Question:
$99%$ of a first order reaction was completed in $32 \mathrm{min}$. When will $99.9%$ of the reaction complete?
$99%$ of a first order reaction was completed in $32 \mathrm{min}$. When will $99.9%$ of the reaction complete?
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For first order, time for $99.9%$ completion is $1.5$ times time for $99%$ completion.
Updated On: May 2, 2026
- $50 \mathrm{min}$
- $46 \mathrm{min}$
- $49 \mathrm{min}$
- $48 \mathrm{min}$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
For first order, $k = \frac{2.303}{t} \log \frac{a}{a-x}$.
Step 2: Detailed Explanation:
For $99%$, $k = \frac{2.303}{32} \log 100 = \frac{2.303 \times 2}{32}$. For $99.9%$, $t = \frac{2.303}{k} \log 1000 = \frac{2.303}{k} \times 3 = 32 \times \frac{3}{2} = 48$ min.
Step 3: Final Answer:
Time required is $48$ min.
For first order, $k = \frac{2.303}{t} \log \frac{a}{a-x}$.
Step 2: Detailed Explanation:
For $99%$, $k = \frac{2.303}{32} \log 100 = \frac{2.303 \times 2}{32}$. For $99.9%$, $t = \frac{2.303}{k} \log 1000 = \frac{2.303}{k} \times 3 = 32 \times \frac{3}{2} = 48$ min.
Step 3: Final Answer:
Time required is $48$ min.
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