A $\to$ B
The rate constants of the above reaction at 200 K and 300 K are 0.03 min$^{-1}$ and 0.05 min$^{-1}$ respectively. The activation energy for the reaction is ___________ J (Nearest integer)
(Given : $\mathrm{ln10=2.3}$
$\mathrm{R=8.3~J~K^{-1}~mol^{-1}}$
$\mathrm{\log5=0.70}$
$\mathrm{\log3=0.48}$
$\mathrm{\log2=0.30}$)
Answer (integer)
2520
Solution
$$
\begin{aligned}
& \log \frac{\mathrm{k}_2}{\mathrm{k}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{2.3 \times 8.3}\left(\frac{1}{200}-\frac{1}{300}\right) \\\\
& \log \frac{0.05}{0.03}=\frac{E_a}{2.3 \times 8.3}\left(\frac{1}{600}\right) \\\\
& \begin{aligned}
(0.70-0.48)=\frac{E_a}{2.3 \times 8.3} \times \frac{1}{600}
\end{aligned} \\\\
& \begin{aligned}
& \Rightarrow 0.22 =\frac{E_a}{2.3 \times 8.3} \times \frac{1}{600} \\\\
& \Rightarrow {E_a} = 2.3 \times 8.3 \times 600 \times 0.22 \\\\
& =2519.88 \\\\
& \approx 2520 \mathrm{~J}
\end{aligned}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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