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The rate of a reaction quadruples when temperature changes from $27^{\circ} \mathrm{C}$ to $57^{\circ} \mathrm{C}$. Calculate the energy o activation.
Given $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \log 4=0.6021$
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Given $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \log 4=0.6021$
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The correct answer is:
$38.04 \mathrm{~kJ} / \mathrm{mol}$
$\begin{aligned} & \log \left(\frac{\mathrm{k}_2}{\mathrm{k}_1}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right) \\ & \log \left(\frac{4}{1}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{1}{300}-\frac{1}{330}\right) \\ & \mathrm{E}_{\mathrm{a}}=\frac{(\log (4)) \times 2.303 \times 8.314 \times 300 \times 330}{30} \\ & =3.804 \times 10^4 \mathrm{~J} / \mathrm{mol} \\ & =38.04 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$
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