Two reactions A _ 1 and A _ 2 have identical pre-exponential factors. The activation energy of A _ 1 is more than A _ 2 by 10kJ mol ^ - 1 . If k _ 1 and k _ 2 are the rate constants for reactions A _ 1 and A _ 2

Correct answer: C. 4 From Arrhenius equation k = Ae ^ - Ea / RT k _ 1 = Ae ^ - Ea _ 1 / RT k _ 2 = Ae ^ - Ea _ 2 / RT ln k _ 2 k _ 1 = e ^ ( Ea _ 1 - Ea _ 2 ) / RT ln k _ 2 k _ 1 = Ea _ 1 - Ea _ 2 RT = (10 × 1000)/(8 .314 × 300) = 4

JEE Main 20172 April

Exam: JEE Main
Year: 2017
Date: 2 April
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2017 · 2 April · Chemistry · Chemical Kinetics

Two reactions A _ 1 and A _ 2 have identical pre-exponential...

Two reactions

{A}_{1} and {A}_{2}

have identical pre-exponential factors. The activation energy of

{A}_{1}

is more than

{A}_{2}

10kJ{mol}^{- 1}

. If

{k}_{1} and {k}_{2}

are the rate constants for reactions

{A}_{1} and {A}_{2}

, respectively at

300K

, then

\ln \left(\dfrac{{k}_{2}}{{k}_{1}}\right) is equal to<br /><br />\left(R = 8 .314 J{mol}^{- 1}{K}^{- 1}\right)

Options

A $12$
B $6$
C $4$
D $8$

Correct answer

C. $4$

Step-by-step solution

From Arrhenius equation

k = {Ae}^{- Ea / RT}

{k}_{1} = {Ae}^{- {Ea}_{1} / RT}

{k}_{2} = {Ae}^{- {Ea}_{2} / RT}

ln\dfrac{{k}_{2}}{{k}_{1}} = {e}^{\left({Ea}_{1} - {Ea}_{2}\right) / RT}

ln \dfrac{{k}_{2}}{{k}_{1}} = \dfrac{{Ea}_{1} - {Ea}_{2}}{RT} = \dfrac{10 \times 1000}{8 .314 \times 300} = 4

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