Question

The half-life for a second order reaction is 50 sec when [A]0=0.84 mol L−1. The time needed for the concentration of A to decrease to one-fourth of its original concentration is


a. 125 sec

b. 75 sec

c. 100 sec

d. 50 sec

e. 150 sec

Answer 1

Half-life of a second-order reaction

t_1/2 = 1/k[A]₀

50 = 1/k*0.84

k = 1/42

Second-order reaction equation

1/[A] = 1/[A]₀ + kt

1/4 of concentration is: 0.84/4 = 0.21

1/0.21 = 1/0.84 + 1/42*t

3/0.84 = 1/42*t

t = 150 s

Answer: e. 150 sec