Decomposition of a hydrocarbon follows the equation k=(5.5×10¹¹ s⁻¹) e^(-28000 K/T). The activation energy of reaction is _ _ _ kJ mol ⁻¹. (Nearest Integer) Given: R=8.3 J K ⁻¹ mol ⁻¹

Chemical Kinetics: JEE Main 2026 Chemistry question

JEE Main 2026 · 6 Apr, Evening shift · Q75

JEE Main 20266 Apr, Evening shiftChemistryMediumChemical Kinetics62× since 2002

Decomposition of a hydrocarbon follows the equation

k=(5.5\times10^{11}\ \mathrm{s^{-1}})\,e^{-28000\,\mathrm{K}/T}

. The activation energy of reaction is ______ kJ mol

^{-1}

. (Nearest Integer) Given :

R=8.3

J K

^{-1}

mol

^{-1}

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2026 Chemical Kinetics Question with Solution (6 Apr Evening shift, Q75) | JEE Arc