1. 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.

2. 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

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

4. ∆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

5. ∆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

6. ∆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.

7. ∆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.

8. ∆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.

9. 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.