JEE Chemistry Chemical Kinetics MCQs Set 29

Chemistry Objective Questions and Answers: Chemical Kinetics

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Multiple Choice Questions

Question 1: For the same two mechanisms (I: rate = 1 × 10-5[A][B]; II: rate = 1 × 10-4[A]), the two rates are equal when [B] equals
(a) 7 M
(b) 10 M
(c) 5 M
(d) 1 M
Show Answer & Explanation

Answer: (b) 10 M

Explanation:
1. Equate: 10-5[A][B] = 10-4[A].
2. [B] = 10 M.

Teacher's Note:
1. [A] cancels.
2. Divide both sides by 10-5[A].

Question 2: The catalytic converter in automobiles mainly serves
(a) to decrease sound pollution
(b) to decrease the rate of combustion of fuel
(c) to increase the rate of combustion of fuel
(d) to control the air pollutants produced by the engine
Show Answer & Explanation

Answer: (d) to control the air pollutants produced by the engine

Explanation:
1. It contains Pt/Pd/Rh catalysts that convert CO, unburnt hydrocarbons and NOx into CO2, H2O and N2.

Teacher's Note:
1. It acts on exhaust gases, not on fuel burning.
2. Heterogeneous catalysis.

Question 3: In a mechanism, an activated molecule A* is formed by A + A → A + A* (rate constant k1) and removed by A* + A → 2A (rate constant k2). Applying the steady state approximation, [A*] is
(a) k2[A]/k1
(b) k1k2[A]
(c) k1[A]/k2
(d) k1k2/[A]
Show Answer & Explanation

Answer: (c) k1[A]/k2

Explanation:
1. Formation rate of A* = k1[A]2; removal rate = k2[A*][A].
2. Steady state: k1[A]2 = k2[A*][A], so [A*] = k1[A]/k2.

Teacher's Note:
1. Set d[A*]/dt = 0.
2. The [A] cancels once.

Question 4: The time for half reaction of an nth order reaction (initial concentration a) is inversely proportional to
(a) a^n
(b) an+1
(c) a
(d) an-1
Show Answer & Explanation

Answer: (d) an-1

Explanation:
1. For order n (n not 1), t1/2 is proportional to a1-n = 1/an-1.

Teacher's Note:
1. Check: n = 2 gives 1/a.
2. n = 1 gives no dependence.

Question 5: The unit of the rate constant of a second order reaction (equal reactant concentrations) is
(a) mol2 dm-6 s-1
(b) s-1
(c) dm3 mol-1 s-1
(d) mol dm-3 s-1
Show Answer & Explanation

Answer: (c) dm3 mol-1 s-1

Explanation:
1. Rate = k c2, so k = rate / c2.
2. Units: (mol dm-3 s-1) / (mol dm-3)2 = dm3 mol-1 s-1.

Teacher's Note:
1. For order n the unit is (mol dm-3)1-n s-1.
2. First order gives s-1.

Question 6: If the activation energy of a reaction equals 2RT, the rate constant is related to the frequency factor A by
(a) k = 2.303 A
(b) k = 0.0135 A
(c) k = 1.35 A
(d) k = 0.135 A
Show Answer & Explanation

Answer: (d) k = 0.135 A

Explanation:
1. k = A e-E/RT = A e-2.
2. e-2 = 0.135, so k = 0.135 A.

Teacher's Note:
1. Substitute E/RT = 2 directly.
2. e-2 is about 1/7.4.

Question 7: For Ag+ + 2NH3 ⇌ [Ag(NH3)2]+, dx/dt = 2 × 107 L2 mol-2 s-1[Ag+][NH3]2 - 1 × 10-2 s-1[Ag(NH3)2+]. The ratio of forward to backward rate constants (equilibrium constant) is
(a) 2 × 107 L2 mol-2
(b) 1 × 10-2 L2 mol-2
(c) 2 × 109 L2 mol-2
(d) 0.5 × 10-9 L2 mol-2
Show Answer & Explanation

Answer: (c) 2 × 109 L2 mol-2

Explanation:
1. K = kf/kb = 2 × 107 / 10-2 = 2 × 109.
2. Units: (L2 mol-2 s-1)/(s-1) = L2 mol-2.

Teacher's Note:
1. At equilibrium forward rate equals backward rate.
2. K is the formation constant of the complex.

Question 8: Iodide is oxidised by acidified dichromate: Cr2O72- + 9I- + 14H+ → 2Cr3+ + 3I3- + 7H2O. At constant pH: Exp 1 [Cr2O72-] = 0.0050 M, [I-] = 0.0125 M, rate 0.00050; Exp 2 [Cr2O72-] = 0.010 M, [I-] = 0.0125 M, rate 0.0010; Exp 3 [Cr2O72-] = 0.0150 M, [I-] = 0.0250 M, rate 0.0060. The orders are
(a) second in Cr2O72- and first in I-
(b) first with respect to both Cr2O72- and I-
(c) first in Cr2O72- and second in I-
(d) second with respect to both
Show Answer & Explanation

Answer: (c) first in Cr2O72- and second in I-

Explanation:
1. Exp 1 to 2: dichromate doubles, rate doubles, so first order in dichromate.
2. Exp 2 to 3: dichromate x 1.5, iodide x 2, rate x 6. So 1.5 × 2^y = 6, 2^y = 4, y = 2.

Teacher's Note:
1. Isolate one species at a time.
2. The remaining factor after removing dichromate gives the iodide order.

Question 9: Consider the decay scheme A → B → C where A and B are radioisotopes with half-lives 10 s and 20 s. Initially there are 10000 atoms of A and none of B or C. At t = 30 s the correct statement is
(a) NA = 2500 and NB more than 7500
(b) NA = 1250 and NC more than 2500
(c) NA = 2500 and NC less than 1000
(d) NA = 1250 and NC more than 7500
Show Answer & Explanation

Answer: (b) NA = 1250 and NC more than 2500

Explanation:
1. 30 s is 3 half-lives of A, so NA = 10000/8 = 1250.
2. With λA = ln2/10 and λB = ln2/20: NB = N0 λA/(λB - λA) (e-λA t - e-λB t) = about 4571.
3. NC = 10000 - 1250 - 4571 = about 4179, which is more than 2500.

Teacher's Note:
1. Use the Bateman solution for a decay chain.
2. Atoms are conserved: NA + NB + NC = 10000.

Question 10: For a first order reversible reaction A ⇌ B with [A]=[A]0 and [B] = 0 at the start, the time taken for [B] to reach half its equilibrium value is (kf and kb are the forward and backward rate constants)
(a) 2 ln 2 / (kf + kb)
(b) ln 2 / (kf + kb)
(c) ln 2 / kf
(d) ln 2 / (kf - kb)
Show Answer & Explanation

Answer: (b) ln 2 / (kf + kb)

Explanation:
1. [B] = [B]eq (1 - e-(kf + kb)t).
2. Setting [B] = [B]eq/2 gives e-(kf+kb)t = 1/2.
3. t = ln 2 / (kf + kb).

Teacher's Note:
1. The approach to equilibrium is governed by the sum of the rate constants.
2. The result is independent of [A]0.

Question 11: For NH4+ + NO2- → N2 + 2H2O at 278 K: Set 1 [NH4+] = 0.24, [NO2-] = 0.10, rate 7.2 × 10-6 M s-1; Set 2 [NH4+] = 0.12, [NO2-] = 0.10, rate 3.6 × 10-6; Set 3 [NH4+] = 0.12, [NO2-] = 0.15, rate 5.4 × 10-6. The rate law and the unit of k are
(a) rate = k[NH4+] and s-1
(b) rate = k[NO2-] and s-1
(c) rate = k[NH4+][NO2-] and M s-1
(d) rate = k[NH4+][NO2-] and M-1 s-1
Show Answer & Explanation

Answer: (d) rate = k[NH4+][NO2-] and M-1 s-1

Explanation:
1. Set 1 to 2: [NH4+] halves and rate halves, so order 1.
2. Set 2 to 3: [NO2-] × 1.5 and rate x 1.5, so order 1.
3. k = 7.2 × 10-6/(0.24 × 0.10) = 3.0 × 10-4 M-1 s-1.

Teacher's Note:
1. Second order overall gives M-1 s-1.
2. Compute k to confirm the unit.

Question 12: The half-time of a second order reaction (single reactant) is
(a) inversely proportional to the initial concentration
(b) inversely proportional to the square of the initial concentration
(c) independent of the initial concentration
(d) proportional to the initial concentration
Show Answer & Explanation

Answer: (a) inversely proportional to the initial concentration

Explanation:
1. For second order, 1/[A] - 1/a = kt.
2. At half reaction, 1/(a/2) - 1/a = 1/a = k t1/2, so t1/2 = 1/(ka).

Teacher's Note:
1. Half-life is a general function of a1-n.
2. For n = 2 this is 1/a.

Question 13: For A + B → C + D, set 1: [A] = 4.00, [B] = 3.00, rate 0.10; set 2: [A] = 12.00, [B] = 3.00, rate 0.90; set 3: [A] = 12.00, [B] = 6.00, rate 0.90 (all concentrations in mol dm-3). The rate expression is
(a) rate = k[A][B]2
(b) rate = k[A]2[B]2
(c) rate = k[A][B]
(d) rate = k[A]2
Show Answer & Explanation

Answer: (d) rate = k[A]2

Explanation:
1. Set 1 to 2: [A] × 3 and rate x 9, so order in A is 2.
2. Set 2 to 3: [B] doubles and the rate is unchanged, so order in B is 0.

Teacher's Note:
1. 3^x = 9 gives x = 2.
2. A zero order species does not appear in the rate law.

Question 14: Match the order of reaction with its rate constant expression, where a is the initial concentration and x the amount reacted. (P) zero, (Q) first, (R) second, (S) third; (1) k = (1/2t)[1/(a-x)2 - 1/a2], (2) k = (1/t)[1/(a-x) - 1/a], (3) k = x/t, (4) k = (1/t) ln(a/(a-x)). The correct matching is
(a) P-3, Q-4, R-2, S-1
(b) P-2, Q-3, R-1, S-4
(c) P-3, Q-1, R-2, S-4
(d) P-1, Q-2, R-4, S-3
Show Answer & Explanation

Answer: (a) P-3, Q-4, R-2, S-1

Explanation:
1. Zero order: x = kt, so k = x/t (3).
2. First order: k = (1/t) ln(a/(a-x)) (4).
3. Second order: k = (1/t)(1/(a-x) - 1/a) (2).
4. Third order: k = (1/2t)(1/(a-x)2 - 1/a2) (1).

Teacher's Note:
1. Recall the integrated laws.
2. Check by dimensions: third order has 1/c2.

Question 15: The initial activity of a radionuclide is 9750 counts per minute and it is 975 counts per minute after 5 minutes. The decay constant (min-1) is about
(a) 0.69
(b) 0.46
(c) 0.99
(d) 0.23
Show Answer & Explanation

Answer: (b) 0.46

Explanation:
1. λ = (2.303/t) log(9750/975) = (2.303/5) log 10.
2. λ = 0.4606, about 0.46 min-1.

Teacher's Note:
1. The ratio of counts is exactly 10.
2. log 10 = 1.

Question 16: 60% of a first order reaction was completed in 60 minutes. The time taken for half of the reactant to decompose is about
(a) 45 min
(b) 30 min
(c) 40 min
(d) 20 min
Show Answer & Explanation

Answer: (a) 45 min

Explanation:
1. k = (2.303/60) log(100/40) = 0.01527 min-1.
2. t1/2 = 0.693/0.01527 = 45.4 min, about 45 min.

Teacher's Note:
1. 40% remains after 60 min.
2. ln 2.5 = 0.916.

Question 17: The correct statement about the order of a reaction is that
(a) it always has positive integral values
(b) it always has positive integral or fractional values
(c) it can be predicted from the stoichiometric coefficients of the reactants
(d) it has to be determined experimentally
Show Answer & Explanation

Answer: (d) it has to be determined experimentally

Explanation:
1. Order is the sum of exponents in the experimental rate law and may be zero, fractional or even negative.
2. It cannot be obtained from the balanced equation, so it must be found by experiment.

Teacher's Note:
1. Only for elementary steps does order equal molecularity.
2. Zero and negative orders do occur.

Question 18: For A + B ⇌ C + D with dx/dt = k1[A][B] - k2[C][D] and k1 = k2, the reaction ceases for the concentration set
(a) [A] = 0.4, [B] = 0.25, [C] = 0.2, [D] = 0.5 M
(b) [A] = 0.1, [B] = 0.2, [C] = 0.3, [D] = 0.4 M
(c) [A] = 0.2, [B] = 0.2, [C] = 0.4, [D] = 0.2 M
(d) [A] = 0.2, [B] = 0.2, [C] = 0.3, [D] = 0.2 M
Show Answer & Explanation

Answer: (a) [A] = 0.4, [B] = 0.25, [C] = 0.2, [D] = 0.5 M

Explanation:
1. Net rate is zero when k1[A][B] = k2[C][D], i.e. [A][B] = [C][D] since k1 = k2.
2. Option 2: 0.4 × 0.25 = 0.10 and 0.2 × 0.5 = 0.10. The others do not match.

Teacher's Note:
1. Compute both products for each option.
2. Equilibrium means net rate zero.

Question 19: Iodine-131 has a half-life of 8.0 days. The percentage remaining in the body after 32 days, with no elimination, is
(a) 2.77
(b) 6.25
(c) 0.0625
(d) 25
Show Answer & Explanation

Answer: (b) 6.25

Explanation:
1. 32 days = 4 half-lives.
2. Remaining = 100 × (1/2)4 = 6.25%.

Teacher's Note:
1. Count half-lives first.
2. (1/2)4 = 1/16.

Question 20: The decomposition of H2O2 is followed by KMnO4 titration and found first order with k = 4.5 × 10-2 min-1. The initial titre is 25 mL. The titre will be 5 mL after
(a) (ln 5)/(4.5 × 10-2) × 5 min
(b) (ln 5)/(4.5 × 10-2) / 2 min
(c) (ln 25)/(4.5 × 10-2) min
(d) (ln 5)/(4.5 × 10-2) min
Show Answer & Explanation

Answer: (d) (ln 5)/(4.5 × 10-2) min

Explanation:
1. Titre is proportional to concentration.
2. t = (1/k) ln(25/5) = ln 5/(4.5 × 10-2) = 35.8 min.

Teacher's Note:
1. Ratio 25:5 = 5.
2. Use first order integrated law.

Question 21: In a multi-step reaction, the rate of the overall reaction is governed by
(a) the rate of the slowest step
(b) the rate of the fastest step
(c) the sum of the rates of all the steps
(d) the average rate of all the steps
Show Answer & Explanation

Answer: (a) the rate of the slowest step

Explanation:
1. Products can form no faster than the slowest step allows.
2. The slowest step is the rate determining step.

Teacher's Note:
1. Think of a bottleneck in a pipeline.
2. Fast steps only set up pre-equilibria.

Question 22: For 3A + B → C + D with mechanism: step 1 A + B → E (k1); step 2 E + 2A → C + D (k2). Using the steady state approximation for E, the overall order of the reaction is
(a) 4
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. Steady state for E: k1[A][B] = k2[E][A]2.
2. Rate = k2[E][A]2 = k1[A][B]. This is first order in A and in B: overall order 2.

Teacher's Note:
1. Set d[E]/dt = 0.
2. Substitute [E] back into the rate.

Question 23: The activation energy of a reaction is obtained from a plot of ln k against 1/T as
(a) slope / R
(b) R / slope
(c) - slope x R
(d) slope x R
Show Answer & Explanation

Answer: (c) - slope x R

Explanation:
1. ln k = ln A - E/(RT), so the slope of ln k versus 1/T is -E/R.
2. Hence E = -(slope) x R.

Teacher's Note:
1. The slope of an Arrhenius plot is negative.
2. E is always positive.

Question 24: A radioactive element has a half-life of 4 minutes. After 20 minutes, 1 g of the element reduces to
(a) 0.0625 g
(b) 0.25 g
(c) 0.125 g
(d) 0.03125 g
Show Answer & Explanation

Answer: (d) 0.03125 g

Explanation:
1. 20 minutes is 5 half-lives.
2. Mass left = 1 × (1/2)5 = 0.03125 g.

Teacher's Note:
1. Number of half-lives = time / half-life.
2. Each half-life halves the mass.

Question 25: For X + Y → P: Set I [A] = 0.1 M, [B] = 0.2 M, rate 0.001; Set II [A] = 0.2 M, [B] = 0.2 M, rate 0.004; Set III [A] = 0.2 M, [B] = 0.8 M, rate 0.008. The overall order is
(a) 1
(b) 2.5
(c) 2
(d) 3
Show Answer & Explanation

Answer: (b) 2.5

Explanation:
1. I to II: [A] doubles, rate x 4, so order 2 in A.
2. II to III: [B] × 4, rate x 2, so order 1/2 in B.
3. Total = 2.5.

Teacher's Note:
1. Isolate one reactant at a time.
2. 4^y = 2 gives y = 0.5.

Question 26: Adsorption of lead from drinking water on activated carbon is first order with k = 1.8 × 10-5 s-1. The time needed to remove 90% of the lead is
(a) 1.3 × 105 s
(b) 2.7 × 106 s
(c) 2.4 × 104 s
(d) 1.8 × 107 s
Show Answer & Explanation

Answer: (a) 1.3 × 105 s

Explanation:
1. t = (2.303/k) log(100/10) = 2.303 / (1.8 × 10-5).
2. t = 1.28 × 105 s, about 1.3 × 105 s.

Teacher's Note:
1. 90% removal means 10% remains.
2. The initial concentration cancels.

Question 27: According to the Arrhenius equation, the rate constant k and activation energy E are related by
(a) k = A e-E/RT
(b) k = -A eE/RT
(c) k = A eE/RT
(d) A = k e-E/RT
Show Answer & Explanation

Answer: (a) k = A e-E/RT

Explanation:
1. The fraction of molecules with energy at least E is e-E/RT.
2. So k = A e-E/RT, with A the frequency (pre-exponential) factor.

Teacher's Note:
1. k increases with temperature and decreases with E.
2. The exponent must be negative.

Question 28: For an elementary rearrangement A → P (with P0 = 0) the rate of conversion of A is 0.100, 0.050 and 0.025 mol L-1 min-1 for [A]0 = 0.340, 0.170 and 0.085 mol L-1. The equilibrium constant is 1.12 at 303 K. The rate constant of the reverse reaction P → A is
(a) 0.526 min-1
(b) 0.588 min-1
(c) 0.263 min-1
(d) 0.294 min-1
Show Answer & Explanation

Answer: (c) 0.263 min-1

Explanation:
1. Rate is proportional to [A]: first order, kf = 0.100/0.340 = 0.294 min-1.
2. K = kf/kb, so kb = 0.294/1.12 = 0.263 min-1.

Teacher's Note:
1. Initially the reverse rate is zero because P0 = 0.
2. K = kf / kb for elementary reversible steps.

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