The Gibb's Energy is used to find the spontaneity of reaction and is connected to any equilibrium process.
Find ΔH°, ΔS°, and ΔG° for the solubilization of calcium phosphate at 25 °C.
Write the reaction:
Ca3(PO4)2(s) → ← 3 Ca2+(aq) + 2 PO43–(aq)
From the Table of Thermodynamic Quantities:
ΔHf° (Ca3(PO4)2(s)) = –4121 kJ/mol
ΔHf°(Ca2+(aq)) = –542.96 kJ/mol
ΔHf°(PO43–(aq)) = –1277 kJ/mol
S°(Ca3(PO4)2(s)) = 236 J/mol·K
S°(Ca2+(aq)) = –55.2 J/mol·K
S°(PO43–(aq)) = –222 J/mol·K
Find the enthalpy and entropy changes for the reaction:
ΔH° = [3(–542.96) + 2(–1277)] – [–4121] = –62 kJ/mol
ΔS° = [3(–55.2) + 2(–222)] – [236] = –846 J/mol·K = –0.846 kJ/mol·K
ΔG° = ΔH° – TΔS°
T = 25 + 273 = 298 K
ΔG° = –62 – (298)(–0.846) = +190. kJ/mol
The reaction is nonspontaneous at room temperature.
What is ΔG° for this reaction at 75 °C?
The Gibb's Energy has obvious temperature dependence, but do the enthalpy and entropy terms change with temperature?
The answer is yes, but not too much around room temperature. Even better, for ΔG° determinations, the errors in ΔH° and ΔS° often tend to cancel each other out. Thus, to a reasonably good approximation, we can find ΔG° at other temperatures by using the defining equation and assuming that ΔH° and ΔS° do not change significantly with temperature.
With this assumption:
ΔG° = ΔH° – TΔS°
T = 75 + 273 = 348 K
ΔG° = –62 – (348)(–0.846) = +232 kJ/mol
The reaction becomes less spontaneous at higher temperature. The salt becomes less soluble at higher temperature.
Is this consistent with LeChatelier's Principle?
Consider the evaporation of water:
H2O(l) → ← H2O(g)
ΔHf°(H2O(l)) = –285.8 kJ/mol
ΔHf°(H2O(g)) = –241.8 kJ/mol
S°(H2O(l)) = 69.91 J/mol·K
S°(H2O(g)) = 188.7 J/mol·K
ΔH° = [–241.8] – [–285.8] = 44.0 kJ
ΔS° = [188.7] – [69.91] = 118.8 J/K
ΔG° = ΔH° – TΔS° = 44.0 – (298)(0.1188) = 8.6 kJ
At the normal boiling point of water:
ΔG° = 44.0 – (373)(0.1188) = –0.3 kJ
Experimentally, the free energy change for any phase transition is 0. Thus, the error in this approximation (in this case) is 0.3 kJ.
At the normal boiling point or the normal melting point of a substance:
ΔH° = TΔS°
For nonpolar substances, the entropy change for vaporization is fairly constant from compound to compound, about 87 J/mol·K.
This gives rise to Trouton's Rule:
ΔS°vapn = ΔH°vapn Tbp ~ 87 J/mol·K
This means that thermodynamic data can be used to estimate normal boiling points.
Compare the estimated normal boiling point for benzene, C6H6, using Trouton's Rule and using the calculated entropy of vaporization.
Using Trouton's Rule:
C6H6(l) → ← C6H6(g)
ΔHf°(C6H6(l)) = 48.99 kJ/mol
ΔHf°(C6H6(g)) = 82.93 kJ/mol
ΔH°vap = [82.93] – [48.99] = 33.94 kJ/mol
Tbp = ΔH°vapn ΔS°vapn ~ ΔH°vapn 87 J/mol·K = 33940 J/mol 87 J/mol·K = 390 K = 117 °C
Using the calculated entropy of vaporization:
S°(C6H6(l)) = 173.3 J/mol·K
S°(C6H6(g)) = 269.2 J/mol·K
ΔS° = [269.2] – [173.3] = 95.9 J/mol·K
Tbp = ΔH°vapn ΔS°vapn = 33940 J/mol 95.9 J/mol·K = 354 K = 81 °C
Experimental value: 80.1 °C – Trouton's Rule is not great!