16.4 Free Energy

Learning Objectives

By the end of this section, you will be able to:

Define Gibbs free energy, and describe its relation to spontaneity
Calculate free energy change for a process using free energies of formation for its reactants and products
Calculate free energy change for a process using enthalpies of formation and the entropies for its reactants and products
Explain how temperature affects the spontaneity of some processes
Relate standard free energy changes to equilibrium constants

One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that we must determine the entropy change for the system and the entropy change for the surroundings. An alternative approach involving a new thermodynamic property defined in terms of system properties only was introduced in the late nineteenth century by American mathematician Josiah Willard Gibbs. This new property is called the Gibbs free energy change (G) (or simply the free energy), and it is defined in terms of a system’s enthalpy and entropy as the following:

[latex]G=H-TS[/latex]

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

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

(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:

[latex]\Delta {S}_{\text{univ}}=\Delta S+\frac{{q}_{\text{surr}}}{T}[/latex]

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:

[latex]\Delta {S}_{\text{univ}}=\Delta S-\frac{\Delta H}{T}[/latex]

ΔH is the enthalpy change of the system. Multiplying both sides of this equation by −T, and rearranging yields the following:

[latex]-T\Delta {S}_{\text{univ}}=\Delta H-T\Delta S[/latex]

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

[latex]\Delta G=-T\Delta {S}_{\text{univ}}[/latex]

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 1 summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.

Table 1. 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)

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 that have undergone some change. 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 1.

[latex]\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}[/latex]

Example 1: Evaluation of ΔG° Change from ΔH° and ΔS°

Use standard enthalpy and entropy data from Standard Thermodynamic Properties for Selected Substances 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?

Show Answer

Substance	[latex]\Delta {H}_{\text{f}}^{\circ }\text{(kJ/mol)}[/latex]	[latex]{S}_{298}^{\circ }\text{(J/K\cdot mol)}[/latex]
H₂O(l)	−286.83	70.0
H₂O(g)	−241.82	188.8

Check Your Learning

Use standard enthalpy and entropy data from Standard Thermodynamic Properties for Selected Substances 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?

[latex]{\text{C}}_{2}{\text{H}}_{6}\left(g\right)\longrightarrow {\text{H}}_{2}\left(g\right)+{\text{C}}_{2}{\text{H}}_{4}\left(g\right)[/latex]

Show Answer

Free energy changes may also use the standard free energy of formation [latex]\left(\Delta {G}_{\text{f}}^{\circ }\right)[/latex], for each 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 enthalpies of formation, [latex]\Delta {G}_{\text{f}}^{\circ }[/latex] is by definition zero for elemental substances under standard state conditions. The approach to computing the free energy change for a reaction using this approach is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction

[latex]m\text{A}+n\text{B}\longrightarrow x\text{C}+y\text{D}[/latex],

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

[latex]\begin{array}{rll}{}\Delta {G}_{298}^{\circ }=\Delta G^{\circ }&=&\sum \nu \Delta {G}_{298}^{\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{298}^{\circ }\left(\text{reactants}\right)\\{}&=&\left[x\Delta {G}_{\text{f}}^{\circ }\left(\text{C}\right)+y\Delta {G}_{\text{f}}^{\circ }\left(\text{D}\right)\right]-\left[m\Delta {G}_{\text{f}}^{\circ }\left(\text{A}\right)+n\Delta {G}_{\text{f}}^{\circ }\left(\text{B}\right)\right]\end{array}[/latex]

Example 2: Calculation of [latex]\Delta {G}_{298}^{\circ }[/latex]

Consider the decomposition of yellow mercury(II) oxide.

[latex]\text{HgO}\left(s,\text{yellow}\right)\longrightarrow \text{Hg}\left(l\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)[/latex]

Calculate the standard free energy change at room temperature, [latex]\Delta {G}_{298}^{\circ }[/latex], 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?

Show Answer

The required data are available in Standard Thermodynamic Properties for Selected Substances and are shown here.

Compound	[latex]\Delta {G}_{\text{f}}^{\circ }\text{(kJ/mol)}[/latex]	[latex]\Delta {H}_{\text{f}}^{\circ }\text{(kJ/mol)}[/latex]	[latex]{S}_{298}^{\circ }\text{(J/K}\cdot\text{mol)}[/latex]
HgO (s, yellow)	−58.43	−90.46	71.13
Hg(l)	0	0	75.9
O₂(g)	0	0	205.2

Using free energies of formation:
[latex]\begin{array}{rll}{}\Delta {G}_{298}^{\circ }&=&\sum \nu G{S}_{298}^{\circ }\text{(products)}-\sum \nu \Delta {G}_{298}^{\circ }\text{(reactants)}\\{}&=&\left[1\Delta {G}_{298}^{\circ }\text{Hg}\left(l\right)+\frac{1}{2}\Delta {G}_{298}^{\circ }{\text{O}}_{\text{2}}\left(g\right)\right]-1\Delta {G}_{298}^{\circ }\text{HgO}\left(s,\text{yellow}\right)\\{}&=&\left[1\text{mol}\text{(0 kJ/mol)}+\frac{1}{2}\text{mol(0 kJ/mol)}\right]-\text{1 mol(-58.43 kJ/mol)}=\text{58.43 kJ/mol}\end{array}[/latex]
Using enthalpies and entropies of formation:
[latex]\begin{array}{rll}{}\Delta {H}_{\text{298}}^{\circ }&=&\sum \nu \Delta {H}_{\text{298}}^{\circ }\text{(products)}-\sum \nu \Delta {H}_{\text{298}}^{\circ }\text{(reactants)}\\{}&=&\left[1\Delta {H}_{298}^{\circ }\text{Hg}\left(l\right)+\frac{1}{2}\Delta {H}_{298}^{\circ }{\text{O}}_{2}\left(g\right)\right]-1\Delta {H}_{298}^{\circ }\text{HgO}\left(s,\text{yellow}\right)\\{}&=&\left[\text{1 mol}\left(\text{0 kJ/mol}\right)+\frac{1}{2}\text{mol}\left(\text{0 kJ/mol}\right)\right]-\text{1 mol}\left(-\text{90.46 kJ/mol}\right)=\text{90.46 kJ/mol}\end{array}[/latex]
[latex]\begin{array}{rll}\\ \Delta {S}_{298}^{\circ }&=&\sum\nu\Delta {S}_{298}^{\circ }\text{(products)}-\sum \nu\Delta {S}_{298}^{\circ }\text{(reactants)}\\{}&=&\left[1\Delta {S}_{298}^{\circ }\text{Hg}\left(l\right)+\frac{1}{2}\Delta {S}_{298}^{\circ }{\text{O}}_{2}\left(g\right)\right]-1\Delta {S}_{298}^{\circ }\text{HgO}\left(s,\text{yellow}\right)\\{}&=&\left[\text{1 mol}\left(\text{75.9 J/mol K}\right)+\frac{1}{2}\text{mol}\left(\text{205.2 J/mol K}\right)\right]-\text{1 mol}\left(\text{71.13 J/mol K}\right)=\text{107.4 J/mol K}\end{array}[/latex]
[latex]\begin{array}{rll}\\ \Delta G^{\circ }&=&\Delta H^{\circ }-T\Delta S^{\circ }=\text{90.46 kJ}-\text{298.15 K}\times \text{107.4 J/K}\cdot \text{mol}\times \frac{\text{1 kJ}}{\text{1000 J}}\\\Delta G^{\circ }&=&\left(90.46 - 32.01\right)\text{kJ/mol}=\text{58.45 kJ/mol}\end{array}[/latex]

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 (Standard Thermodynamic Properties for Selected Substances). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?

[latex]{\text{C}}_{2}{\text{H}}_{4}\left(g\right)\longrightarrow {\text{H}}_{2}\left(g\right)+{\text{C}}_{2}{\text{H}}_{2}\left(g\right)[/latex]

Show Answer

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:

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

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 1.

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

Example 3: Predicting the Temperature Dependence of Spontaneity

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

[latex]\text{2C}\left(s\right)+{\text{O}}_{2}\left(g\right)\longrightarrow \text{2CO}\left(g\right)[/latex]

How does the spontaneity of this process depend upon temperature?

Show Answer

Check Your Learning

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

[latex]\text{4Fe}\left(s\right)+{\text{3O}}_{2}\left(g\right)\longrightarrow {\text{2Fe}}_{2}{\text{O}}_{3}\left(s\right)[/latex]

How does the spontaneity of this process depend upon temperature?

Show Answer

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

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:

[latex]\Delta G=\Delta H-T\Delta S[/latex]
[latex]y=b+mx[/latex]

Such a plot is shown in Figure 2. 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:

[latex]\Delta G=0=\Delta H-T\Delta S[/latex]
[latex]T=\frac{\Delta H}{\Delta S}[/latex]

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, this condition describes a system at equilibrium.

Example 4: 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 solid and liquid phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Standard Thermodynamic Properties for Selected Substances to estimate the boiling point of water.

Show Answer

Check Your Learning

Use the information in Standard Thermodynamic Properties for Selected Substances to estimate the boiling point of CS₂.

Show Answer

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:

[latex]\Delta G=\Delta G^{\circ }+RT\text{ln}Q[/latex]

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 5.

Example 5: Calculating ΔG under Nonstandard Conditions

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

T = 25 °C, [latex]{P}_{{\text{N}}_{2}}=\text{0.870 atm}[/latex], [latex]{P}_{{\text{H}}_{2}}=\text{0.250 atm}[/latex], and [latex]{P}_{{\text{NH}}_{3}}=\text{12.9 atm}[/latex]

[latex]{\text{2NH}}_{3}\left(g\right)\longrightarrow {\text{3H}}_{2}\left(g\right)+{\text{N}}_{2}\left(g\right)\Delta G^{\circ }=\text{33.0 kJ/mol}[/latex]

Show Answer

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?

Show Answer

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

[latex]0=\Delta G^{\circ }+RT\text{ln}K\left(\text{at equilibrium}\right)[/latex]
[latex]\Delta G^{\circ }=-RT\text{ln}K\text{ or }K={e}^{-\frac{\Delta G^{\circ }}{RT}}[/latex]

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 2.

Table 2. 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.

Example 6

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.

Show Answer

Check Your Learning

Use the thermodynamic data provided in Standard Thermodynamic Properties for Selected Substances to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.

[latex]{\text{2NO}}_{2}\left(g\right)\rightleftharpoons {\text{N}}_{2}{\text{O}}_{4}\left(g\right)[/latex]

Show Answer

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 3). If a system is present with reactants and products present 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 3. These plots show the free energy versus reaction progress for systems whose standard free 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.

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
ΔG = ΔG° + RT ln Q
ΔG° = −RT ln K

Exercises

What is the difference between ΔG, ΔG°, and [latex]\Delta {G}_{298}^{\circ }[/latex] for a chemical change?
A reactions has [latex]\Delta {H}_{298}^{\circ }[/latex] = 100 kJ/mol and [latex]\Delta {S}_{298}^{\circ }=\text{250 J/mol}\cdot\text{K}[/latex]. Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it 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 Standard Thermodynamic Properties for Selected Substances 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.
1. [latex]{\text{MnO}}_{2}\left(s\right)\longrightarrow \text{Mn}\left(s\right)+{\text{O}}_{2}\left(g\right)[/latex]
2. [latex]{\text{H}}_{2}\left(g\right)+{\text{Br}}_{2}\left(l\right)\longrightarrow \text{2HBr}\left(g\right)[/latex]
3. [latex]\text{Cu}\left(s\right)+\text{S}\left(g\right)\longrightarrow \text{CuS}\left(s\right)[/latex]
4. [latex]\text{2LiOH}\left(s\right)+{\text{CO}}_{2}\left(g\right)\longrightarrow {\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)+{\text{H}}_{2}\text{O}\left(g\right)[/latex]
5. [latex]{\text{CH}}_{4}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow \text{C}\left(s,\text{graphite}\right)+{\text{2H}}_{2}\text{O}\left(g\right)[/latex]
6. [latex]{\text{CS}}_{2}\left(g\right)+{\text{3Cl}}_{2}\left(g\right)\longrightarrow {\text{CCl}}_{4}\left(g\right)+{\text{S}}_{2}{\text{Cl}}_{2}\left(g\right)[/latex]
Use the standard free energy data in Standard Thermodynamic Properties for Selected Substances 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.
1. [latex]\text{C}\left(s\text{, graphite}\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{CO}}_{2}\left(g\right)[/latex]
2. [latex]{\text{O}}_{2}\left(g\right)+{\text{N}}_{2}\left(g\right)\longrightarrow \text{2NO}\left(g\right)[/latex]
3. [latex]\text{2Cu}\left(s\right)+\text{S}\left(g\right)\longrightarrow {\text{Cu}}_{2}\text{S}\left(s\right)[/latex]
4. [latex]\text{CaO}\left(s\right)+{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow \text{Ca}{\left(\text{OH}\right)}_{2}\left(s\right)[/latex]
5. [latex]{\text{Fe}}_{2}{\text{O}}_{3}\left(s\right)+\text{3CO}\left(g\right)\longrightarrow \text{2Fe}\left(s\right)+{\text{3CO}}_{2}\left(g\right)[/latex]
Given:
[latex]{\text{P}}_{4}\left(s\right)+{\text{5O}}_{2}\left(g\right)\longrightarrow {\text{P}}_{4}{\text{O}}_{10}\left(s\right)\Delta {G}_{298}^{\circ }=-\text{2697.0 kJ/mol}[/latex]
[latex]{\text{2H}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{2H}}_{2}\text{O}\left(g\right)\Delta {G}_{298}^{\circ }=-\text{457.18 kJ/mol}[/latex]
[latex]{\text{6H}}_{2}\text{O}\left(g\right)+{\text{P}}_{4}{\text{O}}_{10}\left(g\right)\longrightarrow {\text{4H}}_{3}{\text{PO}}_{4}\left(l\right)\Delta {G}_{298}^{\circ }=-\text{428.66 kJ/mol}[/latex]
1. Determine the standard free energy of formation, [latex]\Delta {G}_{\text{f}}^{\circ }[/latex], for phosphoric acid.
2. How does your calculated result compare to the value in Standard Thermodynamic Properties for Selected Substances? Explain.
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.
[latex]\text{2HgO}\left(s,\text{red}\right)\longrightarrow \text{2Hg}\left(l\right)+{\text{O}}_{2}\left(g\right)[/latex]
1. Is the decomposition spontaneous under standard state conditions?
2. Above what temperature does the reaction become 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.
1. Ammonia: [latex]{\text{2NH}}_{3}\left(g\right)\longrightarrow {\text{N}}_{2}\left(g\right)+{\text{3H}}_{2}\left(g\right)[/latex]
2. Diborane: [latex]{\text{B}}_{2}{\text{H}}_{6}\left(g\right)\longrightarrow \text{2B}\left(g\right)+{\text{3H}}_{2}\left(g\right)[/latex]
3. Hydrazine: [latex]{\text{N}}_{2}{\text{H}}_{4}\left(g\right)\longrightarrow {\text{N}}_{2}\left(g\right)+{\text{2H}}_{2}\left(g\right)[/latex]
4. Hydrogen peroxide: [latex]{\text{H}}_{2}{\text{O}}_{2}\left(l\right)\longrightarrow {\text{H}}_{2}\text{O}\left(g\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)[/latex]
Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.
1. [latex]{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow \text{2NO}\left(g\right)\,\,\,{;}\,\,\,\text{T}=2000^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=4.1\times {10}^{-\text{4}}[/latex]
2. [latex]{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)\longrightarrow \text{2HI}\left(g\right)\,\,\,{;}\,\,\,\text{T}=400^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=50.0[/latex]
3. [latex]{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\left(g\right)\longrightarrow \text{CO}\left(g\right)+{\text{H}}_{2}\text{O}\left(g\right)\,\,\,{;}\,\,\,\text{T}=980^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=1.67[/latex]
4. [latex]{\text{CaCO}}_{3}\left(s\right)\longrightarrow \text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)\,\,\,{;}\,\,\,\text{T}=900^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=1.04[/latex]
5. [latex]\text{HF}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{F}}^{-}\left(aq\right)\,\,\,{;}\,\,\,\text{T}=25^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=7.2\times {10}^{-\text{4}}[/latex]
6. [latex]\text{AgBr}\left(s\right)\longrightarrow {\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)\,\,\,{;}\,\,\,\text{T}=25^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=3.3\times {10}^{-\text{13}}[/latex]
Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.
1. [latex]{\text{Cl}}_{2}\left(g\right)+{\text{Br}}_{2}\left(g\right)\longrightarrow \text{2BrCl}\left(g\right)\,\,\,{;}\,\,\,\text{T}=25^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=4.7\times {10}^{-\text{2}}[/latex]
2. [latex]{\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightleftharpoons {\text{2SO}}_{3}\left(g\right)\,\,\,{;}\,\,\,\text{T}=500^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=48.2[/latex]
3. [latex]{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{2}\text{O}\left(g\right)\,\,\,{;}\,\,\,\text{T}=60^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=\text{0.196 atm}[/latex]
4. [latex]\text{CoO}\left(s\right)+\text{CO}\left(g\right)\rightleftharpoons \text{Co}\left(s\right)+{\text{CO}}_{2}\left(g\right)\,\,\,{;}\,\,\,\text{T}=550^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=4.90\times {10}^{2}[/latex]
5. [latex]{\text{CH}}_{3}{\text{NH}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow {\text{CH}}_{3}{\text{NH}}_{3}{}^{\text{+}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\,\,\,{;}\,\,\,\text{T}=25^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=4.4\times {10}^{-\text{4}}[/latex]
6. [latex]{\text{PbI}}_{2}\left(s\right)\longrightarrow {\text{Pb}}^{2+}\left(aq\right)+{\text{2I}}^{-}\left(aq\right)\,\,\,{;}\,\,\,\text{T}=25^{\circ}\text{C}\,\,\,{;}\,\,\,{K}_{p}=8.7\times {10}^{-\text{9}}[/latex]
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.
1. [latex]{\text{O}}_{2}\left(g\right)+{\text{2F}}_{2}\left(g\right)\longrightarrow {\text{2OF}}_{2}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\Delta G^{\circ }=-\text{9.2 kJ}[/latex]
2. [latex]{\text{I}}_{2}\left(s\right)+{\text{Br}}_{2}\left(l\right)\longrightarrow \text{2IBr}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\Delta G^{\circ }=\text{7.3 kJ}[/latex]
3. [latex]\text{2LiOH}\left(s\right)+{\text{CO}}_{2}\left(g\right)\longrightarrow {\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)+{\text{H}}_{2}\text{O}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\Delta G^{\circ }=-\text{79 kJ}[/latex]
4. [latex]{\text{N}}_{2}{\text{O}}_{3}\left(g\right)\longrightarrow \text{NO}\left(g\right)+{\text{NO}}_{2}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\Delta G^{\circ }=-\text{1.6 kJ}[/latex]
5. [latex]{\text{SnCl}}_{4}\left(l\right)\longrightarrow {\text{SnCl}}_{4}\left(l\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\Delta G^{\circ }=\text{8.0 kJ}[/latex]
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.
1. [latex]{\text{I}}_{2}\left(s\right)+{\text{Cl}}_{2}\left(g\right)\longrightarrow \text{2ICl}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\Delta G^{\circ }=-\text{10.88 kJ}[/latex]
2. [latex]{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(s\right)\longrightarrow \text{2HI}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\Delta G^{\circ }=\text{3.4 kJ}[/latex]
3. [latex]{\text{CS}}_{2}\left(g\right)+{\text{3Cl}}_{2}\left(g\right)\longrightarrow {\text{CCl}}_{4}\left(g\right)+{\text{S}}_{2}{\text{Cl}}_{2}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\Delta G^{\circ }=-\text{39 kJ}[/latex]
4. [latex]{\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{2SO}}_{3}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\Delta G^{\circ }=-\text{141.82 kJ}[/latex]
5. [latex]{\text{CS}}_{2}\left(g\right)\longrightarrow {\text{CS}}_{2}\left(l\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\Delta G^{\circ }=-\text{1.88 kJ}[/latex]
Calculate the equilibrium constant at the temperature given.
1. [latex]{\text{O}}_{2}\left(g\right)+{\text{2F}}_{2}\left(g\right)\longrightarrow {\text{2F}}_{2}\text{O}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\left(\text{T}=100^{\circ C}\right)[/latex]
2. [latex]{\text{I}}_{2}\left(s\right)+{\text{Br}}_{2}\left(l\right)\longrightarrow \text{2IBr}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\left(\text{T}=0.0^{\circ C}\right)[/latex]
3. [latex]\text{2LiOH}\left(s\right)+{\text{CO}}_{2}\left(g\right)\longrightarrow {\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)+{\text{H}}_{2}\text{O}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\left(\text{T}=575^{\circ C}\right)[/latex]
4. [latex]{\text{N}}_{2}{\text{O}}_{3}\left(g\right)\longrightarrow \text{NO}\left(g\right)+{\text{NO}}_{2}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\left(\text{T}=-10.0^{\circ C}\right)[/latex]
5. [latex]{\text{SnCl}}_{4}\left(l\right)\longrightarrow {\text{SnCl}}_{4}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\left(\text{T}=200^{\circ C}\right)[/latex]
Calculate the equilibrium constant at the temperature given.
1. [latex]{\text{I}}_{2}\left(s\right)+{\text{Cl}}_{2}\left(g\right)\longrightarrow \text{2ICl}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\left(\text{T}=100^{\circ C}\right)[/latex]
2. [latex]{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(s\right)\longrightarrow \text{2HI}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\left(\text{T}=0.0^{\circ C}\right)[/latex]
3. [latex]{\text{CS}}_{2}\left(g\right)+{\text{3Cl}}_{2}\left(g\right)\longrightarrow {\text{CCl}}_{4}\left(g\right)+{\text{S}}_{2}{\text{Cl}}_{2}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\left(\text{T}=125^{\circ C}\right)[/latex]
4. [latex]{\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{2SO}}_{3}\left(g\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\left(\text{T}=675^{\circ C}\right)[/latex]
5. [latex]{\text{CS}}_{2}\left(g\right)\longrightarrow {\text{CS}}_{2}\left(l\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,\left(\text{T}=90^{\circ C}\right)[/latex]
Consider the following reaction at 298 K: [latex]{\text{N}}_{2}{\text{O}}_{4}\left(g\right)\rightleftharpoons {\text{2NO}}_{2}\left(g\right){K}_{P}=0.142[/latex]
What is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.
Determine the normal boiling point (in kelvin) of dichloroethane, CH₂Cl₂. Find the actual boiling point using the Internet or some other source, and calculate the percent error in the temperature. Explain the differences, if any, between the two values.
Under what conditions is [latex]{\text{N}}_{2}{\text{O}}_{3}\left(g\right)\longrightarrow \text{NO}\left(g\right)+{\text{NO}}_{2}\left(g\right)[/latex] spontaneous?
At room temperature, the equilibrium constant (K_w) for the self-ionization of water is 1.00 [latex]\times[/latex] 10⁻¹⁴. Using this information, calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water. (Hint: The reaction is the reverse of the self-ionization reaction.)
Hydrogen sulfide is a pollutant found in natural gas. Following its removal, it is converted to sulfur by the reaction [latex]{\text{2H}}_{2}\text{S}\left(g\right)+{\text{SO}}_{2}\left(g\right)\rightleftharpoons \frac{3}{8}{\text{S}}_{8}\left(s,\text{rhombic}\right)+{\text{2H}}_{2}\text{O}\left(l\right)[/latex]. What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic?
Consider the decomposition of CaCO₃(s) into CaO(s) and CO₂(g). What is the equilibrium partial pressure of CO₂ at room temperature?
In the laboratory, hydrogen chloride (HCl(g)) and ammonia (NH₃(g)) often escape from bottles of their solutions and react to form the ammonium chloride (NH₄Cl(s)), the white glaze often seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure of HCl and NH₃ in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.)
Benzene can be prepared from acetylene. [latex]{\text{3C}}_{2}{\text{H}}_{2}\left(g\right)\rightleftharpoons {\text{C}}_{6}{\text{H}}_{6}\left(g\right)[/latex]. Determine the equilibrium constant at 25 °C and at 850 °C. Is the reaction spontaneous at either of these temperatures? Why is all acetylene not found as benzene?
Carbon dioxide decomposes into CO and O₂ at elevated temperatures. What is the equilibrium partial pressure of oxygen in a sample at 1000 °C for which the initial pressure of CO₂ was 1.15 atm?
Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at 850 K.
[latex]{\text{CH}}_{4}\left(g\right)+{\text{4Cl}}_{2}\left(g\right)\longrightarrow {\text{CCl}}_{4}\left(g\right)+\text{4HCl}\left(g\right)[/latex]
What is the equilibrium constant for the reaction at 850 K? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?
Acetic acid, CH₃CO₂H, can form a dimer, (CH₃CO₂H)₂, in the gas phase.
[latex]{\text{2CH}}_{3}{\text{CO}}_{2}\text{H}\left(g\right)\longrightarrow {\left({\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right)}_{2}\left(g\right)[/latex]
The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer.

At 25 °C, the equilibrium constant for the dimerization is 1.3 [latex]\times[/latex] 10³ (pressure in atm). What is ΔS° for the reaction?
Nitric acid, HNO₃, can be prepared by the following sequence of reactions:
[latex]{\text{4NH}}_{3}\left(g\right)+{\text{5O}}_{2}\left(g\right)\longrightarrow \text{4NO}\left(g\right)+{\text{6H}}_{2}\text{O}\left(g\right)[/latex]
[latex]\text{2NO}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{2NO}}_{2}\left(g\right)[/latex]
[latex]{\text{3NO}}_{2}\left(g\right)+{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow {\text{2HNO}}_{3}\left(l\right)+\text{NO}\left(g\right)[/latex]
How much heat is evolved when 1 mol of NH₃(g) is converted to HNO₃(l)? Assume standard states at 25 °C.
Determine ΔG for the following reactions.
1. Antimony pentachloride decomposes at 448 °C. The reaction is: [latex]{\text{SbCl}}_{5}\left(g\right)\longrightarrow {\text{SbCl}}_{3}\left(g\right)+{\text{Cl}}_{2}\left(g\right)[/latex]
  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.
2. Chlorine molecules dissociate according to this reaction: [latex]{\text{Cl}}_{2}\left(g\right)\longrightarrow 2\text{Cl}\left(g\right)[/latex]
  1.00% of Cl₂ molecules dissociate at 975 K and a pressure of 1.00 atm.
Given that the [latex]\Delta {G}_{\text{f}}^{\circ }[/latex] for Pb²⁺(aq) and Cl⁻(aq) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, K_sp, for PbCl₂(s).
Determine the standard free energy change, [latex]\Delta {G}_{\text{f}}^{\circ }[/latex], for the formation of S²⁻(aq) given that the [latex]\Delta {G}_{\text{f}}^{\circ }[/latex] 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 [latex]\times[/latex] 10⁻⁵¹.
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.
[latex]{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{2}\text{O}\left(g\right)\Delta {G}_{298}^{\circ }=\text{8.58 kJ}[/latex]
1. Is the evaporation of water under standard thermodynamic conditions spontaneous?
2. Determine the equilibrium constant, K_P, for this physical process.
3. By calculating ∆G, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, [latex]{P}_{{\text{H}}_{2}\text{O}}[/latex], is 0.011 atm.
4. 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 [latex]{P}_{{\text{H}}_{2}\text{O}}[/latex] in the air?
In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation: [latex]\text{Glu}+\text{ATP}\longrightarrow \text{G6P}+\text{ADP}\Delta {G}_{298}^{\circ }=-\text{17 kJ}[/latex]
In this process, ATP becomes ADP summarized by the following equation: [latex]\text{ATP}\longrightarrow \text{ADP}\Delta {G}_{298}^{\circ }=-\text{30 kJ}[/latex]
Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process: [latex]\text{Glu}\longrightarrow \text{G6P}\Delta {G}_{298}^{\circ }=?[/latex]
One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P): [latex]\text{G6P}\rightleftharpoons \text{F6P}\Delta {G}_{298}^{\circ }=\text{1.7 kJ}[/latex]
1. Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?
2. 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.
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.
1. [latex]{\text{N}}_{2}\left(g\right)+{\text{3H}}_{2}\left(g\right)\longrightarrow {\text{2NH}}_{3}\left(g\right)[/latex]
2. [latex]\text{HCl}\left(g\right)+{\text{NH}}_{3}\left(g\right)\longrightarrow {\text{NH}}_{4}\text{Cl}\left(s\right)[/latex]
3. [latex]{\left({\text{NH}}_{4}\right)}_{2}{\text{Cr}}_{2}{\text{O}}_{7}\left(s\right)\longrightarrow {\text{Cr}}_{2}{\text{O}}_{3}\left(s\right)+{\text{4H}}_{2}\text{O}\left(g\right)+{\text{N}}_{2}\left(g\right)[/latex]
4. [latex]\text{2Fe}\left(s\right)+{\text{3O}}_{2}\left(g\right)\longrightarrow {\text{Fe}}_{2}{\text{O}}_{3}\left(s\right)[/latex]
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.
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: [latex]{\text{Cu}}_{2}\text{S}\left(s\right)\longrightarrow \text{Cu}\left(s\right)+\text{S}\left(s\right)[/latex]
1. Determine [latex]\Delta {G}_{298}^{\circ }[/latex] for the decomposition of Cu₂S(s).
2. The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine [latex]\Delta {G}_{298}^{\circ }[/latex] for the process.
3. 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 [latex]\Delta {G}_{298}^{\circ }[/latex] (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?
1. [latex]\text{S}\left(s\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{SO}}_{2}\left(g\right)[/latex]
2. [latex]{\text{2SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\longrightarrow {\text{SO}}_{3}\left(g\right)[/latex]
3. [latex]\text{HgO}\left(s\right)\longrightarrow \text{Hg}\left(l\right)+{\text{O}}_{2}\left(g\right)[/latex]

Show Selected Answers

2. [latex]\Delta {G}_{298}^{\circ }=\Delta {H}_{298}^{\circ }-T\Delta {S}_{298}^{\circ }[/latex]

[latex]\Delta {G}_{298}^{\circ }=100 - 298.15\times 250\frac{\text{J}}{\text{mol}\cdot\text{K}}\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)=\text{25.5 kJ/mol}[/latex]

The [latex]\Delta {G}_{298}^{\circ }>0[/latex], so the reaction is nonspontaneous at room temperature.

ΔG = ΔH − TΔS
[latex]0=100-T\left[250\frac{\text{J}}{\text{mol}\cdot\text{K}}\left(\frac{\text{1 kJ}}{\text{1000 J}}\right)\right][/latex]
T = 400 K

Above 400 K, ΔG will become negative, and the reaction will become spontaneous.

4. The reactions are spontaneous or nonspontaneous as follows:

[latex]\Delta {G}_{298}^{\circ }=\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{reactants}\right)[/latex]
[latex]\Delta {G}_{298}^{\circ }=\left[\text{1 mol}\left(\text{0 kJ/mol}\right)+\text{1 mol}\left(\text{0 kJ/mol}\right)\right]-\left[\text{1 mol}\left(-\text{465.1 kJ/mol}\right)\right]=465.1[/latex] nonspontaneous
[latex]\Delta {G}_{298}^{\circ }=\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{reactants}\right)[/latex]
[latex]\Delta {G}_{298}^{\circ }=\left[\text{2 mol}\left(-\text{53.43 kJ/mol}\right)\right]-\left[\text{1 mol}\left(\text{0 kJ/mol}\right)+\text{1 mol}\left(\text{0 kJ/mol}\right)\right]=-\text{106.86 kJ}[/latex] spontaneous
[latex]\Delta {G}_{298}^{\circ }=\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{reactants}\right)[/latex]
[latex]\Delta {G}_{298}^{\circ }=\left[\text{1 mol}\left(-\text{53.6 kJ/mol}\right)\right]-\left[\text{1 mol}\left(\text{0 kJ/mol}\right)+\text{1 mol}\left(\text{0 kJ/mol}\right)\right]=-\text{53.6 kJ}[/latex] spontaneous
[latex]\Delta {G}_{298}^{\circ }=\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{reactants}\right)[/latex]
[latex]\Delta {G}_{298}^{\circ }=\left[\text{1 mol}\left(-\text{1132.19 kJ/mol}\right)+\text{1 mol}\left(-\text{228.59 kJ/mol}\right)\right]-\left[\text{2 mol}\left(-\text{441.5 kJ/mol}\right)+\text{1 mol}\left(-\text{394.36 kJ/mol}\right)\right]=-\text{83.4 kJ}[/latex] spontaneous
[latex]\Delta {G}_{298}^{\circ }=\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{reactants}\right)[/latex]
[latex]\Delta {G}_{298}^{\circ }=\left[\text{1 mol}\left(\text{0 kJ/mol}\right)+\text{2 mol}\left(-\text{228.59 kJ/mol}\right)\right]-\left[\text{1 mol}\left(-\text{50.5 kJ/mol}\right)+\text{1 mol}\left(\text{0 kJ/mol}\right)\right]=-\text{406.7 kJ}[/latex] spontaneous
[latex]\Delta {G}_{298}^{\circ }=\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{reactants}\right)[/latex]
[latex]\Delta {G}_{298}^{\circ }=\left[\text{1 mol}\left(-\text{62.5 kJ/mol}\right)+\text{1 mol}\left(-\text{29.25 kJ/mol}\right)\right]-\left[\text{1 mol}\left(\text{66.8 kJ/mol}\right)+\text{3 mol}\left(\text{0 kJ/mol}\right)\right]=-\text{30.0 kJ}[/latex] spontaneous

6. (a) The standard free energy of formation is the standard free energy change for [latex]\frac{1}{4}{\text{P}}_{4}\left(s\right)+\frac{3}{2}{\text{H}}_{2}\left(g\right)+{\text{2O}}_{2}\left(g\right)\longrightarrow {\text{H}}_{3}{\text{PO}}_{4}\left(l\right)[/latex]. We can use a Hess’s law-like approach. Note that adding the first reaction plus three times the second reaction plus the third reaction gives, after canceling terms:

[latex]{\text{P}}_{4}\left(s\right)+{\text{6H}}_{2}\left(g\right)+{\text{8O}}_{2}\left(g\right)\longrightarrow {\text{4H}}_{3}{\text{PO}}_{4}\left(l\right)[/latex]

[latex]\Delta {G}_{\text{rxn}}^{\circ }=\left[\left(-2697.0\right)+3\left(-457.18\right)+\left(-428.66\right)\right]\text{kJ/mol}=-\text{4497.2 kJ/mol}[/latex]

Dividing this result by four gives the equation of interest, and it gives −1124.3 kJ/mol for the standard free energy change.

(b) The calculation agrees with the value in Standard Thermodynamic Properties for Selected Substances 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.

8. (a) Using the data in Standard Thermodynamic Properties for Selected Substances, determine [latex]\Delta {G}_{298}^{\circ }:[/latex]

[latex]\Delta {G}_{298}^{\circ }=2\Delta {G}_{\text{f}}^{\circ }\left(\text{Hg}\left(l\right)\right)+\Delta {G}_{\text{f}}^{\circ }\left({\text{O}}_{2}\left(g\right)\right)-2\Delta {G}_{\text{f}}^{\circ }\left(\text{HgO}\left(s,\text{red}\right)\right)=\left\{2\left(0\right)+0 - 2\left(-58.5\right)\right\}\text{kJ/mol}=\text{117.0 kJ/mol}[/latex]

From its value at 298.15 K, the reaction is nonspontaneous

(b) requires the ratio of the standard enthalpy change to the standard entropy change:

[latex]\Delta {H}_{298}^{\circ }=2\Delta {H}_{\text{f}}^{\circ }\left(\text{Hg}\left(l\right)\right)+\Delta {H}_{\text{f}}^{\circ }\left({\text{O}}_{2}\left(g\right)\right)-2\Delta {H}_{\text{f}}^{\circ }\left(\text{HgO}\left(s,\text{red}\right)\right)[/latex]
= [2(0) + 0 − 2(−90.83)] kJ/mol = 181.66 kJ/mol
[latex]\Delta {S}_{298}^{\circ }=2{S}_{298}^{\circ }\left(\text{Hg}\left(l\right)\right)+{S}_{298}^{\circ }\left({\text{O}}_{2}\left(g\right)\right)-2{S}_{298}^{\circ }\left(\text{HgO}\left(s,\text{red}\right)\right)[/latex]
= [2(75.9) + 205.0 − 2(70.29)] J/K·mol = 216.42 J/K·mol
[latex]T=\frac{\Delta {H}_{298}^{\circ }}{\Delta {S}_{298}^{\circ }}=\frac{181.66\times {10}^{3}\text{J/mol}}{\text{216.42 J/K\cdot mol}}=\text{839 K}=566^{\circ C}[/latex]

Above 566 °C the process is spontaneous.

10. ΔG° for each is as follows:

ΔG° = −RT ln K_P = −(8.314 J K⁻¹)(2273.15 K)(ln 4.1 [latex]\times[/latex] 10⁻⁴) = 147 kJ = 1.5 [latex]\times[/latex] 10² kJ
ΔG° = −(8.314 J K⁻¹)(673.15 K)(ln 50.0) = −21,893 J = 21.9 kJ
ΔG° = −(8.314 J K⁻¹)(1253.15 K)(ln 1.67) = −5.34 kJ
ΔG° = −(8.314 J K⁻¹)(1173.15 K)(ln 1.04) = −0.383 kJ
ΔG° = −(8.314 J K⁻¹)(298.15 K)(ln 7.2 [latex]\times[/latex] 10⁻⁴) = 17,937 J = 18 kJ
ΔG° = −(8.314 J K⁻¹)(298.15 K)(ln 3.3 [latex]\times[/latex] 10⁻¹³) = 71,240 J = 71 kJ

12. Equilibrium constants are calculated from [latex]\text{ln}K=\frac{-\Delta G^{\circ }}{RT}[/latex]. Note that K is a function of T and thus changes as T changes.

[latex]\text{ln}K=-\left[\frac{-9.2\times {10}^{3}\text{J}}{\left({\text{8.314 J mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\times \text{298.15 K}\right)}\right]=3.71[/latex], K = 41
[latex]\text{ln}K=\frac{-7300\text{J}}{\left(8.314\text{J}{\text{mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\times 298.15\text{K}\right)}=-2.945[/latex], K = 0.053
[latex]\text{ln}K=-\left[\frac{-79\times {10}^{3}\text{J}}{\left({\text{8.314 J mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\times \text{298.15 K}\right)}\right]=31.870[/latex], K = 6.9 [latex]\times[/latex] 10¹³
[latex]\text{ln}K=-\left[\frac{-1.6\times {10}^{3}\text{J}}{\left({\text{8.314 J mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\times \text{298.15 K}\right)}\right]=0.645[/latex], K = 1.9
[latex]\text{ln}K=\frac{-\text{8}\times {10}^{3}\text{J}}{\left(8.314\text{J}{\text{mol}}^{-\text{1}}{\text{K}}^{-\text{1}}\times 298.15\text{K}\right)}=-3.227[/latex], K = 0.04

14. In each of the following, the value of ΔG is not given at the temperature of the reaction. Therefore, we must calculate ΔG from the values ΔH° and ΔS and then calculate ΔG from the relation ΔG = ΔH° − TΔS°.

(a)

[latex]\begin{array}{l}\\ \Delta H^{\circ }=2\Delta {H}_{{\text{f}}_{{\text{OF}}_{2}\left(g\right)}}^{\circ }-\Delta {H}_{{\text{f}}_{{\text{O}}_{2}\left(g\right)}}^{\circ }-\Delta {H}_{{\text{f}}_{{\text{F}}_{2}\left(g\right)}}^{\circ }=\text{2 mol}\left(-22{\text{kJ mol}}^{-\text{1}}\right)-0 - 2\left(0\right)=-\text{44 kJ}=-4.4\times {10}^{4}\text{J}\\ \Delta S^{\circ }=2\Delta {S}_{{\text{OF}}_{2}\left(g\right)}^{\circ }-\Delta {S}_{{\text{O}}_{2}\left(g\right)}^{\circ }-2\Delta {S}_{{\text{F}}_{2}\left(g\right)}^{\circ }=2\left(247.3\right)-205.03 - 2\left(202.7\right)=-115.83\text{J}\\ \Delta G=\Delta H^{\circ }-T\Delta S^{\circ }=-\text{44,000 J}-373.15\left(-115.83\text{J}\right)=-\text{778 J}\\ \text{ln}K=\frac{\Delta G}{-RT}=\frac{-778}{-8.314\times 373.15}=0.2507,K=1.29\end{array}[/latex]

(b)

[latex]\begin{array}{l}\Delta H^{\circ }=2\Delta {H}_{{\text{f}}_{\text{IBr}\left(g\right)}}^{\circ }-\Delta {H}_{{\text{f}}_{{\text{I}}_{2}\left(s\right)}}^{\circ }-\Delta {H}_{{\text{f}}_{{\text{Br}}_{2}\left(l\right)}}^{\circ }=2\left(\text{40,800 J}\right)=\text{81,600 J}\\ \Delta S^{\circ }=2\Delta {S}_{\text{IBr}\left(g\right)}^{\circ }-\Delta {S}_{{\text{I}}_{2}\left(s\right)}^{\circ }-2\Delta {S}_{{\text{Br}}_{2}\left(l\right)}^{\circ }=2\left(258.66\right)-116.14 - 152.23=248.95\\ \Delta G=81,600 - 273.15\left(248.95\right)=1.36\times {10}^{4}\text{J}\\ \text{ln}K=\frac{13,600}{-8.314\left(273.15\right)}=-5.989,K=2.51\times {10}^{-3}\end{array}[/latex]

(c)

[latex]\begin{array}{l}\Delta H^{\circ }=2\Delta {H}_{{\text{f}}_{{\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)}}^{\circ }+\Delta {H}_{{\text{f}}_{{\text{H}}_{2}\text{O}\left(g\right)}}^{\circ }-2\Delta {H}_{{\text{f}}_{\text{LiOH}\left(g\right)}}^{\circ }-\Delta {H}_{{\text{f}}_{\text{C}{\text{O}}_{2}\left(g\right)}}^{\circ }=-1215.6 - 241.82 - 2\left(-487.23\right)-\left(-393.51\right)=-\text{44 kJ}\\ \Delta S^{\circ }=\Delta {S}_{{\text{f}}_{{\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)}}^{\circ }+\Delta {S}_{{\text{f}}_{{\text{H}}_{2}\text{O}\left(g\right)}}^{\circ }-2\left(\Delta {S}_{{\text{f}}_{\text{LiOH}\left(g\right)}}^{\circ }\right)-\Delta {S}_{{\text{f}}_{{\text{CO}}_{2}\left(g\right)}}^{\circ }=90.4+188.71 - 2\left(50.2\right)-213.6=-34.9{\text{J K}}^{-\text{1}}\\ \Delta G=\Delta H^{\circ }-T\Delta S^{\circ },T=575^{\circ C}+273.15=\text{848.15 K}\\ \Delta G=-89,400 - 848.15\left(-34.9\right)=-5.9800\times {10}^{-4}\\ \text{ln}K=\frac{\Delta G}{-RT}=\frac{-5.98\times {10}^{4}}{-8.314\times 848.15}=8.4816,K=4.83\times {10}^{3}\end{array}[/latex]

(d)

[latex]\begin{array}{l}\Delta H^{\circ }=\Delta {H}_{{\text{f}}_{\text{NO}\left(g\right)}}^{\circ }+\Delta {H}_{{\text{f}}_{{\text{NO}}_{2}\left(g\right)}}^{\circ }-\Delta {H}_{{\text{f}}_{{\text{N}}_{2}\text{O}\left(g\right)}}^{\circ }=90.25+33.2 - 83.72=\text{39.73 kJ}\\ \Delta S^{\circ }=\Delta {S}_{\text{NO}\left(g\right)}^{\circ }+\Delta {S}_{{\text{f}}_{{\text{NO}}_{2}\left(g\right)}}^{\circ }-\Delta {S}_{{\text{N}}_{2}{\text{O}}_{3}\left(g\right)}^{\circ }=210.65+239.9 - 312.2={\text{138.35 J K}}^{-1}\\ \Delta G=\Delta H^{\circ }-T\Delta S^{\circ }=39,730 - 263.15\left(138.35\right)=3323\\ \text{ln}K=\frac{\Delta G}{-RT}=\frac{3323}{-8.314\times 263.15}=-1.519,K=0.219\end{array}[/latex]

(e)

[latex]\begin{array}{l}\Delta H^{\circ }=2\Delta {H}_{{\text{f}}_{{\text{SnCl}}_{4}\left(g\right)}}^{\circ }-\Delta {H}_{{\text{f}}_{{\text{SnCl}}_{4}\left(l\right)}}^{\circ }=-471.5{\text{kJ mol}}^{-\text{1}}-\left(-511.2{\text{kJ mol}}^{-\text{1}}\right)={\text{39.7 kJ mol}}^{-1}\\ \Delta S^{\circ }=\Delta {S}_{{\text{SnCl}}_{4}\left(g\right)}^{\circ }-{\Delta }_{S}^{{\text{SnCl}}_{4}\left(l\right)}=-366{\text{J K}}^{-\text{1}}{\text{mol}}^{-\text{1}}\\ \Delta G=\Delta H^{\circ }-T\Delta S^{\circ }=39,700 - 473.15\left(107\right)=-\text{10.927 kJ}\\ \text{ln}K=\frac{-10,927}{-8.314\times 473.15}=2.778,K=16.1\end{array}[/latex]

16. The standard free energy change is [latex]\Delta {G}_{298}^{\circ }=-RT\text{ln}K=\text{4.84 kJ/mol}[/latex].

When reactants and products are in their standard states (1 bar or 1 atm), Q = 1. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): Q < 1, and [latex]\Delta {G}_{298}[/latex] becomes less positive as it approaches zero. At equilibrium, Q = K, and ΔG = 0.

18. This reaction will become spontaneous when ΔG goes from positive to negative, or at the point where ΔG = 0. [latex]\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }T=\frac{\Delta H}{\Delta S}[/latex]

Calculate ΔH° and ΔS°:

[latex]\begin{array}{l}\Delta H^{\circ }=\Delta {H}_{\text{NO}\left(g\right)}^{\circ }+\Delta {H}_{{\text{NO}}_{2}\left(g\right)}^{\circ }-\Delta {H}_{{\text{N}}_{2}{\text{O}}_{3}\left(g\right)}^{\circ }=\left(90.25+33.2 - 83.72\right)\text{kJ}=\text{39.73 kJ}\\ \Delta S^{\circ }={S}_{\text{NO}\left(g\right)}^{\circ }+{S}_{{\text{NO}}_{2}\left(g\right)}^{\circ }-{S}_{{\text{N}}_{2}{\text{O}}_{3}\left(g\right)}^{\circ }=\left(210.65+239.9 - 312.2\right){\text{J K}}^{-\text{1}}=138.35{\text{J K}}^{-\text{1}}\\ \Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }=\text{39.73 kJ}-\left(\text{298.15 K}\right)\left({\text{0.13835 kJ K}}^{-\text{1}}\right)=-\text{1.519 kJ}\end{array}[/latex]

Thus this reaction is spontaneous at 298 K because ΔG is negative at this temperature, and because TΔS° term is negative, ΔG will become progressively more negative as the temperature increases above 298 K. Thus the temperature at which ΔG first becomes negative must lie at a temperature lower than 298 K. We can find this temperature by substituting into [latex]T=\frac{\Delta H}{\Delta S}[/latex].

[latex]T=\frac{\Delta H}{\Delta S}=\frac{\text{39,700 J}}{{\text{138.35 J K}}^{-\text{1}}}=\text{287 K}[/latex]

The reaction will be spontaneous at temperatures greater than 287 K.

20. Find ΔG° for the reaction, and from ΔG° find K.

[latex]\begin{array}{l}\Delta G^{\circ }=3\Delta {G}_{\text{S}\left(s\right)}^{\circ }+2\Delta {G}_{{\text{H}}_{2}\text{O}\left(g\right)}^{\circ }-2\Delta {G}_{{\text{H}}_{2}\text{S}\left(g\right)}^{\circ }-\Delta {G}_{{\text{SO}}_{2}\left(s\right)}^{\circ }=3\left(0\right)+2\left(-228.59\right)-2\left(-33.6\right)-\left(-300.19\right)=-\text{89.79 kJ}\\ \Delta G^{\circ }=-RT\text{ln}K=-2.303RT\text{ log }K\\ \\ K=\text{antilog}\left(\frac{-\Delta G}{2.303RT}\right)=\left[\frac{89,790}{2.303\left(8.314\right)\left(298.15\right)}\right]=\text{antilog }15.729=5.35\times {10}^{15}\end{array}[/latex]

The process is exothermic.

22. The equilibrium may be written as

[latex]{\text{NH}}_{4}\text{Cl}\left(s\right)\rightleftharpoons \text{HCl}\left(g\right)+{\text{NH}}_{3}\left(g\right){K}_{P}={P}_{{\text{NH}}_{3}}{P}_{\text{HCl}}={x}^{2}[/latex]

to help simplify the mathematics. Use the standard free energy change to determine the equilibrium constant for the reaction: [latex]\Delta {G}_{298}^{\circ }=\text{91.1 kJ/mol}[/latex], K_P = 1.1 [latex]\times[/latex] 10⁻¹⁶. At equilibrium, the partial pressures are equal so that [latex]x=\sqrt{{K}_{P}}=1.0\times {10}^{-\text{8}}\text{atm}[/latex]. This is the maximum pressure of the gases under the stated conditions.

24. The reaction is [latex]{\text{2CO}}_{2}\left(g\right)\longrightarrow \text{2CO}\left(g\right)+{\text{O}}_{2}\left(g\right)[/latex].

ΔH° = (2 mol)(−110.52 kJ mol⁻¹) + (1 mol)(0) − (2 mol)(−393.51 kJ mol⁻¹) = 565.98 kJ
ΔS° = (2 mol)(197.56 J K⁻¹ mol⁻¹) + (1 mol)(205.03 J K⁻¹ mol⁻¹) − (2 mol)(−393.51 kJ mol⁻¹) = 172.95 J K⁻¹
ΔG° = (565,980 J) − (1273 K)(172.95 J K⁻¹) = 345,810 J

[latex]\begin{array}{rll}{}\text{log }K&=&\frac{\text{345,810 J}}{\left(-2.303\right)\left(8.314{\text{J K}}^{-\text{1}}\right)\left(1273K\right)}=-14.1975\\{}K&=&6.494\times {10}^{-15}=\frac{{\left({P}_{\text{CO}}\right)}^{2}\left({P}_{{\text{O}}_{2}}\right)}{\left({P}_{{\text{CO}}_{2}}\right)}=\frac{{\left(2x\right)}^{2}\left(x\right)}{{\left(1.15 - 2x\right)}^{2}}\approx \frac{{\left(2x\right)}^{2}\left(x\right)}{{\left(1.15\right)}^{2}}\\ 4{x}^{3}&=&8.59\times {10}^{-15}\\ x&=&1.29\times {10}^{-5}\text{atm}={P}_{{\text{O}}_{2}}\end{array}[/latex]

26. The equilibrium constant allows the calculation of [latex]\Delta {G}_{298}^{\circ }[/latex].

[latex]\begin{array}{rll}\Delta {G}_{298}^{\circ }&=&-RT\text{ln}K=-2.303RT\text{log}K=-2.303\left(8.314{\text{J K}}^{-\text{1}}{\text{mol}}^{-\text{1}}\right)\left(298.2K\right)\left(\text{log}1.3\times {10}^{3}\right)\\{}&=&-17.78{\text{kJ mol}}^{-\text{1}}=1.8{\text{kJ mol}}^{-\text{1}}\left(\text{2 significant figures}\right)\end{array}[/latex]

The strength of the bond (66.5 kJ) means that it requires 66.5 kJ to pull 1 mol of bonds apart. In other words, [latex]\Delta {H}_{298}^{\circ }=-\text{66.5 kJ}[/latex]. The values of [latex]\Delta {G}_{298}^{\circ }[/latex] and [latex]\Delta {H}_{298}^{\circ }[/latex] allow us to calculate ΔS° by use of the equation

[latex]\begin{array}{rll}\Delta {G}_{298}^{\circ }&=&\Delta {H}_{298}^{\circ }-T\Delta {H}_{298}^{\circ }\\{}\Delta S^{\circ }&=&-\frac{\Delta {G}_{298}^{\circ }-\Delta {H}_{298}^{\circ }}{T}=-\frac{\left(-\text{17,780 J}\right)-\left(-\text{66,500 J}\right)}{\text{298.2 K}}=-163{\text{J K}}^{-\text{1}}=-\text{0.16 kJ}\end{array}[/latex]

28. (a) Find the equilibrium constant by first calculating the number of moles of each component.

molar masses: SbCl₅ = 299.11 g mol⁻¹
SbCl₃ = 228.11 g mol⁻¹
Cl₂ = 70.9054 g mol⁻¹

[latex]\begin{array}{rll}{}{\text{mol SbCl}}_{5}&=&\frac{\text{3.85 g}}{299.01g{\text{mol}}^{-\text{1}}}=\text{0.0129 mol}\\ {\text{mol SbCl}}_{3}&=&\frac{\text{9.14 g}}{228.11{\text{g mol}}^{-\text{1}}}=\text{0.0401 mol}\\ {\text{mol Cl}}_{2}&=&\frac{\text{2.84 g}}{70.9054{\text{g mol}}^{-\text{1}}}=\text{0.0401 mol}\end{array}[/latex]

[latex]K=\frac{\left[{\text{SbCl}}_{3}\right]\left[C{l}_{2}\right]}{\left[{\text{SbCl}}_{5}\right]}=\frac{\left(\frac{\text{0.0401 mol}}{\text{5.00 L}}\right)\left(\frac{\text{0.0401 mol}}{\text{5.00 L}}\right)}{\left(\frac{\text{0.0129 mol}}{\text{5.00 L}}\right)}=0.0249{\text{mol L}}^{-\text{1}}[/latex]

At 448 °C = 721.15 K, [latex]\Delta G^\circ =-RT\text{ln}K=-\left(8.314{\text{J K}}^{-\text{1}}\right)\text{(721.15 K)(ln 0.0249)}=-\text{2.21}\times {10}^{4}\text{J}=-\text{22.1 kJ}[/latex]

(b) Determine the value of K and then determine the value of ΔG° from the value of K. Molar concentrations are required to find K. The concentration of Cl₂ at 1.00 atm and 975 K is determined from the ideal gas law, PV = nRT

[latex]\frac{n}{V}=\frac{P}{RT}=\frac{\text{1.00 atm}}{0.08206\frac{\text{L atm}}{\text{K mol}}\left(\text{975 K}\right)}=0.0125M[/latex]

Since the Cl₂ gas is 1.00% dissociated, the concentration of Cl₂ is 0.9900 [latex]\times[/latex] 0.0125 M = 0.0124 M. The concentration of Cl₂ is 2 [latex]\times[/latex] 0.0100 [latex]\times[/latex] 0.0125 = 0.000250 M (we have to use the factor of two since each dissociated Cl₂ molecule produces two Cl particles).

[latex]K=\frac{{\left[\text{Cl}\right]}^{2}}{\left[{\text{Cl}}_{2}\right]}=\frac{{0.00250}^{2}}{0.0125}=0.000500[/latex]

Using this result, ΔG° = −RT ln K = −8.314 J/mol K [latex]\times[/latex] 975 K ln (0.000500) = 61,600 J/mol = 61.6 kJ/mol.

30. The reaction of interest is: [latex]{\text{Ag}}_{2}\text{S}\left(s\right)\rightleftharpoons {\text{2Ag}}^{\text{+}}\left(aq\right)+{\text{S}}^{2-}\left(aq\right){K}_{\text{sp}}=8\times {10}^{-51}[/latex]

[latex]\begin{array}{rll}\Delta {G}_{298}^{\circ }&=&-RT\text{ln}K\\ \Delta {G}_{298}^{\circ }&=&-8.314\times \text{298.15 ln}8\times {10}^{-51}\\ \Delta {G}_{298}^{\circ }&=&\text{286 kJ}\\{}\Delta {G}_{298}^{\circ }&=&\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{products}\right)-\sum \nu \Delta {G}_{\text{f}}^{\circ }\left(\text{reactants}\right)\end{array}[/latex]

[latex]\begin{array}{l}\\ 286=\left[\text{1 mol}\left(\Delta {G}_{\text{f}}^{\circ }{S}^{2-}\text{kJ/mol}\right)+\text{2 mol}\left(\text{77.1 kJ/mol}\right)\right]-\left[\text{1 mol}\left(-\text{39.5 kJ/mol}\right)\right]\\ \Delta {G}_{\text{f}}^{\circ }{S}^{2-}=\text{90 kJ/mol}\end{array}[/latex]

32. The answers are as follows:

Under standard thermodynamic conditions, the temperature is 298 K, and the pressure of water vapor would be 1 atm (or 1 bar). Under these conditions, the evaporation is nonspontaneous as indicated by [latex]\Delta {G}_{298}^{\circ }>0[/latex]
[latex]\Delta {G}_{298}^{\circ }=-RT\text{ln}K[/latex], 8.58 [latex]\times[/latex] 10³ = −8.314 [latex]\times[/latex] 2981.15 ln K, K_p = 0.031
[latex]Q={P}_{{\text{H}}_{2}\text{O}}=\text{0.011 atm}[/latex], [latex]\Delta G=\Delta {G}_{298}^{\circ }+RT\text{ln}Q[/latex], ΔG = 8.58 [latex]\times[/latex] 10³ + (8.314 [latex]\times[/latex] 298 [latex]\times[/latex] ln 0.011) = −2.6 kJ, Under these conditions, the evaporation of water is spontaneous
[latex]{P}_{{\text{H}}_{2}\text{O}}[/latex] must always be less than K_p or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.

34.The answers are as follows:

Nonspontaneous as [latex]\Delta {G}_{298}^{\circ }>0[/latex]
[latex]\Delta {G}_{298}^{\circ }=-RT\text{ln}K[/latex], [latex]\Delta G=1.7\times {10}^{3}+\left(8.314\times 335\times \text{ln}\frac{28}{128}\right)=-\text{2.5 kJ}[/latex]. The forward reaction to produce F6P is spontaneous under these conditions.

36. 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.

38. The answers are as follows:

Increasing [latex]{P}_{{\text{O}}_{2}}[/latex] will shift the equilibrium toward the products, which increases the value of K. [latex]\Delta {G}_{298}^{\circ }[/latex] therefore becomes more negative.
Increasing [latex]{P}_{{\text{O}}_{2}}[/latex] will shift the equilibrium toward the products, which increases the value of K. [latex]\Delta {G}_{298}^{\circ }[/latex] therefore becomes more negative.
Increasing [latex]{P}_{{\text{O}}_{2}}[/latex] will shift the equilibrium the reactants, which decreases the value of K. [latex]\Delta {G}_{298}^{\circ }[/latex] 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 [latex]\left(\Delta {G}_{\text{f}}^{\circ }\right)[/latex]: change in free energy accompanying the formation of one mole of substance from its elements in their standard states

Chapter 16: Thermodynamics

Learning Objectives

Calculating Free Energy Change

Example 1: Evaluation of ΔG° Change from ΔH° and ΔS°

Check Your Learning

Example 2: Calculation of [latex]\Delta {G}_{298}^{\circ }[/latex]

Check Your Learning

Temperature Dependence of Spontaneity

Example 3: Predicting the Temperature Dependence of Spontaneity

Check Your Learning

Example 4: Equilibrium Temperature for a Phase Transition

Check Your Learning

Free Energy and Equilibrium

Example 5: Calculating ΔG under Nonstandard Conditions

Check Your Learning

Example 6

Calculating an Equilibrium Constant using Standard Free Energy Change

Check Your Learning

Key Concepts and Summary

Key Equations

Exercises

Glossary

Candela Citations