Question

The rate of a certain biochemical reaction at physiological temperature (T) occurs ${10}^6$ times faster with enzyme than without. The change in activation energy upon adding enzyme is:

• A. $-6\text{RT}$
• B. $-6\times 2.303\text{RT}$
• C. $+6\text{RT}$
• D. $+6\times 2.303\text{RT}$

Answer 1

The correct option is B $-6\times 2.303\text{RT}$

$K_1=A{e}^{-E_{a1}/RT}....\left(1\right)$
$K_1=A{e}^{-E_{a2}/RT}....\left(1\right)$
Dividing equation 1 with equation 2, we get
$\frac{K_1}{K_2}=e^{E_{a2}-E_{a1}/RT}{10}^{-6}=e^{(E_{a}2-E_{a}1)/RT}$
Taking $lo{g}_e$ on both sides, we get
$\Delta E=E_{a2}-E_{a1}=-6\times 2.303\text{RT}$