JEE Chemistry Chemical Kinetics MCQs Set 03

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Chapter-wise Objective Questions: Chemical Kinetics

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Multiple Choice Questions

Question 1: For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), which relationship between the rates of change of concentrations is correct?
(a) \( \text{Rate} = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt} \)
(b) \( \text{Rate} = -\frac{d[N_2]}{dt} = -\frac{3\,d[H_2]}{dt} = \frac{2\,d[NH_3]}{dt} \)
(c) \( \text{Rate} = \frac{d[N_2]}{dt} = \frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt} \)
(d) \( \text{Rate} = -\frac{d[N_2]}{dt} = -\frac{d[H_2]}{dt} = \frac{d[NH_3]}{dt} \)
Show Answer & Explanation

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

Explanation:
1. Divide each rate of change by its stoichiometric coefficient.
2. Reactants are consumed, so their terms carry a negative sign; products carry a positive sign.
3. Hence Rate = \( -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt} \).

Teacher's Note:
1. The rate of reaction is defined so that it is the same whichever species is followed.
2. Always divide by the coefficient and use a minus sign for reactants.

Question 2: In the contact process, 2SO₂(g) + O₂(g) → 2SO₃(g), the rate is \( -\frac{d[O_2]}{dt} = 2.5\times10^{-4} \) 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. From the equation, 2 mol SO₂ are consumed for every 1 mol O₂.
2. \( -\frac{d[SO_2]}{dt} = 2\times\left(-\frac{d[O_2]}{dt}\right) = 2\times2.5\times10^{-4} = 5\times10^{-4} \) mol L⁻¹ s⁻¹.

Teacher's Note:
1. Rates of disappearance are always quoted as positive numbers.
2. The species with the larger coefficient changes faster.

Question 3: In the Haber process N₂ + 3H₂ → 2NH₃, the rate of appearance of NH₃ is 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹. The rate of disappearance of H₂ is
(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. Rate of reaction = \( \frac{1}{2}\frac{d[NH_3]}{dt} = 1.25\times10^{-4} \).
2. \( -\frac{d[H_2]}{dt} = 3\times1.25\times10^{-4} = 3.75\times10^{-4} \) mol L⁻¹ s⁻¹.

Teacher's Note:
1. First convert to the rate of reaction, then scale by the coefficient of the species you need.
2. Check: H₂ and NH₃ are in the ratio 3 : 2, so 2.5 × 3/2 = 3.75.

Question 4: For N₂O₅ → 2NO₂ + ½O₂, it is given that \( -\frac{d[N_2O_5]}{dt} = K_1[N_2O_5] \), \( \frac{d[NO_2]}{dt} = K_2[N_2O_5] \) and \( \frac{d[O_2]}{dt} = K_3[N_2O_5] \). 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. \( \frac{d[NO_2]}{dt} = 2\times\left(-\frac{d[N_2O_5]}{dt}\right) \), so K₂ = 2K₁.
2. \( \frac{d[O_2]}{dt} = \frac{1}{2}\times\left(-\frac{d[N_2O_5]}{dt}\right) \), so K₃ = K₁/2.
3. Thus K₁ : K₂ : K₃ = 1 : 2 : ½, i.e. 2K₁ = K₂ = 4K₃.

Teacher's Note:
1. Write every rate in terms of the same quantity before comparing constants.
2. The coefficient of each product fixes its constant.

Question 5: 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 (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 have the same coefficient, so moles of SO₂ used per minute = moles of SO₃ formed per minute.
2. Mass ratio = molar mass ratio = 64/80 = 0.8.
3. Rate for SO₂ = 1.6 × 10⁻³ × 0.8 = 1.28 × 10⁻³ kg min⁻¹.

Teacher's Note:
1. The rate is given in mass units here, so convert through moles.
2. Equal moles do not mean equal masses.

Question 6: In 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 (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 = \( \frac{1}{2}\times2.5\times10^{-4} = 1.25\times10^{-4} \).
2. Rate of disappearance of H₂ = 3 × 1.25 × 10⁻⁴ = 3.75 × 10⁻⁴.

Teacher's Note:
1. Rate of reaction is a single number for the whole equation.
2. Species rates are multiples of it.

Question 7: For 4A + B → 2C + 2D, which statement is NOT correct?
(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 of 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. Let the rate of reaction be r. Then A disappears at 4r, B at r, C and D appear at 2r each.
2. B disappears at r, while C appears at 2r, so B disappears at half the rate of appearance of C, not twice.
3. The other three statements agree with this.

Teacher's Note:
1. Assign r to the equation, then write each species as coefficient × r.
2. Read 'twice' and 'half' carefully in such statements.

Question 8: For a gaseous reaction, rate = k[A][B]. If the volume of the container is reduced to \( \frac{1}{4} \) of the initial volume, the rate becomes how many times the initial rate?
(a) \( \frac{1}{8} \)
(b) 8
(c) \( \frac{1}{16} \)
(d) 16
Show Answer & Explanation

Answer: (d) 16

Explanation:
1. Reducing the volume to 1/4 increases each concentration 4 times.
2. Rate ∝ [A][B], so the new rate = 4 × 4 = 16 times.

Teacher's Note:
1. Volume and concentration are inversely related for a fixed amount of gas.
2. Multiply the factors for each reactant according to its order.

Question 9: The rate of a hypothetical reaction is \( r = k[A]^{1/2}[B]^{1/3}[C]^{1/4} \). The order of the reaction is
(a) 1
(b) \( \frac{1}{2} \)
(c) 2
(d) \( \frac{13}{12} \)
Show Answer & Explanation

Answer: (d) \( \frac{13}{12} \)

Explanation:
1. Order = sum of exponents = \( \frac{1}{2}+\frac{1}{3}+\frac{1}{4} \).
2. = \( \frac{6+4+3}{12} = \frac{13}{12} \).

Teacher's Note:
1. Order may be a fraction.
2. Add the powers of all concentration terms in the rate law.

Question 10: Which rate law has an overall order of 0.5 for a reaction of 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 \( C_x C_y^{0} / C_z^{2} \)
Show Answer & Explanation

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

Explanation:
1. Option (a): 1+1+1 = 3. Option (b): 0.5×3 = 1.5.
2. Option (c): 1.5 − 1 + 0 = 0.5. Option (d): 1 + 0 − 2 = −1.

Teacher's Note:
1. Negative exponents are allowed and reduce the order.
2. Sum the exponents for each option.

Question 11: For H₂(g) + Br₂(g) → 2HBr(g), the experimental rate law is rate = k[H₂][Br₂]^{1/2}. The order of the reaction is
(a) 2
(b) \( 1\frac{1}{2} \)
(c) 1
(d) \( 2\frac{1}{2} \)
Show Answer & Explanation

Answer: (b) \( 1\frac{1}{2} \)

Explanation:
1. Order = 1 + ½ = 1.5.

Teacher's Note:
1. The exponents come from experiment, not from the balanced equation.
2. Fractional order usually signals a complex mechanism.

Question 12: The initial rate of A + B → C was measured: Exp 1: [A] = 0.012, [B] = 0.035, rate = 0.10; Exp 2: [A] = 0.024, [B] = 0.070, rate = 1.6; Exp 3: [A] = 0.024, [B] = 0.035, rate = 0.20; Exp 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. Exp 1 to Exp 3: [A] doubles, rate doubles, so order in A is 1.
2. Exp 1 to Exp 4: [B] doubles, rate becomes 8 times, so order in B is 3.
3. Check with Exp 2: 2 × 8 = 16 and 0.10 × 16 = 1.6. So rate = k[A][B]³.

Teacher's Note:
1. Change one concentration at a time to find each order.
2. Always confirm using the remaining experiment.

Question 13: For 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 = k[X]⁴[Y]³
Show Answer & Explanation

Answer: (b) r = k[X]⁰[Y]²

Explanation:
1. Rows 1 and 2: [X] doubles with no change in rate, so order in X is 0.
2. Rows 1 and 3 (X has no effect): [Y] doubles, rate ×4, so order in Y is 2.
3. Row 4 check: [Y] = 3 times, rate = 9 times (0.002 × 9 = 0.018). So r = k[Y]².

Teacher's Note:
1. A zero-order reactant does not appear in the rate law.
2. Use the three-times data point as a check.

Question 14: For A + B → C, three experiments give ([A], [B], rate): (0.01, 0.01, 1.0 × 10⁻⁴); (0.01, 0.03, 9.0 × 10⁻⁴); (0.03, 0.03, 2.70 × 10⁻³). 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] ×3, rate ×9, so order in B = 2.
2. Rows 2 and 3: [A] ×3, rate ×3, so order in A = 1.
3. Rate = k[A][B]².

Teacher's Note:
1. Tripling giving 9 times means second order.
2. Work out each order separately.

Question 15: 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). The orders with respect to A and B are
(a) 1 and 0
(b) 1 and 1
(c) 0 and 1
(d) 2 and 0
Show Answer & Explanation

Answer: (a) 1 and 0

Explanation:
1. Doubling [A] doubles the rate: order in A = 1.
2. Doubling [B] leaves the rate unchanged: order in B = 0.

Teacher's Note:
1. No change in rate means zero order.
2. Orders are always found from experimental data.

Question 16: The time required for half of a reaction to complete is inversely proportional to the concentration of 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, \( t_{1/2} \propto a^{1-n} \).
2. Here \( t_{1/2} \propto a^{-1} \), so 1 − n = −1 and n = 2.

Teacher's Note:
1. Half-life of a second-order reaction is 1/(ka).
2. Memorise the half-life dependence for orders 0, 1 and 2.

Question 17: The rate law for the single-step reaction 2A → 2C is
(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. A single-step reaction is elementary, so the exponent of A equals its coefficient 2.
2. Rate = k[A]².

Teacher's Note:
1. Order equals molecularity only for elementary steps.
2. [B] does not occur in this reaction at all.

Question 18: For A + B → products, doubling the concentration of A makes the rate four times, but doubling 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 4 times: order in A = 2.
2. Doubling B gives no change: order in B = 0.
3. Rate = k[A]².

Teacher's Note:
1. Match 2ⁿ with the observed factor to get the order.
2. Zero order species is omitted.

Question 19: 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 species colliding in an elementary step.
2. 2 molecules of X + 1 of Y = 3.

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

Question 20: For A + B → products, the rate doubles when [A] is doubled. When [A] and [B] are both doubled, the rate again just doubles. 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: order in A = 1.
2. Doubling both gives only the same factor 2, so B contributes a factor 1: order in B = 0.

Teacher's Note:
1. Separate the effect of each reactant.
2. Combined changes multiply the individual factors.

Question 21: The rate law for A + B → product is rate = k[A][B]². Which statement is false?
(a) If [B] is constant and [A] is doubled, the reaction is twice as fast
(b) If [A] is constant and [B] is reduced to one quarter, the rate is halved
(c) If [A] and [B] are both doubled the reaction is 8 times as fast
(d) The reaction is of third order
Show Answer & Explanation

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

Explanation:
1. Rate ∝ [B]², so [B] reduced to ¼ gives (¼)² = 1/16 of the rate, not half.
2. The other statements follow from rate = k[A][B]² (order 3, doubling both gives 2 × 4 = 8).

Teacher's Note:
1. Squared dependence: quarter the concentration gives sixteenth the rate.
2. Check each statement against the rate law.

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

Answer: (c) \( \frac{1}{3} \)

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

Teacher's Note:
1. Take logarithms if the numbers are not obvious.
2. Orders can be fractions.

Question 23: The 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 changes with temperature (and with a catalyst), as in the Arrhenius equation.

Teacher's Note:
1. k is a characteristic of the reaction at a given temperature.
2. Do not confuse rate (depends on concentration) with rate constant.

Question 24: The reaction 2O₃ → 3O₂ proceeds as: O₃ ⇌ O₂ + O (fast); O + O₃ → 2O₂ (slow). The rate law is
(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: r = k₂[O][O₃].
2. From the fast equilibrium, \( [O] = K\frac{[O_3]}{[O_2]} \).
3. So r = k₂K[O₃]²/[O₂] = k′[O₃]²[O₂]⁻¹.

Teacher's Note:
1. Replace intermediates using the fast pre-equilibrium.
2. A negative order for a product signals inhibition by it.

Question 25: 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) \( 1\frac{1}{2} \)
(d) 0
Show Answer & Explanation

Answer: (c) \( 1\frac{1}{2} \)

Explanation:
1. Slow step: rate = k[A][B₂].
2. From the fast equilibrium [A] = K^{1/2}[A₂]^{1/2}.
3. Rate = k′[A₂]^{1/2}[B₂], order = ½ + 1 = 1.5.

Teacher's Note:
1. Pre-equilibria can give fractional orders.
2. Substitute intermediates in terms of reactants.

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Multiple Choice Questions (MCQs) for JEE Chemistry Chemical Kinetics

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