The rate of a reaction doubles when its temperature changes from 300 K to 310 K . Activation energy of such a reaction will be ( R = 8.314 J K^-1 mol^ - Chemistry MCQ JEE Main 2026 | ExamTest.live

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The rate of a reaction doubles when its temperature changes from 300 K to 310 K . Activation energy of such a reaction will be (R = 8.314 J K^{- 1} mol^{- 1} and lo g 2 = 0.301):

Solution & Step-by-step Explanation

Using the Arrhenius Equation: lo g (\frac{k _{2}}{k _{1}}) = \frac{E _{a}}{2.303 R} [\frac{T _{2} - T _{1}}{T _{1} T _{2}}] Given: \frac{k _{2}}{k _{1}} = 2, T_{1} = 300 K, T_{2} = 310 K . lo g (2) = \frac{E _{a}}{2.303 \times 8.314} [\frac{310 - 300}{300 \times 310}] 0.301 = \frac{E _{a}}{19.147} [\frac{10}{93000}] E_{a} = \frac{0.301 \times 19.147 \times 93000}{10} = 53597.1 J/mol \approx 53.6 kJ/mol

Try it yourself before checking the explanation above.

The rate of a reaction doubles when its temperature changes from 300 K to 310 K . Activation energy of such a reaction will be (R = 8.314 J K^{- 1} mol^{- 1} and lo g 2 = 0.301):

48.6 kJ mol^{- 1}

58.5 kJ mol^{- 1}

60.5 kJ mol^{- 1}

53.6 kJ mol^{- 1}

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