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:

ΔG =

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 ΔG also can determine if a system is at equilibrium, it must be related to an equilibrium constant. In general:

ΔG = ΔG° + RTlnQ

R = the gas constant = 8.314 J/mol·K

T = temperature in K

Q = reaction quotient

This equation determines ΔG at any composition or temperature conditions.

At equilibrium, ΔG = 0 and Q = K_{eq}, so

ΔG° = –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)ΔH

_{f}°(Pb^{2+}(aq)) = –1.7 kJ/molΔH

_{f}°(Cl^{–}(aq)) = –167.2 kJ/molΔH

_{f}°(PbCl_{2}(s)) = –359 kJ/molS°(Pb

^{2+}(aq)) = 10.5 J/mol·KS°(Cl

^{–}(aq)) = 56.5 J/mol·KS°(PbCl

_{2}(s)) = 136 J/mol·KΔH° = [–1.7 + 2(–167.2)] – [–359] = 23 kJ/mol

ΔS° = [10.5 + 2(56.5)] – [136] = –13 J/mol·K

T = 90 + 273 = 363 K

ΔG° = ΔH° – TΔS° = 23000 – 363(–13) = 28000 J/mol

ΔG° = –RTlnK

_{eq}= –RTlnK_{sp}K

_{sp}= e– ΔG°RT = e – 28000 J/mol(8.314 J/mol·K)(363 K) = 9.3×10^{–5}

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

Initial 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.

ΔG° = –RTlnK_{eq}

ΔH° – TΔS° = –RTlnK_{eq}

After some algebra:

ln K

_{eq}= –ΔH°RT + ΔS°R

A plot of ln K_{eq} vs. 1/T (T in units of K) should be a straight line with

slope = –ΔH°/R

intercept = ΔS°/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 ΔH° and ΔS° using a van't Hoff plot.

T (°C)K

_{sp}

105.5×10

^{–6}

204.8×10

^{–6}

303.2×10

^{–6}

402.7×10

^{–6}

502.5×10

^{–6}

601.9×10

^{–6}

701.5×10

^{–6}

801.5×10

^{–6}

901.2×10

^{–6}