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The \Delta \mathrm{H} of a reaction is -200 \mathrm{kJ} / \mathrm{mol} and the activation energy \mathrm{\left(E_f\right)} of the forward reaction is 80 \mathrm{~kJ} / \mathrm{mol}. On addition of the catalyst the activation energy\left(\mathrm{E}_{\mathrm{f}}{ }^{\prime}\right) becomes 40 \mathrm{~kJ} / \mathrm{mol}. What is the ratio of activation energy of backward reaction before and after addition of the catalyst is:
 

Option: 1

2


Option: 2

0.6


Option: 3

2.6


Option: 4

1.6


Answers (1)

best_answer

\mathrm{E_{\mathrm{b}}=E_{\mathrm{i}}-\Delta \mathrm{H}=80-(-200) \mathrm{kJ} / \mathrm{mol}=280 \mathrm{kJ} / \mathrm{mol}}

After adding catalyst

\mathrm{E_{\mathrm{b}}{ }{ }^{\prime}=\mathrm{E}_{\mathrm{i}}{ }^{\prime}-\Delta \mathrm{H}=40-(-200) \mathrm{kJ} / \mathrm{mol}=260 \mathrm{kJ} / \mathrm{mol}}

Ratio \mathrm{=280 / 240=1.16}

Hence option 4 is correct.

Posted by

Pankaj

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