Solve for ΔG (in kJ)

Question

Consider the chemical reaction and thermochemical information and initial partial pressures of the reaction components given below:

CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)

	ΔHf° (kJ/mol)	S° (J mol-1 K-1)	P (atm)
CO2	-393.51	213.74	0.427
H2	0.00	130.68	0.522
CO	-110.53	197.66	8.02
H2O	-241.83	188.84	6.61

Determine ΔG (in kJ) for this reaction at 803.43 K. Assume ΔH°_f and S° do not vary as a function of temperature.

Answer 1

CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)

∆H_rxn = ∑products - ∑reactants = (-393.51 + 0) - (-110.53 + -241.83) = -393.51 - (-352.36) = -41.15 kJ

∆S_rxn = ∑products - ∑reactants = (213.74 + 130.68) - (197.66 + 188.84) = 344.42 - 386.5 = -42.08 J = -0.04208kJ

∆G = ∆H - T∆S

∆G = -41.15 kJ - (803.43K)(-0.04208 kJ/K

∆G = -41.15 - (-33.80)

∆G = -7.35 kJ (negative so reaction is spontaneous)