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  • Given reaction below <img alt="\mathrm{\text { A } \stackrel{800 \mathrm{~K}}{\longrightarrow} \text { Product }}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Ctext

Given reaction below

\mathrm{\text { A } \stackrel{800 \mathrm{~K}}{\longrightarrow} \text { Product }}

\mathrm{B \frac{560 K}{\text { Catalyst }} \text { Product }}

it was found that the Ea is decreased by \mathrm{25 \mathrm{~kJ}/mol}  in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is { Assume pre exponential factor is same } : ~

Option: 1

\mathrm{49.8 \mathrm{~kJ} / \mathrm{mol}}


Option: 2

\mathrm{33.6 \mathrm{~kJ} / \mathrm{mol}}


Option: 3

\mathrm{58.3 \mathrm{~kJ} / \mathrm{mol}}


Option: 4

\mathrm{92.6 \mathrm{~kJ} / \mathrm{mol}}


Answers (1)

best_answer

\mathrm{A \stackrel{800\left(k_1\right)}{\longrightarrow} \text { Product }}

\mathrm{\text { A } \frac{\left.560 \mathrm{} \mathrm{~K}_2\right)}{\text { catalyst }} \text { Product }}

\mathrm{\text { Given } k_1=k_2}

\mathrm{A e-\frac{E a_1}{R T_1}=A e-\frac{E a}{ T_2}}

\mathrm{\begin{aligned} & \frac{E a_1}{T_1}=\frac{E a_2}{T_2} \\\\ & E _{a_2}=E a_1-25 \\\\ & E a_1=E a_2+25 \\\\ & \frac{E a_2+25}{T_1}=\frac{E a_2}{T_2} \end{aligned}}

\mathrm{\begin{aligned} & \frac{E a_2+25}{800}=\frac{E a_2}{560} \\\\ & 560 E a_2+14000=800 E a_2 \\\\ & 14000=800 E a_2-560 E a_2 \\\\ & 14000=240 E a_2 \\\\ & 58.365\, kj / \text { mol }=\mathrm{Ea}_2 \end{aligned}}

Posted by

Pankaj Sanodiya

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