For the equilibrium A(g)⇌B(g),ΔH is −40kJ/mol, If the ratio of activation energies of the forward Ef and reverse Eb is 2/3, then

For the equilibrium , If the ratio of activation energies of the forward $\left({E}_{f}\right)$ and reverse $\left({E}_{\mathrm{b}}\right)$ is 2/3, then

1. A

${E}_{\mathrm{f}}=60\mathrm{kJ}/\mathrm{mol};{E}_{\mathrm{b}}=100\mathrm{kJ}/\mathrm{mol}$

2. B

${E}_{\mathrm{f}}=30\mathrm{kJ}/\mathrm{mol};{E}_{\mathrm{b}}=70\mathrm{kJ}/\mathrm{mol}$

3. C

${E}_{\mathrm{f}}=80\mathrm{kJ}/\mathrm{mol};{E}_{\mathrm{b}}=120\mathrm{kJ}/\mathrm{mol}$

4. D

${E}_{\mathrm{f}}=70\mathrm{kJ}/\mathrm{mol}:{E}_{\mathrm{b}}=30\mathrm{kJ}/\mathrm{mol}$

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Solution:

.

The given reaction is,

$\begin{array}{l}\mathrm{A}\left(\mathrm{g}\right)⇌\mathrm{B}\left(\mathrm{g}\right)\\ \mathrm{△}\mathrm{H}=-40\mathrm{KJ}/\mathrm{mol}\\ \mathrm{&}\frac{{\mathrm{E}}_{\mathrm{f}}}{{\mathrm{E}}_{\mathrm{b}}}=\frac{2}{3}⇒{\mathrm{E}}_{\mathrm{f}}=\frac{2}{3}{\mathrm{E}}_{\mathrm{b}}\end{array}$

From graph we have,

$\begin{array}{l}\mathrm{△}\mathrm{H}={\mathrm{E}}_{\mathrm{b}}-{\mathrm{E}}_{\mathrm{f}}\\ ⇒-40={\mathrm{E}}_{\mathrm{b}}-\frac{2}{3}{\mathrm{E}}_{\mathrm{b}}\\ ⇒-40=\frac{1}{3}{\mathrm{E}}_{\mathrm{b}}⇒{\mathrm{E}}_{\mathrm{b}}=-120\mathrm{KJ}/\mathrm{mol}\\ \mathrm{&}{\mathrm{E}}_{\mathrm{f}}={\mathrm{E}}_{\mathrm{b}}-\mathrm{\Delta H}\\ =-120-\left(-40\right)\\ -80\mathrm{KJ}/\mathrm{mol}\end{array}$

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