The first-order reversible reaction $A B$ has $k_{f} = 0.030 mi n^{- 1}$ and $k_{b} = 0.020 mi n^{- 1}$ . The deviation $([A] - [A]_{e q})$ decays exponentially with rate constant $k_{f} + k_{b}$ . Using $ln 2 = 0.693$ , find the time (in min) for the deviation to fall to half its initial value (two decimal places).

Numerical answer type, answer in min, tolerance 0.02.

Answer: 13.86 min

The relaxation obeys

x = x_{0} e^{- (k_{f} + k_{b}) t}

, with

k_{f} + k_{b} = 0.050 mi n^{- 1}

. The time to halve is

0.693/0.050 = 13.86

min.