# An exothermic reaction

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)

$\Delta \mathrm{H}=\mathrm{E}_{\mathrm{a}, \mathrm{f}}-\mathrm{E}_{\mathrm{a}, \mathrm{b}}$

$-20=30-\mathrm{E}_{\mathrm{a}, \mathrm{b}}$

$\mathrm{E}_{\mathrm{a}, \mathrm{b}}=50 \mathrm{~kJ} / \mathrm{mole}$