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

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

T = Temperature in Kelvin Equilibrium R = 8.314 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.

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 = 133.6 J ; ∆H = 131.3 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.

Coupled Reactions: ∆G = ∆G1 + ∆G2 + ...... 2Fe2O3(s) → 4Fe(s) + 3O2(g) ∆Go = 1487 kJ 6CO(g) + 3O2(g) → 6CO2(g) ∆Go = -1543 kJ 2Fe2O3(s) + 6CO(g) → 4Fe(s) + 6CO2(g) ∆Go = -56 kJ