The rate of a certain biochemical reaction at physiological temperature (T) occurs 10 ^ 6 times faster with enzyme than without. The change in the activation energy upon adding enzyme is:

Correct answer: A. - 6(2.303)RT K = Ae ^ - E / RT . . . . . . ( 1 ) 10 ^ 6 K = Ae ^ - E _ c RT . ( 2 ) On dividing eq, 2 by eq, 1 10 ^ 6 = e ^ (E - E _ c ) / RT on taking logarithm both sides 6 10 = (E - E _ c ) RT 2 .303 × 6 = (E - E _ c ) RT E - E _ c = 2 .303 × ( 6RT ) Δ E _ a = E _ c - E = - 2 .303 × 6RT

JEE Main 20208 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 8 January
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2020 · 8 January · Morning Shift · Chemistry · Chemical Kinetics

The rate of a certain biochemical reaction at physiological...

The rate of a certain biochemical reaction at physiological temperature

\left(T\right)

occurs

{10}^{6}

times faster with enzyme than without. The change in the activation energy upon adding enzyme is:

Options

A $- 6\left(2.303\right)RT$
B $- 6RT$
C $+ 6\left(2.303\right)RT$
D $+ 6RT$

Correct answer

A. $- 6\left(2.303\right)RT$

Step-by-step solution

K = {Ae}^{- E / RT} . . . . . . ( 1 )

{10}^{6}\text{K} = {Ae}^{- {E}_{c}RT} \ldots \ldots . ( 2 )

On dividing eq, 2 by eq, 1

{10}^{6} = {e}^{\left(E - {E}_{c}\right) / RT}

on taking logarithm both sides

6\ln 10 = \dfrac{\left(E - {E}_{c}\right)}{RT}

2 .303 \times 6 = \dfrac{\left(E - {E}_{c}\right)}{RT}

E - {E}_{c} = 2 .303 \times ( 6RT )

{\Delta E}_{a} = {E}_{c} - E = - 2 .303 \times 6RT

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