Question

The rate of a certain biochemical reaction at physiological temperature $\left(T\right)$ occurs ${10}^6$ times faster with enzyme than without. The change in the activation energy upon adding enzyme is
$\left(\text{Find}\right(E_a^{\prime }-E_a\left)\right)$
$E_a^{\prime }=$ Activation Energy with enzyme
$E_a=$ Activation Energy without enzyme

• A. $–6\left(2.303\right)RT$
• B. $+6RT$
• C. $-6RT$
• D. $+6\left(2.303\right)RT$

Answer 1

The correct option is A $–6\left(2.303\right)RT$

The rate constant of a reaction is given by
$k=A{e}^{-E_a/RT}$
The rate constant in presence of enzyme is given by
$k^{\prime }=A{e}^{-E_a^{\prime }/RT}$
$\frac{k^{\prime }}{k}=e^{-\left(E_a^{\prime }{E}_a\right)/RT}$
${10}^6=e^{-(E_a^{\prime }-E_a)/RT}$
In ${10}^6=-\frac{(E_a^{\prime }-E_a)}{RT}$
$E_a^{\prime }-E_a=-6\left(2.303\right)RT$