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The rate coefficient $(k)$ for a particular reactions is $1.3 \times 10^{-4} \mathrm{M}^{-1} \mathrm{~s}^{-1}$ at $100^{\circ} \mathrm{C}$, and $1.3 \times 10^{-3}$ $\mathrm{M}^{-1} \mathrm{~s}^{-1}$ at $150^{\circ} \mathrm{C}$. What is the energy of activation $\left(\mathrm{E}_{\mathrm{A}}\right)$ (in $\left.\mathrm{kJ}\right)$ for this reaction? $(\mathrm{R}=$ molar gas constant $=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )
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Verified Answer
The correct answer is:
60
60
According to Arrhenius equation
$$
\begin{aligned}
&\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\
&\log \frac{1.3 \times 10^{-3}}{1.3 \times 10^{-4}}=\frac{E_a}{2.303 \times 8.314}\left[\frac{1}{373}-\frac{1}{423}\right] \\
&1=\frac{E_a}{2.303 \times 8.314}\left[\frac{1}{373}-\frac{1}{423}\right] \\
&E_a=60 \mathrm{~kJ} / \mathrm{mole}
\end{aligned}
$$
$$
\begin{aligned}
&\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\
&\log \frac{1.3 \times 10^{-3}}{1.3 \times 10^{-4}}=\frac{E_a}{2.303 \times 8.314}\left[\frac{1}{373}-\frac{1}{423}\right] \\
&1=\frac{E_a}{2.303 \times 8.314}\left[\frac{1}{373}-\frac{1}{423}\right] \\
&E_a=60 \mathrm{~kJ} / \mathrm{mole}
\end{aligned}
$$
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