JEE Chemistry Chemical Kinetics MCQs Set 34

Chemistry Objective Questions and Answers: Chemical Kinetics

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Download Chemical Kinetics MCQs with Answers

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Multiple Choice Questions

Question 1: For a given first order reaction it takes 20 min for the concentration to drop from 1 M to 0.6 M. The time required for the concentration to drop from 0.6 M to 0.36 M will be:
(a) More than 20 min
(b) Less than 20 min
(c) Equal to 20 min
(d) Infinity
Show Answer & Explanation

Answer: (c) Equal to 20 min

Explanation:
1. Both changes are by the same fraction (0.6 times the initial value).
2. In first order reactions equal fractions take equal times, so the time is 20 min.
3. Answer: (c)

Teacher's Note:
1. Fractional change depends only on time.
2. Compare ratios 0.6/1 and 0.36/0.6.

Question 2: The reaction L → M is started with 10 g/L. After 30 and 90 minutes, 5 g/L and 1.25 g/L are left respectively. The order of reaction is:
(a) 0
(b) 2
(c) 1
(d) 3
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. 10 → 5 in 30 min; 5 → 2.5 in the next 30 and 2.5 → 1.25 in the next 30, so 1.25 g/L at 90 min.
2. The half life is constant (30 min), so the reaction is first order.
3. Answer: (c)

Teacher's Note:
1. Constant half life means first order.
2. Check the 90 min value against repeated halving.

Question 3: If doubling the initial concentration of a reactant doubles t1/2 of the reaction, the order is:
(a) 3
(b) 2
(c) 1
(d) 0
Show Answer & Explanation

Answer: (d) 0

Explanation:
1. t1/2 ∝ a1−n; t1/2 ∝ a gives 1 − n = 1, so n = 0.
2. Answer: (d)

Teacher's Note:
1. Half life proportional to a is a zero-order signature.
2. Use t1/2 ∝ a1−n.

Question 4: The rate constant k for the reaction 2A + B → product was found to be 2.5 × 10−5 L mol−1 s−1 after 15 s, 2.60 × 10−5 after 30 s and 2.55 × 10−5 after 50 s. The order of the reaction is:
(a) 2
(b) 3
(c) Zero
(d) 1
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. The value of k stays practically constant (the small differences are experimental scatter).
2. Its unit L mol−1 s−1 is the unit for a second order constant.
3. Answer: (a)

Teacher's Note:
1. Second order k has units L mol−1 s−1.
2. Order can be read directly from the unit of k.

Question 5: In a first order reaction 75% of the reactant disappeared in 1.388 h. The rate constant of the reaction is:
(a) 1 s−1
(b) 2.8 × 10−4 s−1
(c) 17.2 × 10−3 s−1
(d) 1.8 × 10−3 s−1
Show Answer & Explanation

Answer: (b) 2.8 × 10−4 s−1

Explanation:
1. t = 1.388 h = 4997 s.
2. k = (2.303/t) log(100/25) = 1.386/4997 = 2.77 × 10−4 s−1 ≈ 2.8 × 10−4 s−1.
3. Answer: (b)

Teacher's Note:
1. 75% completion equals two half lives.
2. Convert hours to seconds.

Question 6: For a first order reaction, the ratio of the time required for 99.9% completion to that for 50% completion is:
(a) 2
(b) 5
(c) 10
(d) None of these
Show Answer & Explanation

Answer: (c) 10

Explanation:
1. t99.9% = (2.303/k) log 1000 = 6.909/k.
2. t50% = 0.693/k.
3. Ratio = 6.909/0.693 ≈ 10.
4. Answer: (c)

Teacher's Note:
1. t99.9% ≈ 10 t1/2.
2. A useful result for first order reactions.

Question 7: For a reaction, t1/2 × a (a is the initial concentration) is found to be constant at different initial concentrations. The order of reaction is:
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. t1/2 ∝ 1/a (t1/2 × a constant) means second order.
2. Answer: (c)

Teacher's Note:
1. Product with a constant means inverse proportionality.
2. Second order: t1/2 = 1/(ka).

Question 8: The following data were obtained at a certain temperature for the decomposition of ammonia: p (mm) = 50, 100, 200; t1/2 = 3.64, 1.82, 0.91. The order of the reaction is:
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. t1/2 halves each time p doubles, so t1/2 ∝ 1/p.
2. The order is 2.
3. Answer: (c)

Teacher's Note:
1. Half life inversely proportional to pressure is second order.
2. Check each doubling.

Question 9: A reaction is found to have a rate constant of x s−1. By what factor is the rate increased if the initial concentration of A is tripled?
(a) 3
(b) 9
(c) x
(d) Remains the same
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. Unit s−1 means first order.
2. Rate ∝ [A], so tripling concentration triples the rate.
3. Answer: (a)

Teacher's Note:
1. Read the order from the unit of k.
2. First order: rate is proportional to concentration.

Question 10: Which is incorrect?
(a) Half life of a first order reaction is independent of initial concentration
(b) Rate of reaction is constant for a first order reaction
(c) Unit of k for a second order reaction is mol−1 L s−1
(d) None of these
Show Answer & Explanation

Answer: (b) Rate of reaction is constant for a first order reaction

Explanation:
1. In a first order reaction it is the rate constant that is constant; the rate itself falls as [A] falls.
2. (a) and (c) are correct.
3. Answer: (b)

Teacher's Note:
1. Rate and rate constant are different quantities.
2. Only zero order has a constant rate.

Question 11: Hydrolysis of an ester in alkaline medium is:
(a) First order with molecularity one
(b) Second order with molecularity two
(c) First order with molecularity two
(d) Second order with molecularity one
Show Answer & Explanation

Answer: (b) Second order with molecularity two

Explanation:
1. Rate = k[ester][OH−] is second order.
2. Two species collide in the elementary step, so molecularity is two.
3. Answer: (b)

Teacher's Note:
1. Elementary bimolecular reaction has order 2.
2. Acid hydrolysis, with water in excess, is pseudo first order.

Question 12: The rate constant of a zero order reaction is 0.2 mol dm−3 h−1. If the concentration of the reactant after 30 minutes is 0.05 mol dm−3, then its initial concentration would be:
(a) 6.05 mol dm−3
(b) 0.15 mol dm−3
(c) 0.25 mol dm−3
(d) 4.00 mol dm−3
Show Answer & Explanation

Answer: (b) 0.15 mol dm−3

Explanation:
1. 30 min = 0.5 h.
2. [A]0 = 0.05 + 0.2 × 0.5 = 0.15 mol dm−3.
3. Answer: (b)

Teacher's Note:
1. Convert minutes to hours before using k.
2. Zero order: [A]0 = [A] + kt.

Question 13: According to the collision theory of reaction rates:
(a) Every collision between reactants leads to chemical reaction
(b) Rate of reaction is proportional to the velocity of molecules
(c) All reactions which occur in the gaseous phase are zero order reactions
(d) Rate of reaction is directly proportional to the collision frequency
Show Answer & Explanation

Answer: (d) Rate of reaction is directly proportional to the collision frequency

Explanation:
1. Only collisions with enough energy and proper orientation are effective, so (a) is wrong.
2. Rate = collision frequency × fraction of effective collisions, so it is proportional to collision frequency.
3. Answer: (d)

Teacher's Note:
1. Rate = Z × e−Ea/RT × steric factor.
2. Gas phase reactions have many orders.

Question 14: The rate constant K1 of a reaction is found to be double that of the rate constant K2 of another reaction (the pre-exponential factors being equal). The relationship between the corresponding activation energies E1 and E2 at the same temperature is:
(a) E1 > E2
(b) E1 < E2
(c) E1 = E2
(d) E1 = 4E2
Show Answer & Explanation

Answer: (b) E1 < E2

Explanation:
1. k = Ae−Ea/RT; with equal A a larger k needs a smaller Ea.
2. So E1 < E2.
3. Answer: (b)

Teacher's Note:
1. Lower barrier means faster reaction.
2. Equal pre-exponential factors are assumed.

Question 15: At room temperature the reaction between NO and O2 to give NO2 is fast while that between CO and O2 is slow. It is due to:
(a) CO is smaller in size than NO
(b) CO is poisonous
(c) The activation energy for 2NO + O2 → 2NO2 is less than that for 2CO + O2 → 2CO2
(d) None of these
Show Answer & Explanation

Answer: (c) The activation energy for 2NO + O2 → 2NO2 is less than that for 2CO + O2 → 2CO2

Explanation:
1. Rates at the same temperature differ because of the energy barrier.
2. The reaction with the smaller activation energy is faster.
3. Answer: (c)

Teacher's Note:
1. Size and toxicity do not control rate.
2. Lower Ea gives a larger rate constant.

Question 16: An endothermic reaction A → B has an activation energy of 15 kcal/mol and the heat of reaction is 5 kcal/mol. The activation energy of the reaction B → A is:
(a) 20 kcal/mol
(b) 15 kcal/mol
(c) 10 kcal/mol
(d) Zero
Show Answer & Explanation

Answer: (c) 10 kcal/mol

Explanation:
1. ΔH = Ea(forward) − Ea(reverse).
2. 5 = 15 − Ea(reverse), so Ea(reverse) = 10 kcal/mol.
3. Answer: (c)

Teacher's Note:
1. Endothermic: reverse barrier is lower.
2. Draw the energy profile if in doubt.

Question 17: A large increase in the rate of a reaction for a rise in temperature is due to:
(a) An increase in the number of collisions
(b) An increase in the number of activated molecules
(c) Lowering of activation energy
(d) Shortening of the mean free path
Show Answer & Explanation

Answer: (b) An increase in the number of activated molecules

Explanation:
1. Collision frequency rises only slightly (∝ √T).
2. The fraction of molecules with energy ≥ Ea rises exponentially, which causes the large increase.
3. Answer: (b)

Teacher's Note:
1. Maxwell-Boltzmann curve flattens with temperature.
2. Temperature does not change Ea.

Question 18: The rate of reactions increases with temperature for:
(a) Any reaction
(b) Exothermic reactions only
(c) Endothermic reactions only
(d) None of these
Show Answer & Explanation

Answer: (a) Any reaction

Explanation:
1. The rate constant k = Ae−Ea/RT rises with T for every reaction whose Ea is positive.
2. Answer: (a)

Teacher's Note:
1. Equilibrium yield and rate are different questions.
2. Exothermic reactions also go faster when heated.

Question 19: The rate of a chemical reaction doubles for every 10°C rise in temperature. If the temperature is increased by 60°C, the rate of the reaction increases by:
(a) 20 times
(b) 32 times
(c) 64 times
(d) 128 times
Show Answer & Explanation

Answer: (c) 64 times

Explanation:
1. A 60°C rise is six 10°C steps.
2. Factor = 26 = 64.
3. Answer: (c)

Teacher's Note:
1. Temperature coefficient = 2.
2. Raise it to the number of 10°C steps.

Question 20: Given that K is the rate constant for some order of any reaction at temperature T, the value of limT→∞ log K is:
(a) A/2.303
(b) A
(c) 2.303 A
(d) log A
Show Answer & Explanation

Answer: (d) log A

Explanation:
1. K = Ae−Ea/RT; as T → ∞ the exponential tends to 1, so K → A.
2. log K → log A.
3. Answer: (d)

Teacher's Note:
1. A is the Arrhenius pre-exponential factor.
2. Take the log of the limit.

Question 21: The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25°C are 3 × 10−4 s−1, 104.4 kJ mol−1 and 6.0 × 1014 s−1 respectively. The value of the rate constant as T → ∞ is:
(a) 2 × 108 s−1
(b) 6 × 1014 s−1
(c) Infinity
(d) 3.6 × 1030 s−1
Show Answer & Explanation

Answer: (b) 6 × 1014 s−1

Explanation:
1. As T → ∞, e−Ea/RT → 1 and k → A = 6.0 × 1014 s−1.
2. Answer: (b)

Teacher's Note:
1. k cannot exceed the pre-exponential factor.
2. Other data are not needed.

Question 22: The rate of a reaction increases to 2.3 times when the temperature is raised from 300 K to 310 K. If K is the rate constant at 300 K, the rate constant at 310 K will be:
(a) 2K
(b) K
(c) 2.3K
(d) 3K2
Show Answer & Explanation

Answer: (c) 2.3K

Explanation:
1. The rate (at fixed concentrations) is proportional to k.
2. So k310 = 2.3 K.
3. Answer: (c)

Teacher's Note:
1. Rate scales with k when concentrations are unchanged.
2. Temperature coefficient is 2.3 here.

Question 23: If the concentration of reactants is increased by x, then the rate constant K becomes:
(a) ln(K/x)
(b) K/x
(c) K + x
(d) K
Show Answer & Explanation

Answer: (d) K

Explanation:
1. The rate constant depends on temperature and catalyst, not on concentration.
2. K remains unchanged.
3. Answer: (d)

Teacher's Note:
1. The rate changes, but k does not.
2. k depends on temperature.

Question 24: The energy of activation of a forward reaction is 50 kcal. The energy of activation of its backward reaction is:
(a) Equal to 50 kcal
(b) Greater than 50 kcal
(c) Less than 50 kcal
(d) Either greater or less than 50 kcal
Show Answer & Explanation

Answer: (d) Either greater or less than 50 kcal

Explanation:
1. Ea(reverse) = Ea(forward) − ΔH.
2. It exceeds 50 kcal for an exothermic forward reaction and is below 50 kcal for an endothermic one, so either is possible.
3. Answer: (d)

Teacher's Note:
1. ΔH decides the relation between the two barriers.
2. No sign of ΔH is given.

Question 25: An exothermic reaction X → Y has an activation energy of 30 kJ mol−1. If the energy change (ΔE) during the reaction is −20 kJ, the activation energy of the reverse reaction is:
(a) 10 kJ
(b) 20 kJ
(c) 50 kJ
(d) −30 kJ
Show Answer & Explanation

Answer: (c) 50 kJ

Explanation:
1. Ea(reverse) = Ea(forward) − ΔE = 30 − (−20) = 50 kJ.
2. Answer: (c)

Teacher's Note:
1. Exothermic reactions have a larger reverse barrier.
2. Sketch the energy profile.

Question 26: The rate of a reaction increases with an increase of temperature because:
(a) Collisions increase
(b) Energy of products decreases
(c) The fraction of molecules possessing energy ≥ ET (threshold energy) increases
(d) The mechanism of the reaction changes
Show Answer & Explanation

Answer: (c) The fraction of molecules possessing energy ≥ ET (threshold energy) increases

Explanation:
1. Raising T shifts the Maxwell-Boltzmann distribution so a larger fraction of molecules has energy at least equal to the threshold.
2. Answer: (c)

Teacher's Note:
1. The effect on collision number is small.
2. The exponential term e−Ea/RT grows.

Question 27: For producing effective collisions, the colliding molecules must possess:
(a) A certain minimum amount of energy
(b) Energy equal to or greater than the threshold energy
(c) Proper orientation
(d) Threshold energy as well as proper orientation of collision
Show Answer & Explanation

Answer: (d) Threshold energy as well as proper orientation of collision

Explanation:
1. An effective collision needs both energy at least equal to the threshold and the correct orientation.
2. Answer: (d)

Teacher's Note:
1. Both conditions must be met together.
2. Orientation is covered by the steric factor.

Question 28: For a reaction of fixed order and fixed initial concentration, the half life with increase of temperature:
(a) Is independent of temperature
(b) Increases
(c) Decreases
(d) Increases or decreases
Show Answer & Explanation

Answer: (c) Decreases

Explanation:
1. Raising T increases k, and t1/2 varies inversely with k.
2. So the half life decreases.
3. Answer: (c)

Teacher's Note:
1. t1/2 = 0.693/k for first order.
2. Higher k means faster completion.

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