JEE Chemistry Chemical Kinetics MCQs Set 27

Multiple Choice Questions (MCQs) for JEE Chemistry: Chemical Kinetics

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Multiple Choice Questions

Question 1: In the scheme A + B → 2C (k1), C + B → 2D (k2), C + A → 2E (k3), the rate of disappearance of C is
(a) -d[C]/dt = k2[B][C] + k3[E] - k1[C]
(b) -d[C]/dt = k2[B][C] + k3[A][C] - 2k1[A][B]
(c) -d[C]/dt = k2[D] + k3[E] - k1[C]
(d) -d[C]/dt = 2k1[A][B] - k2[B][C] - k3[A][C]
Show Answer & Explanation

Answer: (b) -d[C]/dt = k2[B][C] + k3[A][C] - 2k1[A][B]

Explanation:
1. C is produced in step 1 at rate 2k1[A][B] and consumed in steps 2 and 3 at rates k2[B][C] and k3[A][C].
2. Net disappearance = k2[B][C] + k3[A][C] - 2k1[A][B].

Teacher's Note:
1. Rate of formation minus rate of consumption gives the net change.
2. The factor 2 comes from the coefficient of C in step 1.

Question 2: For 3A → B + C, the reaction would be of zero order when
(a) the rate doubles if the concentration of B is doubled
(b) the rate remains the same at any concentration of A
(c) the rate remains unchanged at any concentration of B and C
(d) the rate is proportional to the square of the concentration of A
Show Answer & Explanation

Answer: (b) the rate remains the same at any concentration of A

Explanation:
1. A zero order reaction has rate = k, independent of the concentration of the reactant.
2. Dependence on products is not what defines the order with respect to A.

Teacher's Note:
1. Order refers to dependence on reactant concentration.
2. Rate = k[A]0.

Question 3: For A → B + C (first order), only B and C are optically active and dextrorotatory. The total optical rotation 10 minutes after the start is 50° and when the reaction is complete it is 100°. The rate constant is
(a) 0.693 min-1
(b) 0.0693 s-1
(c) 0.00693 s-1
(d) 0.0693 min-1
Show Answer & Explanation

Answer: (d) 0.0693 min-1

Explanation:
1. Rotation is proportional to the extent of reaction, so the reaction is half complete after 10 minutes.
2. t1/2 = 10 min, so k = 0.693/10 = 0.0693 min-1.

Teacher's Note:
1. Rotation 50 out of 100 means half the reactant has been consumed.
2. k = ln 2 / t1/2.

Question 4: A substance decomposes by two parallel first order paths to B (k1 = 1.26 × 10-4 s-1) and C (k2 = 3.8 × 10-5 s-1). The percentage distribution of B and C is
(a) 90% B and 10% C
(b) 60% B and 40% C
(c) 80% B and 20% C
(d) 76.83% B and 23.17% C
Show Answer & Explanation

Answer: (d) 76.83% B and 23.17% C

Explanation:
1. In parallel first order reactions the products are formed in the ratio of their rate constants.
2. % B = 1.26/(1.26 + 0.38) × 100 = 76.83%, and % C = 23.17%.

Teacher's Note:
1. Fraction of a product = its k / total k.
2. Total k = 1.64 × 10-4 s-1.

Question 5: The forward rate constant for the elementary reversible gaseous reaction C2H6 ⇌ 2CH3 is 1.57 × 10-3 s-1 at a given temperature. What is the rate constant of the backward reaction if 10-4 mol of CH3 and 10 mol of C2H6 are present at equilibrium in a 10 L vessel?
(a) 1.57 × 1010 L mol-1 s-1
(b) 1.57 × 1011 L mol-1 s-1
(c) 1.57 × 107 L mol-1 s-1
(d) 1.57 × 109 L mol-1 s-1
Show Answer & Explanation

Answer: (c) 1.57 × 107 L mol-1 s-1

Explanation:
1. [CH3] = 10-5 M and [C2H6] = 1 M.
2. At equilibrium Kc = [CH3]2/[C2H6] = 10-10 mol L-1 = kf/kb.
3. kb = 1.57 × 10-3/10-10 = 1.57 × 107 L mol-1 s-1.

Teacher's Note:
1. At equilibrium forward and backward rates are equal.
2. Convert amounts to concentrations.

Question 6: A reaction is second order in carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else unchanged, the rate
(a) triples
(b) doubles
(c) increases by a factor of 4
(d) remains unchanged
Show Answer & Explanation

Answer: (c) increases by a factor of 4

Explanation:
1. Rate is proportional to [CO]2, so doubling [CO] multiplies the rate by 22 = 4.

Teacher's Note:
1. Raise the factor to the order.
2. 22 = 4.

Question 7: The first order rate constant of a reaction is 2.303 min-1. If the initial concentration is 1 mol L-1, the rate of the reaction after 1 minute is
(a) 0.2303 M min-1
(b) 2.303 M min-1
(c) 0.1 M min-1
(d) none of these
Show Answer & Explanation

Answer: (a) 0.2303 M min-1

Explanation:
1. [A] = e-kt = e-2.303 = 0.1 M after 1 minute.
2. Rate = k[A] = 2.303 × 0.1 = 0.2303 M min-1.

Teacher's Note:
1. Find the concentration first, then the rate.
2. e-2.303 = 0.1.

Question 8: The average life of U-238 with a half-life of 138.5 days (as given in the problem) is
(a) 238 days
(b) none of these
(c) 199.86 days
(d) 138.5 days
Show Answer & Explanation

Answer: (c) 199.86 days

Explanation:
1. Average life = t1/2/0.693 = 138.5/0.693 = 199.86 days.

Teacher's Note:
1. Mean life = 1/λ = 1.44 x half-life.
2. It is greater than the half-life.

Question 9: The rate of a reaction with k = 2.0 × 10-3 s-1 is measured at 25 °C. If it is first order, the half-life is (ln 2 = 0.693)
(a) 346.5 s
(b) 0.693 s
(c) 1386 s
(d) 693 s
Show Answer & Explanation

Answer: (a) 346.5 s

Explanation:
1. t1/2 = 0.693/k = 0.693/(2.0 × 10-3) = 346.5 s.

Teacher's Note:
1. First order half-life is 0.693/k.
2. Units of k show it is first order.

Question 10: For A(g) + 2B(g) → C(g) the initial rates at 300 K are: [A] = 0.01 M, [B] = 0.01 M, 6.930 × 10-6 M s-1; [A] = 0.02 M, [B] = 0.01 M, 1.386 × 10-5 M s-1; [A] = 0.02 M, [B] = 0.02 M, 1.386 × 10-5 M s-1. In another experiment starting with [A] = 0.5 M and [B] = 1 M at 300 K, the rate after 50 minutes is
(a) 3.46 × 10-9 M s-1
(b) 6.93 × 10-4 M s-1
(c) 4.33 × 10-5 M s-1
(d) 0.25 × 10-7 M s-1
Show Answer & Explanation

Answer: (c) 4.33 × 10-5 M s-1

Explanation:
1. Rate doubles when [A] doubles and does not change with [B], so rate = k[A], first order with k = 6.930 × 10-6/0.01 = 6.93 × 10-4 s-1 (half-life 1000 s).
2. After 50 min = 3000 s = 3 half-lives, [A] = 0.5/8 = 0.0625 M.
3. Rate = 6.93 × 10-4 x 0.0625 = 4.33 × 10-5 M s-1.

Teacher's Note:
1. Use half-lives: 3000 s is three of them.
2. Order in B is zero.

Question 11: At room temperature the reaction between NO and O2 to give NO2 is fast while that between CO and O2 is slow. This is because
(a) CO is poisonous
(b) the activation energy of 2NO + O2 → 2NO2 is lower than that of 2CO + O2 → 2CO2
(c) CO is smaller than NO
(d) none of these
Show Answer & Explanation

Answer: (b) the activation energy of 2NO + O2 → 2NO2 is lower than that of 2CO + O2 → 2CO2

Explanation:
1. The rate depends on the fraction of molecules with energy above the activation energy.
2. The NO reaction has the lower activation energy, so it is faster at the same temperature.

Teacher's Note:
1. Lower activation energy means a larger rate constant.
2. Size and toxicity are not relevant.

Question 12: The units of the rate constants of first order and zero order reactions in terms of molarity M are respectively
(a) M, s-1
(b) s-1, M
(c) s-1, M s-1
(d) M s-1, s-1
Show Answer & Explanation

Answer: (c) s-1, M s-1

Explanation:
1. k for order n has units M1-n s-1.
2. First order: s-1; zero order: M s-1.

Teacher's Note:
1. Rate has units M s-1.
2. Divide by concentration^n.

Question 13: The correct differential rate law for H2 + I2 → 2HI is
(a) -(1/2) d[H2]/dt = -(1/2) d[I2]/dt = d[HI]/dt
(b) -d[H2]/dt = -d[I2]/dt = (1/2) d[HI]/dt
(c) -d[H2]/dt = -d[I2]/dt = d[HI]/dt
(d) -2 d[H2]/dt = -2 d[I2]/dt = d[HI]/dt
Show Answer & Explanation

Answer: (b) -d[H2]/dt = -d[I2]/dt = (1/2) d[HI]/dt

Explanation:
1. H2 and I2 are used up at the same rate, and HI forms twice as fast.
2. Rate of reaction = -d[H2]/dt = -d[I2]/dt = (1/2) d[HI]/dt.

Teacher's Note:
1. Divide by the coefficient.
2. HI has coefficient 2.

Question 14: For rate = k[A]^n[B]^m, the concentration of A is doubled and that of B is halved. The ratio of the new rate to the earlier rate is
(a) 2n/m
(b) 2n-m
(c) 1/2m+n
(d) 2m+n
Show Answer & Explanation

Answer: (b) 2n-m

Explanation:
1. New rate = k(2[A])^n([B]/2)^m = 2^n 2-m k[A]^n[B]^m.
2. Ratio = 2n-m.

Teacher's Note:
1. Substitute the modified concentrations.
2. Powers of 2 combine.

Question 15: For aA + bB → products, rate = k[A]^a[B]^b. Doubling the concentration of A increases the rate four times, and increasing the concentration of B four times doubles the rate. The relation between the rates of disappearance of A and B is
(a) -d[A]/dt = -4 d[B]/dt
(b) -4 d[A]/dt = -d[B]/dt
(c) none of these
(d) -d[A]/dt = -d[B]/dt
Show Answer & Explanation

Answer: (a) -d[A]/dt = -4 d[B]/dt

Explanation:
1. Doubling A gives 2^a = 4, so a = 2; quadrupling B gives 4^b = 2, so b = 1/2.
2. -(1/a) d[A]/dt = -(1/b) d[B]/dt, so -d[A]/dt = (a/b)(-d[B]/dt) = 4(-d[B]/dt).

Teacher's Note:
1. Here the exponents are used as stoichiometric coefficients, as the question states.
2. Rate of reaction is the same whichever species is used once divided by its coefficient.

Question 16: The half-life of a radioactive element is 20 days for a 10 g sample. The half-life for a 100 g sample of the same element is
(a) 200 days
(b) 10 days
(c) 20 days
(d) 100 days
Show Answer & Explanation

Answer: (c) 20 days

Explanation:
1. Radioactive decay is first order, so the half-life does not depend on the amount present.

Teacher's Note:
1. First order half-life is independent of initial concentration.
2. Decay constant is a property of the nucleus.

Question 17: Statements: (1) a second order reaction must be a bimolecular elementary reaction; (2) a bimolecular elementary reaction must be a second order reaction; (3) a zero order reaction must be a complex reaction; (4) a first order reaction may be complex or elementary. How many are correct?
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (c) 3

Explanation:
1. (1) is false: order is experimental and complex reactions can be second order.
2. (2) is true because an elementary step has order equal to its molecularity.
3. (3) is true because an elementary step cannot be zero order, and (4) is true. So three are correct.

Teacher's Note:
1. Order equals molecularity only for elementary reactions.
2. A zero order rate is found only for multistep (for example surface catalysed) reactions.

Question 18: For the first order reaction A → B + 2C + 3D (A optically inactive, B, C and D dextrorotatory), rt and r∞ are the optical rotations at time t and at completion. The rate constant is
(a) k = (1/t) ln(r∞ rt)
(b) none of these
(c) k = (1/t) ln(r∞/(r∞ - rt))
(d) k = (1/t) ln(rt/(r∞ - rt))
Show Answer & Explanation

Answer: (c) k = (1/t) ln(r∞/(r∞ - rt))

Explanation:
1. Rotation is proportional to the amount of A reacted: rt is proportional to x and r∞ to a.
2. Amount of A left is proportional to r∞ - rt.
3. k = (1/t) ln[r∞/(r∞ - rt)].

Teacher's Note:
1. Initial rotation is zero as A is inactive.
2. Reactant remaining is proportional to the rotation still to come.

Question 19: For the first order reaction A(g) → B(g) + C(g) + D(g) started with pure A in a rigid vessel (initial pressure Pi, total pressure Pt at time t), the correct expression for k is
(a) k = (2.303/t) log(Pi/Pt)
(b) k = (2.303/t) log(Pt/Pi)
(c) k = (2.303/t) log[3Pi/(2Pi - 3Pt)]
(d) k = (2.303/t) log[2Pi/(3Pi - Pt)]
Show Answer & Explanation

Answer: (d) k = (2.303/t) log[2Pi/(3Pi - Pt)]

Explanation:
1. If x of A reacts, Pt = (Pi - x) + 3x = Pi + 2x, so x = (Pt - Pi)/2.
2. Pressure of A = Pi - x = (3Pi - Pt)/2.
3. k = (2.303/t) log[Pi/(Pi - x)] = (2.303/t) log[2Pi/(3Pi - Pt)].

Teacher's Note:
1. Three moles of gas are formed from each mole of A.
2. Express P(A) in terms of Pt.

Question 20: For Cl2(aq) + H2S(aq) → S(s) + 2H+ + 2Cl-, the rate equation is rate = k[Cl2][H2S]. Mechanism A: Cl2 + H2S → H+ + Cl- + Cl+ + HS- (slow), then Cl+ + HS- → H+ + Cl- + S (fast). Mechanism B: H2S ⇌ H+ + HS- (fast equilibrium), Cl2 + HS- → 2Cl- + H+ + S (slow). Which is consistent with the rate equation?
(a) A only
(b) neither A nor B
(c) both A and B
(d) B only
Show Answer & Explanation

Answer: (a) A only

Explanation:
1. Mechanism A: the slow step is bimolecular in Cl2 and H2S, so rate = k[Cl2][H2S], as observed.
2. Mechanism B: rate = k[Cl2][HS-] with [HS-] = K[H2S]/[H+], so rate = kK[Cl2][H2S]/[H+], which has an extra inverse [H+] term.
3. Only mechanism A fits.

Teacher's Note:
1. Write the rate law of the slow step and remove intermediates.
2. An extra concentration term means inconsistency.

Question 21: The half-life of a first order reaction is 6.93 minutes. The time required for 99% completion is (log 2 = 0.301)
(a) 46.06 minutes
(b) 460.6 minutes
(c) 230.3 minutes
(d) 23.03 minutes
Show Answer & Explanation

Answer: (a) 46.06 minutes

Explanation:
1. k = 0.693/6.93 = 0.1 min-1.
2. t = (2.303/k) log(100/1) = 2.303 × 2/0.1 = 46.06 minutes.

Teacher's Note:
1. 99% completion means 1% is left.
2. t = 6.64 half-lives.

Question 22: The elementary reaction A(g) + 2B(g) → C(g) + D(g) starts with PA = 0.60 atm and PB = 0.80 atm. When PC = 0.2 atm, the rate relative to the initial rate is
(a) 1/24
(b) 9/16
(c) 1/6
(d) 1/48
Show Answer & Explanation

Answer: (c) 1/6

Explanation:
1. Initial rate = k(0.60)(0.80)2 = 0.384 k.
2. When PC = 0.2 atm, PA = 0.40 and PB = 0.80 - 0.40 = 0.40 atm, rate = k(0.40)(0.40)2 = 0.064 k.
3. Ratio = 0.064/0.384 = 1/6.

Teacher's Note:
1. B is consumed twice as fast as A.
2. Use the elementary rate law.

Question 23: For A + 2B → C the rate is R = k[A][B]2. The order of the reaction is
(a) 5
(b) 7
(c) 6
(d) 3
Show Answer & Explanation

Answer: (d) 3

Explanation:
1. The order is the sum of the exponents in the rate law: 1 + 2 = 3.

Teacher's Note:
1. Order is not the sum of the stoichiometric coefficients unless the step is elementary.
2. Here they happen to coincide.

Question 24: The half-life of a radioisotope is four hours. If the initial mass was 200 g, the mass remaining undecayed after 24 hours is
(a) 4.167 g
(b) 2.084 g
(c) 3.125 g
(d) 1.042 g
Show Answer & Explanation

Answer: (c) 3.125 g

Explanation:
1. 24 h = 6 half-lives.
2. Mass left = 200/64 = 3.125 g.

Teacher's Note:
1. n = 24/4 = 6.
2. 26 = 64.

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