Q21 / 120 · VITEEE 2016 · Chemistry

For a reaction $\mathrm{A} \rightarrow \mathrm{B} ; \Delta \mathrm{H}=20 \mathrm{~kJ} \mathrm{~mol}^{-1}$ the activation energy of the forward reaction is $85 \mathrm{~kJ} / \mathrm{mol}$. The activation energy of the backward reaction will be

  • A
    $105 \mathrm{~kJ} / \mathrm{mol}$
  • Correct answer
    B
    $65 \mathrm{~kJ} / \mathrm{mol}$
  • C
    $45 \mathrm{~kJ} / \mathrm{mol}$
  • D
    $75 \mathrm{~kJ} / \mathrm{mol}$

Solution

Energy of activation for forward reaction $\left(\mathrm{E}_{\mathrm{a}}\right)=85 \mathrm{~kJ} / \mathrm{mol}$ $\therefore \quad$ Energy of activation for backward reaction $=\mathrm{E}_{\mathrm{a}}-\Delta \mathrm{H}=85-20=65 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Hence, the correct answer is option: 'B'.

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