Chemistry - Chemical Kinetics Question with Solution | TestHub

For a first-order reaction, the integrated rate law is $kt=\ln \frac{[A{]}_0}{[A{]}_t}$.

When decomposition is to $1/8$ of initial concentration, $[A{]}_t=[A{]}_0/8$. So, $k{t}_{1/8}=\ln \frac{[A{]}_0}{[A{]}_0/8}=\ln 8$.

When decomposition is to $1/10$ of initial concentration, $[A{]}_t=[A{]}_0/10$. So, $k{t}_{1/10}=\ln \frac{[A{]}_0}{[A{]}_0/10}=\ln 10$.

The ratio $\frac{t_{1/8}}{t_{1/10}}=\frac{\ln 8}{\ln 10}=\frac{3\ln 2}{\ln 10}=\frac{3\times 0.3}{1}=0.9$.

Therefore, $\frac{t_{1/8}}{t_{1/10}}\times 10=0.9\times 10=9$.