The rate of a reaction quadruples when temperature changes from 27^ to 57^ . Calculate the energy of activation. (Given R = 8.314 J/mol K ) - Chemistry MCQ NEET (UG) 2024 2026 | ExamTest.live

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The rate of a reaction quadruples when temperature changes from 2 7^{\circ} C to 5 7^{\circ} C . Calculate the energy of activation. (Given R = 8.314 J/mol K)

Solution & Step-by-step Explanation

Using Arrhenius Equation: ln (\frac{k _{2}}{k _{1}}) = \frac{E _{a}}{R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}}) k_{2} = 4 k_{1} \Rightarrow \frac{k _{2}}{k _{1}} = 4 . T_{1} = 300 K, T_{2} = 330 K . ln 4 = \frac{E _{a}}{8.314} (\frac{330 - 300}{300 \times 330}) 2 \times 0.693 = \frac{E _{a}}{8.314} \times \frac{30}{99000} 1.386 = \frac{E _{a}}{8.314} \times \frac{1}{3300} E_{a} = 1.386 \times 8.314 \times 3300 \approx 38038 J/mol \approx 38.04 kJ/mol .

Try it yourself before checking the explanation above.

The rate of a reaction quadruples when temperature changes from 2 7^{\circ} C to 5 7^{\circ} C . Calculate the energy of activation. (Given R = 8.314 J/mol K)

380.4 kJ/mol

3.80 kJ/mol

3804 kJ/mol

38.04 kJ/mol

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