ΔG° = ΔH° - TΔS°

Question

For the reaction

2 NO(g) + O₂(g)  2 NO₂(g)

G° = -64.1 kJ and S° = -146.5 J/K at 342 K and 1 atm.

This reaction is (reactant, product) _____ favored under standard conditions at 342 K.

The standard enthalpy change for the reaction of 1.88 moles of NO(g) at this temperature would be ____  kJ.

Answer 1

∆Gº = ∆Hº - T∆Sº

Solving for ∆Hº we have...

∆Hº = ∆Gº + T∆Sº = -64.1 kJ + (298K)(-0.1465 kJ/K) = -64.1 kJ - 43.7 kJ

∆Hº = -107.8 kJ = standard enthalpy change

Solving for ∆G we have...

∆G = ∆H = T∆S

∆G = -107.8 kJ - (342K * -0.1465kJ/K) = -107.8 kJ + 50.1 kJ

∆G = -57.7 kJ

This reaction is product favored