JEE PYQ: Chemical Kinetics Question 24
Question 24 - 2020 (08 Jan Shift 1)
The rate of a certain biochemical reaction at physiological temperature ($T$) occurs $10^6$ times faster with enzyme than without. The change in the activation energy upon adding enzyme is:
(1) $-6(2.303)\text{RT}$
(2) $-6\text{RT}$
(3) $+6(2.303)\text{RT}$
(4) $+6\text{RT}$
Show Answer
Answer: (1)
Solution
The rate constant of a reaction is given by $k = Ae^{-E_a/RT}$. The rate constant in presence of catalyst is $k’ = Ae^{-E_a’/RT}$. $\frac{k’}{k} = e^{-(E_a’ - E_a)/RT}$. $10^6 = e^{-(E_a’ - E_a)/RT}$. $\ln 10^6 = -\frac{(E_a’ - E_a)}{RT}$. $E_a’ - E_a = -6(2.303)\text{RT}$.
BETA
