74 Free Energy

[latexpage]

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:

\(\text{Δ}G=\text{Δ}H-T\text{Δ}S\)

(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:

\(\text{Δ}{S}_{\text{univ}}=\text{Δ}S+\phantom{\rule{0.2em}{0ex}}\frac{{q}_{\text{surr}}}{T}\)

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

\(\text{Δ}{S}_{\text{univ}}=\text{Δ}S-\phantom{\rule{0.1em}{0ex}}\frac{\text{Δ}H}{T}\)

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

\(\text{−}T\text{Δ}{S}_{\text{univ}}=\text{Δ}H-T\text{Δ}S\)

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

\(\text{Δ}G=\text{−}T\text{Δ}{S}_{\text{univ}}\)

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. (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
ΔSuniv > 0 ΔG < 0 spontaneous
ΔSuniv < 0 ΔG > 0 nonspontaneous
ΔSuniv = 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:

\(\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}\)

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:

\({\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(g\right)\)

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

\(\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}\)

From Appendix G:

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

\(\begin{array}{l}\text{Δ}H\text{°}=\text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(g\right)\right)\phantom{\rule{0.2em}{0ex}}-\text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(l\right)\right)\\ =\left[\text{−241.82 kJ/mol}-\left(\text{−286.83}\right)\right]\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}=\text{45.01 kJ}\end{array}\)
\(\begin{array}{c}\text{Δ}S\text{°}=\text{1 mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}S\text{°}\left({\text{H}}_{2}\text{O}\left(g\right)\right)\phantom{\rule{0.2em}{0ex}}-\text{1 mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}S\text{°}\left({\text{H}}_{2}\text{O}\left(l\right)\right)\\ =\left(1 mol\right)188.8\phantom{\rule{0.2em}{0ex}}\text{J/mol·K}-\left(1\phantom{\rule{0.2em}{0ex}}\text{mol)}70.0\phantom{\rule{0.2em}{0ex}}\text{J/mol K}=118.8\phantom{\rule{0.2em}{0ex}}\text{J/mol·K}\end{array}\)
\(\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}\)

Substitution into the standard free energy equation yields:

\(\begin{array}{c}\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}\\ \\ =\text{45.01 kJ}-\left(\text{298 K}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}118.8\phantom{\rule{0.2em}{0ex}}\text{J/K}\right)\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\frac{\text{1 kJ}}{\text{1000 J}}\end{array}\)
\(\text{45.01 kJ}-\text{35.4 kJ}=\text{9.6 kJ}\)

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

\({\text{C}}_{2}{\text{H}}_{6}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\left(g\right)+{\text{C}}_{2}{\text{H}}_{4}\left(g\right)\)
Answer:

\(\text{Δ}G\text{°}=\text{102.0 kJ/mol};\) the reaction is nonspontaneous (not spontaneous) at 25 °C.

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

\(m\text{A}+n\text{B}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}x\text{C}+y\text{D},\)

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

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

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

\(\text{HgO}\left(s,\phantom{\rule{0.2em}{0ex}}\text{yellow}\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Hg}\left(l\right)\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\)

Calculate the standard free energy change at room temperature, \(\text{Δ}G\text{°},\) 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 \(\text{Δ}{G}_{\text{f}}^{°}\phantom{\rule{0.2em}{0ex}}\text{(kJ/mol)}\) \(\text{Δ}{H}_{\text{f}}^{°}\phantom{\rule{0.2em}{0ex}}\text{(kJ/mol)}\) \(S\text{°}\phantom{\rule{0.2em}{0ex}}\text{(J/K·mol)}\)
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:

\(\text{Δ}G\text{°}=\sum \nu G{S}_{f}^{°}\text{(products)}\phantom{\rule{0.2em}{0ex}}-\sum \nu \text{Δ}{G}_{f}^{°}\text{(reactants)}\)
\(=\left[1\text{Δ}{G}_{f}^{°}\text{Hg}\left(l\right)+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\text{Δ}{G}_{f}^{°}{\text{O}}_{\text{2}}\left(g\right)\right]\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}1\text{Δ}{G}_{f}^{°}\text{HgO}\left(s,\phantom{\rule{0.2em}{0ex}}\text{yellow}\right)\)
\(=\left[1\phantom{\rule{0.2em}{0ex}}\text{mol}\text{(0 kJ/mol)}+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{mol(0 kJ/mol)}\right]\phantom{\rule{0.2em}{0ex}}-\text{1 mol(−58.43 kJ/mol)}=\text{58.43 kJ/mol}\)

(b) Using enthalpies and entropies of formation:

\(\text{Δ}H\text{°}=\sum \nu \text{Δ}{H}_{\text{f}}^{°}\text{(products)}\phantom{\rule{0.2em}{0ex}}-\sum \nu \text{Δ}{H}_{\text{f}}^{°}\text{(reactants)}\)
\(=\left[1\text{Δ}{H}_{f}^{°}\text{Hg}\left(l\right)+\frac{1}{2}\text{Δ}{H}_{f}^{°}{\text{O}}_{2}\left(g\right)\right]\phantom{\rule{0.2em}{0ex}}-1\text{Δ}{H}_{f}^{°}\text{HgO}\left(s,\phantom{\rule{0.2em}{0ex}}\text{yellow}\right)\)
\(=\left[\text{1 mol}\left(\text{0 kJ/mol}\right)\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{mol}\left(\text{0 kJ/mol}\right)\right]\phantom{\rule{0.2em}{0ex}}-\text{1 mol}\left(\text{−90.46 kJ/mol}\right)=\text{90.46 kJ/mol}\)
\(\text{Δ}S\text{°}=\sum \text{ν}\text{Δ}S\text{°}\text{(products)}-\sum \text{ν}\text{Δ}S\text{°}\text{(reactants)}\)
\(=\left[1\text{Δ}S\text{°}\text{Hg}\left(l\right)\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\text{Δ}S\text{°}{\text{O}}_{2}\left(g\right)\right]\phantom{\rule{0.2em}{0ex}}-1\text{Δ}S\text{°}\text{HgO}\left(s,\phantom{\rule{0.2em}{0ex}}\text{yellow}\right)\)
\(=\left[\text{1 mol}\phantom{\rule{0.2em}{0ex}}\left(\text{75.9 J/mol K}\right)\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{mol}\left(\text{205.2 J/mol K}\right)\right]\phantom{\rule{0.2em}{0ex}}-\text{1 mol}\left(\text{71.13 J/mol K}\right)=\text{107.4 J/mol K}\)
\(\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}=\text{90.46 kJ}-\text{298.15 K}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{107.4 J/K·mol}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\frac{\text{1 kJ}}{\text{1000 J}}\)
\(\text{Δ}G\text{°}=\left(90.46-32.01\right)\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}=\text{58.45 kJ/mol}\)

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:

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:

\(\begin{array}{c}\underset{¯}{\begin{array}{cccc}{\text{H}}_{2}\left(g\right)+1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}\to \phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(g\right)\hfill & & & \phantom{\rule{1em}{0ex}}\text{Δ}{G}_{\text{f},\text{gas}}^{°}\hfill \\ {\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}\to {\text{H}}_{2}\left(g\right)+1}{2}{\text{O}}_{2}\left(g\right)\hfill & & & \phantom{\rule{1em}{0ex}}-{\text{Δ}G}_{\text{f},\text{liquid}}^{°}\phantom{\rule{5em}{0ex}}\hfill \end{array}}\hfill \\ \begin{array}{cccc}{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}\to {\text{H}}_{2}\text{O}\left(g\right)\hfill & & & \phantom{\rule{5em}{0ex}}\text{Δ}G={\text{Δ}G}_{\text{f},\text{gas}}^{°}-{\text{Δ}G}_{\text{f},\text{liquid}}^{°}\hfill \end{array}\hfill \end{array}\)

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°:

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

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

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

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

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

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?

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

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

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?

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

–199.7 kJ; spontaneous

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 number of approaches to the computation of free energy changes are possible.

Key Equations

  • ΔG = ΔHTΔS
  • ΔG = ΔG° + RT ln Q
  • ΔG° = −RT ln K

Chemistry End of Chapter Exercises

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

A reaction has \(\text{Δ}H\text{°}\) = 100 kJ/mol and \(\text{Δ}S\text{°}=\text{250 J/mol·K.}\) Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?

The reaction is nonspontaneous at room temperature.
Above 400 K, ΔG will become negative, and the reaction will become spontaneous.

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

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

(b) \({\text{H}}_{2}\left(g\right)+{\text{Br}}_{2}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2HBr}\left(g\right)\)

(c) \(\text{Cu}\left(s\right)+\text{S}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CuS}\left(s\right)\)

(d) \(\text{2LiOH}\left(s\right)+{\text{CO}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)+{\text{H}}_{2}\text{O}\left(g\right)\)

(e) \({\text{CH}}_{4}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{C}\left(s,\phantom{\rule{0.2em}{0ex}}\text{graphite}\right)+{\text{2H}}_{2}\text{O}\left(g\right)\)

(f) \({\text{CS}}_{2}\left(g\right)+{\text{3Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CCl}}_{4}\left(g\right)+{\text{S}}_{2}{\text{Cl}}_{2}\left(g\right)\)

(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

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

(b) \({\text{O}}_{2}\left(g\right)+{\text{N}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2NO}\left(g\right)\)

(c) \(\text{2Cu}\left(s\right)+\text{S}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cu}}_{2}\text{S}\left(s\right)\)

(d) \(\text{CaO}\left(s\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Ca}{\left(\text{OH}\right)}_{2}\left(s\right)\)

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

(f) \({\text{CaSO}}_{4}\text{·}{\text{2H}}_{2}\text{O}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CaSO}}_{4}\left(s\right)+{\text{2H}}_{2}\text{O}\left(g\right)\)

Given:
\(\begin{array}{cccc}{\text{P}}_{4}\left(s\right)+{\text{5O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{P}}_{4}{\text{O}}_{10}\left(s\right)\hfill & & & \text{Δ}G\text{°}=\text{−2697.0 kJ/mol}\hfill \\ {\text{2H}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2H}}_{2}\text{O}\left(g\right)\hfill & & & \text{Δ}G\text{°}=\text{−457.18 kJ/mol}\hfill \\ {\text{6H}}_{2}\text{O}\left(g\right)+{\text{P}}_{4}{\text{O}}_{10}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{4H}}_{3}{\text{PO}}_{4}\left(l\right)\hfill & & & \text{Δ}G\text{°}=\text{−428.66 kJ/mol}\hfill \end{array}\)

(a) Determine the standard free energy of formation, \(\text{Δ}{G}_{\text{f}}^{°},\) for phosphoric acid.

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

(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.

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

Consider the decomposition of red mercury(II) oxide under standard state conditions.
\(\text{2HgO}\left(s,\phantom{\rule{0.2em}{0ex}}\text{red}\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2Hg}\left(l\right)+{\text{O}}_{2}\left(g\right)\)

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

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

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

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: \({\text{2NH}}_{3}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}\left(g\right)+{\text{3H}}_{2}\left(g\right)\)

(b) Diborane: \({\text{B}}_{2}{\text{H}}_{6}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2B}\left(g\right)+{\text{3H}}_{2}\left(g\right)\)

(c) Hydrazine: \({\text{N}}_{2}{\text{H}}_{4}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}\left(g\right)+{\text{2H}}_{2}\left(g\right)\)

(d) Hydrogen peroxide: \({\text{H}}_{2}{\text{O}}_{2}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(g\right)+\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\)

Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.

\(\begin{array}{ccccccc}\text{(a)}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2NO}\left(g\right)\hfill & & & \text{T}=2000\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=4.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−4}}\hfill \\ \text{(b)}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2HI}\left(g\right)\hfill & & & \text{T}=400\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=50.0\hfill \\ \text{(c)}\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)\hfill & & & \text{T}=980\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=1.67\hfill \\ \text{(d)}\phantom{\rule{0.2em}{0ex}}{\text{CaCO}}_{3}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)\hfill & & & \text{T}=900\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=1.04\hfill \\ \text{(e)}\phantom{\rule{0.2em}{0ex}}\text{HF}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{F}}^{\text{−}}\left(aq\right)\hfill & & & \text{T}=25\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=7.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−4}}\hfill \\ \text{(f)}\phantom{\rule{0.2em}{0ex}}\text{AgBr}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Br}}^{\text{−}}\left(aq\right)\hfill & & & \text{T}=25\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=3.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−13}}\hfill \end{array}\)

(a) 1.5 \(×\) 102 kJ; (b) −21.9 kJ; (c) −5.34 kJ; (d) −0.383 kJ; (e) 18 kJ; (f) 71 kJ

Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.

\(\begin{array}{ccccccc}\text{(a)}\phantom{\rule{0.2em}{0ex}}{\text{Cl}}_{2}\left(g\right)+{\text{Br}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2BrCl}\left(g\right)\hfill & & & \text{T}=25\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=4.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−2}}\hfill \\ \text{(b)}\phantom{\rule{0.2em}{0ex}}{\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{2SO}}_{3}\left(g\right)\hfill & & & \text{T}=500\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=48.2\hfill \\ \text{(c)}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(g\right)\hfill & & & \text{T}=60\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=\text{0.196 atm}\hfill \\ \text{(d)}\phantom{\rule{0.2em}{0ex}}\text{CoO}\left(s\right)+\text{CO}\left(g\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}\text{Co}\left(s\right)+{\text{CO}}_{2}\left(g\right)\hfill & & & \text{T}=550\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=4.90\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\hfill \\ \text{(e)}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{NH}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{NH}}_{3}{}^{\text{+}}\left(aq\right)+{\text{OH}}^{\text{−}}\left(aq\right)\hfill & & & \text{T}=25\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=4.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−4}}\hfill \\ \text{(f)}\phantom{\rule{0.2em}{0ex}}{\text{PbI}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Pb}}^{2+}\left(aq\right)+{\text{2I}}^{\text{−}}\left(aq\right)\hfill & & & \text{T}=25\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill & & & {K}_{p}=8.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−9}}\hfill \end{array}\)

Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.

\(\begin{array}{cccc}\text{(a)}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{2}\left(g\right)+{\text{2F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2OF}}_{2}\left(g\right)\hfill & & & \text{Δ}G\text{°}=\text{−9.2 kJ}\hfill \\ \text{(b)}\phantom{\rule{0.2em}{0ex}}{\text{I}}_{2}\left(s\right)+{\text{Br}}_{2}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2IBr}\left(g\right)\hfill & & & \text{Δ}G\text{°}=\text{7.3 kJ}\hfill \\ \text{(c)}\phantom{\rule{0.2em}{0ex}}\text{2LiOH}\left(s\right)+{\text{CO}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)+{\text{H}}_{2}\text{O}\left(g\right)\hfill & & & \text{Δ}G\text{°}=\text{−79 kJ}\hfill \\ \text{(d)}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}{\text{O}}_{3}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{NO}\left(g\right)+{\text{NO}}_{2}\left(g\right)\hfill & & & \text{Δ}G\text{°}=\text{−1.6 kJ}\hfill \\ \text{(e)}\phantom{\rule{0.2em}{0ex}}{\text{SnCl}}_{4}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{SnCl}}_{4}\left(l\right)\hfill & & & \text{Δ}G\text{°}=\text{8.0 kJ}\hfill \end{array}\)

(a) K = 41; (b) K = 0.053; (c) K = 6.9 \(×\) 1013; (d) K = 1.9; (e) K = 0.04

Determine ΔGº for the following reactions.

(a) Antimony pentachloride decomposes at 448 °C. The reaction is:
\({\text{SbCl}}_{5}\left(g\right)⟶{\text{SbCl}}_{3}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\)

An equilibrium mixture in a 5.00 L flask at 448 °C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2.

(b) Chlorine molecules dissociate according to this reaction:
\({\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2\text{Cl}\left(g\right)\)

1.00% of Cl2 molecules dissociate at 975 K and a pressure of 1.00 atm.

(a) 22.1 kJ;
(b) 61.6 kJ/mol

Given that the \(\text{Δ}{G}_{\text{f}}^{°}\) for Pb2+(aq) and Cl(aq) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, Ksp, for PbCl2(s).

Determine the standard free energy change, \(\text{Δ}{G}_{\text{f}}^{°},\) for the formation of S2−(aq) given that the \(\text{Δ}{G}_{\text{f}}^{°}\) for Ag+(aq) and Ag2S(s) are 77.1 k/mole and −39.5 kJ/mole respectively, and the solubility product for Ag2S(s) is 8 \(×\) 10−51.

90 kJ/mol

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.

The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.
\({\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(g\right)\phantom{\rule{5em}{0ex}}\text{Δ}G\text{°}=\text{8.58 kJ}\)

(a) Is the evaporation of water under standard thermodynamic conditions spontaneous?

(b) Determine the equilibrium constant, KP, for this physical process.

(c) By calculating ∆G, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, \({P}_{{\text{H}}_{2}\text{O}},\) is 0.011 atm.

(d) If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of \({P}_{{\text{H}}_{2}\text{O}}\) in the air?

(a) Under standard thermodynamic conditions, the evaporation is nonspontaneous; (b) Kp = 0.031; (c) The evaporation of water is spontaneous; (d) \({P}_{{\text{H}}_{2}\text{O}}\) must always be less than Kp or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.

In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:
\(\text{Glu}+\text{ATP}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{G6P}+\text{ADP}\phantom{\rule{5em}{0ex}}\text{Δ}G\text{°}=\text{−17 kJ}\)

In this process, ATP becomes ADP summarized by the following equation:
\(\text{ATP}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{ADP}\phantom{\rule{5em}{0ex}}\text{Δ}G\text{°}=\text{−30 kJ}\)

Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:
\(\text{Glu}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{G6P}\phantom{\rule{5em}{0ex}}\text{Δ}G\text{°}=?\)

One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
\(\text{G6P}\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}\text{F6P}\phantom{\rule{5em}{0ex}}\text{Δ}G\text{°}=\text{1.7 kJ}\)

(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?

(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M, however, in a typical cell, they are not even close to these values. Calculate ΔG when the concentrations of G6P and F6P are 120 μM and 28 μM respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.

(a) Nonspontaneous as \(\text{Δ}G\text{°}>0;\) (b) \(\text{Δ}G\text{°}=\text{Δ}{G}^{°}\text{+}RT\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}Q,\) \(\text{Δ}G=1.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}+\phantom{\rule{0.2em}{0ex}}\left(8.314\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}335\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.4em}{0ex}}\frac{28}{128}\right)\phantom{\rule{0.2em}{0ex}}=\text{−2.5 kJ}.\) The forward reaction to produce F6P is spontaneous under these conditions.

Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.

(a) \({\text{N}}_{2}\left(g\right)+{\text{3H}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2NH}}_{3}\left(g\right)\)

(b) \(\text{HCl}\left(g\right)+{\text{NH}}_{3}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{4}\text{Cl}\left(s\right)\)

(c) \({\left({\text{NH}}_{4}\right)}_{2}{\text{Cr}}_{2}{\text{O}}_{7}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cr}}_{2}{\text{O}}_{3}\left(s\right)+{\text{4H}}_{2}\text{O}\left(g\right)+{\text{N}}_{2}\left(g\right)\)

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

When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of ΔG, ΔH, and ΔS for this process, and justify your choices.

ΔG is negative as the process is spontaneous. ΔH is positive as with the solution becoming cold, the dissolving must be endothermic. ΔS must be positive as this drives the process, and it is expected for the dissolution of any soluble ionic compound.

An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the Cu2S decomposes to form copper and sulfur described by the following equation:
\({\text{Cu}}_{2}\text{S}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Cu}\left(s\right)+\text{S}\left(s\right)\)

(a) Determine \(\text{Δ}G\text{°}\) for the decomposition of Cu2S(s).

(b) The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine \(\text{Δ}G\text{°}\) for the process.

(c) The production of copper from chalcocite is performed by roasting the Cu2S in air to produce the Cu. By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.

What happens to \(\text{Δ}G\text{°}\) (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?

(a) \(\text{S}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{SO}}_{2}\left(g\right)\)

(b) \({\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{SO}}_{3}\left(g\right)\)

(c) \(\text{HgO}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Hg}\left(l\right)+{\text{O}}_{2}\left(g\right)\)

(a) Increasing \({P}_{{\text{O}}_{\text{2}}}\) will shift the equilibrium toward the products, which increases the value of K. ΔG° therefore becomes more negative.
(b) Increasing \({P}_{{\text{O}}_{\text{2}}}\) will shift the equilibrium toward the products, which increases the value of K. ΔG° therefore becomes more negative.
(c) Increasing \({P}_{{\text{O}}_{\text{2}}}\) will shift the equilibrium the reactants, which decreases the value of K. ΔG° therefore becomes more positive.

Glossary

Gibbs free energy change (G)
thermodynamic property defined in terms of system enthalpy and entropy; all spontaneous processes involve a decrease in G
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 \(\left(\text{Δ}{G}_{\text{f}}^{°}\right)\)
change in free energy accompanying the formation of one mole of substance from its elements in their standard states

License

Icon for the Creative Commons Attribution 4.0 International License

Atoms First / OpenStax by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Share This Book