Q36 / 120 · VITEEE 2009 · Chemistry
For a first order reaction at $27^{\circ} \mathrm{C}$, the ratio of time required for $75 \%$ completion to $25 \%$ completion of reaction is
- A$3.0$
- B$2.303$
- Correct answerC$4.8$
- D$0.477$
Solution
For a first order reaction,
$$
t=\frac{2.303}{\lambda} \log _{10} \frac{a}{a-x}
$$
Let initial amount of reactant is 100 .
$$
\begin{aligned}
\frac{t_{1}}{t_{2}} &=\frac{\log \frac{100}{100-75}}{\log \frac{100}{100-25}}[\because \lambda \text { remains constant }] \\
&=\frac{\log \frac{100}{25}}{\log \frac{100}{75}}=\frac{\log 4}{\log 4 / 3} \\
&=\frac{\log 4}{\log 4-\log 3} \\
&=\frac{2 \times 0.3010}{2 \times 0.3010-0.4771} \\
&=\frac{0.6020}{0.1249}=4.81
\end{aligned}
$$
Hence, the correct answer is option: 'C'.