Chemistry 112

Using the Gibb's Energy.

The Gibb's Energy is used to find the spontaneity of reaction and is connected to any equilibrium process.

Example

Find Ho, So, and Go 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:

Hfo (Ca3(PO4)2(s)) = –4121 kJ/mol

Hfo(Ca2+(aq)) = –542.96 kJ/mol

Hfo(PO43–(aq)) = –1277 kJ/mol

So(Ca3(PO4)2(s)) = 236 J/mol•K

So(Ca2+(aq)) = –55.2 J/mol•K

So(PO43–(aq)) = –222 J/mol•K

Find the enthalpy and entropy changes for the reaction:

Ho = [3(–542.96) + 2(–1277)] – [–4121] = –62 kJ/mol

So = [3(–55.2) + 2(–222)] – [236] = –846 J/mol·K = –0.846 kJ/mol·K

Go = Ho – TSo

T = 25 + 273 = 298 K

Go = –62 – (298)(–0.846) = +190. kJ/mol

The reaction is nonspontaneous at room temperature.

What is Go 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 Go determinations, the errors in Ho and So often tend to cancel each other out. Thus, to a reasonably good approximation, we can find Go at other temperatures by using the defining equation and assuming that Ho and So do not change significantly with temperature.

With this assumption:

Go = Ho – TSo

T = 75 + 273 = 348 K

Go = –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?



Phase Changes:

Consider the evaporation of water:

H2O(l) H2O(g)

Hfo(H2O(l)) = –285.8 kJ/mol

Hfo(H2O(g)) = –241.8 kJ/mol

So(H2O(l)) = 69.91 J/mol•K

So(H2O(g)) = 188.7 J/mol•K

Ho = [–241.8] – [–285.8] = 44.0 kJ

So = [188.7] – [69.91] = 118.8 J/K

Go = Ho – TSo = 44.0 – (298)(0.1188) = 8.6 kJ

At the normal boiling point of water:

Go = 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:

Ho = TSo

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:

This means that thermodynamic data can be used to estimate normal boiling points.


Example

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)

Hfo(C6H6(l)) = 48.99 kJ/mol

Hfo(C6H6(g)) = 82.93 kJ/mol

Hovap = [82.93] – [48.99] = 33.94 kJ

Using the calculated entropy of vaporization:

So(C6H6(l)) = 173.3 J/mol•K

So(C6H6(g)) = 269.2 J/mol•K

So = [269.2] – [173.3] = 95.9 J/K

Experimental value: 80.1 oC – Trouton's Rule is not great!