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In a first order reaction, $10 \%$ of the reactant is consumed in 25 minutes. Calculate :
(1) The half-life period of the reaction.
Options:
(1) The half-life period of the reaction.
Solution:
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The correct answer is:
164 min
$t=10 \%$ reactant $\rightarrow 25$ minute
$\begin{aligned} & \text { (1) } \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \\ & \mathrm{~K}=\frac{2.303}{\mathrm{t}} \log \frac{100}{90} \\ & \mathrm{~K}=0.004215 \\ & \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \\ & \mathrm{t}_{1 / 2}=164.405 \text { minute. }\end{aligned}$
(2) $\mathrm{t}=\frac{2.303}{\mathrm{~K}} \log \frac{100}{12.5}$
$t=493.43$ minute.
$\begin{aligned} & \text { (1) } \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \\ & \mathrm{~K}=\frac{2.303}{\mathrm{t}} \log \frac{100}{90} \\ & \mathrm{~K}=0.004215 \\ & \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \\ & \mathrm{t}_{1 / 2}=164.405 \text { minute. }\end{aligned}$
(2) $\mathrm{t}=\frac{2.303}{\mathrm{~K}} \log \frac{100}{12.5}$
$t=493.43$ minute.
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