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ExamsJEE MainChemistry

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

  1. 58.5 kJ mol−1
  2. 60.5 kJ mol−1
  3. 53.6 kJ mol−1
  4. 48.5 kJ mol−1

Correct answer: 53.6 kJ mol−1

Solution

The correct option is derived using the Arrhenius equation, which relates the rate of reaction to temperature and activation energy. Given that the rate doubles with a temperature increase of 10 K, we can apply the formula to calculate the activation energy, leading to the value of 53.6 kJ mol−1.

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