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JEE Main 20266 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
6 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2026 · 6 April · Evening Shift · Chemistry · Chemical Kinetics

Decomposition of a hydrocarbon follows the equation k = (5.5 × 10^...

Decomposition of a hydrocarbon follows the equation k=(5.5×1011 s−1) e−28000 KTk = (5.5 \times 10^{11}\,\text{s}^{-1})\,e^{\frac{-28000\,\text{K}}{T}}. The activation energy of reaction is __________ kJ mol−1^{-1}. (Nearest Integer) Given : R =8.3= 8.3 J K−1^{-1} mol−1^{-1}

Correct answer

0

Step-by-step solution

According to the Arrhenius equation, k=Ae−EaRTk = A e^{-\dfrac{E_a}{RT}}.

The given equation is k=(5.5×1011 s−1)e−28000Tk = (5.5 \times 10^{11} \text{ s}^{-1}) e^{-\dfrac{28000}{T}}.

Comparing the exponent of ee in both equations:

EaRT=28000T\dfrac{E_a}{RT} = \dfrac{28000}{T}

Ea=28000×RE_a = 28000 \times R

Substituting the value of R=8.3 J K−1 mol−1R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}:

Ea=28000×8.3 J mol−1E_a = 28000 \times 8.3 \text{ J mol}^{-1}

Ea=232400 J mol−1E_a = 232400 \text{ J mol}^{-1}

Ea=232.4 kJ mol−1E_a = 232.4 \text{ kJ mol}^{-1}

Rounding to the nearest integer, the activation energy is 232232.

Answer: 232232
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