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)

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

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

$\frac{t_{1}}{t_{2}}=\frac{8}{1} \times \frac{\log (4 / 3)}{\log (4)}$

$\frac{t_{1}}{t_{2}}=\frac{8 \times 0.125}{0.602}=1: 602$