For $A k X_{1} B k X_{2} C$ (first order, $k_{1} = 0.20 mi n^{- 1}$ , $k_{2} = 0.05 mi n^{- 1}$ , only A present initially), the time at which $[B]$ is maximum is $t_{m a x} = ln (k_{1} / k_{2}) / (k_{1} - k_{2})$ . Using $ln 4 = 1.386$ , find $t_{m a x}$ in min (two decimal places).

Numerical answer type, answer in min, tolerance 0.02.

Answer: 9.24 min

At the maximum

d [B] / d t = 0

, i.e.

k_{1} [A] = k_{2} [B]

, which gives

t_{m a x} = ln (k_{1} / k_{2}) / (k_{1} - k_{2})

. Here

k_{1} / k_{2} = 4

, so

t_{m a x} = 1.386/ (0.20 - 0.05) = 1.386/0.15 = 9.24

min.