Free Energy Changes for Coupled Reactions

The use of free energies of formation to compute free energy changes for reactions as described above is possible because ΔG is a state function, and the approach is analogous to the use of Hess’ Law in computing enthalpy changes (see the chapter on thermochemistry). Consider the vaporization of water as an example:

$$H_2O(l)⟶H_2O(g)$$

An equation representing this process may be derived by adding the formation reactions for the two phases of water (necessarily reversing the reaction for the liquid phase). The free energy change for the sum reaction is the sum of free energy changes for the two added reactions:

$$H_2(g)+\frac{1}{2}O_2(g)⟶H_2O(g)\qquad ΔG_f^°gas$$ $$H_2O(l)⟶H_2(g)+\frac{1}{2}O_2(g)\qquad ΔG_f^°liquid$$ $$H_2O(l)⟶H_2O(g)\qquad ΔG^°gas=ΔG_f^°gas-ΔG_f^°liquid$$

This approach may also be used in cases where a nonspontaneous reaction is enabled by coupling it to a spontaneous reaction. For example, the production of elemental zinc from zinc sulfide is thermodynamically unfavorable, as indicated by a positive value for ΔG°:

$$ZnS(s)⟶Zn(s)+S(s)\qquad ΔG_1^°=201.3\;kJ$$

The industrial process for production of zinc from sulfidic ores involves coupling this decomposition reaction to the thermodynamically favorable oxidation of sulfur:

$$S(s)+O_2(g)⟶SO_2(g)\qquad ΔG_2^°=-300.1\;kJ$$

The coupled reaction exhibits a negative free energy change and is spontaneous:

$$ZnS(s)+O_2(g)⟶Zn(s)+SO_2(g)\qquad ΔG^°=201.3\;kJ+-300.1\;kJ=-98.8\;kJ$$

This process is typically carried out at elevated temperatures, so this result obtained using standard free energy values is just an estimate. The gist of the calculation, however, holds true.

Calculating Free Energy Change for a Coupled Reaction
Is a reaction coupling the decomposition of ZnS to the formation of H2S expected to be spontaneous under standard conditions?

Solution
Following the approach outlined above and using free energy values from the Appendix:

$$\text{Decomposition of zinc sulfide:}\quad ZnS(s)⟶Zn(s)+S(s)\quad ΔG_1^°=201.3\;kJ$$ $$\text{Formation of hydrogen sulfide:}\quad S(s)+H_2(g)⟶H_2S(g)\quad ΔG_2^°=-33.4\;kJ$$ $$\text{Coupled reaction:}\quad ZnS(s)+H_2(g)⟶Zn(s)+H_2S(g)\quad ΔG^°=201.3\;kJ+-33.4\;kJ=167.9\;kJ$$

The coupled reaction exhibits a positive free energy change and is thus nonspontaneous.

Check Your Learning
What is the standard free energy change for the reaction below? Is the reaction expected to be spontaneous under standard conditions?

$$FeS(s)+O_2(g)⟶Fe(s)+SO_2(g)$$
Answer:

−199.7 kJ; spontaneous