JEE Chemistry Chemical Kinetics MCQs Set 21

Chemistry Objective Questions and Answers: Chemical Kinetics

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Multiple Choice Questions

Question 1: For the reaction xA + yB → zC, if −d[A]/dt = −d[B]/dt = 1.5 d[C]/dt, then x, y and z are:
(a) 1, 1, 1
(b) 3, 2, 3
(c) 3, 3, 2
(d) 2, 2, 3
Show Answer & Explanation

Answer: (c) 3, 3, 2

Explanation:
1. Rate of reaction r = −(1/x)d[A]/dt = −(1/y)d[B]/dt = (1/z)d[C]/dt.
2. So −d[A]/dt = x r, −d[B]/dt = y r and d[C]/dt = z r.
3. x r = y r = 1.5 z r gives x = y and x = 1.5 z, so x : y : z = 3 : 3 : 2.

Teacher's Note:
1. Each rate of change equals its coefficient times the rate of reaction.
2. Pick the smallest whole numbers.

Question 2: For the reaction xA → yB, log[−d[A]/dt] = log[d[B]/dt] + 0.3. Then x : y is:
(a) 1 : 2
(b) 2 : 1
(c) 3 : 1
(d) 3 : 10
Show Answer & Explanation

Answer: (b) 2 : 1

Explanation:
1. log(−d[A]/dt) − log(d[B]/dt) = 0.3, so (−d[A]/dt)/(d[B]/dt) = 10^0.3 = 2.
2. −d[A]/dt = x r and d[B]/dt = y r, so x/y = 2.
3. x : y = 2 : 1.

Teacher's Note:
1. 10^0.3 = 2 (since log 2 = 0.3).
2. Rates of change are proportional to the coefficients.

Question 3: The rate of formation of SO₃ in the reaction 2SO₂ + O₂ → 2SO₃ is 100 g min⁻¹. The rate of disappearance of O₂ is (S = 32, O = 16):
(a) 50 g min⁻¹
(b) 40 g min⁻¹
(c) 200 g min⁻¹
(d) 20 g min⁻¹
Show Answer & Explanation

Answer: (d) 20 g min⁻¹

Explanation:
1. Rate of formation of SO₃ = 100/80 = 1.25 mol min⁻¹.
2. From the equation, moles of O₂ used = half the moles of SO₃ formed = 0.625 mol min⁻¹.
3. Mass of O₂ = 0.625 × 32 = 20 g min⁻¹.

Teacher's Note:
1. Convert grams to moles before using stoichiometric ratios.
2. Convert back to grams at the end.

Question 4: The rate of a reaction is expressed in different ways as +(1/2)d[C]/dt = −(1/3)d[D]/dt = +(1/4)d[A]/dt = −d[B]/dt. The reaction is:
(a) 4A + B → 2C + 3D
(b) B + 3D → 4A + 2C
(c) 4A + 2B → 2C + 3D
(d) B + (1/2)D → 4A + B
Show Answer & Explanation

Answer: (b) B + 3D → 4A + 2C

Explanation:
1. Species with a negative sign (B and D) are reactants; those with a positive sign (C and A) are products.
2. The coefficients are the numbers dividing each derivative: B 1, D 3, C 2, A 4.
3. Reaction: B + 3D → 4A + 2C.

Teacher's Note:
1. The sign shows whether a species is consumed or formed.
2. The divisor of each term is its stoichiometric coefficient.

Question 5: For the reaction A + B → C, starting with different initial concentrations of A and B, the initial rates were: (1) [A] = 1.6 × 10⁻³, [B] = 5 × 10⁻², rate = 10⁻³ M s⁻¹; (2) [A] = 3.2 × 10⁻³, [B] = 5 × 10⁻², rate = 4 × 10⁻³; (3) [A] = 1.6 × 10⁻³, [B] = 10⁻¹, rate = 2 × 10⁻³; (4) [A] = 3.2 × 10⁻³, [B] = 10⁻¹, rate = 8 × 10⁻³. The rate law is:
(a) r = k[A]²[B]²
(b) r = k[A]²[B]
(c) r = k[A][B]²
(d) r = k[A][B]
Show Answer & Explanation

Answer: (b) r = k[A]²[B]

Explanation:
1. Experiments 1 and 2: [A] doubled with [B] constant, rate × 4, so the order in A is 2.
2. Experiments 1 and 3: [B] doubled with [A] constant, rate × 2, so the order in B is 1.
3. Check with experiment 4: 2² × 2 = 8 times, 8 × 10⁻³ ✓. Rate = k[A]²[B].

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

Question 6: For a reaction aA + bB → product, dx/dt = k[A]^a[B]^b. If the concentration of A is doubled, the rate is four times; if the concentration of B is made four times, the rate is doubled. The relation between the rates of disappearance of A and B is:
(a) −d[A]/dt = −d[B]/dt
(b) −d[A]/dt = −4 d[B]/dt
(c) −4 d[A]/dt = −d[B]/dt
(d) None of these
Show Answer & Explanation

Answer: (b) −d[A]/dt = −4 d[B]/dt

Explanation:
1. Doubling [A] gives 4 times rate, so a = 2. Making [B] 4 times gives 2 times rate, so 4^b = 2 and b = 1/2.
2. −d[A]/dt = a(dx/dt) = 2(dx/dt) and −d[B]/dt = b(dx/dt) = (1/2)(dx/dt).
3. Hence −d[A]/dt = 4(−d[B]/dt).

Teacher's Note:
1. The rate of disappearance of a reactant is its coefficient times dx/dt.
2. Here dx/dt is the rate of reaction.

Question 7: For the reaction 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g), the rate expression can be written as d[N₂]/dt = k₁[NO][H₂]; d[H₂O]/dt = k[NO][H₂]; −d[NO]/dt = k′[NO][H₂]; −d[H₂]/dt = k″[NO][H₂]. The relationship between k, k₁, k′ and k″ is:
(a) k = k₁ = k′ = k″
(b) k = 2k₁ = k′ = k″
(c) k = 2k′ = k₁ = 2k″
(d) k = k₁ = k′ = 2k″
Show Answer & Explanation

Answer: (b) k = 2k₁ = k′ = k″

Explanation:
1. Rate of formation of N₂ = r. Then d[H₂O]/dt = 2r, −d[NO]/dt = 2r and −d[H₂]/dt = 2r.
2. Writing each as (constant)[NO][H₂]: k₁ ∝ r, and k = k′ = k″ ∝ 2r.
3. So k = k′ = k″ = 2k₁.

Teacher's Note:
1. Compare the stoichiometric coefficients: N₂ has 1, the rest have 2.
2. The rate constants carry the same ratio as the coefficients.

Question 8: In acidic medium the rate of reaction between BrO₃⁻ and Br⁻ ions is given by −d[BrO₃⁻]/dt = k[BrO₃⁻][Br⁻][H⁺]². It means:
(a) Rate constant of overall reaction is 4 s⁻¹
(b) Rate of reaction is independent of the concentration of acid
(c) Changes in pH of the solution will not affect the rate
(d) Doubling the concentration of H⁺ ions will increase the reaction rate by 4 times
Show Answer & Explanation

Answer: (d) Doubling the concentration of H⁺ ions will increase the reaction rate by 4 times

Explanation:
1. The rate is second order in H⁺.
2. Doubling [H⁺] multiplies the rate by 2² = 4.
3. The rate depends on pH, and the overall order is 4, so the unit of k is M⁻³ s⁻¹, not s⁻¹.

Teacher's Note:
1. The exponent of a concentration is its order.
2. Overall order = 1 + 1 + 2 = 4.

Question 9: The rate of production of NH₃ in N₂ + 3H₂ → 2NH₃ is 3.4 kg min⁻¹. The rate of consumption of H₂ is (N = 14, H = 1):
(a) 5.1 kg min⁻¹
(b) 0.01 kg s⁻¹
(c) 0.6 kg h⁻¹
(d) None of these
Show Answer & Explanation

Answer: (b) 0.01 kg s⁻¹

Explanation:
1. NH₃ formed = 3400/17 = 200 mol min⁻¹.
2. H₂ consumed = (3/2) × 200 = 300 mol min⁻¹ = 600 g min⁻¹ = 0.6 kg min⁻¹.
3. 0.6 kg per minute = 0.01 kg s⁻¹.

Teacher's Note:
1. Convert mass to moles first.
2. 1 kg min⁻¹ is 1/60 kg s⁻¹.

Question 10: At the point of intersection of the two curves for the reaction A → nB (concentration of A falling from A₀ and that of B rising), the concentration of B is given by:
(a) nA₀/2
(b) A₀/(n − 1)
(c) nA₀/(n + 1)
(d) (n − 1)A₀/(n + 1)
Show Answer & Explanation

Answer: (c) nA₀/(n + 1)

Explanation:
1. Let x be the amount of A consumed at the intersection. Then [A] = A₀ − x and [B] = n x.
2. At the intersection [A] = [B]: A₀ − x = n x, so x = A₀/(n + 1).
3. [B] = n x = nA₀/(n + 1).

Teacher's Note:
1. The curves cross where the two concentrations are equal.
2. Express both in terms of the amount reacted x.

Question 11: The reaction A + B → C + D follows the rate law: rate = k[A]^(1/2)[B]^(1/2). Starting with an initial concentration of 1 M each of A and B, the time taken for the concentration of A to become 0.25 M is (k = 2.303 × 10⁻³ s⁻¹):
(a) 300 s
(b) 600 s
(c) 900 s
(d) None of these
Show Answer & Explanation

Answer: (b) 600 s

Explanation:
1. A and B are consumed in equal amounts and start equal, so [A] = [B] at all times.
2. Rate = k[A]^(1/2)[A]^(1/2) = k[A], a first order law.
3. t = (2.303/k) log(1/0.25) = (2.303/2.303 × 10⁻³) × 0.602 = 602 s ≈ 600 s.

Teacher's Note:
1. Equal concentrations of A and B at the start and 1 : 1 stoichiometry mean they stay equal.
2. Then the overall order is 1.

Question 12: In presence of HCl, sucrose is hydrolysed into glucose and fructose. The concentration of sucrose was found to reduce from 0.4 M to 0.2 M in 1 hour and to 0.1 M in a total of 2 hours. The order of the reaction is:
(a) Zero
(b) One
(c) Two
(d) None of these
Show Answer & Explanation

Answer: (b) One

Explanation:
1. The concentration halves from 0.4 to 0.2 in 1 h and from 0.2 to 0.1 in the next hour.
2. A constant half-life independent of the initial concentration is the signature of a first order reaction.

Teacher's Note:
1. Zero order: half-life decreases as concentration decreases. Second order: half-life doubles each time.
2. Check successive half-lives.

Question 13: In a first order reaction the reacting substance has a half-life of ten minutes. The fraction of the substance left after an hour is:
(a) 1/6 of the initial concentration
(b) 1/64 of the initial concentration
(c) 1/12 of the initial concentration
(d) 1/32 of the initial concentration
Show Answer & Explanation

Answer: (b) 1/64 of the initial concentration

Explanation:
1. 60 minutes = 6 half-lives.
2. Fraction left = (1/2)⁶ = 1/64.

Teacher's Note:
1. Fraction left = (1/2)^n for n half-lives.
2. Do not divide the time by the half-life and then use the result as a fraction.

Question 14: For a given reaction of first order, it takes 20 min for the concentration to drop from 1.0 M to 0.60 M. The time required for the concentration to drop from 0.60 M to 0.36 M will be:
(a) More than 20 min
(b) 20 min
(c) Less than 30 min
(d) Cannot tell
Show Answer & Explanation

Answer: (b) 20 min

Explanation:
1. For a first order reaction the time depends only on the ratio of concentrations.
2. 1.0 → 0.60 is a ratio 0.6, and 0.60 → 0.36 is also a ratio 0.6.
3. So the time is the same, 20 min.

Teacher's Note:
1. t = (2.303/k) log(C₀/C): equal ratios give equal times.
2. This property is why first order reactions have a constant half-life.

Question 15: In the first order reactions (a) → product (k₁) and (b) → product (k₂), the ratio k₁/k₂ is, if 90% of (a) has reacted in time t while 99% of (b) has reacted in time 2t:
(a) 1
(b) 2
(c) 1/2
(d) None of these
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. (a): k₁t = 2.303 log(100/10) = 2.303.
2. (b): k₂(2t) = 2.303 log(100/1) = 2 × 2.303, so k₂t = 2.303.
3. So k₁ = k₂ and the ratio is 1.

Teacher's Note:
1. Write 2.303 log(a/(a − x)) for each reaction.
2. 99% completion needs exactly twice the time of 90% for the same k.

Question 16: If a first order reaction is completed to the extent of 60% and 20% in time intervals t₁ and t₂, the ratio t₁ : t₂ is:
(a) 6.32
(b) 5.58
(c) 4.11
(d) 8.33
Show Answer & Explanation

Answer: (c) 4.11

Explanation:
1. t₁ = (2.303/k) log(100/40) = (2.303/k)(0.398).
2. t₂ = (2.303/k) log(100/80) = (2.303/k)(0.0969).
3. t₁/t₂ = 0.398/0.0969 = 4.11.

Teacher's Note:
1. 60% completed leaves 40%; 20% completed leaves 80%.
2. k cancels in the ratio.

Question 17: Two first order reactions have half-lives in the ratio 3 : 2. The ratio of the time intervals t₁ : t₂, where t₁ and t₂ are the times for 25% and 75% completion of the first and second reaction respectively, is (log 2 = 0.301, log 3 = 0.477):
(a) 0.311 : 1
(b) 0.420 : 1
(c) 0.273 : 1
(d) 0.119 : 1
Show Answer & Explanation

Answer: (a) 0.311 : 1

Explanation:
1. k ∝ 1/t₁/₂, so k₁ : k₂ = 2 : 3.
2. t₁ = (1/k₁) ln(1/0.75) = 0.2877/k₁ and t₂ = (1/k₂) ln(1/0.25) = 1.386/k₂.
3. t₁/t₂ = (0.2877/1.386)(k₂/k₁) = 0.2075 × 1.5 = 0.311.

Teacher's Note:
1. 25% completion leaves 75%; 75% completion leaves 25%.
2. Convert half-life ratios to rate constant ratios by inverting.

Question 18: For the reaction A → products, −d[A]/dt = k, and at different time intervals [A] values are: 0 min, 20 mol; 5 min, 18 mol; 10 min, 16 mol; 15 min, 14 mol. At 20 minutes the rate will be:
(a) 12 mol min⁻¹
(b) 10 mol min⁻¹
(c) 8 mol min⁻¹
(d) 0.4 mol min⁻¹
Show Answer & Explanation

Answer: (d) 0.4 mol min⁻¹

Explanation:
1. −d[A]/dt = k means the reaction is zero order and the rate is constant.
2. [A] falls by 2 mol every 5 min, so the rate = 2/5 = 0.4 mol min⁻¹ at all times, including 20 min.

Teacher's Note:
1. A constant rate (independent of [A]) means zero order.
2. Rate = −Δ[A]/Δt.

Question 19: At a certain temperature the half-life period in the thermal decomposition of a gaseous substance is: P = 500 mm Hg, t₁/₂ = 235 min; P = 250 mm Hg, t₁/₂ = 950 min. The order of the reaction is (log 23.5 = 1.37, log 95 = 1.97):
(a) 1
(b) 2
(c) 2.5
(d) 3
Show Answer & Explanation

Answer: (d) 3

Explanation:
1. For order n, t₁/₂ ∝ P^(1 − n). So t₂/t₁ = (P₁/P₂)^(n − 1).
2. 950/235 = 4.04 = 2^(n − 1), that is (n − 1) log 2 = log 95 − log 23.5 = 0.60.
3. n − 1 = 0.60/0.301 = 2, so n = 3.

Teacher's Note:
1. Half-life of an nth order reaction is inversely proportional to the (n − 1)th power of the initial concentration.
2. Take logs of the ratio to find n.

Question 20: The data for the reaction A + B → C: Exp 1: [A]₀ = 0.012, [B]₀ = 0.035, initial rate 0.10; Exp 2: [A]₀ = 0.024, [B]₀ = 0.035, initial rate 0.80; Exp 3: [A]₀ = 0.012, [B]₀ = 0.070, initial rate 0.10; Exp 4: [A]₀ = 0.024, [B]₀ = 0.070, initial rate 0.80. The rate law is:
(a) r = k[B]³
(b) r = k[A]³
(c) r = k[A][B]⁴
(d) r = k[A]²[B]²
Show Answer & Explanation

Answer: (b) r = k[A]³

Explanation:
1. Exp 1 → 2: [A] doubles with [B] constant and the rate becomes 8 times, so the order in A is 3.
2. Exp 1 → 3: [B] doubles with [A] constant and the rate is unchanged, so the order in B is 0.
3. Rate = k[A]³.

Teacher's Note:
1. A rate unchanged on doubling a concentration means zero order in that species.
2. 2ⁿ = 8 gives n = 3.

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