For following reactions $\mathrm{A} \xrightarrow{700 \mathrm{~K}}$ Product A $\xrightarrow[\text{catalyst}]{500 \mathrm{~K}}$ Product it was found that the $\mathrm{E}_{\mathrm{a}}$ is decreased by $30 \mathrm{~kJ/mol}$ in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Chemical Kinetics

For following reactions

A 700 K

Product
A

500 K catalyst

Product
it was found that the

E_{a}

is decreased by

30 kJ/mol

in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):

Answer: (C)

Because the rate is unchanged and the pre-exponential factor is the same for both paths, the two rate constants are equal, which forces

E_{a} / T

to be the same for the uncatalysed and catalysed reactions. Writing

\frac{E _{a}}{700} = \frac{E _{a} - 30}{500}

for the catalysed energy being

30 kJ/mol

lower,

500 E_{a} = 700 (E_{a} - 30)

gives

E_{a} = 105 kJ/mol

for the uncatalysed reaction. The catalysed activation energy is therefore

105 - 30 = 75 kJ/mol

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option