Chemistry: Atoms First 2e

# 12.4Free Energy

One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that it requires measurements of the entropy change for the system and the entropy change for the surroundings. An alternative approach involving a new thermodynamic property defined in terms of system properties only was introduced in the late nineteenth century by American mathematician Josiah Willard Gibbs. This new property is called the Gibbs free energy (G) (or simply the free energy), and it is defined in terms of a system’s enthalpy and entropy as the following:

$G=H−TSG=H−TS$

Free energy is a state function, and at constant temperature and pressure, the free energy change (ΔG) may be expressed as the following:

$ΔG=ΔH−TΔSΔG=ΔH−TΔS$

(For simplicity’s sake, the subscript “sys” will be omitted henceforth.)

The relationship between this system property and the spontaneity of a process may be understood by recalling the previously derived second law expression:

$ΔSuniv=ΔS+qsurrTΔSuniv=ΔS+qsurrT$

The first law requires that qsurr = −qsys, and at constant pressure qsys = ΔH, so this expression may be rewritten as:

$ΔSuniv=ΔS−ΔHTΔSuniv=ΔS−ΔHT$

Multiplying both sides of this equation by −T, and rearranging yields the following:

$−TΔSuniv=ΔH−TΔS−TΔSuniv=ΔH−TΔS$

Comparing this equation to the previous one for free energy change shows the following relation:

$ΔG=−TΔSunivΔG=−TΔSuniv$

The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, ΔSuniv. Table 12.3 summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.

Relation between Process Spontaneity and Signs of Thermodynamic Properties
ΔSuniv > 0 ΔG < 0 spontaneous
ΔSuniv < 0 ΔG > 0 nonspontaneous
ΔSuniv = 0 ΔG = 0 at equilibrium
Table 12.3

In addition to indicating spontaneity, the free energy change also provides information regarding the amount of useful work (w) that may be accomplished by a spontaneous process. Although a rigorous treatment of this subject is beyond the scope of an introductory chemistry text, a brief discussion is helpful for gaining a better perspective on this important thermodynamic property.

For this purpose, consider a spontaneous, exothermic process that involves a decrease in entropy. The free energy, as defined by

$ΔG=ΔH−TΔS ΔG=ΔH−TΔS$

may be interpreted as representing the difference between the energy produced by the process, ΔH, and the energy lost to the surroundings, TΔS. The difference between the energy produced and the energy lost is the energy available (or “free”) to do useful work by the process, ΔG. If the process somehow could be made to take place under conditions of thermodynamic reversibility, the amount of work that could be done would be maximal:

$ΔG= w max ΔG= w max$

However, as noted previously in this chapter, such conditions are not realistic. In addition, the technologies used to extract work from a spontaneous process (e.g., automobile engine, steam turbine) are never 100% efficient, and so the work done by these processes is always less than the theoretical maximum. Similar reasoning may be applied to a nonspontaneous process, for which the free energy change represents the minimum amount of work that must be done on the system to carry out the process.

### Calculating Free Energy Change

Free energy is a state function, so its value depends only on the conditions of the initial and final states of the system. A convenient and common approach to the calculation of free energy changes for physical and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes, ΔG°, according to the following relation:

$ΔG°=ΔH°−TΔS°ΔG°=ΔH°−TΔS°$

### Example 12.7

Using Standard Enthalpy and Entropy Changes to Calculate ΔG°Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the vaporization of water at room temperature (298 K). What does the computed value for ΔG° say about the spontaneity of this process?

Solution The process of interest is the following:

$H2O(l)⟶H2O(g)H2O(l)⟶H2O(g)$

The standard change in free energy may be calculated using the following equation:

$ΔG°=ΔH°−TΔS°ΔG°=ΔH°−TΔS°$

From Appendix G:

Substance $ΔHf°(kJ/mol)ΔHf°(kJ/mol)$ $S°(J/K·mol)S°(J/K·mol)$
H2O(l) −286.83 70.0
H2O(g) −241.82 188.8

Using the appendix data to calculate the standard enthalpy and entropy changes yields:

$Δ H ° = Δ H f ° ( H 2 O ( g ) ) − Δ H f ° ( H 2 O ( l ) ) = [ −241.82 kJ/mol − ( −286.83 ) ] kJ/mol = 45.01 kJ Δ H ° = Δ H f ° ( H 2 O ( g ) ) − Δ H f ° ( H 2 O ( l ) ) = [ −241.82 kJ/mol − ( −286.83 ) ] kJ/mol = 45.01 kJ$
$Δ S ° = 1 mol × S ° ( H 2 O ( g ) ) − 1 mol × S ° ( H 2 O ( l ) ) = (1 mol)188.8 J/mol·K − (1 mol) 70.0 J/mol K = 118.8 J/mol·K Δ S ° = 1 mol × S ° ( H 2 O ( g ) ) − 1 mol × S ° ( H 2 O ( l ) ) = (1 mol)188.8 J/mol·K − (1 mol) 70.0 J/mol K = 118.8 J/mol·K$
$ΔG°=ΔH°−TΔS°ΔG°=ΔH°−TΔS°$

Substitution into the standard free energy equation yields:

$ΔG°=ΔH°−TΔS°=45.01 kJ−(298 K×118.8J/K)×1 kJ1000 JΔG°=ΔH°−TΔS°=45.01 kJ−(298 K×118.8J/K)×1 kJ1000 J$
$45.01 kJ−35.4 kJ=9.6 kJ45.01 kJ−35.4 kJ=9.6 kJ$

At 298 K (25 °C) $ΔG°>0,ΔG°>0,$ so boiling is nonspontaneous (not spontaneous).

Check Your Learning Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for ΔG° say about the spontaneity of this process?

$C2H6(g)⟶H2(g)+C2H4(g)C2H6(g)⟶H2(g)+C2H4(g)$

$ΔG°=102.0 kJ/mol;ΔG°=102.0 kJ/mol;$ the reaction is nonspontaneous (not spontaneous) at 25 °C.

The standard free energy change for a reaction may also be calculated from standard free energy of formation ΔG°f values of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpy of formation, $ΔGf°ΔGf°$ is by definition zero for elemental substances under standard state conditions. The approach used to calculate $ΔG°ΔG°$ for a reaction from $ΔGf°ΔGf°$ values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction

$mA+nB⟶xC+yD,mA+nB⟶xC+yD,$

the standard free energy change at room temperature may be calculated as

$ΔG°=∑νΔG°(products)−∑νΔG°(reactants)=[xΔGf°(C)+yΔGf°(D)]−[mΔGf°(A)+nΔGf°(B)].ΔG°=∑νΔG°(products)−∑νΔG°(reactants)=[xΔGf°(C)+yΔGf°(D)]−[mΔGf°(A)+nΔGf°(B)].$

### Example 12.8

Using Standard Free Energies of Formation to Calculate ΔG° Consider the decomposition of yellow mercury(II) oxide.

$HgO(s,yellow)⟶Hg(l)+12O2(g)HgO(s,yellow)⟶Hg(l)+12O2(g)$

Calculate the standard free energy change at room temperature, $ΔG°,ΔG°,$ using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions?

Solution The required data are available in Appendix G and are shown here.

Compound $ΔGf°(kJ/mol)ΔGf°(kJ/mol)$ $ΔHf°(kJ/mol)ΔHf°(kJ/mol)$ $S°(J/K·mol)S°(J/K·mol)$
HgO (s, yellow) −58.43 −90.46 71.13
Hg(l) 0 0 75.9
O2(g) 0 0 205.2

(a) Using free energies of formation:

$ΔG°=∑νGf°(products)−∑νΔGf°(reactants)ΔG°=∑νGf°(products)−∑νΔGf°(reactants)$
$=[1ΔGf°Hg(l)+12ΔGf°O2(g)]−1ΔGf°HgO(s,yellow)=[1ΔGf°Hg(l)+12ΔGf°O2(g)]−1ΔGf°HgO(s,yellow)$
$=[1mol(0 kJ/mol)+12mol(0 kJ/mol)]−1 mol(−58.43 kJ/mol)=58.43 kJ/mol=[1mol(0 kJ/mol)+12mol(0 kJ/mol)]−1 mol(−58.43 kJ/mol)=58.43 kJ/mol$

(b) Using enthalpies and entropies of formation:

$ΔH°=∑νΔHf°(products)−∑νΔHf°(reactants)ΔH°=∑νΔHf°(products)−∑νΔHf°(reactants)$
$=[1ΔHf°Hg(l)+12ΔHf°O2(g)]−1ΔHf°HgO(s,yellow)=[1ΔHf°Hg(l)+12ΔHf°O2(g)]−1ΔHf°HgO(s,yellow)$
$=[1 mol(0 kJ/mol)+12mol(0 kJ/mol)]−1 mol(−90.46 kJ/mol)=90.46 kJ/mol=[1 mol(0 kJ/mol)+12mol(0 kJ/mol)]−1 mol(−90.46 kJ/mol)=90.46 kJ/mol$
$ΔS°=∑νΔS°(products)−∑νΔS°(reactants)ΔS°=∑νΔS°(products)−∑νΔS°(reactants)$
$=[1ΔS°Hg(l)+12ΔS°O2(g)]−1ΔS°HgO(s,yellow)=[1ΔS°Hg(l)+12ΔS°O2(g)]−1ΔS°HgO(s,yellow)$
$=[1 mol(75.9 J/mol K)+12mol(205.2 J/mol K)]−1 mol(71.13 J/mol K)=107.4 J/mol K=[1 mol(75.9 J/mol K)+12mol(205.2 J/mol K)]−1 mol(71.13 J/mol K)=107.4 J/mol K$
$ΔG°=ΔH°−TΔS°=90.46 kJ−298.15 K×107.4 J/K·mol×1 kJ1000 JΔG°=ΔH°−TΔS°=90.46 kJ−298.15 K×107.4 J/K·mol×1 kJ1000 J$
$ΔG°=(90.46−32.01)kJ/mol=58.45 kJ/molΔG°=(90.46−32.01)kJ/mol=58.45 kJ/mol$

Both ways to calculate the standard free energy change at 25 °C give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (not spontaneous) at room temperature.

Check Your Learning Calculate ΔG° using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?

$C2H4(g)⟶H2(g)+C2H2(g)C2H4(g)⟶H2(g)+C2H2(g)$

(a) 140.8 kJ/mol, nonspontaneous
(b) 141.5 kJ/mol, nonspontaneous

### 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:

$H 2 O(l)→ H 2 O(g) H 2 O(l)→ H 2 O(g)$

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:

$H2(g)+12O2(g)→H2O(g)ΔGf,gas°H2O(l)→H2(g)+12O2(g)−ΔGf,liquid°¯ H2O(l)→H2O(g)ΔG=ΔGf,gas°−ΔGf,liquid°H2(g)+12O2(g)→H2O(g)ΔGf,gas°H2O(l)→H2(g)+12O2(g)−ΔGf,liquid°¯ H2O(l)→H2O(g)ΔG=ΔGf,gas°−ΔGf,liquid°$

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°:

$ZnS(s)→Zn(s)+S(s)ΔG1° =201.3kJ ZnS(s)→Zn(s)+S(s)ΔG1° =201.3kJ$

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

$S(s)+ O 2 (g)→ SO 2 (g)ΔG2° =−300.1kJ S(s)+ O 2 (g)→ SO 2 (g)ΔG2° =−300.1kJ$

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

$ZnS(s)+ O 2 (g)→Zn(s)+SO(g)ΔG°=201.3kJ+−300.1kJ=−98.8kJ ZnS(s)+ O 2 (g)→Zn(s)+SO(g)ΔG°=201.3kJ+−300.1kJ=−98.8kJ$

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.

### Example 12.9

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?

SolutionFollowing the approach outlined above and using free energy values from Appendix G:

$Decomposition of zinc sulfide: Zn(s)→Zn(s)+S(s) ΔG1° =201.3kJ Formation of hydrogen sulfide: S(s)+ H 2 (g)→ H 2 S(g) ΔG2° =−33.4kJ Coupled reaction: ZnS(s)+ H 2 (g)→Zn(s)+ H 2 S(g) ΔG°=201.3kJ+−33.4kJ=167.9kJ Decomposition of zinc sulfide: Zn(s)→Zn(s)+S(s) ΔG1° =201.3kJ Formation of hydrogen sulfide: S(s)+ H 2 (g)→ H 2 S(g) ΔG2° =−33.4kJ Coupled reaction: ZnS(s)+ H 2 (g)→Zn(s)+ H 2 S(g) ΔG°=201.3kJ+−33.4kJ=167.9kJ$

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?

$FeS(s)+ O 2 (g)→Fe(s)+ SO 2 (g) FeS(s)+ O 2 (g)→Fe(s)+ SO 2 (g)$

–199.7 kJ; spontaneous