The standard free energy of activation of a reaction A is 83.9 kJ mol–1 (20.1 kc
ID: 888444 • Letter: T
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