The standard free energy of activation of a reaction A is 83.9 kJ mol–1 (20.1 kc

Question

The standard free energy of activation of a reaction A is 83.9 kJ mol–1 (20.1 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

K(B) / K(A) = 10 x 106

Ea(A) = 83.9 kJ/mol = 83900 J /mol

(a). Arrhenius equation - K = A * e(-Ea / RT)

K(B) / K(A) = e[ (-Ea(B) + Ea(A)) / RT ]

10 x 106 = e[ (-Ea(B) + 83900) / 8.314 * 298 ]

(-Ea(B) + 83900) / 8.314 * 298 = ln (10 x 106)

(-Ea(B) + 83900) / 2477.572 = 16.118

- Ea(B) + 83900 = 39933.505

- Ea(B) = 39933.505 + 83900

Ea(B) = 43966.495 J/mol

Ea(B) = 43.9 kJ/mol

(b). standard free energy of activation of the reverse of reaction A -

Ea (reverse A) = 83.9 + 10.0

Ea (reverse A) = 93.9 kJ/mol

(c). standard free energy of activation of the reverse of reaction B -

Ea (reverse B) = 43.9 + 10.0

Ea (reverse B) = 53.9 kJ/mol

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