From this value and the values of the standard free-energy of formation of liquid ethanol, acetic acid, and water, calculate ΔG°f for ethyl acetate.

Question

The standard free-energy change in the reaction of water with ethyl acetate (CH₃CO₂C₂H₅) to give ethanol (C₂H₅OH) and acetic acid (CH₃COOH) is –19.7 kJ. From this value and the values of the standard free-energy of formation of liquid ethanol, acetic acid, and water, calculate ΔG°_f for ethyl acetate.

ΔG°f (kJ/mol)
Ethanol(ℓ)	-174.9
Acetic acid(ℓ)	-389.9
Water(ℓ)	-237.2

Answer 1

CH₃CO₂C₂H₅ + H₂O ==> C₂H₅OH + CH₃COOH .. balanced equation

ethyl acetate + water ==> ethyl alc. + acetic acid

∆G_rxn = ∑∆G_f products - ∑∆G_f reactants

-19.7 = (-174.9 + -389.9) - (-237.2 + X) where X = ∆Gf ethyl acetate

-19.7 = -564.8 + 237.2 - X

X = ∆Gf ethyl acetate = -307.9 kJ/mol

(be sure to check the math)