MCQ
Equilibrium constant at $27\,^oC$ and $127\,^oC$ is $30$ and $40$ respectively, then the ratio of activation energy $E_a$ (forward) / $E_a$ (backward) will be
- ✓$>\,1$
- B$-1$
- C$<\,1$
- DData is insufficient
$\mathrm{T}_{2}=127\,^{\circ} \mathrm{C} ;\left(\mathrm{k}_{\mathrm{eq}}\right)_{2}=40$
reaction is endothermic $\left(\mathrm{k}_{\mathrm{eq}} \propto \mathrm{T}\right)$
${\Delta \mathrm{H}=\left(\mathrm{E}_{0}\right)_{\mathrm{f}}-\left(\mathrm{E}_{2}\right)_{0}=+\mathrm{ve}} $
${\left(\mathrm{E}_{\mathrm{q}}\right)_{\mathrm{t}}>\left(\mathrm{E}_{\mathrm{s}}\right)_{\mathrm{b}}} $
${\frac{\left(\mathrm{E}_{\mathrm{a}}\right)_{\mathrm{f}}}{\left(\mathrm{E}_{\mathrm{a}}\right)_{\mathrm{b}}}>1}$
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Given: $\Delta H ^0=-54.07\,kJ\,mol ^{-1}$
$\Delta S ^{\circ}=10\,JK ^{-1}\,mol ^{-1}$
(Take $2.303 \times 8.314 \times 298=5705$ )