Thermodynamics and Keq FACT: ∆Gorxn is the change in free energy when pure reactants convert COMPLETELY to pure products. FACT: Product-favored systems have Keq > 1. Therefore, both ∆G˚rxn and Keq are related to reaction favorability.

Thermodynamics and Keq Keq is related to reaction favorability and so to ∆Gorxn. The larger the value of K the more negative the value of ∆Gorxn ∆Gorxn = - RT lnK where R = 8.31 J/K•mol

Thermodynamics and Keq ∆Gorxn = - RT lnK Calculate K for the reaction N2O4 --->2 NO2 ∆Gorxn = +4.8 kJ ∆Gorxn = +4800 J = - (8.31 J/K)(298 K) ln K K = 0.14 When ∆Gorxn > 0, then K < 1

∆G, ∆G˚, and Keq ∆G is change in free energy at non-standard conditions. ∆G is related to ∆G˚ ∆G = ∆G˚ + RT ln Q where Q = reaction quotient When Q < K or Q > K, reaction is spontaneous. When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K

∆G, ∆G˚, and Keq But systems can reach equilibrium when reactants have NOT converted completely to products. In this case ∆Grxn is < ∆Gorxn , so state with both reactants and products present is MORE STABLE than complete conversion. Figure 19.10

∆G, ∆G˚, and Keq Product favored reaction –∆Go and K > 1 In this case ∆Grxn is < ∆Gorxn , so state with both reactants and products present is MORE STABLE than complete conversion.

∆G, ∆G˚, and Keq Product-favored reaction. 2 NO2 ---> N2O4 ∆Gorxn = – 4.8 kJ Here ∆Grxn is less than ∆Gorxn , so the state with both reactants and products present is more stable than complete conversion.

∆G, ∆G˚, and Keq Reactant-favored reaction. N2O4 --->2 NO2 ∆Gorxn = +4.8 kJ Here ∆Gorxn is greater than ∆Grxn , so the state with both reactants and products present is more stable than complete conversion.

Thermodynamics and Keq Keq is related to reaction favorability. When ∆Gorxn < 0, reaction moves energetically “downhill” ∆Gorxn is the change in free energy when reactants convert COMPLETELY to products.