Exercise 7 | Chemical Thermodynamics

Calculate ΔG0rxn for the following reaction at 25°C:

C2H4 (g) + H2O (l) C2H5OH (l)

What is the direction in which the reaction is spontaneous at 25°C and PC2H4 = 20.0 bar?


Data: ΔH0rxn = -44.2 kJ.mol-1

S0[C2H4 (g)] = 219.3 J.mol-1.K-1

S0[C2H5OH (l)] = 160.7 J.mol-1.K-1

S0[H2O (l)] = 70.00 J.mol-1.K-1

ΔS0rxn = S0[C2H5OH (l)] – S0[C2H4 (g)] – S0[H2O (l)]

ΔS0rxn = 160.7 – 219.3 – 70.00

ΔS0rxn = -128.6 J.mol-1.K-1


ΔG0rxn = ΔH0rxn - TΔS0rxn and 25°C = 298K

ΔG0rxn = - 44.2 x 103 + 298 x 128.6

ΔG0rxn = - 5.88 kJ.mol-1

ΔG0rxn < 0: at standard conditions, the reaction is spontaneous from left to right.


ΔGrxn = ΔG0rxn + RT ln Q

with Q = reaction quotient = 1PC2H4 = 120.0 = 5.00 x 10-2


ΔGrxn = - 5.88 x 103 + 8.314 x 298 x ln Q

ΔGrxn = - 13.3 kJ.mol-1

ΔGrxn < 0 the reaction is spontaneous from left to right as written