# Physical Chemistry Questions and Answers – Kinetics – Activation Energy

This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Kinetics – Activation Energy”.

1. The rate of the certain equation depends on the concentration according to the equation: (-dC/dt = K1C/1+K2C). What will be the order of reaction, when concentration (C) is very high?
a) Second-order reaction
b) C = C0
c) First-order reaction
d) Zero-order reaction

Explanation: [-dC/dt] = [K1C/1+K2C] = [K1/ (1/C + K2], if C is very high then 1/C being small may be neglected.

2. Calculate t0.75/t0.50 for a 1st order reaction.
a) Negative
b) 0
c) 1
d) 2

Explanation: The solution for 1st order reaction is 2. k = 2.303/t3/4 (log C0/(1/4C0)) = 2.303/t1/2 (log C0/(C0/2)) = t3/4/t1/2 = log 4/log 2 = 2log 2/log 2 = 2.

3. At least how many half-lives should elapse for 1st order reaction A → products so that the reaction is at least 95% completed?
a) 3
b) 4
c) 5
d) 6

Explanation: 5 half-lives should have elapsed for 1st order reaction. 100 → 50 → 25 → 12.5 → 6.25 → 3.125. At 3.125 the reaction is 96.875% completed.

4. The rate of the certain equation depends on the concentration according to the equation: (-dC/dt = K1C/1+K2C). What will be the order of reaction, when concentration (C) is very low?
a) First-order reaction
b) Second-order reaction
c) Zero-order reaction
d) C = C0

Explanation: If C is very low 1 + K2C = K’, (-dC/dt) = K1C/K’ = (K1/K’) * concentration.

5. In the reduction of nitric gas with hydrogen, the reaction was found to be 50% complete in 210 seconds when the initial pressure of the mixture was 200 mm. In a second experiment the time of half- reaction was 140 seconds when the initial pressure was 300 mm. Calculate the total order of the reaction.
a) 1
b) 2
c) 0
d) 4

Explanation: The total order of the reaction is n = 2. For a nth order reaction (n ≠ 1), t1/2 ∝ 1/ [c0n-1] = 210/140 = (300/200)n-1 = n = 2.

6. Find the expression for K in terms of P0, Pt and n.
a) K = 2.303/t
b) K = P0(n – 1)
c) K = P0(n – 1)/n P0 – Pt
d) K = (2.303/t) log (P0(n – 1)/n P0 – Pt)

Explanation: x = Pt – P0/n – 1, PA = P0 – (Pt – P0/n – 1) = P0n – Pt/n – 1, a ∝ p0 and a – x ∝ PA = n P0 – Pt/n -1 = K = (2.303/t) log (P0(n – 1)/n P0 – Pt).
7. For a reaction T.C. = 2, Calculate k40℃/k25℃ for this reaction.
a) √10
b) √8
c) √4
d) √6
Explanation: k2/k1 = (2)15/10 = (2)3/2 = √8. But the method of temperature coefficient was not exact and to explain the effect of temperature on reaction rate new theory was evolved.

8. Calculate t0.5/t0.25 for 1st order reaction.
a) 5.5
b) 4.5
c) 3.5
d) 2.5

Explanation: The value for 1st order reaction is 2.5. t0.5/t0.25 = [2.303/K log (a/a-a/2)]/ [2.303/K log (a/a-a/4)] = (log 2)/ (log 4/3) = 2.5.

9. Gaseous cyclobutane isomerizes to butadiene in a first order process which has k = 3.3 * 10-4 s-1 at 153℃. How many minutes would it take for the isomerization to proceed 40% completion at this temperature.
a) 25.80 minutes
b) 34.98 minutes
c) 78.90 minutes
d) 55.66 minutes

Explanation: For the first order reaction t = 2.303/K log a/ (a – x), t = 2.303/3.3 * 10-4 log (100/60), t = 1.54 * 10+3 second = 25.80 minutes.

10. The temperature coefficient of the rate of a reaction is 3. How many times the rate of reaction would increase if the temperature is raised by 30 K:
a) 3
b) 9
c) 27
d) 81

Explanation: For same concentration R2/R1 = K2/K1 = 330/10 = 27. The rate of the reaction would increase by 27 times if the temperature is raised by 30 K.

Sanfoundry Global Education & Learning Series – Physical Chemistry.

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