The standard free energy of activation of a reaction A is 81.8 kJ mol-1 (19.6 kc

Question

The standard free energy of activation of a reaction A is 81.8 kJ mol-1 (19.6 kcal mol-1) at 298 K. Reaction B is one 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? Number kJ mol-1 (b) What is the standard free energy of activation of the reverse of reaction A? Number kJ mol-1 (c) What is the standard free energy of activation of the reverse of reaction B? Number kJ mol-1

Explanation / Answer

(a) From the Arrhenius equation: Rate constant k = A exp(-Ea/RT)

Rate(B)/Rate(A) = k(B)/k(A) = exp(-(Ea(B) - Ea(A))/RT)

Substituting values into equation gives:

1 x 10^6 = exp(-(Ea(B) - 81800)/(8.314 x 298))

-(Ea(B) - 81800)/(8.314 x 298) = ln(1 x 10^6) = 13.816

Activation energy Ea(B) = 81800 - 13.816 x 8.314 x 298

= 47571 J/mol = 47.6 kJ/mol

(b) Activation energy Ea(reverse of A) = Ea(A) + stability of products relative to reactants

= 81.8 + 10.0 = 91.8 kJ/mol

(c) Activation energy Ea(reverse of B) = Ea(B) + stability of products relative to reactants

= 47.6 + 10.0 = 57.6 kJ/mol

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