Paul Flowers; William R. Robinson, PhD; Richard Langley; Klaus Theopold

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 - T S

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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:

Δ G = Δ H - T Δ S

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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:

Δ S_{univ} = Δ S + \frac{q_{surr}}{T}

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The first law requires that q_surr = −q_sys, and at constant pressure q_sys = ΔH, and so this expression may be rewritten as the following:

Δ S_{univ} = Δ S - \frac{Δ H}{T}

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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:

− T Δ S_{univ} = Δ H - T Δ S

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

Δ G = − T Δ S_{univ}

16.32

The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, ΔS_univ. 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
ΔS_univ > 0	ΔG < 0	spontaneous
ΔS_univ < 0	ΔG > 0	nonspontaneous
ΔS_univ = 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.

Δ G ° = Δ H ° - T Δ S °

16.33

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:

H_{2} O (l) ⟶ H_{2} O (g)

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

Δ G_{298}^{°} = Δ H ° - T Δ S °

16.35

From Appendix G, here is the data:

Substance	$Δ H_{f}^{°} (kJ/mol)$	$S_{298}^{°} (J/K·mol)$
H₂O(l)	−286.83	70.0
H₂O(g)	−241.82	188.8

Combining at 298 K:

\begin{array}{l} Δ H ° = Δ H_{298}^{°} = Δ H_{f}^{°} (H_{2} O (g)) - Δ H_{f}^{°} (H_{2} O (l)) \\ = [−241.82 kJ - (−286.83)] kJ/mol = 44.01 kJ/mol \end{array}

16.36

\begin{matrix} Δ S ° = Δ S_{298}^{°} = S_{298}^{°} (H_{2} O (g)) - S_{298}^{°} (H_{2} O (l)) \\ = 188.8 J/mol·K - 70.0 J/K = 118.8 J/mol·K \end{matrix}

16.37

Δ G ° = Δ H ° - T Δ S °

16.38

Converting everything into kJ and combining at 298 K:

\begin{matrix} Δ G_{298}^{°} = Δ H ° - T Δ S ° \\ = 44.01 kJ/mol - (298 K \times 118.8 J/mol·K) \times \frac{1 kJ}{1000 J} \end{matrix}

16.39

44.01 kJ/mol - 35.4 kJ/mol = 9.6 kJ/mol

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At 298 K (25 °C) $Δ G_{298}^{°} > 0,$ and 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?

C_{2} H_{6} (g) ⟶ H_{2} (g) + C_{2} H_{4} (g)

16.41

Answer:

$Δ G_{298}^{°} = 102.0 kJ/mol;$ 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 $(Δ G_{f}^{°}),$ 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, $Δ G_{f}^{°}$ is by definition zero for elemental substances under standard state conditions. The approach used to calculate $Δ G^{°}$ for a reaction from $Δ G_{f}^{°}$ values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction

m A + n B ⟶ x C + y D,

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

\begin{matrix} Δ G_{298}^{°} = Δ G ° = \sum ν Δ G_{298}^{°} (products) - \sum ν Δ G_{298}^{°} (reactants) \\ = [x Δ G_{f}^{°} (C) + y Δ G_{f}^{°} (D)] - [m Δ G_{f}^{°} (A) + n Δ G_{f}^{°} (B)] . \end{matrix}

16.43

Example 16.8

Calculation of $Δ G_{298}^{°}$

Consider the decomposition of yellow mercury(II) oxide.

HgO (s, yellow) ⟶ Hg (l) + \frac{1}{2} O_{2} (g)

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Calculate the standard free energy change at room temperature, $Δ G_{298}^{°},$ 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	$Δ G_{f}^{°} (kJ/mol)$	$Δ H_{f}^{°} (kJ/mol)$	$S_{298}^{°} (J/K·mol)$
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:

Δ G_{298}^{°} = \sum ν G S_{f}^{°} (products) - \sum ν Δ G_{f}^{°} (reactants)

16.45

= [1 Δ G_{f}^{°} Hg (l) + \frac{1}{2} Δ G_{f}^{°} O_{2} (g)] - 1 Δ G_{f}^{°} HgO (s, yellow)

16.46

= [1 mol (0 kJ/mol) + \frac{1}{2} mol(0 kJ/mol)] - 1 mol(−58.43 kJ/mol) = 58.43 kJ/mol

16.47

(b) Using enthalpies and entropies of formation:

Δ H_{298}^{°} = \sum ν Δ H_{f}^{°} (products) - \sum ν Δ H_{f}^{°} (reactants)

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= [1 Δ H_{f}^{°} Hg (l) + \frac{1}{2} Δ H_{f}^{°} O_{2} (g)] - 1 Δ H_{f}^{°} HgO (s, yellow)

16.49

= [1 mol (0 kJ/mol) + \frac{1}{2} mol (0 kJ/mol)] - 1 mol (−90.46 kJ/mol) = 90.46 kJ/mol

16.50

Δ S_{298}^{°} = \sum ν Δ S_{298}^{°} (products) - \sum ν Δ S_{298}^{°} (reactants)

16.51

= [1 Δ S_{298}^{°} Hg (l) + \frac{1}{2} Δ S_{298}^{°} O_{2} (g)] - 1 Δ S_{298}^{°} HgO (s, yellow)

16.52

= [1 mol (75.9 J/mol K) + \frac{1}{2} mol (205.2 J/mol K)] - 1 mol (71.13 J/mol K) = 107.4 J/mol K

16.53

Δ G ° = Δ H ° - T Δ S ° = 90.46 kJ - 298.15 K \times 107.4 J/K·mol \times \frac{1 kJ}{1000 J}

16.54

Δ G ° = (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.

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?

C_{2} H_{4} (g) ⟶ H_{2} (g) + C_{2} H_{2} (g)

16.56

Answer:

(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:

Δ G = Δ H - T Δ S

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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:

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.
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.
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.
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.

A table with three columns and four rows is shown. The first column has the phrase, “Delta S greater than zero ( increase in entropy ),” in the third row and the phrase, “Delta S less than zero ( decrease in entropy),” in the fourth row. The second and third columns have the phrase, “Summary of the Four Scenarios for Enthalpy and Entropy Changes,” written above them. The second column has, “delta H greater than zero ( endothermic ),” in the second row, “delta G less than zero at high temperature, delta G greater than zero at low temperature, Process is spontaneous at high temperature,” in the third row, and “delta G greater than zero at any temperature, Process is nonspontaneous at any temperature,” in the fourth row. The third column has, “delta H less than zero ( exothermic ),” in the second row, “delta G less than zero at any temperature, Process is spontaneous at any temperature,” in the third row, and “delta G less than zero at low temperature, delta G greater than zero at high temperature, Process is spontaneous at low temperature.”

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) + O_{2} (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.

Check Your Learning

Popular chemical hand warmers generate heat by the air-oxidation of iron:

4Fe (s) + {3O}_{2} (g) ⟶ {2Fe}_{2} O_{3} (s)

16.59

How does the spontaneity of this process depend upon temperature?

Answer:

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:

Δ G = Δ H - T Δ S

16.60

y = b + m x

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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:

Δ G = 0 = Δ H - T Δ S

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T = \frac{Δ H}{Δ S}

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.

A graph is shown where the y-axis is labeled, “Free energy,” and the x-axis is labeled, “Increasing temperature ( K ).” The value of zero is written midway up the y-axis with the label, “delta G greater than 0,” written above this line and, “delta G less than 0,” written below it. The bottom half of the graph is labeled on the right as, “Spontaneous,” and the top half is labeled on the right as, “Nonspontaneous.” A green line labeled, “delta H less than 0, delta S greater than 0,” extends from a quarter of the way up the y-axis to the bottom right of the graph. A yellow line labeled, “delta H less than 0, delta S less than 0,” extends from a quarter of the way up the y-axis to the middle right of the graph. A second yellow line labeled, “delta H greater than 0, delta S greater than 0,” extends from three quarters of the way up the y-axis to the middle right of the graph. A red line labeled, “delta H greater than 0, delta S less than 0,” extends from three quarters of the way up the y-axis to the top right of the graph.

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:

H_{2} O (l) ⟶ H_{2} O (g)

16.64

When this process is at equilibrium, ΔG = 0, so the following is true:

0 = Δ H ° - T Δ S ° or T = \frac{Δ H °}{Δ S °}

16.65

Using the standard thermodynamic data from Appendix G,

\begin{matrix} Δ H ° & = & Δ H_{f}^{°} (H_{2} O (g)) - Δ H_{f}^{°} (H_{2} O (l)) \\ = & - 241.82 kJ/mol - (- 286.83 kJ/mol) = 44.01 kJ/mol \end{matrix}

16.66

\begin{matrix} Δ S ° & = & Δ S_{298}^{°} (H_{2} O (g)) - Δ S_{298}^{°} (H_{2} O (l)) \\ = & 188.8 J/K·mol - 70.0 J/K·mol = 118.8 J/K·mol \end{matrix}

16.67

T = \frac{Δ H °}{Δ S °} = \frac{44.01 \times 10^{3} J/mol}{118.8 J/K·mol} = 370.5 K = 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.

Check Your Learning

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

Answer:

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:

Δ G = Δ G ° + R T ln Q

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, $P_{N_{2}} = 0.870 atm,$ $P_{H_{2}} = 0.250 atm,$ and $P_{{NH}_{3}} = 12.9 atm$

{2NH}_{3} (g) ⟶ {3H}_{2} (g) + N_{2} (g) Δ G ° = 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:

Δ G = Δ G ° + R T ln Q = 33.0 \frac{kJ}{mol} + (8.314 \frac{J}{mol K} \times 298 K \times ln \frac{({0.250}^{3}) \times 0.870}{{12.9}^{2}}) = 9680 \frac{J}{mol} or 9.68 kJ/mol

16.71

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

Check Your Learning

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?

Answer:

ΔG = −47 kJ; yes

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

0 = Δ G ° + R T ln K (at equilibrium)

16.72

Δ G ° = − R T ln K or K = e^{- \frac{Δ G °}{R T}}

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
K	ΔG°	Comments
> 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, K_sp, for AgCl.

Solution

The reaction of interest is the following:

AgCl (s) ⇌ {Ag}^{+} (a q) + {Cl}^{−} (a q) K_{sp} = [{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:

\begin{matrix} Δ G ° = Δ G_{298}^{°} = [Δ G_{f}^{°} ({Ag}^{+} (a q)) + Δ G_{f}^{°} ({Cl}^{−} (a q))] - [Δ G_{f}^{°} (AgCl (s))] \\ = [77.1 kJ/mol - 131.2 kJ/mol] - [- 109.8 kJ/mol] = 55.7 kJ/mol \end{matrix}

16.75

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

K_{sp} = e^{- \frac{Δ G °}{R T}} = exp (- \frac{Δ G °}{R T}) = exp (- \frac{55.7 \times 10^{3} J/mol}{8.314 J/mol·K \times 298.15 K}) = exp (- 22.470) = e^{- 22.470} = 1.74 \times 10^{−10}

16.76

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

Check Your Learning

Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.

{2NO}_{2} (g) ⇌ N_{2} O_{4} (g)

16.77

Answer:

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 (Q ≠ K), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.

Three graphs, labeled, “a,” “b,” and “c” are shown where the y-axis is labeled, “Gibbs free energy ( G ),” and, “G superscript degree sign ( reactants ),” while the x-axis is labeled, “Reaction progress,” and “Reactants,” on the left and, “Products,” on the right. In graph a, a line begins at the upper left side and goes steadily down to a point about halfway up the y-axis and two thirds of the way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is slightly higher than halfway up the y-axis. The distance between the beginning and ending points of the graph is labeled as, “delta G less than 0,” while the lowest point on the graph is labeled, “Q equals K greater than 1.” In graph b, a line begins at the middle left side and goes steadily down to a point about two fifths up the y-axis and one third of the way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is near the top of the y-axis. The distance between the beginning and ending points of the graph is labeled as, “delta G greater than 0,” while the lowest point on the graph is labeled, “Q equals K less than 1.” In graph c, a line begins at the upper left side and goes steadily down to a point near the bottom of the y-axis and half way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is equal to the starting point on the y-axis which is labeled, “G superscript degree sign ( reactants ).” The lowest point on the graph is labeled, “Q equals K equals 1.” At the top of the graph is the label, “Delta G superscript degree sign equals 0.”

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.

16.4 Free Energy

Calculating Free Energy Change

Evaluation of ΔG° from ΔH° and ΔS°

Solution

Check Your Learning

Calculation of ΔG298°ΔG298°

Solution

Check Your Learning

Temperature Dependence of Spontaneity

Predicting the Temperature Dependence of Spontaneity

Solution

Check Your Learning

Equilibrium Temperature for a Phase Transition

Solution

Check Your Learning

Free Energy and Equilibrium

Calculating ΔG under Nonstandard Conditions

Solution

Check Your Learning

Calculating an Equilibrium Constant using Standard Free Energy Change

Solution

Check Your Learning

Calculation of $Δ G_{298}^{°}$