JEE MAIN - Chemistry (2024 - 29th January Morning Shift - No. 26)
For a reaction taking place in three steps at same temperature, overall rate constant $$\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_3}$$. If $$\mathrm{Ea}_1, \mathrm{Ea}_2$$ and $$\mathrm{Ea}_3$$ are 40, 50 and $$60 \mathrm{~kJ} / \mathrm{mol}$$ respectively, the overall $$\mathrm{Ea}$$ is ________ $$\mathrm{kJ} / \mathrm{mol}$$.
Answer
30
Explanation
$$\begin{aligned} & \mathrm{K}=\frac{\mathrm{K}_1 \cdot \mathrm{K}_2}{\mathrm{~K}_3}=\frac{\mathrm{A}_1 \cdot \mathrm{A}_2}{\mathrm{~A}_3} \cdot \mathrm{e}^{-\frac{\left(\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}\right)}{R T}} \\ & \mathrm{~A} \cdot \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}=\frac{\mathrm{A}_1 \mathrm{~A}_2}{\mathrm{~A}_3} \cdot \mathrm{e}^{-\frac{\left(\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}\right)}{R T}}\end{aligned}$$
$$\mathrm{E}_{\mathrm{a}}=\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}=40+50-60=30 \mathrm{~kJ} / \mathrm{mole} .$$
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