Free Energy

D. Latimer; D. Vanderwel; J. Hollett

73 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

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:

ΔG = ΔH – TΔS

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:

$\Delta {S}_{\text{univ}}=\Delta S+\phantom{\rule{0.2em}{0ex}}\frac{{q}_{\text{surr}}}{T}$

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

$\Delta {S}_{\text{univ}}=\Delta S-\phantom{\rule{0.1em}{0ex}}\frac{\Delta H}{T}$

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

$\text{−}T\Delta {S}_{\text{univ}}=\Delta H-T\Delta S$

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

$\Delta G=\text{−}T\Delta{S}_{\text{univ}}$

The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, ΔS_univ. (Figure) 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	at equilibrium

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:

$\Delta G={w}_{\mathrm{max}}$

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.

$\Delta G\text{°}=\Delta H\text{°}-T\Delta S\text{°}$

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)

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

$\Delta G\text{°}=\Delta H\text{°}-T\Delta S\text{°}$

From Appendix G:

Substance	$\Delta {H}_{\text{f}}^{°}\text{(kJ/mol)}$	$S\text{°}\text{(J/K·mol)}$
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:

$\begin{array}{l}\Delta H\text{°}=\Delta {H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(g\right)\right)\phantom{\rule{0.2em}{0ex}}-\Delta {H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(l\right)\right)\\ =\left[\text{−241.82 kJ/mol}-\left(\text{−286.83}\right)\right]\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}=\text{45.01 kJ}\end{array}$

ΔS^o = 1 mol x S^o(H₂O(g)) – 1 mol x S^o(H₂O(l))

=(1 mol)(188.8 J mol^-1 K^-1) – (1 mol)(70.0 J mol^-1 K^-1) = 118.8 J K^-1

Substitution into the standard free energy equation yields:

$\begin{array}{c}\Delta G\text{°}=\Delta H\text{°}-T\Delta S\text{°}\\ \\ =\text{45.01 kJ}-\left(\text{298 K}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}118.8\phantom{\rule{0.2em}{0ex}}\text{J/K}\right)\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\frac{\text{1 kJ}}{\text{1000 J}}\end{array}$

$\text{45.01 kJ}-\text{35.4 kJ}=\text{9.6 kJ}$

At 298 K (25 °C) $\Delta G\text{°}>0,$ 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)

Answer: ΔG^o = 102.0 kJ mol^-1; 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, ΔG_f° is by definition zero for elemental substances under standard state conditions. The approach used to calculate ΔG^o for a reaction from ΔG_f° values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction

mA + nB → xC + yD,

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

$\begin{array}{}\\ \\ \Delta G\text{°}=\sum \nu \Delta G\text{°}\left(\text{products}\right)\phantom{\rule{0.2em}{0ex}}-\sum \nu \text{Δ}G\text{°}\left(\text{reactants}\right)\\ \\ =\left[x\Delta {G}_{\text{f}}^{°}\left(\text{C}\right)+y\Delta {G}_{\text{f}}^{°}\left(\text{D}\right)\right]\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\left[m\Delta {G}_{\text{f}}^{°}\left(\text{A}\right)+n\Delta {G}_{\text{f}}^{°}\left(\text{B}\right)\right].\end{array}$

Using Standard Free Energies of Formation to Calculate ΔG°

Consider the decomposition of yellow mercury(II) oxide.

HgO(s, yellow) → Hg(l) +½ O₂(g)

Calculate the standard free energy change at room temperature, ΔG^o, 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	ΔG_f^o (kJ mol^-1)	ΔH_f^o (kJ mol^-1)	S^o (J K^-1 mol^-1)
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:

$\Delta G\text{°}=\sum \nu G{S}_{f}^{°}\text{(products)}\phantom{\rule{0.2em}{0ex}}-\sum \nu \Delta {G}_{f}^{°}\text{(reactants)}$

$=\left[1\Delta {G}_{f}^{°}\text{Hg}\left(l\right)+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\Delta {G}_{f}^{°}{\text{O}}_{\text{2}}\left(g\right)\right]\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}1\Delta {G}_{f}^{°}\text{HgO}\left(s,\phantom{\rule{0.2em}{0ex}}\text{yellow}\right)$

$=\left[1\phantom{\rule{0.2em}{0ex}}\text{mol}\text{(0 kJ/mol)}+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{mol(0 kJ/mol)}\right]\phantom{\rule{0.2em}{0ex}}-\text{1 mol(−58.43 kJ/mol)}=\text{58.43 kJ/mol}$

(b) Using enthalpies and entropies of formation:

$\Delta H\text{°}=\sum \nu \Delta H}_{\text{f}}^{°}\text{(products)}\phantom{\rule{0.2em}{0ex}}-\sum \nu \Delta H}_{\text{f}}^{°}\text{(reactants)}$

$=\left[1\Delta {H}_{f}^{°}\text{Hg}\left(l\right)+\frac{1}{2}\Delta {H}_{f}^{°}{\text{O}}_{2}\left(g\right)\right]\phantom{\rule{0.2em}{0ex}}-1\Delta {H}_{f}^{°}\text{HgO}\left(s,\phantom{\rule{0.2em}{0ex}}\text{yellow}\right)$

$=\left[\text{1 mol}\left(\text{0 kJ/mol}\right)\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{mol}\left(\text{0 kJ/mol}\right)\right]\phantom{\rule{0.2em}{0ex}}-\text{1 mol}\left(\text{−90.46 kJ/mol}\right)=\text{90.46 kJ/mol}$

$\Delta S\text{°}=\sum \text{ν}\Delta S\text{°}\text{(products)}-\sum \text{ν}\Delta S\text{°}\text{(reactants)}$

$=\left[1\Delta S\text{°}\text{Hg}\left(l\right)\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\Delta S\text{°}{\text{O}}_{2}\left(g\right)\right]\phantom{\rule{0.2em}{0ex}}-1\Delta S\text{°}\text{HgO}\left(s,\phantom{\rule{0.2em}{0ex}}\text{yellow}\right)$

$=\left[\text{1 mol}\phantom{\rule{0.2em}{0ex}}\left(\text{75.9 J/mol K}\right)\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{mol}\left(\text{205.2 J/mol K}\right)\right]\phantom{\rule{0.2em}{0ex}}-\text{1 mol}\left(\text{71.13 J/mol K}\right)=\text{107.4 J/mol K}$

$\Delta G\text{°}=\Delta H\text{°}-T\Delta S\text{°}=\text{90.46 kJ}-\text{298.15 K}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{107.4 J/K·mol}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\frac{\text{1 kJ}}{\text{1000 J}}$

$\Delta G\text{°}=\left(90.46-32.01\right)\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}=\text{58.45 kJ/mol}$

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?

${\text{C}}_{2}{\text{H}}_{4}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\left(g\right)+{\text{C}}_{2}{\text{H}}_{2}\left(g\right)$

Answer: (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:

${\text{H}}_{2}\text{O}\left(l\right)\to {\text{H}}_{2}\text{O}\left(g\right)$

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:

$\begin{array}{c}\underset{¯}{\begin{array}{cc}{\text{H}}_{2}\left(g\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\to {\text{H}}_{2}\text{O}\left(g\right)\phantom{\rule{6.6em}{0ex}}\Delta {G}_{\text{f}}^{°}\text{gas}\hfill & \\ {\text{H}}_{2}\text{O}\left(l\right)\to {\text{H}}_{2}\left(g\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{5.6em}{0ex}}-\Delta {G}_{\text{f}}^{°}\text{liquid}\hfill \end{array}}\hfill \\ \begin{array}{c}\hfill \\ {\text{H}}_{2}\text{O}\left(l\right)\to {\text{H}}_{2}\text{O}\left(g\right)\phantom{\rule{10em}{0ex}}\Delta G°=\Delta {G}_{\text{f}}^{°}\text{gas}-\Delta {G}_{\text{f}}^{°}\text{liquid}\hfill & \end{array}\hfill \end{array}$

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°:

$\text{ZnS}\left(s\right)\to \text{Zn}\left(s\right)+\text{S}\left(s\right)\phantom{\rule{5em}{0ex}}\Delta {G}_{1}^{°}=201.3\phantom{\rule{0.2em}{0ex}}\text{kJ}$

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

$\text{S}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{SO}}_{\text{2}}\left(g\right)\phantom{\rule{5em}{0ex}}\Delta {G}_{2}^{°}=\text{−}300.1\phantom{\rule{0.2em}{0ex}}\text{kJ}$

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

$\text{ZnS}\left(s\right)+{\text{O}}_{2}\left(g\right)\to \text{Zn}\left(s\right)+\text{S}{\text{O}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}\Delta G°=201.3\phantom{\rule{0.2em}{0ex}}\text{kJ}+-300.1\phantom{\rule{0.2em}{0ex}}\text{kJ}=98.8\phantom{\rule{0.2em}{0ex}}\text{kJ}$

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 Appendix G:

$\begin{array}{l}\text{Decomposition of zinc sulfide:}\phantom{\rule{5em}{0ex}}\text{ZnS}\left(s\right)\to \text{Zn}\left(s\right)+\text{S}\left(s\right)\phantom{\rule{5em}{0ex}}\Delta {G}_{1}^{°}=201.3\phantom{\rule{0.2em}{0ex}}\text{kJ}\\ \text{Formation of hydrogen sulfide:}\phantom{\rule{5em}{0ex}}\text{S}\left(s\right)+{\text{H}}_{2}\left(g\right)\to {\text{H}}_{\text{2}}\text{S}\left(g\right)\phantom{\rule{5em}{0ex}}\Delta {G}_{2}^{°}=-33.4\phantom{\rule{0.2em}{0ex}}\text{kJ}\\ \text{Coupled reaction:}\phantom{\rule{2em}{0ex}}\text{ZnS}\left(s\right)+{\text{H}}_{2}\left(g\right)\to \text{Zn}\left(s\right)+{\text{H}}_{\text{2}}\text{S}\left(g\right)\phantom{\rule{4em}{0ex}}\Delta G\text{°}=201.3\phantom{\rule{0.2em}{0ex}}\text{kJ}+-33.4\phantom{\rule{0.2em}{0ex}}\text{kJ}=167.9\phantom{\rule{0.2em}{0ex}}\text{kJ}\end{array}$

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?

$\text{FeS}\left(s\right)+{\text{O}}_{2}\left(g\right)\to \text{Fe}\left(s\right)+{\text{SO}}_{\text{2}}\left(g\right)$

Answer: −199.7 kJ; spontaneous

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:

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).

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

Predicting the Temperature Dependence of Spontaneity

The incomplete combustion of carbon is described by the following equation:

$\text{2C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2CO}\left(g\right)$

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:

$\text{4Fe}\left(s\right)+{\text{3O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2Fe}}_{2}{\text{O}}_{3}\left(s\right)$

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:

$\Delta G=\Delta H-T\Delta S$

$y=b+mx$

Such a plot is shown in (Figure). 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:

$\Delta G=0=\Delta H-T\Delta S$

$T=\phantom{\rule{0.2em}{0ex}}\frac{\Delta H}{\Delta S}$

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.

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

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)

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

$0=\Delta H\text{°}-T\Delta S\text{°}\phantom{\rule{5em}{0ex}}\text{or}\phantom{\rule{5em}{0ex}}T=\phantom{\rule{0.2em}{0ex}}\frac{\Delta H\text{°}}{\Delta S\text{°}}$

Using the standard thermodynamic data from Appendix G,

$\begin{array}{ccc}\hfill \Delta H\text{°}& =& 1\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\Delta {H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(g\right)\right)\phantom{\rule{0.2em}{0ex}}-1\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(l\right)\right)\hfill \\ & =& \text{(1 mol)}-\text{241.82 kJ/mol}-\text{(1 mol)}\left(-\text{286.83 kJ/mol}\right)=\text{44.01 kJ}\end{array}$

ΔS^o = 1 mol x S^o(H₂O(g)) – 1 mol x ΔS^o(H₂O(l))

= (1 mol) (188.8 J K^-1 mol^-1) – (1 mol) (70.0 J K^-1 mol^-1) = 118.8 J K^-1

$T=\phantom{\rule{0.2em}{0ex}}\frac{\Delta H\text{°}}{\Delta S\text{°}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{44.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}}{118.8\phantom{\rule{0.2em}{0ex}}\text{J/K}}\phantom{\rule{0.2em}{0ex}}=370.5\phantom{\rule{0.2em}{0ex}}\text{K}=97.3\phantom{\rule{0.2em}{0ex}}\text{°C}$

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)

Key Concepts and Summary

Gibbs free energy (G) is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A negative value for ΔG indicates a spontaneous process; a positive ΔG indicates a nonspontaneous process; and a ΔG of zero indicates that the system is at equilibrium. A number of approaches to the computation of free energy changes are possible.

Key Equations

ΔG = ΔH − TΔS
For the reaction mA + nB → xC + yD, the standard free energy change at room temperature may be calculated as

$\begin{array}{}\\ \\ \Delta G\text{°}=\sum \nu \Delta G\text{°}\left(\text{products}\right)\phantom{\rule{0.2em}{0ex}}-\sum \nu \text{Δ}G\text{°}\left(\text{reactants}\right)\\ \\ =\left[x\Delta {G}_{\text{f}}^{°}\left(\text{C}\right)+y\Delta {G}_{\text{f}}^{°}\left(\text{D}\right)\right]\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\left[m\Delta {G}_{\text{f}}^{°}\left(\text{A}\right)+n\Delta {G}_{\text{f}}^{°}\left(\text{B}\right)\right].\end{array}$

Chemistry End of Chapter Exercises

1. What is the difference between ΔG and ΔG° for a chemical change?

2. A reaction has ΔH^o = 100 kJ mol^-1 and ΔS^o= 250 J mol^-1 K^-1. Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?

Answer(s): The reaction is nonspontaneous at room temperature.
Above 400 K, ΔG will become negative, and the reaction will become spontaneous.

3. Explain what happens as a reaction starts with ΔG < 0 (negative) and reaches the point where ΔG = 0.

4. Use the standard free energy of formation data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.

(a) MnO₂ (s) → Mn (s) + O₂ (g)

(b) H₂ (g) + Br₂ (l) → 2HBr (g)

(c) Cu (s) + S (g) → CuS (s)

(d) 2 LiOH (s) + CO₂ (g) → Li₂CO₃ (s) + H₂O (g)

(e) CH₄ (g) + O₂ (g) → C (s, graphite) + 2 H₂O (g)

(f) CS₂ (g) + 3 Cl₂ (g) → CCl₄ (g) + S₂Cl₂ (g)

Answer(s): (a) 465.1 kJ nonspontaneous; (b) −106.86 kJ spontaneous; (c) −53.6 kJ spontaneous; (d) −83.4 kJ spontaneous; (e) −406.7 kJ spontaneous; (f) −30.0 kJ spontaneous

5. Use the standard free energy data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.

(a) C (s, graphite) + O₂ → CO₂ (g)
(b) O₂ (g) + N₂ (g) → 2NO (g)
(c) 2 Cu (s) + S (g) → Cu₂S (s)
(d) CaO (s) + H₂O (l) → Ca(OH)₂ (s)
(e) Fe₂O₃ (s) + 3 CO (g) → 2 Fe (s) + 3 CO₂ (g)
(f) CaSO₄•2H₂O (s) → CaSO₄ (s) + 2 H₂O (g)

6. Given:

P₄ (s) + 5 O₂ (g) → P₄O₁₀ (s), ΔG^o = -2697.0 kJ mol^-1.

2 H₂ (g) + O₂ (g) → 2 H₂O (g), ΔG^o = -457.18 kJ mol^-1.

6 H₂O (g) + P₄O₁₀ (s) → 4 H₃PO₄ (l), ΔG^o = -428.66 kJ mol^-1.

(a) Determine the standard free energy of formation, ΔG_f^o, for phosphoric acid.

(b) How does your calculated result compare to the value in Appendix G? Explain.

Answer(s): (a) The standard free energy of formation is –1124.3 kJ/mol. (b) The calculation agrees with the value in Appendix G because free energy is a state function (just like the enthalpy and entropy), so its change depends only on the initial and final states, not the path between them.

7. Is the formation of ozone (O₃(g)) from oxygen (O₂(g)) spontaneous at room temperature under standard state conditions?

8. Consider the decomposition of red mercury(II) oxide under standard state conditions.

2 HgO (s, red) → 2 Hg (l) + O₂ (g)

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

Answer(s): (a) The reaction is nonspontaneous; (b) Above 566 °C the process is spontaneous.

9. Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.

(a) Ammonia: 2 NH₃ (g) → N₂ (g) + 3 H₂ (g)

(b) Diborane: B₂H₆ (g) → 2 B (g) + 3 H₂ (g)

(c) Hydrazine: N₂H₄ (g) → N₂ (g) + 2 H₂ (g)

(d) Hydrogen peroxide: H₂O₂ (l) → H₂O (g) + 1/2 O₂ (g)

10. Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.

: thermodynamic property defined in terms of system enthalpy and entropy. All spontaneous processes involve a decrease in G (i.e., a negative value for ΔG.)

standard free energy change (ΔG°): change in free energy for a process occurring under standard conditions (1 bar pressure for gases, 1 M concentration for solutions)

standard free energy of formation (ΔG_f°): change in free energy accompanying the formation of one mole of substance from its elements in their standard states

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What’s “Free” about ΔG?

Calculating Free Energy Change

Using Standard Enthalpy and Entropy Changes to Calculate ΔG°

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Using Standard Free Energies of Formation to Calculate ΔG°

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Free Energy Changes for Coupled Reactions

Calculating Free Energy Change for a Coupled Reaction

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Temperature Dependence of Spontaneity

Predicting the Temperature Dependence of Spontaneity

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Equilibrium Temperature for a Phase Transition

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