JEE MAIN - Chemistry (2020 - 8th January Morning Slot - No. 14)
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 :
+ 6RT
– 6 (2.303)RT
– 6RT
+ 6(2.303)RT
Explanation
The rate constant of a reaction without catalyst is
$$k = A{e^{ - {{{E_a}} \over {RT}}}}$$
The rate constant in presence of catalyst is given by
$$k' = A{e^{ - {{E{'_a}} \over {RT}}}}$$
$$ \therefore $$ $${{k'} \over k} = {e^{ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}}$$
$$ \Rightarrow $$ 106 = $${e^{ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}}$$
$$ \Rightarrow $$ ln 106 = $${ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}$$
$$ \Rightarrow $$ $${E{'_a} - {E_a}}$$ = -RTln 106
$$ \Rightarrow $$ $${E{'_a} - {E_a}}$$ = -6RT$$ \times $$2.303
$$k = A{e^{ - {{{E_a}} \over {RT}}}}$$
The rate constant in presence of catalyst is given by
$$k' = A{e^{ - {{E{'_a}} \over {RT}}}}$$
$$ \therefore $$ $${{k'} \over k} = {e^{ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}}$$
$$ \Rightarrow $$ 106 = $${e^{ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}}$$
$$ \Rightarrow $$ ln 106 = $${ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}$$
$$ \Rightarrow $$ $${E{'_a} - {E_a}}$$ = -RTln 106
$$ \Rightarrow $$ $${E{'_a} - {E_a}}$$ = -6RT$$ \times $$2.303
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