The standard free energy of activation of a reaction A is 71.3kJ mol^-1 (17.o kc

Question

The standard free energy of activation of a reaction A is 71.3kJ mol^-1 (17.o kcal mol^-1) at 298K. Reaction B is one hundred million times faster than reaction A at the same temperature. The products of each reaction are 10.0kJ mol^-1 (2.39kcal 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

let us take rate constant of reaction A = 1 s-1.

k1 = 1

A = Frequency factor = ?

Ea = 71.3*1000 j/mol

R = 8.314 j.k-.mol-1

T = 298 k

K1 = A*e(-Ea/RT)

1 = A*e^((-71.3*1000)/(8.314*298))

A = 3.15 s-1.

Reaction B , at same temperature A = 3.15 s-1.

rate of reaction B = 100*10^6 rate of reaction A

thefore rate constant k2 = 10^8 s-1

10^8 = 3.15*e^((-Ea)/(8.314*298))

Ea = 42.795 kj/mol

a) activation of a reaction B = Ea = 42.795 kj/mol

b) activation of the reverse of reaction A = -71.3 kj/mol

c) activation of the reverse of reaction B = -42.795 kj/mol

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