Question

The standard free energy of activation of a reaction A is 77.3 kJ mol–1 (18.5 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?

Answer 1

A)

According to the Arrhenius equation,

K = A e^{-E_a / RT}

For reaction A,

Rate Constant = K_a

For reaction B,

Rate Constant = Kb

Since, Reaction B is 10 million (10⁷) times faster than A

K_b = 10⁷ Ka

Therefore,

K_b / K_a = 10⁷

=> A e^{-E_a(B) / RT} / A e^{-E_a(A) / RT} = 10⁷

=> e^{E_a(A) -E_a(B) / RT} = 10⁷

=> e^{(77300 - E_a(B)) / 8.314 x 298} = 10⁷

Taking log on both sides,

(77300 - Ea(B)) / 2477.57 = 7 ln 10

=> (77300 - Ea(B)) / 2477.57 = 16.12

=> Ea(B) = 37366.3 J/mol = 37.37 KJ/mol = Activation Energy for B

B)

The reaction is exothermic since products of each reaction are 10.0 kJ mol–1 (2.39 kcal mol–1) more stable than the reactants

Hence delta H = -10 KJ/mol

Activation energy for Reverse reaction of A = 77.3 - (delta H) = 77.3 - (-10) = 87.3 KJ/mol

C)

Activation energy for reverse reaction of B = 37.37 + 10 = 47.37 KJ/mol