The time required for $10 \%$ completion of a first order reaction at $298 \mathrm{~K}$ is equal to that required for its $25 \%$ completion at $308 \mathrm{~K}$. If the preexponential factor for the reaction is $3.56 imes 10^{9} \mathrm{~s}^{-1}$, the activation energy at $318 \mathrm{~K}$ is:

Question

The time required for $10 \%$ completion of a first order reaction at $298 \mathrm{~K}$ is equal to that required for its $25 \%$ completion at $308 \mathrm{~K}$. If the preexponential factor for the reaction is $3.56 	imes 10^{9} \mathrm{~s}^{-1}$, the activation energy at $318 \mathrm{~K}$ is:

Aicharya · Accepted Answer

Let the initial concentration $(A)=100$
Final concentration at $298 \mathrm{~K}=100-10=90$
Final concentration at $308 \mathrm{~K}=100-25=75$
Substituting the values in the 1 st order rate reaction
$t=\frac{2.303}{k_{298}} \log \frac{100}{90}$ ...(i)
$t=\frac{2.303}{k_{308}} \log \frac{100}{75} \quad$...(ii)
From (i) and (ii) $\frac{k_{308}}{k_{208}}=2.73$
Substituting the value in the following relation
$E_{a}=\frac{2.303 R \times T_{1} \times T_{2}}{T_{2}-T_{1}} \log \frac{k_{2}}{k_{1}}$
$=\frac{2.303 \times 8.314 \times 298 \times 308}{308-298} \log 2.73$
$E_{o}=76.62 \mathrm{~kJ} \mathrm{~mol}^{-1}=18.39 \mathrm{kcal} \mathrm{mol}^{-1}$

The time required for $10 \%$ completion of a first order reaction at $298 \mathrm{~K}$ is equal to that…

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