Q26 / 160 · TS EAMCET 2021 (05 Aug Shift 1) · Chemistry

Calculate the activation energy of a reaction, whose rate constant doubles on raising the temperature from $300 K$ to $600 K .$

Correct answer
A
$3.45 kJ / mol$
B
$6.90 kJ / mol$
C
$9.68 kJ / mol$
D
$19.6 kJ / mol$

Solution

Relation between temperature and activation energy is given by Arrhenius equation.

$k = {Ae}^{- (\frac{E_{a}}{RT})}$

$\frac{k_{2}}{k_{1}} = e^{- (\frac{E_{a}}{{RT}_{2}} + \frac{E_{a}}{{RT}_{1}})} \ln (\frac{k_{2}}{k_{1}}) = - (\frac{E_{a}}{{RT}_{2}} + \frac{E_{a}}{{RT}_{1}}) \ln (2) = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \log (2) = \frac{E_{a}}{2.303 R} (\frac{1}{300} - \frac{1}{600}) 0.3010 = \frac{E_{a}}{2.303 \times 8.314} (\frac{1}{600}) E_{a} = 3.45 kJ / mol$

Hence, the correct answer is option: 'A'.

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Calculate the activation energy of a reaction, whose rate constant doubles on raising the temperature from 300 K to 600 K.

Solution

Relevant Questions

Calculate the activation energy of a reaction, whose rate constant doubles on raising the temperature from $300 K$ to $600 K .$