Question:
An exothermic reaction $X \rightarrow Y$ has an activation energy $30 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If energy change $\Delta \mathrm{E}$ during the reaction is $-20 \mathrm{~kJ}$, then the activation energy for the reverse reaction in $\mathbf{k J}$ is
Solution:
(50)
$\Delta \mathrm{H}=\mathrm{E}_{\mathrm{a}, \mathrm{f}}-\mathrm{E}_{\mathrm{a}, \mathrm{b}}$
$-20=30-\mathrm{E}_{\mathrm{a}, \mathrm{b}}$
$\mathrm{E}_{\mathrm{a}, \mathrm{b}}=50 \mathrm{~kJ} / \mathrm{mole}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.