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JEE Main 202126 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 26 February
Shift: Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2021 · 26 February · Evening Shift · Chemistry · Chemical Kinetics

If the activation energy of a reaction is 80 . 9kJ mol ^ - 1 , the...

If the activation energy of a reaction is

80 . 9kJ{mol}^{- 1}

, the fraction of molecules at

700K

, having enough energy to react to form products is

{e}^{- x}

. The value of

x

is (Rounded off to the nearest integer) [Use

R = 8 . 31J{K}^{- 1}{mol}^{- 1}

]

Correct answer

0

Step-by-step solution

Fraction of molecules to have enough energy to react $= {e}^{- Ea / RT}$

So, $x = \dfrac{{E}_{a}}{RT}$

$= \dfrac{80 . 9 \times {10}^{3}}{8 . 31 \times 700}$

$= 13 . 9 \approx 14$

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