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← JEE Main 2022 · 27 Jun · Shift 1
ChemistryInteger+4 / -1Question 86 of 90

JEE Main 2022 · 27 Jun · Shift 1, Chemistry question 86

The rate constant for a first order reaction is given by the following equation :
ln⁡k=33.24−2.0×104 KT\ln k=33.24-\frac{2.0\times10^4\ \mathrm K}T
The Activation energy for the reaction is given by ______ kJ mol−1^{-1}. (In Nearest integer)

(Given : R=8.3R=8.3 J K−1^{-1} mol−1^{-1})

Answer and worked solution

Official answer: 166

Arrhenius. EaR=2.0×104\dfrac{E_a}R=2.0\times10^4 K, so Ea=2.0×104×8.3=166E_a=2.0\times10^4\times8.3=166 kJ mol−1^{-1}.

Answer: 166