Question

Two first order reaction have half-lives in the ratio $8:1.$ Calculate the ratio of time intervals $t_1:t_2.$The time $t_1$ and $t_2$ are the time period for ${\left(\frac{1}{4}\right)}^{th}$ and ${\left(\frac{3}{4}\right)}^{\text{th }}$

• A. 1: 0.301
• B. 0.125: 0.602
• C. 1: 602
• D. None of these

Answer 1

Answer: (C)

$t_1=\frac{{\left(t_{1/2}\right)}_1}{0.693}\times 2.303{\log }_{10}\frac{1}{1-\frac{1}{4}}$

$t_2=\frac{{\left(t_{1/2}\right)}_2}{0.693}\times 2.303{\log }_{10}\frac{1}{1-\frac{3}{4}}$

$\frac{t_1}{t_2}=\frac{8}{1}\times \frac{\log (4/3)}{\log (4)}$

$\frac{t_1}{t_2}=\frac{8\times 0.125}{0.602}=1:602$