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The rate of the chemical reaction doubles for an increase of \(10 \mathrm{~K}\) in absolute temperature from \(298 \mathrm{~K}\). Calculate \(\mathbf{E}_{\mathrm{a}}\).
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Verified Answer
Here, \(T_1=298 \mathrm{~K}, T_2=308 \mathrm{~K}, k_1=k, k_2=2 k\)
We know,
\(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 \mathrm{R}}\left(\frac{T_2-T_1}{T_1 T_2}\right)\)
\(\begin{aligned}
\log \frac{2 k}{k} &=\frac{E_a}{2.303 \times 8.314}\left(\frac{308-298}{308 \times 298}\right) \\
\log 2 &=\frac{E_a}{2.303 \times 8.314} \times \frac{10}{308 \times 298} \\
\Rightarrow \quad E_a &=\frac{(\log 2)(2.303 \times 8.314)(308 \times 298)}{10} \\
&=52897.7 \mathrm{~J} \mathrm{~mol}^{-1}=52.8 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}\)
We know,
\(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 \mathrm{R}}\left(\frac{T_2-T_1}{T_1 T_2}\right)\)
\(\begin{aligned}
\log \frac{2 k}{k} &=\frac{E_a}{2.303 \times 8.314}\left(\frac{308-298}{308 \times 298}\right) \\
\log 2 &=\frac{E_a}{2.303 \times 8.314} \times \frac{10}{308 \times 298} \\
\Rightarrow \quad E_a &=\frac{(\log 2)(2.303 \times 8.314)(308 \times 298)}{10} \\
&=52897.7 \mathrm{~J} \mathrm{~mol}^{-1}=52.8 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}\)
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