For $A B$ (first order both ways, $k_{f} = 0.020 mi n^{- 1}$ , $k_{b} = 0.005 mi n^{- 1}$ ) starting with $[A]_{0} = 1.00 M$ and no B, find the time (in min) at which $[B] = 0.400 M$ . Use $ln 2 = 0.693$ and give the answer to one decimal place.

Numerical answer type, answer in min, tolerance 0.1.

Answer: 27.7 min

The equilibrium constant is

K = k_{f} / k_{b} = 4

, so

[B]_{e q} = K / (1 + K) = 0.80 M

. Approach to equilibrium follows

[B] = [B]_{e q} (1 - e^{- (k_{f} + k_{b}) t})

. Setting

0.400 = 0.80 (1 - e^{- x t})

gives

e^{- x t} = 0.5

, so

t = 0.693/0.025 = 27.72 \approx 27.7

min.