teacher's Note:
1. Rate depends on the slow step only.
2. C is an intermediate, so it should not appear.

Question 25: The reaction \( \ce{2O3 -> 3O2} \) proceeds as \( \ce{O3 <=> O2 + O} \) (fast) and \( \ce{O + O3 -> 2O2} \) (slow). The rate law should be :
(a) r = k[O₃]²
(b) r = k[O₃]²[O₂]⁻¹
(c) r = k[O₃][O₂]
(d) unpredictable
Show Answer & Explanation

Answer: (b) r = k[O₃]²[O₂]⁻¹

Explanation:
1. Slow step: rate = k₂[O][O₃].
2. From the fast equilibrium, \( [O] = K[O_3]/[O_2] \).
3. So rate = k[O₃]²[O₂]⁻¹.

Teacher's Note:
1. Eliminate the intermediate using the fast equilibrium.
2. The product appears with a negative power.

Question 26: The reaction \( \ce{A2 + B2 -> 2AB} \) follows: \( \ce{A2 <=> A + A} \) (fast), \( \ce{A + B2 -> AB + B} \) (slow), \( \ce{A + B -> AB} \) (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 step \( [A] = K^{1/2}[A_2]^{1/2} \).
3. Rate = k'[A₂]^{1/2}[B₂], so the order is 1/2 + 1 = 1.5.

Teacher's Note:
1. Replace the intermediate by reactants.
2. Add the exponents.

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