Q14 / 50 · Manipal (MET) 2024 · Chemistry
For a first order reaction, time required for $99 \%$ completion of a reaction is $x$ times the time required for completion of $90 \%$. The value of $x$ is .......
Correct answer: 2
Solution
$\begin{aligned}
t_{99 \%} & =\frac{2.303}{k} \log \frac{(a)}{a-0.99 a} \\
& =\frac{2.303}{k} \log 10^2 \Rightarrow 2 \times \frac{2.303}{k} ...(i)\\
t_{90 \%} & =\frac{2.303}{k} \log \frac{a}{a-0.90 a} \\
& =\frac{2.303}{k} \log 10=\frac{2.303}{k}...(ii)
\end{aligned}$
Take the ratio of (i) and (ii)
$\begin{aligned}
& \frac{t_{99 \%}}{t_{90 \%}}=2 \\
& t_{99 \%}=2 \times t_{90 \%}
\end{aligned}$
Hence, 2 times is the final answer.
Hence, the correct answer is 2.