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Chemical Kinetics

Section: Chemistry

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Question : 2 of 18

Marks: +1, -0

The following equation is obtained for a first order reaction at 300 K .

\log_{10} \frac{k}{A}=0.00174

What is the activation energy (in

\mathrm{J} \ \mathrm{mol}^{-1}

) of the reaction?

(R=8.314 \ \mathrm{J} \ \mathrm{mol}^{-1} \ \mathrm{K}^{-1})

[AP EAPCET 23 May 2025 S2]

$10.0$
$100.0$
$0.1$
$1.0$

Solution:

The given equation is

\log_{10} \frac{k}{A}=0.00174

Multiply 2.303 on both side

\ln \left(\frac{k}{A}\right)=2.303 \times 0.00714

\Rightarrow \ \ln \left(\frac{k}{A}\right)=0.00400722 \ \cdots(i)

The Arrhenius equation is

k=A e^{-E_e / R T} \Rightarrow \frac{k}{A}=e^{-E a / R T}

Taking natural log

\ln \left(\frac{k}{A}\right)=-\frac{E_a}{R T}\ \cdots(ii)

Compare Eqs. (i) and (ii)

-\frac{E_a}{R T}=0.00400722

Substituting all values

E_a \simeq 10.0 \text{ J/mol}

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