Catalyst A reduces the activation energy for a reaction by $10 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at 300 K. The ratio of rate constants, $\frac{k_{\mathrm{T}, \text{Catalysed}}}{k_{\mathrm{T}, \text{Uncatalysed}}}$ is $\mathrm{e}^{\mathrm{x}}$. The value of x is $\_\_\_\_$. [nearest integer] [Assume that the pre-exponential factor is same in both the cases. Given $\mathrm{R}=8.31 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$]

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Chemical Kinetics

Catalyst A reduces the activation energy for a reaction by

10 kJ mol^{- 1}

at 300 K. The ratio of rate constants,

\frac{k _{T, Catalysed}}{k _{T, Uncatalysed}}

e^{x}

. The value of x is

____

. [nearest integer]
[Assume that the pre-exponential factor is same in both the cases.

Given

R = 8.31 J K^{- 1} mol^{- 1}

]

Answer: 4

Using the Arrhenius equation

k = A e^{- E_{a} / R T}

\frac{k _{cat}}{k _{uncat}} = \frac{A e ^{- E_{a}^{'} / R T}}{A e ^{- E_{a} / R T}} = e^{(E_{a} - E_{a}^{'}) / R T}

.

Given that the catalyst reduces

E_{a}

by 10 kJ/mol, so

E_{a} - E_{a}^{'} = 10

kJ/mol

= 10000

J/mol.

\frac{k _{cat}}{k _{uncat}} = e^{10000/ (8.31 \times 300)} = e^{4.009} \approx e^{4}

.

Therefore,

x = 4

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Integer answer