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 = K_{1}C/1+K_{2}C). 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

View Answer

Explanation: [-dC/dt] = [K

_{1}C/1+K

_{2}C] = [K

_{1}/ (1/C + K

_{2}], if C is very high then 1/C being small may be neglected.

2. Calculate t_{0.75}/t_{0.50} for a 1^{st} order reaction.

a) Negative

b) 0

c) 1

d) 2

View Answer

Explanation: The solution for 1

^{st}order reaction is 2. k = 2.303/t

_{3/4}(log C

_{0}/(1/4C

_{0})) = 2.303/t

_{1/2}(log C

_{0}/(C

_{0/2})) = t

_{3/4}/t

_{1/2}= log 4/log 2 = 2log 2/log 2 = 2.

3. At least how many half-lives should elapse for 1^{st} order reaction A → products so that the reaction is at least 95% completed?

a) 3

b) 4

c) 5

d) 6

View Answer

Explanation: 5 half-lives should have elapsed for 1

^{st}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 = K_{1}C/1+K_{2}C). 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

View Answer

Explanation: If C is very low 1 + K

_{2}C = K’, (-dC/dt) = K

_{1}C/K’ = (K

_{1}/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

View Answer

Explanation: The total order of the reaction is n = 2. For a nth order reaction (n ≠ 1), t

_{1/2}∝ 1/ [c0

^{n-1}] = 210/140 = (300/200)

^{n-1}= n = 2.

6. Find the expression for K in terms of P_{0}, P_{t} and n.

a) K = 2.303/t

b) K = P_{0}(n – 1)

c) K = P_{0}(n – 1)/n P_{0} – P_{t}

d) K = (2.303/t) log (P_{0}(n – 1)/n P_{0} – P_{t})

View Answer

Explanation: x = P

_{t}– P

_{0}/n – 1, P

_{A}= P

_{0}– (P

_{t}– P

_{0}/n – 1) = P

_{0}n – P

_{t}/n – 1, a ∝ p

_{0}and a – x ∝ P

_{A}= n P

_{0}– P

_{t}/n -1 = K = (2.303/t) log (P

_{0}(n – 1)/n P

_{0}– P

_{t}).

_{40℃}/k

_{25℃}for this reaction.

a) √10

b) √8

c) √4

d) √6

View Answer

Explanation: k

_{2}/k

_{1}= (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 t_{0.5}/t_{0.25} for 1^{st} order reaction.

a) 5.5

b) 4.5

c) 3.5

d) 2.5

View Answer

Explanation: The value for 1

^{st}order reaction is 2.5. t

_{0.5}/t

_{0.25}= [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

View Answer

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

View Answer

Explanation: For same concentration R

_{2}/R

_{1}= K

_{2}/K

_{1}= 3

^{30/10}= 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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