How long (in seconds) will it take for the concentration of AB to reach one-third of its initial concentration of 1.12?

Question

For the simple decomposition reaction:  AB(g)  A(g) + B(g), the rate = k{AB}2   ({ = [) and k = 0.47 1/Ms.  How long (in seconds) will it take for the concentration of AB to reach one-third of its initial concentration of 1.12?  Enter to 2 decimal places.

Answer 1

Since the rate equation tells us the decomposition is 2nd order in [AB], we can use the 2nd order integrated rate law to solve the problem.

1/[A] = 1/[A]_o + kt

[A] = 1/3 x 1.12 = 0.373 M

[A]_o = 1.12 M

k = 0.47 M^-1s^-1

t = ?

1/0.373 = 1/1.12 + 0.047t

2.68 = 0.893 + 0.047t

t = 38.02 sec