A first-order reaction is 40% complete in 20 min at 300 K. The rate constant doubles when the temperature is raised to 310 K. The time required for…

For a first-order reaction $k=\frac{1}{t}\ln\frac{[A]_0}{[A]}$. At $300\ \mathrm{K}$, $40\%$ completion leaves $60\%$, so $k_{300}=\frac{\ln(1/0.6)}{20}=2.55\times10^{-2}\ \mathrm{min^{-1}}$. The statement that the rate constant doubles gives $k_{310}=2k_{300}=5.11\times10^{-2}\ \mathrm{min^{-1}}$ (the corresponding activation energy from the Arrhenius equation is about $53.6\ \mathrm{kJ\,mol^{-1}}$, but it is not needed because the ratio is given directly). For $75\%$ completion $[A]_0/[A]=4$, so $t=\frac{\ln 4}{k_{310}}=\frac{1.386}{0.0511}=27.1\ \mathrm{min}$. The value $54\ \mathrm{min}$ is the time at $300\ \mathrm{K}$, which is twice the correct answer, and $18\ \mathrm{min}$ results from using $60\%$ completion instead of $75\%$.