The Gibb's Energy is a critical function for understanding chemical reactivity. As we have seen, it universally determines the spontaneity of reaction (at constant pressure).
The Gibb's Energy also is the measure of the maximum amount of useful work available from a reaction:
G = w_{max}
Because nothing is 100% efficient (there is always a loss to entropy, which we detect by a heat change in the system), w_{max} is never achieved in any real system.
Since G also can determine if a system is at equilibrium, it must be related to an equilibrium constant. In general:
G = G^{o} + RTlnQ
R = the gas constant = 8.314 J/mol·K
T = temperature in K
Q = reaction quotient
This equation determines G at any composition or temperature conditions.
At equilibrium, G = 0 and Q = K_{eq}, so
G^{o} = –RTlnK_{eq}
The equilibrium constant, K_{eq}, in this equation is a thermodynamic equilibrium constant. The units of the terms in the mass action expression for K_{eq} must be atm for gases and molarity for concentrations of dissolved species. Pure liquids and solids do not contribute.
All of the labeled equilibrium constants that we have looked at are thermodynamic equilibrium constants: K_{p}, K_{a}, K_{b}, K_{sp}, K_{f}.
This equation allows us to use thermochemical data to find equilibrium constants and vice versa.
Compare the molar solubility of lead chloride at room temperature (25 °C) and 90 °C.
The room temperature solubility was done previously: 0.016 mol/L.
To find the solubility at 90 °C, we need to find K_{sp} at 90 °C using thermodynamic data.
PbCl_{2}(s)Pb^{2+}(aq) + 2 Cl^{–}(aq)
H_{f}^{o}(Pb^{2+}(aq)) = –1.7 kJ/mol
H_{f}^{o}(Cl^{–}(aq)) = –167.2 kJ/mol
H_{f}^{o}(PbCl_{2}(s)) = –359 kJ/mol
S^{o}(Pb^{2+}(aq)) = 10.5 J/mol•K
S^{o}(Cl^{–}(aq)) = 56.5 J/mol•K
S^{o}(PbCl_{2}(s)) = 136 J/mol•K
H^{o} = [–1.7 + 2(–167.2)] – [–359] = 23 kJ
S^{o} = [10.5 + 2(56.5)] – [136] = –13 J/K
T = 90 + 273 = 363 K
G^{o} = H^{o} – TS^{o} = 23000 – 363(–13) = 28000 J
G^{o} = –RTlnK_{eq} = –RTlnK_{sp}
Now we can do a standard equilibrium problem to find the molar solubility.

PbCl_{2}(s) 
Pb^{2+}(aq) 
+ 
2 Cl^{–}(aq) 


K_{sp} = [Pb^{2+}]_{e}[Cl^{–}]_{e}^{2} = 9.3×10^{–5} 

Initial 


0 

0 
Change 


+ x 

+ 2x 
Equilibrium 


x 

2x 
9.3×10^{–5} = [x][2x]^{2} = 4x^{3}
x = 2.9×10^{–2} = the molar solubility of lead chloride at 90 °C.
Equilibrium constants can be used to evaluate thermodynamic parameters.
G^{o} = –RTlnK_{eq}
H^{o} – TS^{o} = –RTlnK_{eq}
After some algebra:
A plot of ln K_{eq} vs. 1/T should be a straight line with
slope = –H^{o}/R
intercept = S^{o}/R
This is called the van't Hoff equation.
The solubility product constant of calcium hydroxide was measured at several temperatures, as given below. Find H^{o} and S^{o} using a van't Hoff plot.
T (^{o}C) 
K_{sp} 
T (K) 
1/T (K^{–1}) 
ln K_{sp} 
10 
5.5×10^{–6} 
283 
0.00353 
–12.11 
20 
4.8×10^{–6} 
293 
0.00341 
–12.25 
30 
3.2×10^{–6} 
303 
0.00330 
–12.65 
40 
2.7×10^{–6} 
313 
0.00319 
–12.82 
50 
2.5×10^{–6} 
323 
0.00310 
–12.90 
60 
1.9×10^{–6} 
333 
0.00300 
–13.17 
70 
1.5×10^{–6} 
343 
0.00292 
–13.41 
80 
1.5×10^{–6} 
353 
0.00283 
–13.41 
90 
1.2×10^{–6} 
363 
0.00275 
–13.63 
slope = 1970
intercept = –19.1
H^{o} = –slope×R = –(1970)×(8.314) = –16400 J/mol = –16.4 kJ/mol
S^{o} = intercept×R = (–19.1)×(8.314) = –159 J/mol•K
Intercepts have a larger inherent error, so this is not a preferred method to find S^{o}