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9.3: Free Energy

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    431429
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    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 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:

    \[ G=H - TS \]

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

    \[ \Delta G= \Delta H - T \Delta S \]

    (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:

    \[ \Delta S_{\text {univ }}=\Delta S+\frac{q_{\mathrm{surr}}}{T} \]

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

    \[ \Delta S_{\text {univ }}=\Delta S - \frac{\Delta H}{T} \]

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

    \[ -T \Delta S_{\text {univ }}=\Delta H - T \Delta S \]

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

    \[ \Delta G = - T \Delta S_{\text {univ }} \]

    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 at equilibrium
    Table 16.3

    What’s “Free” about ΔG?

    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

    \[ \Delta G = \Delta H- T \Delta S \]

    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:

    \[ \Delta G = w_{\text {max}} \]

    where \( w_{\text {max}} \) refers to all types of work except expansion (pressure-volume) work.

    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., batteries) 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 16.7

    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) −285.83 70.0
    H2O(g) −241.82 188.8

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

    ΔH°=ΔHf°(H2O(g))ΔHf°(H2O(l))=[−241.82 kJ/mol(−285.83)]kJ/mol=44.01 kJΔH°=ΔHf°(H2O(g))ΔHf°(H2O(l))=[−241.82 kJ/mol(−285.83)]kJ/mol=44.01 kJ
    ΔS°=1mol×S°(H2O(g))1mol×S°(H2O(l))=(1 mol)188.8J/mol·K(1 mol)70.0J/mol K=118.8J/KΔS°=1mol×S°(H2O(g))1mol×S°(H2O(l))=(1 mol)188.8J/mol·K(1 mol)70.0J/mol K=118.8J/K
    ΔG°=ΔH°TΔS°ΔG°=ΔH°TΔS°

    Substitution into the standard free energy equation yields:

    ΔG°=ΔH°TΔS°=44.01 kJ(298 K×118.8J/K)×1 kJ1000 JΔG°=ΔH°TΔS°=44.01 kJ(298 K×118.8J/K)×1 kJ1000 J
    44.01 kJ35.4 kJ=8.6 kJ44.01 kJ35.4 kJ=8.6 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)

    Answer:

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

    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:

    \[ \Delta G = \Delta H- T \Delta S \]

    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.

    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 16.10

    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)

    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)+3O2(g)2Fe2O3(s)4Fe(s)+3O2(g)2Fe2O3(s)

    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:

    \begin{aligned}
    &\Delta G=\Delta H-T \Delta S\\
    &y=b+m x
    \end{aligned}

    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

    \begin{aligned}
    &\Delta G=0=\Delta H-T \Delta S\\
    &T=\frac{\Delta H}{\Delta S}
    \end{aligned}

    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 16.11

    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)

    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°

    Using the standard thermodynamic data from Appendix G,

    ΔH°=1mol×ΔHf°(H2O(g))1mol×ΔHf°(H2O(l))=(1 mol)241.82 kJ/mol(1 mol)(286.83 kJ/mol)=44.01 kJΔH°=1mol×ΔHf°(H2O(g))1mol×ΔHf°(H2O(l))=(1 mol)241.82 kJ/mol(1 mol)(286.83 kJ/mol)=44.01 kJ
    ΔS°=1mol×ΔS°(H2O(g))1mol×ΔS°(H2O(l))=(1 mol)188.8 J/K·mol(1 mol)70.0 J/K·mol=118.8 J/KΔS°=1mol×ΔS°(H2O(g))1mol×ΔS°(H2O(l))=(1 mol)188.8 J/K·mol(1 mol)70.0 J/K·mol=118.8 J/K
    T=ΔH°ΔS°=44.01×103J118.8J/K=370.5K=97.3°CT=ΔH°ΔS°=44.01×103J118.8J/K=370.5K=97.3°C

    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)

    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° Composition of an Equilibrium Mixture
    > 1 < 0 Products are more abundant
    < 1 > 0 Reactants are more abundant
    = 1 = 0 Reactants and products are comparably abundant
    Table 16.4

    Example 16.13

    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]

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

    ΔG°=[ΔGf°(Ag+(aq))+ΔGf°(Cl(aq))][ΔGf°(AgCl(s))]=[77.1 kJ/mol131.2 kJ/mol][109.8 kJ/mol]=55.7 kJ/molΔG°=[ΔGf°(Ag+(aq))+ΔGf°(Cl(aq))][ΔGf°(AgCl(s))]=[77.1 kJ/mol131.2 kJ/mol][109.8 kJ/mol]=55.7 kJ/mol

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

    Ksp=eΔG°RT=exp(ΔG°RT)=exp(55.7×103J/mol8.314J/mol·K×298.15K)=exp(22.470)=e22.470=1.74×1010Ksp=eΔG°RT=exp(ΔG°RT)=exp(55.7×103J/mol8.314J/mol·K×298.15K)=exp(22.470)=e22.470=1.74×10−10

    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.
    2NO2(g)N2O4(g)2NO2(g)N2O4(g)

    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 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.

    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.

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