The standard free energy of activation of a reaction A is 84.0 kJ mol^-1(20.1 kc

Question

The standard free energy of activation of a reaction A is 84.0 kJ mol^-1(20.1 kcal mol^-1)l 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)l more stable than the reactants. (a) What is the standard free energy of activation of reaction B? kJ mol^-1 (b) What is the standard free energy of activation of the reverse of reaction A? kJ mol^-1 (c) What is the standard free energy of activation of the reverse of reaction B? kJ mol^-1

Explanation / Answer

a)Reaction B is 107 times faster

So, its activation energy will be 7 times lesser ie 84/7 kJ/mole = 12.0 kJ/mole

b)For reverse of reaction A, activation energy =10 + 84 kJ/mole = 94.0 kJ/mole

c)For reverse of reaction B, activation energy = 10 +12 kJ/mole= 22.0 kJ/mole

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