10.4: Free Energy

Learning Objectives

• Define Gibbs free energy, and describe its relation to spontaneity
• Calculate the free energy change of a process using standard free energies of formation
• Calculate the free energy change of a process using enthalpy and entropy values
• Explain how temperature affects the spontaneity of some processes

The Second Law of Thermodynamics states that a spontaneous process increases the total entropy of the universe. This involves calculating the entropy change for both the system and the surroundings. To reduce calculations to just the system, Josiah Willard Gibbs introduced a new thermodynamic quantity: the Gibbs free energy, \(G\). This quantity depends only on system properties, enthalpy, entropy, and temperature, and allows us to predict spontaneity of a process at constant temperature and pressure without considering the surroundings.

The Gibbs free energy (\(G\)) is defined as:

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

At constant temperature and pressure, the free energy change (\(\Delta G\)) is:

\[\Delta G=\Delta H-T \Delta S \label{Gibbs} \]

This equation gives us a simple way to predict spontaneity at constant temperature and pressure, as shown in Table \(\PageIndex{1}\):

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

\(\Delta S_{univ} > 0\)	\(\Delta G < 0\)	Spontaneous in the forward direction; Nonspontaneous in the reverse direction
\(\Delta S_{univ} < 0\)	\(\Delta G > 0\)	Nonspontaneous in the forward direction;  Spontaneous in the reverse direction
\(\Delta S_{univ} = 0\)	\(\Delta G = 0\)	At equilibrium;  Both directions are spontaneous

As seen in Table \(\PageIndex{1}\), while a positive value of \(\Delta S_{univ}\) indicates a spontaneous process, a negative value of \(\Delta G\) is an indicator of spontaneity. This stems from the relationship between \(\Delta G\) and \(\Delta S_{univ}\), defined at constant temperature and pressure as:

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

What’s “Free” about \(\Delta G \)?

In addition to predicting spontaneity at constant temperature and pressure, the free energy change (\(\Delta G\)) tells us about the amount of useful work a process can do.

Let’s revisit the definition:
\[\Delta G=\Delta H-T \Delta S \nonumber \]

In this equation:

• \(\Delta H\) is the total heat released or absorbed by the system.
• \(T \Delta S \) represents the energy of the system that is lost to disorder and is no longer available to do work.
• The remaining energy, \(\Delta G\) , is “free” in the sense that it is the amount of energy left that can be harnessed to do useful work.

If a process is spontaneous and could occur reversibly (idealized conditions), then:

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

That is, the free energy change equals the maximum amount of non-expansion work that can be obtained from a system under reversible conditions.

In real systems, however, no process is perfectly reversible, and devices like batteries or engines are never 100% efficient. So the actual work done is always less than \(\Delta G\).

Similarly, for a nonspontaneous process, \(\Delta G > 0\) represents the minimum amount of work that must be done on the system to make the process happen.

Calculating Free Energy Change

Free energy (\( G\)) combines three state functions, enthalpy (\(H\)), entropy (\(S\)), and temperature (\(T\)), so it is also a state function. As with enthalpy, we focus on the free energy changes (\(\Delta G\)). As we saw with other thermodynamic parameters, we can define the standard free energy change (\(\Delta G^\circ\)) as the free energy change that occurs when all components of the system are at standard state: 1 atm total pressure and 1 M concentration for solutions. We can write the following relationship:

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

We have seen that standard enthalpy and entropy changes can be calculated from tabulated values. These allow us to determine \(\Delta G^\circ\) for a reaction under standard conditions.

The sign of \(\Delta G^\circ\) tells us which direction a reaction will proceed under standard conditions, and therefore whether the products or reactants are favoured. 

If \(\Delta G^\circ < 0\), the reaction is spontaneous and products are favoured.

If \(\Delta G^\circ > 0\), the reaction is nonspontaneous and reactants are favoured.

If \(\Delta G^\circ = 0\), the system is at equilibrium under standard conditions.

Example \(\PageIndex{1}\)

Calculate the standard free-energy change (\(\Delta G^\circ\)) at 25°C for the reaction

\[ \ce{ H2(g) + O2(g) \rightleftharpoons H2O2(l)}\nonumber \]

At 25°C, the standard enthalpy change (\(\Delta H^\circ\)) is -187.78 kJ/mol, and the absolute entropies of the products and reactants are:

• \(S^\circ\)(H₂O₂) = 109.6 J/(mol‧K),
• \(S^\circ\)(O₂) = 205.2 J/(mol‧K), and
• \(S^\circ\)(H₂) = 130.7 J/(mol‧K).

Is the reaction spontaneous as written under standard state conditions and at 25°C?

Given: balanced chemical equation, \(\Delta H^\circ\) and \(\Delta S^\circ\) for reactants and products

Asked for: spontaneity of reaction as written

Strategy:

1. To calculate \(\Delta G^\circ\) for the reaction, we need to know \(\Delta H^\circ\), \(\Delta S^\circ\), and \(T\). We are given \(\Delta H^\circ\), and we know that T = 298.15 K. We can calculate \(\Delta S^\circ\) from the absolute molar entropy values provided.
2. Use Equation \(\ref{Gibbs}\), the calculated value of \(\Delta S^\circ\), and the given \(\Delta H^\circ\) and temperature to calculate \(\Delta G^\circ\) for the reaction. Use the value of \(\Delta G^\circ\) to determine whether the reaction is spontaneous under standard state conditions.

Solution

A. Calculate \(\Delta S^\circ\) from the absolute molar entropy values provided using the formula: sum of entropies of products minus sum of entropies of reactants:

\[
\begin{aligned} \Delta S^\circ &= \sum m S^\circ(\text{products}) - \sum n S^\circ(\text{reactants})\\
\Delta S^\circ &= S^\circ(\ce{H2O2(l)}) - \left[ S^\circ(\ce{H2(g)}) + S^\circ(\ce{O2(g)}) \right] \\[4pt]
&= 109.6\ \mathrm{J/(mol\cdot K)} - \left[130.7\ \mathrm{J/(mol\cdot K)} + 205.2\ \mathrm{J/(mol\cdot K)} \right] \\[4pt]
&= -226.3\ \mathrm{J/(mol\cdot K)}
\end{aligned}
\]

Given the reaction consumes gas to form a liquid, \(\Delta S^\circ\) is very negative for this reaction as expected.

B. Substituting the appropriate quantities into Equation \(\ref{Gibbs}\),

\[
\begin{aligned}
\Delta G^\circ &= \Delta H^\circ - T\Delta S^\circ \\[6pt]
&= -187.78\ \mathrm{kJ/mol} - \left(298.15\ \mathrm{K}\right)\left(-226.3\ \mathrm{\frac{J}{mol\cdot K}} \right)\left(\frac{ \mathrm{kJ}}{1000\ \mathrm{J}}\right) \\[6pt]
&= -120.31\ \mathrm{kJ/mol}
\end{aligned} \]

The negative value of \(\Delta G^\circ\) indicates that the reaction is spontaneous in the forward direction under standard state conditions (1 atm H₂(g), 1 atm O₂(g) and 1 M H₂O₂(l) present).

Because both \(\Delta S^\circ\) and \(\Delta H^\circ\) are negative, the sign of \(\Delta G^\circ\) depends on the relative magnitudes of the \(\Delta H^\circ\) and \(T \Delta S^\circ\) terms. At 298 K, the enthalpy term dominates, so the forward reaction is spontaneous.

Exercise \(\PageIndex{1}\)

Calculate the standard free-energy change (\(\Delta G^\circ\)) at 25°C for the reaction

\[\ce{2H2 (g) + N2 (g) \rightleftharpoons N2H4 (l)} \nonumber \]

Is this reaction spontaneous under standard state conditions (1 atm, 1 M) and at 25°C?

At 25°C, the standard enthalpy change (\(\Delta H^\circ\)) is 50.6 kJ/mol, and the absolute entropies of the products and reactants are

•  \(S^\circ\)(N₂H₄) = 121.2 J/(mol‧K),
• \(S^\circ\)(N₂) = 191.6 J/(mol‧K), and
• \(S^\circ\)(H₂) = 130.7 J/(mol‧K).

 

Answer

\[\begin{aligned} \Delta S^\circ &= \sum m S^\circ(\text{products}) - \sum n S^\circ(\text{reactants})\\
\Delta S^\circ &= S^\circ(\ce{N2H4(l)}) - \left[2 \times S^\circ(\ce{H2(g)}) + S^\circ(\ce{N2(g)}) \right] \\[4pt]
&= 121.2\ \mathrm{J/(mol\cdot K)} - \left[2(130.7\ \mathrm{J/(mol\cdot K)}) + 191.6\ \mathrm{J/(mol\cdot K)} \right] \\[4pt]
&= -331.8\ \mathrm{J/(mol\cdot K)}
\end{aligned}\]

The change in entropy is negative as expected because there are gases present in the reactants and not in the products.

\[\begin{aligned}
\Delta G^\circ &= \Delta H^\circ - T\Delta S^\circ \\[6pt]
&= 50.6\ \mathrm{kJ/mol} - \left(298.15\ \mathrm{K}\right)\left(-331.8\ \mathrm{\frac{J}{mol\cdot K}} \right)\left(\frac{ \mathrm{kJ}}{1000\ \mathrm{J}}\right) \\[6pt]
&= 149.5\ \mathrm{kJ/mol}
\end{aligned} \]

\(\Delta G^\circ\) is positive so the reaction is non spontaneous in the forward direction (the reverse reaction is spontaneous) under standard state conditions.

Because the reaction is endothermic and the entropy change is negative, the free energy change is positive at all temperatures, so the reaction is never spontaneous under standard conditions.

We can also calculate the standard free energy change (\(\Delta G^\circ\)) for a reaction using standard free energies of formation \(\Delta G_f^\circ\) of the reactants and products. The standard free energy of formation is the free energy change of the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpy of formation, \(\Delta G_f^\circ\) is by definition zero for elemental substances in their standard states. The approach used to calculate \(\Delta G^\circ\) for a reaction from \(\Delta G_f^\circ\) values is the same as that demonstrated previously for enthalpy:

\[ \Delta G^\circ = \sum m G_f^\circ(\text{products}) - \sum n G_f^\circ(\text{reactants}) \nonumber\]

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

Consider the decomposition of yellow mercury(II) oxide.

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

Calculate the standard free energy change at room temperature at 25°C, using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. 

Is this reaction spontaneous under standard state conditions (1 atm, 1 M) and at 25°C?

The required data are shown here.

Compound	\(\Delta G_{ f }^{\circ}( \text{kJ/mol} )\)	\(\Delta H_{ f }^{\circ}( \text{kJ/mol} )\)	\(S^{\circ}( \text{J/(mol‧K} )\)
HgO (s, yellow)	−58.43	−90.46	71.13
Hg(l)	0	0	75.9
O2(g)	0	0	205.2

Solution

(a) Using free energies of formation:

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

(b) Using enthalpies and entropies of formation:

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

Both methods give the same \(\Delta G^\circ\) value at 25 °C (to three significant figures), confirming the reliability of either approach. The positive value means that the reaction is nonspontaneous under standard state conditions.

Exercise \(\PageIndex{2}\)

Calculate \(\Delta G^\circ\) using (a) free energies of formation and (b) enthalpies of formation and entropies (values in the reference table). Is the reaction spontaneous or nonspontaneous at 25 °C under standard state conditions?

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

Answer

(a) Using free energies of formation:

\[\begin{align*}
\Delta G^{\circ} & =\sum m G_{ f }^{\circ}(\text { products })-\sum n \Delta G_{ f }^{\circ}(\text { reactants }) \\[4pt]
&=\left[1 \Delta G_{ f }^{\circ}~( \ce{H2 (g)}) + 1\Delta G_{f}^{\circ} ~(\ce{C2H2 (g)}) \right]-1 \Delta G_{ f }^{\circ} ~(\ce{C2H4 (g)}) \\[4pt]
&=\left[1 (0 ~\text{kJ/mol} )+1(209.9 ~\text{kJ/mol} )\right] - 1 (68.4 ~\text{kJ / mol} )=141.5 ~\text{kJ/mol}
\end{align*} \nonumber \]

(b) Using enthalpies and entropies of formation:

\[\begin{align*}
\Delta H^{\circ}&=\sum m \Delta H_{ f }^{\circ}(\text { products })-\sum n \Delta H_{ f }^{\circ}(\text { reactants }) \\[4pt]
&=\left[1 \Delta H_{ f }^{\circ}~(\ce{H2(g)}) + 1\Delta H_{ f }^{\circ}~(\ce{C2H2(g)}) \right]-1 \Delta H_{ f }^{\circ} ~(\ce{C2H4 (g)}) \\[4pt]
&=\left[1(0~ \text{kJ/mol} ) + (227.4 ~\text{kJ/mol} )\right]-1(52.4~\text{kJ/mol} ) = 175.0 ~\text{kJ/mol} \\[8pt]
\Delta S^{\circ} &=\sum m \Delta S^{\circ}(\text { products })-\sum n \Delta S^{\circ}(\text { reactants }) \\[4pt]
&=\left[1 ~\Delta S^{\circ} ~(\ce{H2 (g)}) + 1 \Delta S^{\circ} ~(\ce{C2H2(g)}) \right]-1 \Delta S^{\circ} ~(\ce{C2H4 (g)}) \\[4pt]
& =\left[1(130.7~\text{J /(mol‧K)} ) + 1(200.9~\text{J/(mol‧K)} )\right] -1(219.3 ~\text{J/(mol‧K)} )=112.3 ~\text{J/(mol‧K)} \\[8pt]
\Delta G^{\circ} &=\Delta H^{\circ}-T \Delta S^{\circ}=175.0 ~\text{kJ} - (298.15 ~\text{K})(112.3 ~\text{J/(mol‧K)})\left( \frac{1 ~\text{ kJ} }{1000 ~\text{J} }\right) \\[4pt]
&=141.5 ~\text{kJ/mol}
\end{align*} \nonumber \]

\( \Delta G^\circ\) is positive, so the reaction is nonspontaneous under standard state conditions at 25°C.

 \(\Delta G^\circ\) predicts spontaneity at constant temperature and pressure and under standard conditions (1 M and 1 atm). Non-standard conditions (e.g., different concentrations or pressures) will be addressed later in the chapter.

Free Energy Changes for Coupled Reactions

In multi-step reactions, a nonspontaneous step can be driven by a spontaneous step if the overall free energy change is negative. This is how many biological and industrial reactions proceed. In these coupled reactions, one step supplies sufficient free energy for the other step to occur. Because free energy is a state function, we can combine \(\Delta G\) values just as we combined enthalpy changes using Hess's Law (Ch. 5).

For example, the production of elemental zinc from zinc sulfide is nonspontaneous under standard conditions, as indicated by a positive value for \( \Delta G^\circ\):

\[\ce{ZnS (s) \rightleftharpoons Zn(s) + S(s)} \quad \Delta G_1^{\circ}=201.3\;\text{kJ/mol} \nonumber \]

The industrial process involves coupling this decomposition reaction to the thermodynamically favourable oxidation of sulfur:

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

The coupled reaction, the sum of the two equations, has a negative free energy change and is therefore spontaneous:

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

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

The overall negative value of \( \Delta G^\circ\) means that the coupled process, decomposing zinc sulfide while oxidizing sulfur, is spontaneous under standard state conditions.

Let’s consider another possible coupling pathway and assess its spontaneity.

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

An industrial chemist considers coupling the decomposition of \(\ce{ZnS}\) to the formation of \(\ce{H2S}\):

\[\ce{ZnS (s) \rightleftharpoons Zn(s) + S(s)} \quad \Delta G_1^{\circ}=201.3\;\text{kJ/mol} \nonumber \]

\[\ce{H2(g) + S(s) \rightleftharpoons H2S(g)} \quad \Delta G_2^{\circ}=-33.4\;\text{kJ/mol} \nonumber \]

Is the coupled reaction spontaneous under standard conditions?

Solution

\[\begin{align*} &\text{Decomposition of zinc sulfide:} & \ce{ZnS (s) \rightleftharpoons Zn(s) + S(s)} && \Delta G_1^{\circ}=201.3 ~\text{kJ/mol}\\[4pt]
&\text{Formation of hydrogen sulfide:} & \ce{S(s) + H2(g) \rightleftharpoons H2S(g)} && \Delta G_2^{\circ}=-33.4 ~\text{kJ/mol} \\[4pt]
\hline &\text{Coupled reaction:} & \ce{ZnS(s) + H2(g) \rightleftharpoons Zn(s) + H2S(g)} && \Delta G^{\circ}=201.3 ~\text{kJ/mol} +(-33.4 ~\text{kJ/mol}) =167.9 ~\text{kJ/mol}
\end{align*} \]

Because the total \( \Delta G^\circ\) is positive (+167.9 kJ/mol), the overall reaction is nonspontaneous under standard conditions and would require input of external energy to proceed.

Temperature Dependence of Spontaneity

As we saw earlier in this chapter, the spontaneity of a process can depend on temperature. For example, phase transitions proceed spontaneously in one direction or the other depending on the temperature. Similarly, some chemical reactions are spontaneous in the forward direction at certain temperatures and spontaneous in the reverse direction at others.

To illustrate this, consider the expression for standard free energy change:

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

When \( \Delta G^\circ\) is negative, the forward reaction is spontaneous under standard state conditions, and products are favoured. When \(\Delta G^\circ\) is positive, the reverse reaction is spontaneous, and reactants are favoured. Since temperature (in Kelvin) is always positive, the sign of \( \Delta G^\circ\) depends on the signs and relative magnitudes of \( \Delta H^\circ\) and \( \Delta S^\circ\). Depending on the signs of \( \Delta H^\circ\) and \( \Delta S^\circ\), there are four possible combinations:

1. Both \( \Delta H^\circ\) and \( \Delta S^\circ\) are positive:
   • This describes an endothermic process with an increase in entropy. \( \Delta G^\circ\) will be negative if the magnitude of the \(T \Delta S^\circ\) term is greater than \( \Delta H^\circ\). At higher temperatures, \( \Delta G^\circ\) will be negative and the forward reaction spontaneous. At lower temperatures, the reverse will be true: \( \Delta G^\circ\) will be positive and the reverse reaction spontaneous.
2. Both \( \Delta H^\circ\) and \( \Delta S^\circ\) are negative:
   • This describes an exothermic process with a decrease in entropy. \( \Delta G^\circ\) will be negative if the magnitude of the \(T\Delta S^\circ\) term is less than \( \Delta H^\circ\). At lower temperatures, the forward reaction is spontaneous. At higher temperatures, the \(T \Delta S^\circ\) term is greater than \( \Delta H^\circ\), giving a positive \( \Delta G^\circ\).
3. \( \Delta H^\circ\) is positive and \( \Delta S^\circ\) is negative:
   • This describes an endothermic process with a decrease in system entropy. \( \Delta G^\circ\) will be positive at all temperatures. The reaction is never spontaneous in the forward direction under standard conditions.
4. \( \Delta H^\circ\) is negative and \( \Delta S^\circ\) is positive:
   • This describes an exothermic process with an increase in system entropy. \( \Delta G^\circ\) is negative at all temperatures. The forward reaction is always spontaneous under standard conditions.

These scenarios are summarized in Table \(\PageIndex{2}\)

Table \(\PageIndex{2}\)

\( \Delta H^\circ\)	\( \Delta S^\circ\)	\( \Delta G^\circ\)	Spontaneity of Forward Reaction
−	+	always −	spontaneous (at all temperatures)
+	−	always +	nonspontaneous (at all temperatures)
−	−	− (lower T) + (higher T)	spontaneous at lower temperatures
+	+	+ (lower T) − (higher T)	spontaneous at higher temperatures

Example \(\PageIndex{4}\): Predicting the Temperature Dependence of Spontaneity

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

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

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

Solution

Combustion processes are exothermic, so \(\Delta H^\circ < 0\). This reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, \(\Delta S^\circ > 0\)). \(\Delta G^\circ < 0\) is negative at all temperatures, and the forward reaction is always spontaneous.

Exercise \(\PageIndex{3}\)

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

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

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

Answer

Since the reaction generates heat, \( \Delta H^\circ\) is negative. \( \Delta S^\circ\) is negative because gas is consumed in the reaction. \( \Delta G^\circ\) will be negative at lower temperatures.

For a process in which the sign of \( \Delta G^\circ\) depends on temperature, there is a specific temperature at which \( \Delta G^\circ=0\). At this point, the system is at equilibrium under standard state conditions. The “higher” and “lower” temperatures mentioned earlier refer to whether the system temperature is above or below this equilibrium temperature. To visualize this, consider a plot of \( \Delta G^\circ\) vs. \(T\), based on the equation:

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

This equation has the same form as a linear function, with an intercept \(b=\Delta H^\circ\) a slope \(m=-\Delta S^\circ\):

\[y=b+mx \nonumber \]

Figure \(\PageIndex{1}\) shows this relationship for the four possible combinations of signs for \( \Delta H^\circ\) and \( \Delta S^\circ\):

• Green line: Exothermic (\( \Delta H^\circ < 0\)) with increasing entropy (\( \Delta S^\circ > 0\)). \( \Delta G^\circ\) is negative at all temperatures, so the forward reaction is always spontaneous.
• Red line: Endothermic reaction (\( \Delta H^\circ > 0\)) with decreasing entropy (\( \Delta S^\circ < 0\)). \( \Delta G^\circ\) is positive at all temperatures, so the forward reaction is never spontaneous.
• Yellow lines: \( \Delta H^\circ\) and \( \Delta S^\circ\) have the same sign, so \( \Delta G^\circ\) depends on temperature:
  • For \( \Delta H^\circ > 0\) and \( \Delta S^\circ > 0\) (upper yellow line), \( \Delta G^\circ\) is positive at lower temperatures and becomes negative at higher temperatures. The forward reaction becomes spontaneous above a certain temperature.
  • or \( \Delta H^\circ < 0\) and \( \Delta S^\circ < 0\) (lower yellow line), \( \Delta G^\circ\) is negative at lower temperatures and becomes positive at higher temperatures. The forward reaction is spontaneous only below a certain temperature.

Figure \(\PageIndex{1}\): Plot of standard free energy change (\( \Delta G^\circ\)) versus temperature (\(T\)) for different combinations of enthalpy (\( \Delta H^\circ\)) and entropy (\( \Delta S^\circ\)). The green and red lines represent processes where \( \Delta G^\circ\) is always negative or always positive, respectively, and spontaneity is temperature-independent. The yellow lines show processes where \( \Delta G^\circ\) changes sign with temperature. The temperature at which each yellow line crosses the x-axis (\( \Delta G^\circ = 0\)) marks the point where the reaction changes from spontaneous to nonspontaneous, or vice versa, under standard state conditions.

The "cross-over" temperature, the x-intercept of each yellow line, is the temperature at which \( \Delta G^\circ\) is zero:

\[\Delta G^{\circ}=0=\Delta H^{\circ}-T \Delta S^{\circ} \nonumber \]

Solving for \(T\):

\[T=\dfrac{ΔH^{\circ}}{ΔS^{\circ}} \nonumber \]

This temperature marks the point where the direction of spontaneity changes under standard state conditions.

Using Standard Enthalpy and Entropy Values at Other Temperatures

As long as no phase change occurs, the values of \( \Delta H^\circ\) and \( \Delta S^\circ\) typically do not change much with temperature. This allows us to treat them as constants over moderate temperature ranges. We can use the tabulated standard values (typically reported at 298 K) to estimate \( \Delta G^\circ\) at other temperatures using the expression:

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

This approximation is reasonable as long as the temperature range is moderate and no phase transitions (e.g., melting, boiling) occur for any substances involved in the reaction.

Example \(\PageIndex{5}\)

Calculate \( \Delta G^\circ\) at 25°C and at 300°C for the reaction:

\[\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)} \nonumber \]

Assume that \( \Delta H^\circ\) and \( \Delta S^\circ\) do not change between 25°C and 300°C. 

Use these data:

• \(S^\circ\)(N₂) = 191.6 J/(mol⋅K),
• \(S^\circ\)(H₂) = 130.7 J/(mol⋅K),
• \(S^\circ\)(NH₃) = 192.8 J/(mol⋅K)
• \( \Delta H_f^\circ \)(NH₃) = −45.9 kJ/mol.

Given: balanced chemical equation, temperatures, \( \Delta S^\circ\) values, and \( \Delta H_f^\circ\) for NH₃

Asked for: \( \Delta G^\circ\) at 25°C and 300°C

Strategy:

1. Calculate \( \Delta S^\circ\) for the reaction. Calculate \( \Delta H^\circ\) for the reaction, recalling that \( \Delta H_f^\circ\) for any element in its standard state is zero.
2. Substitute the appropriate values into Equation \(\ref{Gibbs}\) to obtain \( \Delta G^\circ\) for the reaction at 298 K.
3. Assuming that \( \Delta H^\circ\) and \( \Delta S^\circ\) are independent of temperature, substitute values into Equation \(\ref{Gibbs}\) to obtain \( \Delta G^\circ\) for the reaction at 300°C.

Solution

A. Calculate \( \Delta S^\circ\):

\[\begin{align*}\Delta S^\circ_{\textrm{rxn}}&=2S^\circ(\mathrm{NH_3})-[S^\circ(\mathrm{N_2})+3S^\circ(\mathrm{H_2})]
\nonumber\\ &=[2(192.8\;\mathrm{J/(mol\cdot K))}]-[1(191.6\;\mathrm{J/(mol\cdot K))}+ 3(130.7\;\mathrm{J/(mol\cdot K))}]\nonumber\\ &=-198.1\;\mathrm{J/(mol\cdot K))}\end{align*} \]

Calculate \( \Delta H^\circ\):

\[\begin{align*}\Delta H^\circ_{\textrm{rxn}}&=2\Delta H^\circ_\textrm f(\mathrm{NH_3})-[\Delta H^\circ_\textrm f(\mathrm{N_2})+3\Delta H^\circ_\textrm f(\mathrm{H_2})]\nonumber \\ &=[2(-45.9\textrm{ kJ/mol})]-[(1(0\textrm{ kJ/mol})+3(0 \textrm{ kJ/mol})]\nonumber \\ &=-91.8\textrm{ kJ/mol}\nonumber\end{align*} \]

B. Calculate \( \Delta G^\circ\) at 298 K:

\[
\begin{aligned}
\Delta G^\circ_{\text{rxn}} &= \Delta H^\circ - T\Delta S^\circ \\
&= \left( -91.8\ \mathrm{\frac{kJ}{mol}} \right) - \left( 298\ \mathrm{K} \right) \left( -198.1\ \mathrm{\frac{J}{mol \cdot K}} \right) \left( \frac{1\ \mathrm{kJ}}{1000\ \mathrm{J}} \right) \\
&= -32.7\ \mathrm{\frac{kJ}{mol}}
\end{aligned}
\]

C. Calculate \( \Delta G^\circ\) at 300°C (573 K):

\[\begin{align*}\Delta G^\circ &=(-\textrm{91.8 kJ/mol})-(\textrm{573 K})(-198.1\; \mathrm{J/(mol\cdot K)})(\textrm{1 kJ/1000 J}) \\[4pt]&=21.7\textrm{ kJ/mol} \end{align*} \nonumber \]

Changing the temperature has a major effect on the spontaneity of this reaction. At 298 K, the reaction is spontaneous under standard conditions, though in practice it is extremely slow. Although raising the temperature speeds up the reaction (beneficial for kinetics), it also makes the process thermodynamically unfavourable by increasing \(\Delta G^\circ\). This tradeoff is central to designing industrial processes like ammonia synthesis.

Exercise \(\PageIndex{4}\)

Calculate \( \Delta G^\circ\) at (a) 25°C and (b) 750 °C for the reaction important in the formation of urban smog:

\[\ce{ 2NO(g) + O2 (g) \rightleftharpoons 2NO2 (g)}\nonumber \]

Assume that \(\Delta H\) and \(\Delta S\) do not change between 25.0°C and 750°C and use these data:

• \(S^\circ\)(NO) = 210.8 J/(mol⋅K),
• \(S^\circ\)(O₂) = 205.2 J/(mol⋅K),
• \(S^\circ\)(NO₂) = 240.1 J/(mol⋅K),
• \( \Delta H_f^\circ \)(NO₂) = 33.2 kJ/mol, and
• \( \Delta H_f^\circ \)(NO) = 91.3 kJ/mol.

Answer

Calculate \( \Delta S^\circ\):

\[\begin{align*}\Delta S^\circ_{\textrm{rxn}}&=2S^\circ(\mathrm{NO_2})-[2S^\circ(\mathrm{NO})+1S^\circ(\mathrm{O_2})]
\\ &=[2(240.1\;\mathrm{J/(mol\cdot K))}]-[2(210.8\;\mathrm{J/(mol\cdot K))}+ 1(205.2\;\mathrm{J/(mol\cdot K))}]\\ &=-146.6\;\mathrm{J/(mol\cdot K))}\end{align*} \]

Calculate \( \Delta H^\circ\):

\[\begin{align*}\Delta H^\circ_{\textrm{rxn}}&=2\Delta H^\circ_\textrm f(\mathrm{NO_2})-[2\Delta H^\circ_\textrm f(\mathrm{NO})+1\Delta H^\circ_\textrm f(\mathrm{O_2})] \\ &=[2(33.2\textrm{ kJ/mol})]-[(2(91.3\textrm{ kJ/mol})+1(0 \textrm{ kJ/mol})] \\ &=-116.2\textrm{ kJ/mol} \end{align*} \]

Calculate \( \Delta G^\circ\) at 25°C (298 K):

\[\Delta G^\circ_{\textrm{rxn}}=\Delta H^\circ-T\Delta S^\circ=(-\textrm{116.2 kJ/mol})-(\textrm{298 K})(-\textrm{146.6 J/(mol\cdot K)})(\textrm{1 kJ/1000 J})=-\textrm{72.5 kJ/mol}\nonumber \]

Calculate \( \Delta G^\circ\) at 750°C (1023 K):

\[\Delta G^\circ_{\textrm{rxn}}=\Delta H^\circ-T\Delta S^\circ=(-\textrm{116.2 kJ/mol})-(\textrm{1023 K})(-146.6\;J/(mol\cdot K))(\textrm{1 kJ/1000 J})=\textrm{33.8 kJ/mol}\nonumber \]

The effect of temperature on the spontaneity of a reaction, which is an important factor in experimental design and industrial processes, depends on both the sign and magnitude of \( \Delta H^\circ\) and \( \Delta S^\circ\). The temperature at which a reaction is at equilibrium under standard state conditions can be calculated by setting \( \Delta G^\circ\) = 0. This is known as the crossover temperature, the temperature at which a reaction switches from being spontaneous to nonspontaneous (or vice versa) under standard conditions.

Example \(\PageIndex{6}\)

Consider the reaction from Example \(\PageIndex{5}\):

\[\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)} \nonumber \]

From the calculations in the earlier example:

\( \Delta S^\circ\)= -198.1 J/(mol⋅K)

\( \Delta H^\circ\)= -91.8 kJ/mol

Because both \( \Delta H^\circ\) and \( \Delta S^\circ\) are negative, the reaction is spontaneous at lower temperatures and nonspontaneous at higher temperatures. 

Calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that \( \Delta H^\circ\) and \( \Delta S^\circ\) are independent of temperature.

Given: \( \Delta H^\circ\) and \( \Delta S^\circ\)

Asked for: temperature at which reaction changes from spontaneous to nonspontaneous

Strategy:

Set \( \Delta G^\circ\) equal to zero in Equation \(\ref{Gibbs}\) and solve for \(T\), the temperature at which the reaction becomes nonspontaneous.

Solution

The temperature at which the reaction becomes nonspontaneous is found by setting \( \Delta G^\circ\) equal to zero and rearranging Equation \(\ref{Gibbs}\) to solve for \(T\):

\[
\begin{aligned}
\Delta G^\circ &= \Delta H^\circ - T\Delta S^\circ = 0 \\[6pt]
\Delta H^\circ &= T\Delta S^\circ \\[6pt]
T &= \frac{\Delta H^\circ}{\Delta S^\circ}
= \frac{-91.8\ \mathrm{kJ/mol}\left( \frac{1000\ \mathrm{J}}{1\ \mathrm{kJ}}\right)}{-198.1\ \mathrm{J/(mol \cdot K)}}
= 463\ \mathrm{K}
\end{aligned}
\]

The reaction is spontaneous under standard state conditions below 463 K and nonspontaneous above 463 K.

This example highlights a key challenge in industrial chemistry: increasing the temperature speeds up the reaction (good for kinetics) but also makes it thermodynamically less favorable (bad for yield). The synthesis of ammonia is thus a classic case of balancing kinetics and thermodynamics.

Exercise \(\PageIndex{5}\)

Consider the reaction from Exercise \(\PageIndex{4}\):

\[\ce{ 2NO(g) + O2 (g) \rightleftharpoons 2NO2 (g)}\nonumber \]

From the calculations in the earlier example:

\( \Delta S^\circ\)= -146.6 J/(mol⋅K)

\( \Delta H^\circ\)= -116.2 kJ/mol

Because both \( \Delta H^\circ\) and \( \Delta S^\circ\) are negative, the reaction is spontaneous at lower temperatures and nonspontaneous at higher temperatures. 

Calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that \( \Delta H^\circ\) and \( \Delta S^\circ\) are independent of temperature.

Answer

The temperature at which the reaction becomes nonspontaneous is found by setting \( \Delta G^\circ\) equal to zero and rearranging Equation \(\ref{Gibbs}\) to solve for \(T\):

\[
\begin{aligned}
\Delta G^\circ &= \Delta H^\circ - T\Delta S^\circ = 0 \\[6pt]
\Delta H^\circ &= T\Delta S^\circ \\[6pt]
T &= \frac{\Delta H^\circ}{\Delta S^\circ}
= \frac{-116.2\ \mathrm{kJ/mol}\left( \frac{1000\ \mathrm{J}}{1\ \mathrm{kJ}}\right)}{-146.6\ \mathrm{J/(mol \cdot K)}}
= 792.6\ \mathrm{K}
\end{aligned}
\]

At temperatures below 793 K, the forward reaction is spontaneous under standard conditions. At higher temperatures, the reverse reaction becomes spontaneous.

Summary

The change in Gibbs free energy (\(\Delta G\)), a state function, allows us to predict the spontaneity of a process using only system properties:

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

A negative value of \(\Delta G\) corresponds to a positive \(\Delta S_{universe}\), indicating a spontaneous process.

In addition to indicating spontaneity, \(\Delta G\) also tells us the maximum amount of useful work (\(w\)) a system can perform:

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

The standard free-energy change (\(\Delta G^\circ\)) is the change in free energy when all reactants and products are in their standard states (1 M for solutions, 1 atm for gases and pure substances for solids and liquids). The standard free energy of formation (\(\Delta G_f^\circ\)), is the change in free energy when one mol of a substance in its standard state forms from its elements in their standard states. These values can be used to calculate \(\Delta G^\circ\) for a reaction:

\[ \Delta G^\circ = \sum m G_f^\circ(\text{products}) - \sum n G_f^\circ(\text{reactants}) \nonumber\]

A nonspontaneous reaction (\(\Delta G > 0\)) can be driven by coupling it to a spontaneous reaction (\(\Delta G < 0\)), as long as the overall \(\Delta G\) for the combine process is negative.

The value of \(\Delta G\) for a reaction depends on temperature. For processes with the same sign of \(\Delta H\) and \(\Delta S\), either an exothermic reaction (\(\Delta H < 0\)) with a decrease in entropy (\(\Delta S < 0\)) or an endothermic reaction (\(\Delta H > 0\)) with an increase in entropy (\(\Delta S > 0\)), the spontaneity of the reaction is temperature-dependent. In such cases, the temperature at which the reaction switches from spontaneous to nonspontaneous can be found by setting:

\[\Delta G = 0 \Rightarrow T = \frac{\Delta H}{\Delta S} \nonumber\]

This crossover temperature represents the point where the direction of spontaneity changes under standard state conditions.