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In a first order reaction, $10 \%$ of the reactant is consumed in 25 minutes. Calculate: The time required for completing $17 \%$ of the reaction.
Solution:
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Verified Answer
The correct answer is:
44.21
Again, for the given reaction
$\begin{aligned} & \mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \\ & \mathrm{t}=\frac{2.303}{0.0042145} \log \frac{100}{100-17} \\ & \mathrm{t}=44.21 \text { minutes. }\end{aligned}$
$\therefore$ time required for completing $17 \%$ of the reaction is 44.21 minutes.
$\begin{aligned} & \mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \\ & \mathrm{t}=\frac{2.303}{0.0042145} \log \frac{100}{100-17} \\ & \mathrm{t}=44.21 \text { minutes. }\end{aligned}$
$\therefore$ time required for completing $17 \%$ of the reaction is 44.21 minutes.
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