Print Cakulatr Periodic Tabe uestion 22 of 23 Map d Organic Chemistry5 Roberts &

Question

Print Cakulatr Periodic Tabe uestion 22 of 23 Map d Organic Chemistry5 Roberts & The standard free energy of activation of a reaction A is 79.5 kJ mol- (19.0 kcal mol) at 298 K. Reaction B is one hundred million times faster than reaction A at the same temperature. The products of each reaction are 10.0 kJ mor" (2.39 kcal mol) more stable than the reactants. (a) What is the standard free energy of activation of reaction B? Number kJ mol (b) What is the standard free energy of activation of the reverse of reaction A? Number kJ mol B7 kJ mol 5

Explanation / Answer

The standard free energy of activation of a reaction A is 79.5 kJ/mol at 298 K.
EA = 79.5kJ/mol = 79500 J/mol

Reaction B is one hundred million times faster than reaction A at the same

temperature.
rate constant of reaction B, kB = 1E8*kA

The products of each reaction are 10 kJ/mol more stable than the reactants.
Change in gibbs free energy of reaction A, DGA = -10kJ/mol
Change in gibbs free energy of reaction B, DGB = -10kJ/mol

(a) what is the standard free energy of activation of reaction, B?
Arrhenius equation, kB/kA = exp((EB-EA)/(R*T))
EB = EA-2.303*R*T*Log(kB/kA)
R = 8.314 J/mol*K, T = 298 K, kB/kA = 1E8, EA = 79500 J/mol
EB = 79500-2.303*8.314*298*Log(1E8)= 33853.2 J/mol = 33.85 kJ/mol

(b) what is the standard free energy of activation of the reverse of reaction, A?
Change in gibbs free energy of reaction A, DGA = -10kJ/mol
Change in gibbs free energy of reverse of reaction A, DGAr = -DGA = 10kJ/mol

(c) what is the standard free energy of activation of the reverse of reaction, B?
Change in gibbs free energy of reaction B, DGB = -10kJ/mol
Change in gibbs free energy of reverse of reaction B, DGBr = -DGB = 10kJ/mol

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