The standard free energy of activation of a reaction A is 76.5 kJ mol–1 (18.3 kc

Question

The standard free energy of activation of a reaction A is 76.5 kJ mol–1 (18.3 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

a.

Use the Arrhenius equation

ln k1 = lnA – Ea1/RT

ln k2 = lnA – Ea2/RT

Assuming that lnA is the same in the two cases

lnk2 – lnk2 = (Ea1 – Ea2)/RT

Assume T is about 298 K (ambient temperature)

ln (k2/k1) x(RT) = Ea1 – Ea2

ln (107 ) x(8.314 x 10-3 kJ/mol.K x 298 K) = 76.5 kJ/mol - Ea2

1.95 x 2.49 = 76.5 kJ/mol - Ea2

Ea2 = 76.5 kJ/mol – 4.8 kJ/mol = 71.7 kJ/mol

b.

In the reverse reaction, Ea,A is 10 kJ/mol higher

76.5 kJ/ mol + 10 kJ/mol = 86.5 kJ/mol

c.

In the reverse reaction, Ea,B is 10 kJ/mol higher

71.7 kJ/ mol + 10 kJ/mol = 81.7 kJ/mol

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