Chemistry - Chemical Kinetics Question with Solution | TestHub

For a first order reaction, the half-life $t_{1/2}$ is independent of the initial concentration.

The concentration drops from $0.8\mathrm{mol}{\mathrm{L}}^{-1}$ to $0.4\mathrm{mol}{\mathrm{L}}^{-1}$ (which is exactly half of the initial concentration) in $15$ minutes. Thus, $t_{1/2}=15$ minutes.

The time required for the concentration to drop from $0.1\mathrm{mol}{\mathrm{L}}^{-1}$ to $0.025\mathrm{mol}{\mathrm{L}}^{-1}$ can be calculated by finding the number of half-lives.

$0.1\stackrel{t_{1/2}}{\to }0.05\stackrel{t_{1/2}}{\to }0.025$

This process takes $2$ half-lives.

Total time required $=2\times t_{1/2}=2\times 15=30$ minutes.

Answer: 30 minute