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← JEE Main 2026 · 6 Apr · Shift 2
ChemistryInteger+4 / -1Question 75 of 75

JEE Main 2026 · 6 Apr · Shift 2, Chemistry question 75

Decomposition of a hydrocarbon follows the equation k=(5.5×1011 s−1)e−28000 K/Tk=\left(5.5\times10^{11}\text{ s}^{-1}\right)e^{-28000\,\mathrm K/T}. The activation energy of reaction is ________ kJ mol−1^{-1}. (Nearest Integer)

Given : R = 8.3 J K−1^{-1} mol−1^{-1}

Answer and worked solution

Official answer: 232

Arrhenius. k=Ae−Ea/RTk=Ae^{-E_a/RT}, so EaR=28000\dfrac{E_a}R=28000 K and
Ea=28000×8.3=232 400 J mol−1≈232 kJ mol−1.E_a=28000\times8.3=232\,400\text{ J mol}^{-1}\approx232\text{ kJ mol}^{-1}.

Answer: 232

Topics:Arrhenius equation
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