1. HReaction = ΣnH(products) - ΣnH(reactants)
exothermic, ∆H = - points toward spontaneous
endothermic, ∆H = + points toward non-spon.
SReaction = ΣnS(products) - ΣnS(reactants)
more disorder, ∆S = + points toward spontaneous
more order, ∆S = - points toward non-spon.
GReaction = ΣnG(products) - ΣnG(reactants)
∆G = - reaction IS spontaneous as written
∆G = + reaction IS non-spon.
∆G = 0 reaction is at equilibrium

2. ΔG = ΔH – TΔS = 0 @Equilibrium
and..
This allows us to perform phase transition calculations (melting “fusion” and boiling “vaporization”.

3. T = Temperature in Kelvin
Equilibrium
R = J∙mol-1∙K-1
T = Temperature in Kelvin
Keq is the thermodynamic equilibrium constant. We will ignore activities. Gases will use partial pressure values in atmospheres and solutes will have concentrations expressed in molarity.

4. Use ΔG = ΔH – TΔS to solve for T. Since ΔS
Assuming ∆S and ∆H do not vary with temperature, at what temperature will the reaction shown below become spontaneous? C(s) + H2O(g) → H2(g) + CO(s) ∆S = J ; ∆H = kJ
Use ΔG = ΔH – TΔS to solve for T. Since ΔS
Points toward spontaneous (+ value) and ΔH points toward non-spontaneous (+ value), the temperature at which the term TΔS dominates is when the reaction becomes spontaneous.

5. Coupled Reactions: ∆G = ∆G1 + ∆G2 + ......
2Fe2O3(s) → 4Fe(s) + 3O2(g) ∆Go = 1487 kJ
6CO(g) + 3O2(g) → 6CO2(g) ∆Go = kJ
2Fe2O3(s) + 6CO(g) → 4Fe(s) + 6CO2(g) ∆Go = -56 kJ