For the reaction of H _ 2 with I _ 2 , the rate constant is 2.5 × 10 ^ - 4 d m ^ 3 mo l ^ - 1 s ^ - 1 at 327 ^ o C and 1.0d m ^ 3 mol- 1 s ^ - 1 at 527 ° C . The activation energy for the reaction, in kJmol- 1 is

Correct answer: A. 166 T _ 2 = 527 ^ o C ⇒ 800k; k _ 1 = 2.5 × 10 ^ - 4 T _ 1 = 327 ^ o C ⇒ 600k; k _ 2 = 1 According to Arrhenius Equation; ln k _ 2 k _ 1 = - (Ea)/(R)( 1 T _ 2 - 1 T _ 1 ) ( 10 ^ 4 2.5 ) = (Ea)/(R)((1)/(600) - (1)/(800)) (4000) = (Ea)/(8.314)((1)/(2400)) Ea = 165.431kJ, Ea 166kJ / mol

JEE Main 201910 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 10 April
Shift: Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2019 · 10 April · Evening Shift · Chemistry · Chemical Kinetics

For the reaction of H _ 2 with I _ 2 , the rate constant is 2.5 ×...

For the reaction of

{H}_{2}

with

{I}_{2}

, the rate constant is

2.5 × {10}^{- 4}d{m}^{3}mo{l}^{- 1}{s}^{- 1}

{327}^{o}C and 1.0d{m}^{3}mol- 1{s}^{- 1}

527 &deg; C

. The activation energy for the reaction, in

kJmol- 1

is:

\left(R = 8.314J{K}^{- 1}mo{l}^{- 1}\right)

Options

A $166$
B $59$
C $72$
D $150$

Correct answer

A. $166$

Step-by-step solution

{T}_{2} = {527}^{o}C ⇒ 800k; {k}_{1} = 2.5 \times {10}^{- 4}

{T}_{1} = {327}^{o}C ⇒ 600k; {k}_{2} = 1

According to Arrhenius Equation;

ln \dfrac{{k}_{2}}{{k}_{1}} = - \dfrac{Ea}{R}\left(\dfrac{1}{{T}_{2}} - \dfrac{1}{{T}_{1}}\right)

\ln \left(\dfrac{{10}^{4}}{2.5}\right) = \dfrac{Ea}{R}\left(\dfrac{1}{600} - \dfrac{1}{800}\right)

\ln \left(4000\right) = \dfrac{Ea}{8.314}\left(\dfrac{1}{2400}\right)

Ea = 165.431kJ, Ea \approx 166kJ / mol

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