For $\mathrm{A}_2+\mathrm{B}_2 \rightleftharpoons 2 \mathrm{AB}$
$\mathrm{E}_{\mathrm{a}}$ for forward and backward reaction are 180 and $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively
If catalyst lowers $\mathrm{E}_{\mathrm{a}}$ for both reaction by $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
Which of the following statement is correct?
Solution
<p>$\mathrm{A}_2+\mathrm{B}_2 \rightleftharpoons 2 \mathrm{AB}$</p>
<p>$$\begin{aligned}
&\begin{aligned}
& \mathrm{E}_{\mathrm{f}}=180 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{E}_{\mathrm{b}}=200 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{b}}=-20 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}\\
&\text { In presence of catalyst : }\\
&\begin{aligned}
& \mathrm{E}_{\mathrm{f}}=180-100=80 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{E}_{\mathrm{b}}=200-100=100 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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