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# 20.3: Free Energy and Equilibrium

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Determine the temperature at which a reversible reaction will achieve equilibrium by using the Gibbs free energy equation.
• Describe the relationship between standard free energy change (ΔG°) and the equilibrium constant (Keq) for reversible reactions.
• Convert between Keq and ΔG° for a reaction at a given temperature.

### Recalling Prior Knowledge

• What is the equation that relates enthalpy change, entropy change, and free energy change?
• How is the value of the equilibrium constant for a reaction related to the equilibrium position?

A system is at equilibrium when the rates of the forward and reverse reactions are equal. In this case, neither reaction is spontaneous. In this lesson, you will learn about the relationship of free energy to equilibrium and the equilibrium constant.

## Temperature and Free Energy

In the last lesson, you learned how to calculate the free energy change (ΔG) for a reaction when the enthalpy change (ΔH) and the entropy change (ΔS) are known. Consider the reversible reaction in which calcium carbonate decomposes into calcium oxide and carbon dioxide gas. The production of CaO (called quicklime) has been an important reaction for centuries, as shown by the lime kiln below (Figure below).

\begin{align*}\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)\end{align*}

The ΔH° value for this reaction is 177.8 kJ/mol, and the ΔS° value is 160.5 J/K•mol. The reaction is endothermic with an increase in entropy due to the production of a gas. We can first calculate the ΔG° at 25°C in order to determine if the reaction is spontaneous at room temperature.

ΔG° = ΔH° - TΔS°
ΔG° = 177.8 kJ/mol - 298 K(0.1605 kJ/K•mol) = 130.0 kJ/mol

Since ΔG° is a large positive quantity, the reaction strongly favors the reactants, and very little products would be formed. In order to determine a temperature at which ΔG° will become negative, we can first solve the equation for the temperature when ΔG° is equal to zero.

0=ΔH°-TΔS°
\begin{align*}\text{T}=\dfrac{\Delta \text{H}^\circ}{\Delta \text{S}^\circ}=\dfrac{177.8 \ \text{kJ/mol}}{0.1605 \ \text{kJ/K} \cdot \text{mol}}=1108 \ \text{K}=835^\circ \text{C}\end{align*}

At any temperature higher than 835°C, the value of ΔG° will be negative, and the decomposition reaction will be spontaneous.

This lime kiln in Utah was used to produce quicklime (calcium oxide), an important ingredient in mortar and cement.

Recall that we are assuming the values for ΔH and ΔS are independent of temperature. Because they actually vary slightly at non-standard temperatures, the point at which the sign of ΔG° switches from being positive to negative (835°C) is an approximation. Also, it is not correct to assume that absolutely no products are formed below 835°C and decomposition suddenly begins once that temperature is reached. Rather, at lower temperatures, the reactants are simply favored to some extent over the products when the reaction is at equilibrium. When this reaction is performed, the amount of products can be detected by monitoring the pressure of the CO2 gas that is produced. Above about 700°C, measurable amounts of CO2 are produced. The pressure of CO2 at equilibrium gradually increases with increasing temperature. Above 835°C, the pressure of CO2 at equilibrium begins to exceed 1 atm, the standard-state pressure. This is an indication that the products of the reaction are now favored above that temperature. When quicklime is manufactured, the CO2 is constantly removed from the reaction mixture as it is produced. This causes the reaction to be driven toward the products according to Le Châtelier’s principle.

### Changes of State

At the temperature at which a change of state occurs, the two states are in equilibrium with one another. For an ice-water system, equilibrium takes place at 0°C, so ΔG° is equal to 0 at that temperature. The heat of fusion of water is known to be equal to 6.01 kJ/mol, and so the Gibbs free energy equation can be solved for the entropy change that occurs during the melting of ice. The symbol ΔSfus represents the entropy change during the melting process, while Tf is the freezing point of water.

0 = ΔHfus - TfΔSfus
\begin{align*}\mathrm{\Delta S_{fus}=\dfrac{\Delta H_{fus}}{T_f}=\dfrac{6.01 \ kJ/mol}{273 \ K}=0.0220 \ kJ/K \cdot mol=22.0 \ J/K \cdot mol}\end{align*}

The entropy change is positive as the solid state changes into the liquid state. A similar calculation can be performed for the vaporization of liquid to gas.

## Equilibrium Constant and ΔG

The ΔG value of a reaction tells us whether reactants or products are favored at equilibrium. We also know that the equilibrium constant, Keq, relates the concentrations of all substances in the reaction at equilibrium. It stands to reason that there is a relationship between the values of ΔG° and Keq. A more advanced treatment of thermodynamics yields the following equation:

ΔG°=-RT ln(Keq)

The variable R is the ideal gas constant (8.314 J/K•mol), T is the Kelvin temperature, and ln(Keq) is the natural logarithm of the equilibrium constant.

When Keq is large, the products of the reaction are favored, so ΔG° should be negative. When Keq is small, the reactants are favored. The natural logarithm of a number less than one is negative, so the sign of ΔG° is positive when Keq < 1. The table below (Table below) summarizes the relationship between ΔG° and Keq.

Relationship between ΔG° and K
Keq ln(Keq) ΔG° Description
>1 positive negative Products are favored at equilibrium.
1 0 0 Reactants and products are equally favored.
<1 negative positive Reactants are favored at equilibrium.

Knowledge of either the standard free energy change or the equilibrium constant for a reaction allows for the calculation of the other. The following two sample problems illustrate each case.

Sample Problem 20.2: Gibbs Free Energy and the Equilibrium Constant

The formation of nitrogen monoxide from nitrogen and oxygen gases is a reaction that strongly favors the reactants at 25°C.

N2(g) + O2(g) 2NO(g)

The actual concentrations of each gas would be difficult to measure, and so the Keq for the reaction can more easily be calculated from the ΔG°, which is equal to 173.4 kJ/mol.

Step 1: List the known values and plan the problem.

Known

• ΔG° = +173.4 kJ/mol
• R = 8.314 J/K•mol
• T = 25°C = 298 K

Unknown

• Keq = ?

In order to make the units agree, the value of ΔG° will need to be converted to J/mol (173,400 J/mol). To solve for Keq, the inverse of the natural logarithm, ex, will be used.

Step 2: Solve.

\begin{align*}\Delta \text{G}^\circ=-\text{RT} \ln{\text{K}_{\text{eq}}}\end{align*}
\begin{align*}\ln{\text{K}_{\text{eq}}} = \dfrac{- \Delta \text{G}^\circ}{\text{RT}}\end{align*}
\begin{align*}\text{K}_{\text{eq}}=\text{e}^\frac{- \Delta \text{G}^\circ}{\text{RT}}=\text{e}^\frac{-173400 \: \text{J/mol}}{8.314 \: \text{J/K} \cdot \text{mol}(298 \: \text{K})}=4.0 \times 10^{-31}\end{align*}

The large positive free energy change leads to a Keq value that is extremely small. Both lead to the conclusion that the reactants are highly favored, and very few product molecules are present at equilibrium.

Sample Problem 20.3: Free Energy from Ksp

The solubility product constant (Ksp) of lead(II) iodide is 1.4 × 10−8 at 25°C. Calculate ΔG° for the dissociation of lead(II) iodide in water.

\begin{align*}\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq)\end{align*}

Step 1: List the known values and plan the problem.

Known

• Keq = Ksp = 1.4 × 10−8
• R = 8.314 J/K•mol
• T = 25°C = 298 K

Unknown

• ΔG° = ? kJ/mol

The equation relating ΔG° to Keq can be solved directly.

Step 2: Solve.

ΔG° = -RT ln(Keq)
ΔG° = -(8.314 J/K•mol)(298 K)ln(1.4 × 10-8) = 45,000 J/mol = 45 kJ/mol

The large, positive ΔG° indicates that the solid lead(II) iodide is nearly insoluble and is mostly present as a solid at equilibrium.

Practice Problem
1. For the Haber-Bosch process at 25°C, ΔH° = −92.6 kJ/mol and ΔS° = −198.5 J/K•mol.
\begin{align*}\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\end{align*}
a. Calculate ΔG° for the reaction.
b. Calculate the value of Keq.

## Lesson Summary

• The value of the free energy change (ΔG) is zero when a reaction is at equilibrium. The Gibbs free energy equation can be used to solve for the temperature, which equals ΔH/ΔS.
• Two states of a substance are in equilibrium with one another during a change of state.
• The standard free energy change for a reversible reaction is related to the equilibrium constant by the equation ΔG° = −RT ln(Keq). A large Keq value corresponds to a negative ΔG°, while a small Keq value corresponds to a positive ΔG°.

## Lesson Review Questions

### Reviewing Concepts

1. What is the general value of ΔG° for a reaction that strongly favors the formation of products? That strongly favors the reactants?
2. What is the value of ΔG when a reaction is at equilibrium?
3. A certain reaction has a free energy change of ΔG = +10 kJ/mol at a certain temperature. Does this mean that no products are formed in this reaction? Explain.
4. A reaction has a negative ΔH°, a negative ΔS°, and a positive ΔG° at a given temperature. In order to make formation of the products more favorable, should the temperature be raised or lowered? Explain.

### Problems

1. For a certain reaction, ΔH° = −29.2 kJ/mol and ΔS° = −141 J/K•mol. Is the reaction spontaneous at 25°C? If not, at what temperatures would it become spontaneous?
2. Find the Celsius temperatures at which reactions with the following ΔH° and ΔS° values would be spontaneous.
1. ΔH° = +76.5 kJ/mol; ΔS° = +231 J/K•mol
2. ΔH° = −50.2 kJ/mol; ΔS° = +97.7 J/K•mol
3. ΔH° = +19.1 kJ/mol; ΔS° = −73.0 J/K•mol
3. For the nearly insoluble compound silver iodide, Ksp = 8.3 × 10−17. Calculate ΔG° (in kJ/mol) for the following reaction at 25°C: \begin{align*}\text{AgI}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{I}^-(aq)\end{align*}.
4. Calculate the equilibrium constant at 25°C for the following reaction in which hydrogen and bromine combine to form hydrogen bromide. \begin{align*}\text{H}_2(g) + \text{Br}_2(g) \rightleftharpoons 2\text{HBr}(g) \ \ \ \ \Delta \text{G}^\circ = -106.4 \ \text{kJ}\end{align*}
5. Aqueous ammonia combines with water in an equilibrium reaction to form the ammonium ion and the hydroxide ion according to the following equation: \begin{align*}\text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}^+_4(aq) + \text{OH}^-(aq)\end{align*}. For this reaction at 25°C, ΔH° = 3.36 kJ/mol and ΔS° = −78.9 J/K•mol.
1. Calculate ΔG° at 25°C.
2. Calculate Keq at 25°C.
6. Carbon in the form of diamond can form graphite according to the following equation: \begin{align*}\text{C(diamond)} \rightleftharpoons \text{C(graphite)}\end{align*}. Use the thermodynamic data below (Table below).
1. Calculate the ΔG° for the reaction in which diamond changes to graphite. (Hint: first calculate ΔH° and ΔS° from the data table.)
2. Calculate the equilibrium constant, Keq.
3. Is the reaction spontaneous at 25°C? If so, do we need to be concerned about diamonds turning into graphite? Explain.
Graphite and Diamond Thermodynamic Data
Graphite Diamond
heat of formation (ΔHf°) 0 1.90 kJ/mol
standard entropy (S°) 5.69 J/K•mol 2.4 J/K•mol

## Points to Consider

Acids and bases are extremely important types of chemical compounds that we use virtually every day.

• What are the physical and chemical properties of acids and bases?
• What types of reactions do acids and bases undergo?

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