For $A k X_{1} B k X_{2} C$ (first order, $k_{1} = 0.020 mi n^{- 1}$ , $k_{2} = 0.100 mi n^{- 1}$ , pure A at $t = 0$ ), the ratio $[B] / [A]$ at long times (transient equilibrium) approaches

Answer: B

For

k_{2} > k_{1}

the solution is

[B] = \frac{k _{1} [ A ] _{0}}{k _{2} - k _{1}} (e^{- k_{1} t} - e^{- k_{2} t})

. At long times

e^{- k_{2} t}

is negligible and

[A] = [A]_{0} e^{- k_{1} t}

, so

[B] / [A] = k_{1} / (k_{2} - k_{1}) = 0.020/0.080 = 0.25

. The simple steady-state value

k_{1} / k_{2} = 0.20

ignores the decay of A.