5.5: Free Energy

Learning Objectives

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

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

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

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

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

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

(For simplicity’s sake, the subscript “sys” will be omitted henceforth.)

The relationship between this system property and the spontaneity of a process may be understood by recalling the previously derived second law expression:

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

The first law requires that \(q_{surr} = −q_{sys}\), and at constant pressure \(q_{sys} = ΔH\), so this expression may be rewritten as:

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

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

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

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

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

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 \(\PageIndex{1}\) summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.

Table \(\PageIndex{1}\): Relation between Process Spontaneity and Signs of Thermodynamic Properties

\(ΔS_{univ} > 0\)	ΔG < 0	moves spontaneously in the forward direction, as written, to reach equilibrium
\(ΔS_{univ} < 0\)	ΔG > 0	nonspontaneous in the forward direction, as written, but moves spontaneously in the reverse direction, as written, to reach equilibrium
\(ΔS_{univ} = 0\)	ΔG = 0	reversible (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

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

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:

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

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

However, as noted previously in this chapter, such conditions are not realistic. In addition, the technologies used to extract work from a spontaneous process (e.g., batteries) are never 100% efficient, and so the work done by these processes is always less than the theoretical maximum. Similar reasoning may be applied to a nonspontaneous process, for which the free energy change represents the minimum amount of work that must be done on the system to carry out the process.

Calculating Free Energy Change

Free energy is a state function, so its value depends only on the conditions of the initial and final states of the system. A convenient and common approach to the calculation of free energy changes for physical and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes, \(ΔG^o\), according to the following relation.

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

It is important to understand that for phase changes, \(\Delta G^º\) tells you if the phase change is spontaneous or not; will it happen, or not happen. For chemical reactions, \(\Delta G^º\) tells you the extent of a reaction. In other words, \(\Delta G^º\) for a reaction tells you how much product will be present at equilibrium. A reaction with \(\Delta G^º\) < 0 is considered product-favored at equilibrium; there will be more products than reactants when the reaction reaches equilibrium. A reaction with \(\Delta G^º\) > 0 is considered reactant-favored at equilibrium; there will be more reactants than products when the reaction reaches equilibrium.

Example \(\PageIndex{1}\): 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 \(ΔG^o\) say about the spontaneity of this process?

Solution

The process of interest is the following:

\[\ce{H_2O(l) \longrightarrow H_2O(g)} \nonumber \]

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

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

From Appendix G:

Substance	\(\Delta H_{ f }^{\circ} (\text{kJ/mol})\)	\(S^{\circ} (\text{kJ/K mol})\)
H2O(l)	−285.83	70.0
H2O(g)	−241.82	188.8

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

\[\begin{align*}
\Delta H^{\circ} &=\Delta H_{ f }^{\circ}\left( \ce{H2O(g)} \right)-\Delta H_{ f }^{\circ}\left( \ce{H2O(l)} \right) \\[4pt]
&=[-241.82 ~\text{kJ/mol} -(-285.83)] ~\text{kJ/mol} =44.01 ~\text{kJ/mol} \\[4pt]
\Delta S^{\circ} &=1 ~\text{mol} \times S^{\circ}\left( \ce{H2O(g)} \right)-1 ~\text{mol} \times S^{\circ}\left( \ce{H2O(l)} \right) \\[4pt]
&=(1 ~\text{mol}) ~188.8 J / mol \cdot K -(1 ~\text{mol}) ~ 70.0 ~ \text{J} / \text{mol K} =118.8 ~ \text{J / K} \\[4pt]
\Delta G^{\circ} &=\Delta H^{\circ}-T \Delta S^{\circ}
\end{align*} \nonumber \]

Substitution into the standard free energy equation yields:

\[\begin{align*}
\Delta G^{\circ} &=\Delta H^{\circ}-T \Delta S^{\circ} \\[4pt]
&=44.01~\text{kJ} -(298 K \times 118.8~\text{J/K}) \times \frac{1 ~\text{kJ}}{1000~\text{J} } \\[4pt]
&=44.01~\text{kJ} -35.4~\text{kJ} \\[4pt] &=8.6~\text{kJ}
\end{align*} \nonumber \]

At 298 K (25 °C) so boiling is nonspontaneous (not spontaneous).

Exercise \(\PageIndex{1}\)

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?

\[\ce{C2H6(g) \longrightarrow H2(g) + C2H4(g)} \nonumber \]

Answer

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 \(ΔG_f^o\) values of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpy of formation, \(ΔG_f^o\) is by definition zero for elemental substances in their standard states. The approach used to calculate \(ΔG^o\) for a reaction from \(ΔG_f^o\) values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction

\[m A +n B \longrightarrow x C +y D \nonumber \]

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

\[\begin{align*}
\Delta G^{\circ} &= \sum \nu \Delta G^{\circ}(\text { products })-\sum \nu \Delta G^{\circ}(\text { reactants }) \\[4pt]
&= \left[x \Delta G_{ f }^{\circ}( C )+y \Delta G_{ f }^{\circ}( D )\right]-\left[m \Delta G_{ f }^{\circ}( A )+n \Delta G_{ f }^{\circ}( B )\right] .
\end{align*} \nonumber \]

Example \(\PageIndex{2}\): Using Standard Free Energies of Formation to Calculate ΔG°

Consider the decomposition of yellow mercury(II) oxide.

\[\ce{HgO (s, \text { yellow }) -> Hg (l) + 1/2 O2(g)} \nonumber \]

Calculate the standard free energy change at room temperature, 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	\(\Delta G_{ f }^{\circ}( \text{kJ / mol} )\)	\(\Delta H_{ f }^{\circ}( \text{kJ / mol} )\)	\(S^{\circ}( \text{kJ / K mol} )\)
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:

\[\begin{align*}
\Delta G^{\circ} & =\sum \nu G_{ f }^{\circ}(\text { products })-\sum \nu \Delta G_{ f }^{\circ}(\text { reactants }) \\[4pt]
&=\left[1 \Delta G_{ f }^{\circ}~ \ce{Hg(l)} + \dfrac{1}{2} \Delta G_{f}^{\circ} ~\ce{O2(g)} \right]-1 \Delta G_{ f }^{\circ} ~\ce{HgO(s, yellow)} \\[4pt]
&=\left[1 ~\text{mol} (0 ~\text{kJ / mol} )+\frac{1}{2} ~\text{mol} (0 ~\text{kJ / mol} )\right] - 1 ~\text{mol} (-58.43 ~\text{kJ / mol} )=58.43 ~\text{kJ / mol}
\end{align*} \nonumber \]

(b) Using enthalpies and entropies of formation:

\[\begin{align*}
\Delta H^{\circ}&=\sum \nu \Delta H_{ f }^{\circ}(\text { products })-\sum \nu \Delta H_{ f }^{\circ}(\text { reactants }) \\[4pt]
&=\left[1 \Delta H_{ f }^{\circ}~\ce{Hg(l)} +\frac{1}{2} \Delta H_{ f }^{\circ}~\ce{O2(g)} \right]-1 \Delta H_{ f }^{\circ} ~\ce{HgO (s, yellow )} \\[4pt]
&=\left[1 ~\text{mol} (0~ \text{kJ / mol} ) + \frac{1}{2} ~\text{mol} (0 ~\text{kJ / mol} )\right]-1 ~\text{mol} (-90.46~\text{kJ / mol} ) = 90.46 ~\text{kJ / mol} \\[8pt]
\Delta S^{\circ} &=\sum \nu \Delta S^{\circ}(\text { products })-\sum \nu \Delta S^{\circ}(\text { reactants }) \\[4pt]
&=\left[1 ~\Delta S^{\circ} ~\ce{Hg (l)} + \frac{1}{2} \Delta S^{\circ} ~\ce{O2(g)} \right]-1 \Delta S^{\circ} ~\ce{HgO (s, yellow )} \\[4pt]
& =\left[1 ~\text{mol} (75.9~\text{J / mol K} ) + \frac{1}{2} ~\text{mol} (205.2~\text{J / mol K} )\right] -1 \text{mol} (71.13 ~\text{J / mol K} )=107.4 ~\text{ J / mol K} \\[8pt]
\Delta G^{\circ} &=\Delta H^{\circ}-T \Delta S^{\circ}=90.46 ~\text{kJ} - 298.15 ~\text{K} \times 107.4 ~\text{J / K} \cdot \text{mol} \times \frac{1 ~\text{ kJ} }{1000 ~\text{J} } \\[4pt]
& =(90.46 - 32.01) ~\text{kJ / mol} =58.45 ~\text{kJ / mol}
\end{align*} \nonumber \]

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.

Exercise \(\PageIndex{1}\)

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?

\[\ce{C2H4(g) \longrightarrow H2(g) + C2H2(g)} \nonumber \]

Answer

1. 140.8 kJ/mol, nonspontaneous
2. 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:

\[\ce{H_2O(l) -> H_2O(g)} \nonumber \]

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:

\[\begin{align*} \ce{H2(g) + 1/2 O2(g) &-> H2O(g)} && \Delta G_{ f }^{\circ} \text { gas } \\[4pt]
\ce{H2O(l) &-> H2(g) + 1/2 O2(g)} && - \Delta G_{ f }^{ o } \text { liquid } \\[4pt]
\hline \ce{H2O(l) &-> H2O(g)} && \Delta G^{\circ}=\Delta G_{ f }^{\circ} ~\text{gas} - \Delta G_{ f }^{\circ} ~\text{liquid}
\end{align*} \]

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 \(ΔG^o\):

\[\ce{ZnS (s) \rightarrow Zn(s) + S(s)} \quad \Delta G_1^{\circ}=201.3 kJ \nonumber \]

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

\[\ce{S(s) + O2(g) \rightarrow SO2(g)} \quad \Delta G_2^{\circ}=-300.1 kJ \nonumber \]

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

\[\ce{ZnS(s) + O2(g) \rightarrow Zn(s) + SO2(g)} \nonumber \]

\[\Delta G^{\circ}=201.3 ~\text{kJ} + -300.1 ~\text{kJ} =-98.8 ~\text{kJ} \nonumber \]

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.

Example \(\PageIndex{3}\): Calculating Free Energy Change for a Coupled Reaction

Is a reaction coupling the decomposition of \(\ce{ZnS}\) to the formation of \(\ce{H2S}\) expected to be spontaneous under standard conditions?

Solution

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

\[\begin{align*} &\text{Decomposition of zinc sulfide:} & \ce{ZnS (s) \rightarrow Zn(s) + S(s)} && \Delta G_1^{\circ}=201.3 ~\text{kJ}\\[4pt]
&\text{Formation of hydrogen sulfide:} & \ce{S(s) + H2(g) \rightarrow H2S(g)} && \Delta G_1^{\circ}=-33.4 ~\text{kJ} \\[4pt]
\hline &\text{Coupled reaction:} & \ce{ZnS(s) + H2(g) \rightarrow Zn(s) + H2S(g)} && \Delta G^{\circ}=201.3 ~\text{kJ} +-33.4 ~\text{kJ} =167.9 ~\text{kJ}
\end{align*} \]

The coupled reaction exhibits a positive free energy change and is thus nonspontaneous.

Exercise \(\PageIndex{3}\)

What is the standard free energy change for the reaction below? Is the reaction expected to be spontaneous under standard conditions?

\[\ce{FeS(s) \, + O2(g) \rightarrow Fe(s) \, + SO2(g)} \nonumber \]

Answer

−199.7 kJ; spontaneous

Temperature Dependence of Spontaneity and Extent of Reaction

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. In a similar, but not identical fashion, some chemical reactions can switch from being product-favored at equilibrium, to being reactant-favored at equilibrium, depending on the temperature.

Note

The numerical value of \(\Delta G^º\) is always dependent on the temperature. In this section we are determining whether or not the sign of \(\Delta G^º\) is dependent on the temperature.

To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:

\[ ΔG^º=ΔH^º−TΔS^º \nonumber \]

The extent of a process, as reflected in the arithmetic sign of its standard free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (Kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:

1. Both ΔHº and ΔSº are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔGº will be negative if the magnitude of the TΔSº term is greater than ΔHº. If the TΔSº term is less than ΔHº, the free energy change will be positive. Such a process is product-favored at equilibrium at high temperatures and reactant-favored at equilibrium at low temperatures.
2. Both ΔHº and ΔSº are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔGº will be negative if the magnitude of the TΔSº term is less than ΔHº. If the TΔSº term’s magnitude is greater than ΔHº, the free energy change will be positive. Such a process is product-favored at equilibrium at low temperatures and reactant-favored at equilibrium at high temperatures.
3. ΔHº is positive and ΔSº is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔGº will be positive regardless of the temperature. Such a process is reactant-favored at equilibrium at all temperatures.
4. ΔHº is negative and ΔSº is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔGº will be negative regardless of the temperature. Such a process is product-favored at equilibrium at all temperatures.

These four scenarios are summarized in Table \(\PageIndex{1}\)

Table \(\PageIndex{1}\)

Sign of \(\Delta H^o\)	Sign of \(\Delta S^o\)	Sign of \(\Delta G^o\)	Temperature Dependence of \(\Delta G^o\)
-	+	-	The sign of \(\Delta G^o\) does not depend on the temperature.The reaction is product-favored at equilibrium at all temperatures.
+	-	+	The sign of \(\Delta G^o\) does not depend on the temperature.The reaction is reactant-favored at equilibrium at all temperatures.
-	-	- or +	The sign of \(\Delta G^o\) does  depend on the temperature. The reaction will be product-favored at equilibrium at lower temperatures.
+	+	- or +	The sign of \(\Delta G^o\) does  depend on the temperature. The reaction will be product-favored at equilibrium at lower temperatures.

 

Example \(\PageIndex{3}\): Predicting the Temperature Dependence of Spontaneity

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

\[\ce{2C}(s)+\ce{O2}(g)⟶\ce{2CO}(g) \nonumber \]

Does the sign of \(\Delta G^º\) of this process depend upon temperature?

Solution

Combustion processes are exothermic (\(ΔH^º < 0\)). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, \(ΔS^º > 0\)). The reaction is therefore product-favored at equilibrium (\(ΔG^º < 0\)) at all temperatures.

Exercise \(\PageIndex{3}\)

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

\[\ce{4Fe}(s)+\ce{3O2}(g)⟶\ce{2Fe2O3}(s) \nonumber \]

Does the sign of \(\Delta G^o\) of this process depend upon temperature?

Answer

ΔHº and ΔSº are both negative; the reaction is product-favored at equilibrium at low temperatures.

When considering the conclusions drawn regarding the temperature dependence of the sign of ΔGº, 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 reactant-favored at equilibrium at one temperature but product-favored at equilibrium at another temperature will necessarily undergo a change in “extent” (as reflected by its ΔGº) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which ΔGº is plotted on the y axis versus T on the x axis:

\[ΔG^º=ΔH^º−TΔS^º \nonumber \]

\[y=b+mx \nonumber \]

Such a plot is shown in Figure \(\PageIndex{2}\). A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependence for the sign of ΔGº as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative ΔGº) 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 ΔGº is zero:

\[ΔG^º=0=ΔH^º−TΔS^º \nonumber \]

\[T=\dfrac{ΔH^º}{ΔS^º} \nonumber \]

Thus, saying a process is product-favored at equilibrium at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which ΔGº for the process is zero. 

Note

In this discussion, we have used two different descriptions for the meaning of the sign of ΔGº. You should be aware of the meaning of each description. 

a) Extent of Reaction: This description is used to predict the ratio of the product and reactant concentrations at equilibrium. In this description, we use the thermodynamic term ΔGº to tell us the same information as the equilibrium constant, K. When ΔGº < 0, K > 1, and the reaction will be product-favored at equilibrium. When ΔGº > 0, K< 1, and the reaction is reactant-favored at equilibrium. When ΔGº = 0, K =1, and the reaction will have roughly equal amounts of products and reactants at equilibrium. In all cases, the reaction will form a mixture of products and reactants at equilibrium. We use the sign and magnitude of ΔGº to tell us how much product will be made if the reaction is allowed to reach equilibrium.

b) Spontaneity: This description is much more complicated because it involves two different interpretations of how a reaction at standard state occurs. One interpretation involves the hypothetical process in which the reaction proceeds from a starting point of pure reactants to a finishing point of pure products, with all substances isolated in their own containers under standard state conditions. In the second, more realistic interpretation, the reaction starts with all reactants and all products in their standard state in one container. We then allow this specific mixture to react an infinitesimally small amount so that we can obtain a rate of change in free energy with respect to the extent of reaction when all reactants and products are mixed and (essentially) in their standard states. Although each interpretation describes a different reaction scenario, the value of the difference in free energy and the value of the rate of change in free energy are the same number. If ΔGº < 0, we say that the reaction is spontaneous, meaning that the reaction would proceed in the forward direction, as written, to form pure products in their standard state. If ΔGº > 0, we say that the reaction is nonspontaneous, meaning that the reaction would proceed in the reverse direction, as written, to form pure reactants in their standard state. If ΔGº = 0, we say that the neither the reactants nor the products are favored to be formed.

A detailed treatment of the meaning of ΔGº can be found in the paper, "Free Energy versus Extent of Reaction" by Richard S. Treptow, Journal of Chemical Education, 1996, Volume 73 (1), 51-54.

Figure \(\PageIndex{2}\): These plots show the variation in ΔGº with temperature for the four possible combinations of arithmetic sign for ΔHº and ΔSº. Note that in this graph of ΔGº, "spontaneous" is synonymous with "product-favored at equilibrium", and "nonspontaneous" is synonymous with "reactant-favored at equilibrium".

Example \(\PageIndex{4}\): 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 Tables T1 or T2 to estimate the boiling point of water.

Solution

The process of interest is the following phase change:

\[\ce{H2O}(l)⟶\ce{H2O}(g) \nonumber \]

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

\[0=ΔH°−TΔS°\hspace{40px}\ce{or}\hspace{40px}T=\dfrac{ΔH°}{ΔS°} \nonumber \]

Using the standard thermodynamic data from Tables T1 or T2,

\[\begin{align*}
ΔH°&=ΔH^\circ_\ce{f}(\ce{H2O}(g))−ΔH^\circ_\ce{f}(\ce{H2O}(l)) \nonumber\\
&=\mathrm{−241.82\: kJ/mol−(−285.83\: kJ/mol)=44.01\: kJ/mol} \nonumber
\end{align*} \nonumber \]

\[\begin{align*}
ΔS°&=ΔS^\circ_{298}(\ce{H2O}(g))−ΔS^\circ_{298}(\ce{H2O}(l)) \nonumber\\
&=\mathrm{188.8\: J/K⋅mol−70.0\: J/K⋅mol=118.8\: J/K⋅mol} \nonumber
\end{align*} \nonumber \]

\[T=\dfrac{ΔH°}{ΔS°}=\mathrm{\dfrac{44.01×10^3\:J/mol}{118.8\:J/K⋅mol}=370.5\:K=97.3\:°C} \nonumber \]

The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Tables T1 or T2.). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.

Exercise \(\PageIndex{4}\)

Use the information in Tables T1 or T2 to estimate the boiling point of CS₂.

Answer

313 K (accepted value 319 K).