The rate of a certain biochemical reaction at physiological temperature (T) occurs 106 times faster with enzyme than without. The change in the activation energy upon adding enzyme is :
Solution
The rate constant of a reaction without catalyst is
<br><br>$k = A{e^{ - {{{E_a}} \over {RT}}}}$
<br><br>The rate constant in presence of catalyst is
given by
<br><br>$k' = A{e^{ - {{E{'_a}} \over {RT}}}}$
<br><br>$\therefore$ ${{k'} \over k} = {e^{ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}}$
<br><br>$\Rightarrow$ 10<sup>6</sup> = ${e^{ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}}$
<br><br>$\Rightarrow$ ln 10<sup>6</sup> = ${ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}$
<br><br>$\Rightarrow$ ${E{'_a} - {E_a}}$ = -RTln 10<sup>6</sup>
<br><br>$\Rightarrow$ ${E{'_a} - {E_a}}$ = -6RT$\times$2.303
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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