Chemistry

16.4Free Energy

Chemistry16.4 Free Energy

Learning Objectives

By the end of this section, you will be able to:
• Define Gibbs free energy, and describe its relation to spontaneity
• Calculate free energy change for a process using free energies of formation for its reactants and products
• Calculate free energy change for a process using enthalpies of formation and the entropies for its reactants and products
• Explain how temperature affects the spontaneity of some processes
• Relate standard free energy changes to equilibrium constants

One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that we must determine 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 change (G) (or simply the free energy), and it is defined in terms of a system’s enthalpy and entropy as the following:

$G=H−TSG=H−TS$
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Free energy is a state function, and at constant temperature and pressure, the standard free energy change (ΔG°) may be expressed as the following:

$ΔG=ΔH−TΔSΔG=ΔH−TΔS$
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(For simplicity’s sake, the subscript “sys” will be omitted henceforth.)

We can understand the relationship between this system property and the spontaneity of a process by recalling the previously derived second law expression:

$ΔSuniv=ΔS+qsurrTΔSuniv=ΔS+qsurrT$
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The first law requires that qsurr = −qsys, and at constant pressure qsys = ΔH, and so this expression may be rewritten as the following:

$ΔSuniv=ΔS−ΔHTΔSuniv=ΔS−ΔHT$
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ΔH is the enthalpy change of the system. Multiplying both sides of this equation by −T, and rearranging yields the following:

$−TΔSuniv=ΔH−TΔS−TΔSuniv=ΔH−TΔS$
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Comparing this equation to the previous one for free energy change shows the following relation:

$ΔG=−TΔSunivΔG=−TΔSuniv$
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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 16.3 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 reversible (at equilibrium)
Table 16.3

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 according to the following relation as demonstrated in Example 16.7.

$ΔG°=ΔH°−TΔS°ΔG°=ΔH°−TΔS°$
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Example 16.7

Evaluation of ΔG° from ΔH° and ΔS°

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:
$H2O(l)⟶H2O(g)H2O(l)⟶H2O(g)$
16.34

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

$ΔG298°=ΔH°−TΔS°ΔG298°=ΔH°−TΔS°$
16.35

From Appendix G, here is the data:

Substance $ΔHf°(kJ/mol)ΔHf°(kJ/mol)$ $S298°(J/K·mol)S298°(J/K·mol)$
H2O(l) −286.83 70.0
H2O(g) −241.82 188.8

Combining at 298 K:

$ΔH°=ΔH298°=ΔHf°(H2O(g))−ΔHf°(H2O(l))=[−241.82 kJ−(−286.83)]kJ/mol=44.01 kJ/molΔH°=ΔH298°=ΔHf°(H2O(g))−ΔHf°(H2O(l))=[−241.82 kJ−(−286.83)]kJ/mol=44.01 kJ/mol$
16.36
$ΔS°=ΔS298°=S298°(H2O(g))−S298°(H2O(l))=188.8J/mol·K−70.0J/K=118.8J/mol·KΔS°=ΔS298°=S298°(H2O(g))−S298°(H2O(l))=188.8J/mol·K−70.0J/K=118.8J/mol·K$
16.37
$ΔG°=ΔH°−TΔS°ΔG°=ΔH°−TΔS°$
16.38

Converting everything into kJ and combining at 298 K:

$ΔG298°=ΔH°−TΔS°=44.01 kJ/mol−(298 K×118.8J/mol·K)×1 kJ1000 JΔG298°=ΔH°−TΔS°=44.01 kJ/mol−(298 K×118.8J/mol·K)×1 kJ1000 J$
16.39
$44.01 kJ/mol−35.4 kJ/mol=9.6 kJ/mol44.01 kJ/mol−35.4 kJ/mol=9.6 kJ/mol$
16.40

At 298 K (25 °C) $ΔG298°>0,ΔG298°>0,$ and so boiling is nonspontaneous (not spontaneous).

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?
$C2H6(g)⟶H2(g)+C2H4(g)C2H6(g)⟶H2(g)+C2H4(g)$
16.41

$ΔG298°=102.0 kJ/mol;ΔG298°=102.0 kJ/mol;$ 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 $(ΔGf°),(ΔGf°),$ 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

$mA+nB⟶xC+yD,mA+nB⟶xC+yD,$
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the standard free energy change at room temperature may be calculated as

$ΔG298°=ΔG°=∑νΔG298°(products)−∑νΔG298°(reactants)=[xΔGf°(C)+yΔGf°(D)]−[mΔGf°(A)+nΔGf°(B)].ΔG298°=ΔG°=∑νΔG298°(products)−∑νΔG298°(reactants)=[xΔGf°(C)+yΔGf°(D)]−[mΔGf°(A)+nΔGf°(B)].$
16.43

Example 16.8

Calculation of $ΔG298°ΔG298°$

Consider the decomposition of yellow mercury(II) oxide.
$HgO(s,yellow)⟶Hg(l)+12O2(g)HgO(s,yellow)⟶Hg(l)+12O2(g)$
16.44

Calculate the standard free energy change at room temperature, $ΔG298°,ΔG298°,$ 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 $ΔGf°(kJ/mol)ΔGf°(kJ/mol)$ $ΔHf°(kJ/mol)ΔHf°(kJ/mol)$ $S298°(J/K·mol)S298°(J/K·mol)$
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:

$ΔG298°=∑νGSf°(products)−∑νΔGf°(reactants)ΔG298°=∑νGSf°(products)−∑νΔGf°(reactants)$
16.45
$=[1ΔGf°Hg(l)+12ΔGf°O2(g)]−1ΔGf°HgO(s,yellow)=[1ΔGf°Hg(l)+12ΔGf°O2(g)]−1ΔGf°HgO(s,yellow)$
16.46
$=[1mol(0 kJ/mol)+12mol(0 kJ/mol)]−1 mol(−58.43 kJ/mol)=58.43 kJ/mol=[1mol(0 kJ/mol)+12mol(0 kJ/mol)]−1 mol(−58.43 kJ/mol)=58.43 kJ/mol$
16.47

(b) Using enthalpies and entropies of formation:

$ΔH298°=∑νΔHf°(products)−∑νΔHf°(reactants)ΔH298°=∑νΔHf°(products)−∑νΔHf°(reactants)$
16.48
$=[1ΔHf°Hg(l)+12ΔHf°O2(g)]−1ΔHf°HgO(s,yellow)=[1ΔHf°Hg(l)+12ΔHf°O2(g)]−1ΔHf°HgO(s,yellow)$
16.49
$=[1 mol(0 kJ/mol)+12mol(0 kJ/mol)]−1 mol(−90.46 kJ/mol)=90.46 kJ/mol=[1 mol(0 kJ/mol)+12mol(0 kJ/mol)]−1 mol(−90.46 kJ/mol)=90.46 kJ/mol$
16.50
$ΔS298°=∑νΔS298°(products)−∑νΔS298°(reactants)ΔS298°=∑νΔS298°(products)−∑νΔS298°(reactants)$
16.51
$=[1ΔS298°Hg(l)+12ΔS298°O2(g)]−1ΔS298°HgO(s,yellow)=[1ΔS298°Hg(l)+12ΔS298°O2(g)]−1ΔS298°HgO(s,yellow)$
16.52
$=[1 mol(75.9 J/mol K)+12mol(205.2 J/mol K)]−1 mol(71.13 J/mol K)=107.4 J/mol K=[1 mol(75.9 J/mol K)+12mol(205.2 J/mol K)]−1 mol(71.13 J/mol K)=107.4 J/mol K$
16.53
$ΔG°=ΔH°−TΔS°=90.46 kJ−298.15 K×107.4 J/K·mol×1 kJ1000 JΔG°=ΔH°−TΔS°=90.46 kJ−298.15 K×107.4 J/K·mol×1 kJ1000 J$
16.54
$ΔG°=(90.46−32.01)kJ/mol=58.45 kJ/molΔG°=(90.46−32.01)kJ/mol=58.45 kJ/mol$
16.55

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.

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?
$C2H4(g)⟶H2(g)+C2H2(g)C2H4(g)⟶H2(g)+C2H2(g)$
16.56

(a) 140.8 kJ/mol, nonspontaneous

(b) 141.5 kJ/mol, nonspontaneous

Temperature Dependence of Spontaneity

As was previously demonstrated in this chapter’s section on entropy, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:

$ΔG=ΔH−TΔSΔG=ΔH−TΔS$
16.57

The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:

1. Both ΔH and ΔS are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is greater than ΔH. If the TΔS term is less than ΔH, the free energy change will be positive. Such a process is spontaneous at high temperatures and nonspontaneous at low temperatures.
2. Both ΔH and ΔS are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is less than ΔH. If the TΔS term’s magnitude is greater than ΔH, the free energy change will be positive. Such a process is spontaneous at low temperatures and nonspontaneous at high temperatures.
3. ΔH is positive and ΔS is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔG will be positive regardless of the temperature. Such a process is nonspontaneous at all temperatures.
4. ΔH is negative and ΔS is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔG will be negative regardless of the temperature. Such a process is spontaneous at all temperatures.

These four scenarios are summarized in Figure 16.12.

Figure 16.12 There are four possibilities regarding the signs of enthalpy and entropy changes.

Example 16.9

Predicting the Temperature Dependence of Spontaneity

The incomplete combustion of carbon is described by the following equation:
$2C(s)+O2(g)⟶2CO(g)2C(s)+O2(g)⟶2CO(g)$
16.58

How does the spontaneity of this process depend upon temperature?

Solution

Combustion processes are exothermic (ΔH < 0). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, ΔS > 0). The reaction is therefore spontaneous (ΔG < 0) at all temperatures.

Popular chemical hand warmers generate heat by the air-oxidation of iron:
$4Fe(s)+3O2(g)⟶2Fe2O3(s)4Fe(s)+3O2(g)⟶2Fe2O3(s)$
16.59

How does the spontaneity of this process depend upon temperature?

ΔH and ΔS are negative; the reaction is spontaneous at low temperatures.

When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms “high” and “low” mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is nonspontaneous at one temperature but spontaneous at another will necessarily undergo a change in “spontaneity” (as reflected by its ΔG) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which ΔG is plotted on the y axis versus T on the x axis:

$ΔG=ΔH−TΔSΔG=ΔH−TΔS$
16.60
$y=b+mxy=b+mx$
16.61

Such a plot is shown in Figure 16.13. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative ΔG) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the x-intercept of the line, that is, the value of T for which ΔG is zero:

$ΔG=0=ΔH−TΔSΔG=0=ΔH−TΔS$
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$T=ΔHΔST=ΔHΔS$
16.63

And so, saying a process is spontaneous at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which ΔG for the process is zero. As noted earlier, the condition of ΔG = 0 describes a system at equilibrium.

Figure 16.13 These plots show the variation in ΔG with temperature for the four possible combinations of arithmetic sign for ΔH and ΔS.

Example 16.10

Equilibrium Temperature for a Phase Transition

As defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its liquid and gaseous phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Appendix G to estimate the boiling point of water.

Solution

The process of interest is the following phase change:
$H2O(l)⟶H2O(g)H2O(l)⟶H2O(g)$
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When this process is at equilibrium, ΔG = 0, so the following is true:

$0=ΔH°−TΔS°orT=ΔH°ΔS°0=ΔH°−TΔS°orT=ΔH°ΔS°$
16.65

Using the standard thermodynamic data from Appendix G,

$ΔH°=ΔHf°(H2O(g))−ΔHf°(H2O(l))=−241.82 kJ/mol−(−286.83 kJ/mol)=44.01 kJ/molΔH°=ΔHf°(H2O(g))−ΔHf°(H2O(l))=−241.82 kJ/mol−(−286.83 kJ/mol)=44.01 kJ/mol$
16.66
$ΔS°=ΔS298°(H2O(g))−ΔS298°(H2O(l))=188.8 J/K·mol−70.0 J/K·mol=118.8 J/K·molΔS°=ΔS298°(H2O(g))−ΔS298°(H2O(l))=188.8 J/K·mol−70.0 J/K·mol=118.8 J/K·mol$
16.67
$T=ΔH°ΔS°=44.01×103J/mol118.8J/K·mol=370.5K=97.3°CT=ΔH°ΔS°=44.01×103J/mol118.8J/K·mol=370.5K=97.3°C$
16.68

The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Appendix G). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.

Use the information in Appendix G to estimate the boiling point of CS2.

313 K (accepted value 319 K)

Free Energy and Equilibrium

The free energy change for a process may be viewed as a measure of its driving force. A negative value for ΔG represents a finite driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When ΔG is zero, the forward and reverse driving forces are equal, and so the process occurs in both directions at the same rate (the system is at equilibrium).

In the chapter on equilibrium the reaction quotient, Q, was introduced as a convenient measure of the status of an equilibrium system. Recall that Q is the numerical value of the mass action expression for the system, and that you may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When Q is lesser than the equilibrium constant, K, the reaction will proceed in the forward direction until equilibrium is reached and Q = K. Conversely, if Q > K, the process will proceed in the reverse direction until equilibrium is achieved.

The free energy change for a process taking place with reactants and products present under nonstandard conditions, ΔG, is related to the standard free energy change, ΔG°, according to this equation:

$ΔG=ΔG°+RTlnQΔG=ΔG°+RTlnQ$
16.69

R is the gas constant (8.314 J/K mol), T is the kelvin or absolute temperature, and Q is the reaction quotient. We may use this equation to predict the spontaneity for a process under any given set of conditions as illustrated in Example 16.11.

Example 16.11

Calculating ΔG under Nonstandard Conditions

What is the free energy change for the process shown here under the specified conditions?

T = 25 °C, $PN2=0.870 atm,PN2=0.870 atm,$ $PH2=0.250 atm,PH2=0.250 atm,$ and $PNH3=12.9 atmPNH3=12.9 atm$

$2NH3(g)⟶3H2(g)+N2(g)ΔG°=33.0 kJ/mol2NH3(g)⟶3H2(g)+N2(g)ΔG°=33.0 kJ/mol$
16.70

Solution

The equation relating free energy change to standard free energy change and reaction quotient may be used directly:
$ΔG=ΔG°+RTlnQ=33.0kJmol+(8.314Jmol K×298 K×ln(0.2503)×0.87012.92)=9680Jmolor 9.68 kJ/molΔG=ΔG°+RTlnQ=33.0kJmol+(8.314Jmol K×298 K×ln(0.2503)×0.87012.92)=9680Jmolor 9.68 kJ/mol$
16.71

Since the computed value for ΔG is positive, the reaction is nonspontaneous under these conditions.

Calculate the free energy change for this same reaction at 875 °C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?

ΔG = −47 kJ; yes

For a system at equilibrium, Q = K and ΔG = 0, and the previous equation may be written as

$0=ΔG°+RTlnK(at equilibrium)0=ΔG°+RTlnK(at equilibrium)$
16.72
$ΔG°=−RTlnKorK=e−ΔG°RTΔG°=−RTlnKorK=e−ΔG°RT$
16.73

This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in Table 16.4.

Relations between Standard Free Energy Changes and Equilibrium Constants
> 1 < 0 Products are more abundant at equilibrium.
< 1 > 0 Reactants are more abundant at equilibrium.
= 1 = 0 Reactants and products are equally abundant at equilibrium.
Table 16.4

Example 16.12

Calculating an Equilibrium Constant using Standard Free Energy Change

Given that the standard free energies of formation of Ag+(aq), Cl(aq), and AgCl(s) are 77.1 kJ/mol, −131.2 kJ/mol, and −109.8 kJ/mol, respectively, calculate the solubility product, Ksp, for AgCl.

Solution

The reaction of interest is the following:
$AgCl(s)⇌Ag+(aq)+Cl−(aq)Ksp=[Ag+][Cl−]AgCl(s)⇌Ag+(aq)+Cl−(aq)Ksp=[Ag+][Cl−]$
16.74

The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products:

$ΔG°=ΔG298°=[ΔGf°(Ag+(aq))+ΔGf°(Cl−(aq))]−[ΔGf°(AgCl(s))]=[77.1 kJ/mol−131.2 kJ/mol]−[−109.8 kJ/mol]=55.7 kJ/molΔG°=ΔG298°=[ΔGf°(Ag+(aq))+ΔGf°(Cl−(aq))]−[ΔGf°(AgCl(s))]=[77.1 kJ/mol−131.2 kJ/mol]−[−109.8 kJ/mol]=55.7 kJ/mol$
16.75

The equilibrium constant for the reaction may then be derived from its standard free energy change:

$Ksp=e−ΔG°RT=exp(−ΔG°RT)=exp(−55.7×103J/mol8.314J/mol·K×298.15K)=exp(−22.470)=e−22.470=1.74×10−10Ksp=e−ΔG°RT=exp(−ΔG°RT)=exp(−55.7×103J/mol8.314J/mol·K×298.15K)=exp(−22.470)=e−22.470=1.74×10−10$
16.76

This result is in reasonable agreement with the value provided in Appendix J.

Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.
$2NO2(g)⇌N2O4(g)2NO2(g)⇌N2O4(g)$
16.77

K = 6.9

To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of Q), equilibrium is established when the system’s free energy is minimized (Figure 16.14). If a system consists of reactants and products in nonequilibrium amounts (QK), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.

Figure 16.14 These plots show the free energy versus reaction progress for systems whose standard free energy changes are (a) negative, (b) positive, and (c) zero. Nonequilibrium systems will proceed spontaneously in whatever direction is necessary to minimize free energy and establish equilibrium.
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