The standard free energy of activation of a reaction A is 89.0 kJ mol-1 (21.3 kc

Question

The standard free energy of activation of a reaction A is 89.0 kJ mol-1 (21.3 kcal mol-1) at 298 K. Reaction B is ten million times faster than reaction A at the same temperature. The products of each reaction are 10.0 kJ mol-1 (2.39 kcal mol-1) more stable than the reactants. (a) What is the standard free energy of activation of reaction B? (b) What is the standard free energy of activation of the reverse of reaction A? (c) What is the standard free energy of activation of the reverse of reaction B?

Explanation / Answer

As reaction B is 10^7 times faster than reaction A, 10^7 = e^[(Ea-Eb)/RT] On solving, Eb = 49.08 KJ mol-1 = (a) As the products are 10 KJ mol-1 more stable than the reactants, The activation energy of the reverse of the reactions will be Ea + 10 = 99.0 KJ mol-1 = (b) and Eb + 10 = 59.08 KJ mol-1 = (c) respectively.

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