Decomposition of a hydrocarbon follows the equation $k=(5.5 \times 10^{11} \mathrm{~s}^{-1}) e^{\frac{-28000 \mathrm{~K}}{T}}$. The activation energy of reaction is $\_\_\_\_$ $\mathrm{kJ~mol}^{-1}$. (Nearest Integer) Given : $R=8.3 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}$

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Chemical Kinetics

Decomposition of a hydrocarbon follows the equation

k = (5.5 \times 1 0^{11} s^{- 1}) e^{\frac{- 28000 K}{T}}

. The activation energy of reaction is

____

kJ mol^{- 1}

. (Nearest Integer)
Given :

R = 8.3 JK^{- 1} mol^{- 1}

Answer: 232

Writing the Arrhenius form

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

beside the given

k = (5.5 \times 1 0^{11} s^{- 1}) e^{- 28000 K / T}

identifies

\frac{E _{a}}{R} = 28000

K. Hence

E_{a} = 28000 \times 8.3 = 2.324 \times 1 0^{5}

J mol

^{- 1}

, which to the nearest integer is 232 kJ mol

^{- 1}

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Integer answer