ΔG° = ΔH° - TΔS°

Question

For the reaction

C(s,graphite) + O₂(g)  CO₂(g)

G° = -394.3 kJ and S° = 2.9 J/K at 259 K and 1 atm.

This reaction is (reactant, product) ______favored under standard conditions at 259 K.

The standard enthalpy change for the reaction of 1.89 moles of C(s,graphite) at this temperature would be  ______kJ.

Answer 1

C(s, graphite) + O₂(g) ==> CO₂(g)

The reaction PRODUCT is favored (based on the positive value of ∆Gº)

Finding standard enthalpy change:

∆Gº = ∆Hº - T∆Sº

-394.3 kJ = ∆H - (259K)(0.0029 J/K)

∆Hº = -393.5 kJ / mol x 1.89 mol = - 744 kJ