JEE Main 2021 · Chemistry · Chemical Kinetics · hard

A substance A decomposes simultaneously by two parallel first-order paths, AtoB with k1=1.0×10^-3 s^-1 and Ea1=120 kJ mol^-1, and AtoC with…

Single correct option

A substance A decomposes simultaneously by two parallel first-order paths,

\mathrm{A}\to\mathrm{B}

with

k_1=1.0\times10^{-3}\ \mathrm{s^{-1}}

and

E_{a1}=120\ \mathrm{kJ\,mol^{-1}}

, and

\mathrm{A}\to\mathrm{C}

with

k_2=3.0\times10^{-3}\ \mathrm{s^{-1}}

and

E_{a2}=40\ \mathrm{kJ\,mol^{-1}}

(both values measured at the same temperature). The activation energy for the overall disappearance of A, in

\mathrm{kJ\,mol^{-1}}

, is

Show answer and solution

Answer: (C) $60$

For parallel paths the overall rate constant is

k=k_1+k_2

, so

\dfrac{d\ln k}{dT}=\dfrac{1}{k}\left(k_1\dfrac{d\ln k_1}{dT}+k_2\dfrac{d\ln k_2}{dT}\right)

. Since

\dfrac{d\ln k_i}{dT}=\dfrac{E_{ai}}{RT^2}

, the overall activation energy is the rate-constant-weighted mean:

E_a=\dfrac{k_1E_{a1}+k_2E_{a2}}{k_1+k_2}=\dfrac{1(120)+3(40)}{4}=60\ \mathrm{kJ\,mol^{-1}}

. The simple average

80

ignores the weights and the value

40

would apply only if the faster path alone existed.

Original question written in the JEE Main 2021 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake