ΔG° = ΔH° - TΔS°

Question

For the reaction

2CO₂(g) + 5H₂(g)C₂H₂(g) + 4H₂O(g)

H° = 46.5 kJ and S° = -124.8 J/K

The standard free energy change for the reaction of 1.62 moles of CO₂(g) at 279 K, 1 atm would be _____ kJ.

This reaction is (reactant, product) _______favored under standard conditions at 279 K.

Assume that H° and S° are independent of temperature.

Answer 1

NOTE: Because ∆H is positive and ∆S is negative, this reaction is not spontaneous at any temperature and ∆G will be positive

2CO₂(g) + 5H₂(g) ==> C₂H₂(g) + 4H₂O(g)

∆Gº = ∆Hº - T∆Sº

∆Gº = 46.5 kJ - (273K )(-0.1248 kJ/K)

∆Gº = 46.5 kJ - (- 34.1)

∆Gº = 80.6 kJ for 2 moles of CO2 (as the equation is written)

∆G for 1.62 moles CO2 @ 279K:

80.6 kJ/2 mol CO2 x 1.62 = 65.3 kJ = ∆G for 1.62 moles of CO2

The reaction is REACTANT favored under standard conditions because of the positive ∆G