Question

For the reaction $A\to B$, the rate constant $k\left({\text{s}}^{-1}\right)$ is given by

${\text{log}}_{10}k=20.35-\frac{(2.47\times {10}^3)}{T}$

The energy of activation in ${\text{kJ mol}}^{-1}$ is
(Nearest integer)

[Given : R=8.314 ${\text{JK}}^{-1}{\text{mol}}^{-1}$]

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• A. 47
• B. 47.00
• C. 47.0

Answer 1

Answer: (A), (B), (C)

${\text{log}}_{10}k=20.35-\frac{(2.47\times {10}^3)}{T}....\left(1\right)$

$\log k=\log A-\frac{E_a}{2.303RT}....\left(ii\right)$

Comparing (i) and (ii),

$\frac{E_a}{2.303RT}=\frac{2.47\times {10}^3}{T}$

$E_a=2.47\times {10}^3\times 2.303\times 8.314$

$E_a=47293.44{\text{J mol}}^{-1}$
$E_a=47.2934{\text{kJ mol}}^{-1}$
Nearest integer is 47