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For the following reactions:

\mathrm{A \overset{700K}{\rightarrow}product}

\mathrm{A \xrightarrow[catalyst]{500K}product}

it was found that Ea is decreased by 30 kJ/mol in the presence of a catalyst. If the rate remains unchanged, the activation energy (in KJ/mol)  for catalyzed reaction is ( Assume pre-exponential factor is same):

Option: 1

75


Option: 2

105


Option: 3

198


Option: 4

135


Answers (1)

The formula of the rate of reaction

        \mathrm{k=A e^{-E a / R T}}

Ea is decreased by 30 kJ/mol in the presence of a catalyst. If the rate remains unchanged

\mathrm{E_{a}{(catalyst)} -E_a=-30\ kJ/mol}

and  

\mathrm{k}_{\text {catalyst }}=\mathrm{k}

So,

\large \mathrm{Ae}^ \frac{-\mathrm{Ea(catalyst)}}{\mathrm{RT_{catalyst}}} =\mathrm{Ae}^{-\frac{\mathrm{Ea}}{\mathrm{RT}}}

\\ {\frac{\mathrm{E_a(catalyst)}}{\mathrm{T_{(catalyst)}}}=\frac{\mathrm{E_a}}{\mathrm{T}} }

\\ {\frac{\mathrm{E_a(catalyst)}}{\mathrm{500}}=\frac{\mathrm{E_a}}{\mathrm{700}} }

\\ {\frac{\mathrm{E_a(catalyst)}}{\mathrm{E_a}}=\frac{\mathrm{500}}{\mathrm{700}} }

\\ {\frac{\mathrm{E_a(catalyst)}}{\mathrm{E_a(catalyst)-E_a}}=\frac{\mathrm{500}}{\mathrm{500-700}} }

\\ {\frac{\mathrm{E_a(catalyst)}}{\mathrm{-30}}=\frac{\mathrm{500}}{\mathrm{-200}} }

\mathrm{E_a(catalyst)} =\frac{\mathrm{-30\times500}}{\mathrm{-200}}

\mathrm{E_a(catalyst) = 75\ kJ/mol}

 

Posted by

Ramraj Saini

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