# Chemistry Questions and Answers – Chemical Kinetics – Integrated Rate Equations

«
»

This set of Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Chemical Kinetics – Integrated Rate Equations ”.

1. The rate constant of a reaction is 0.01s-1, how much time does it take for 2.4 mol L-1 concentration of reactant reduced to 0.3 mol L-1?
a) 108.3s-1
b) 207.9s-1
c) 248.2s-1
d) 164.8s-1

Explanation: Given,
K = 0.01s-1
t1/2 = 0.693/0.01
t1/2 = 69.3s
[R] = [R]0/2n
2n = [R]0/[R]
2n = 2.4/0.3
2n = 8
n = 3 (number of half-lives)
For 1 half-life t1/2 = 69.3s
For 3 half-life 3t1/2 = 3 x 69.3s = 207.9s.

2. What time does it take for reactants to reduce to 3/4 of initial concentration if the rate constant is 7.5 x 10-3 s-1?
a) 38.4s
b) 40.2s
c) 39.3s
d) 36.8s

Explanation: Given,
K=7.5 x 10-3 s-1
K = (2.303/t) x log([R]0/[R])
t = (2.303/7.5 x 10-3) x log([100]/[25])
t = (2.303/7.5 x 10-3) x 0.6
t = 38.4s.

3. A zero-order reaction is 25% complete in 30seconds. What time does it take for 50% completion?
a) 40s
b) 70s
c) 50s
d) 60s

Explanation:
t25% = (1-0.75)/K
K = 0.25/30
t50% = (1 x 30)/(2 x 0.25)
t50% = 15/0.25
t50% = 60s.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

4. What is the integrated rate equation for a first order reaction?
a) [A] = [A]0e-kt
b) [A] = [A]0/e-kt
c) [A] = [A]0e-t
d) [A] = [A]0e-k

Explanation: A reaction is said to be of the first order if the rate of the reaction depends upon one concentration term only. The integrated rate equation for a first order reaction in exponential form is [A] = [A]0e-kt.

5. For a certain reaction the values of Arrhenius factor and Activation energy are 4 x 1013 collision/sec and 98.6KJ/mol at 303K. Calculate the rate constant if reaction is 1st order?( R=8.341mol-1K-1)
a) 6.07 x 10-3
b) 3.02 x 10-5
c) 4.07 x 10-4
d) 7.42 x 10-3

Explanation: Given,
Arrhenius factor(A) = 4 x 1013 collisions/sec
Activation energy (Ea)=98.6KJ/mol=98.6 x 103J/mol, T=303 K
log K = log A – (Ea/2.303RT)
log K = log (4 x 1013) – (98.6 x 103)/(2.303 x 8.314 x 303)
log K = 13.6020 – (98.6 x 103/5801.584)
log k = -3.39
K = 10-3.39
K = 4.07 x 10-4.

6. The decomposition of N2O5 in CCl4 solution was studied. N2O5 → 2NO2 + 1/2O2. The rate constant of the reaction is 6.2 x 10-4 sec-1. Calculate the rate when the concentration of N2O5 is 1.25 molar.
a) 6.45 x 10-4
b) 7.45 x 10-4
c) 6.75 x 10-4
d) 7.75 x 10-4

Explanation: Given,
N2O5 → 2N02 + 1/2 O2
Rate = k[N2O5] 1
Rate =6.25 x 10-4 x [1.25]
Rate = 7.75 x 10-4.

7. What is the formula to calculate the time taken for the completion of a zero-order reaction?
a) t100% = [A]0/k
b) t100% = [A]0/2k
c) t100% = 2[A]0/k
d) t100% = [A]0/3k

Explanation: The time taken for the zero-order reaction to complete can be calculated as follows:
When the reaction is complete, [A]0 = 0
Therefore, k = [A]0/t or t100% = [A]0/k.

8. The unit of rate constant of a first-order reaction is s-1.
a) True
b) False

Explanation: The rate constant is defined as the proportionality constant which explains the relationship between the molar concentration of the reactants and the rate of a chemical reaction. It is denoted by k.
For a first-order reaction, the unit of k is s-1.

9. The unit of the rate constant of a zero-order reaction and second-order reaction is same.
a) True
b) False

Explanation: The unit of the rate constant k for a zero-order reaction is mol L-1 s-1. For a first-order reaction, the unit of k is s-1. The unit of the rate constant k for a second-order reaction is mol-1 L s-1.

10. The half-life of a given reaction is doubled if the initial concentration of the reactant is doubled. What is the order of the reaction?
a) 0
b) 1
c) 2
d) 3

Explanation: Half-life (t1⁄2) is the time required for a quantity to reduce to half of its initial value. The half-life of a zero-order reaction is directly proportional to its initial concentration. They are related as:
t1/2 = [R]0/2k.

11. A first-order reaction was 70 percent complete in 20 minutes. What is the rate constant of the reaction?
a) 0.07 min-1
b) 0.06 min-1
c) 0.08 min-1
d) 0.09 min-1

Explanation: Given, the reaction is 70 percent complete so, a = 70
Time t = 20 minutes
First order integrated rate equation, k = $$\frac{2.303}{t}$$log$$\frac{100}{100-a}$$
k= $$\frac{2.303}{20}$$log$$\frac{100}{100-70}$$ = 0.06 min-1.

12. What is the time taken to complete 75 percent of the reaction if the rate of the first-order reaction is 0.023 min-1?
a) 60.28 minutes
b) 69.28 minutes
c) 50.37 minutes
d) 65.97 minutes

Explanation: Given, the reaction is 75 complete so, a=75
Rate of the reaction k=0.023 min-1
First-order integrated rate equation, k = $$\frac{2.303}{t}$$log$$\frac{100}{100-a}$$
t=$$\frac{2.303}{0.023}$$log$$\frac{100}{100-75}$$ = 60.28 minutes.

13. For the reaction X → Y + Z, the rate constant is 0.00058 s-1. What percentage of X will be decomposed in 50 minutes?
a) 90.02 percent
b) 82.44 percent
c) 88.82 percent
d) 82.67 percent

Explanation: Given, rate constant k= 0.00058 s-1
Time t= 50 minutes = 50 × 60 = 3000 seconds
First-order integrated rate equation, k = $$\frac{2.303}{t}$$log$$\frac{100}{100-a}$$
log$$\frac{100}{100-a}$$ = $$\frac{kt}{2.303}$$
log$$\frac{100}{100-a}$$ = $$\frac{0.00058 \times 3000}{2.303}$$ = 0.756
a=82.44 percent.

14. What is the time required for 75 percent completion of a first-order reaction?
a) 3 × t50
b) 4 × t50
c) 4.5 × t50
d) 2 × t50

Explanation: Given, the reaction is 75 percent complete so, a = 75
We know, t = $$\frac{2.303}{k}$$log$$\frac{100}{100-a}$$
T75 = $$\frac{2.303}{k}$$log$$\frac{100}{100-75}$$ = $$\frac{1.386}{k}$$
t50 = $$\frac{2.303}{k}$$log$$\frac{100}{100-50}$$ = $$\frac{0.693}{k}$$
t75 = 2 × t50.

15. A first-order reaction is 50 percent complete in 30 minutes. Calculate the time taken for completion of 87.5 percent of the reaction.
a) 30 minutes
b) 60 minutes
c) 90 minutes
d) 120 minutes

Explanation: Reaction is 50 percent complete in 30 minutes. Hence, t1/2 = 30 minutes
75 percent of the reaction is completed in two half-lives. Hence, t = 2 × 30 = 60 minutes
87.5 percent of the reaction is completed in three half-lives. Hence, t = 3 × 30 = 90 minutes.

Sanfoundry Global Education & Learning Series – Chemistry – Class 12.

To practice all areas of Chemistry, here is complete set of 1000+ Multiple Choice Questions and Answers.