# 2.2: Free Energy and the Equilibrium Constant

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We have identified three criteria for whether a given reaction will occur spontaneously: ΔS_{univ} > 0, ΔG_{sys} < 0, and the relative magnitude of the reaction quotient Q versus the equilibrium constant K. Recall that if Q < K, then the reaction proceeds spontaneously to the right as written, resulting in the net conversion of reactants to products. Conversely, if Q > K, then the reaction proceeds spontaneously to the left as written, resulting in the net conversion of products to reactants. If Q = K, then the system is at equilibrium, and no net reaction occurs. Table \(\PageIndex{1}\) summarizes these criteria and their relative values for spontaneous, nonspontaneous, and equilibrium processes.

Spontaneous |
Equilibrium |
Nonspontaneous* |
---|---|---|

*Spontaneous in the reverse direction. | ||

ΔS_{univ} > 0 |
ΔS_{univ} = 0 |
ΔS_{univ} < 0 |

ΔG_{sys} < 0 |
ΔG_{sys} = 0 |
ΔG_{sys} > 0 |

Q < K | Q = K | Q > K |

Because all three criteria are assessing the same thing—the spontaneity of the process—it would be most surprising indeed if they were not related. In this section, we explore the relationship between the standard free energy of reaction (ΔG°) and the equilibrium constant (K).

## Free Energy and the Equilibrium Constant

Because ΔH° and ΔS° determine the magnitude of ΔG° and because K is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of ΔG° and vice versa. "Free Energy", ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating ΔH from the equation for ΔG. The general relationship can be shown as follow (derivation not shown):

\[ \Delta G = V \Delta P − S \Delta T \label{18.29}\]

If a reaction is carried out at constant temperature (ΔT = 0), then Equation \(\ref{18.29}\) simplifies to

\[\Delta{G} = V\Delta{P} \label{18.30}\]

Under normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important.

Assuming ideal gas behavior, we can replace the \(V\) in Equation \(\ref{18.30}\) by nRT/P (where n is the number of moles of gas and R is the ideal gas constant) and express \(\Delta{G}\) in terms of the initial and final pressures (\(P_i\) and \(P_f\), respectively):

\[\Delta G=\left(\dfrac{nRT}{P}\right)\Delta P=nRT\dfrac{\Delta P}{P}=nRT\ln\left(\dfrac{P_\textrm f}{P_\textrm i}\right) \label{18.31}\]

If the initial state is the standard state with P_{i} = 1 atm, then the change in free energy of a substance when going from the standard state to any other state with a pressure P can be written as follows:

\[G − G^° = nRT\ln{P}\]

This can be rearranged as follows:

\[G = G^° + nRT\ln {P} \label{18.32}\]

As you will soon discover, Equation \(\ref{18.32}\) allows us to relate ΔG° and K_{p}. Any relationship that is true for \(K_p\) must also be true for \(K\) because \(K_p\) and \(K\) are simply different ways of expressing the equilibrium constant using different units.

Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species:

Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for ΔG:

Substituting Equation \(\ref{18.32}\) for each term into Equation \(\ref{18.34}\),

Combining terms gives the following relationship between ΔG and the reaction quotient Q:

where ΔG° indicates that all reactants and products are in their standard states. For gases at equilibrium (\(Q = K_p\),), and as you’ve learned in this chapter, ΔG = 0 for a system at equilibrium. Therefore, we can describe the relationship between ΔG° and K_{p} for gases as follows:

\[ 0 = ΔG° + RT\ln K_p \label{18.36a}\]

\[ΔG°= −RT\ln K_p \label{18.36b}\]

If the products and reactants are in their standard states and ΔG° < 0, then K_{p} > 1, and products are favored over reactants when the reaction is at equilibrium. Conversely, if ΔG° > 0, then K_{p} < 1, and reactants are favored over products when the reaction is at equilibrium. If ΔG° = 0, then \(K_p = 1\), and neither reactants nor products are favored when the reaction is at equilibrium.

For a spontaneous process under standard conditions, \(K_{eq}\) and \(K_p\) are greater than 1.

Although K_{p} is defined in terms of the partial pressures of the reactants and the products, the equilibrium constant K is defined in terms of the concentrations of the reactants and the products. We described the relationship between the numerical magnitude of K_{p} and K in Chapter 15 and showed that they are related:

\[K_p = K(RT)^{Δn} \label{18.37}\]

where Δn is the number of moles of gaseous product minus the number of moles of gaseous reactant. For reactions that involve only solutions, liquids, and solids, Δn = 0, so K_{p} = K. For all reactions that do not involve a change in the number of moles of gas present, the relationship in Equation \(\ref{18.36b}\) can be written in a more general form:

\[ΔG° = −RT \ln K \label{18.38}\]

Only when a reaction results in a net production or consumption of gases is it necessary to correct Equation \(\ref{18.38}\) for the difference between K_{p} and K.

Combining Equation \(\ref{18.38}\) with \(ΔG^o = ΔH^o − TΔS^o\) provides insight into how the components of ΔG° influence the magnitude of the equilibrium constant:

\[ΔG° = ΔH° − TΔS° = −RT \ln K \label{18.39}\]

Notice that \(K\) becomes larger as ΔS° becomes more positive, indicating that the magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder. Moreover, K increases as ΔH° decreases. Thus the magnitude of the equilibrium constant is also directly influenced by the tendency of a system to seek the lowest energy state possible.

The magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum entropy and seek the lowest energy state possible.

## Temperature Dependence of the Equilibrium Constant

The fact that ΔG° and K are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship is shown explicitly in Equation \(\ref{18.39}\), which can be rearranged as follows:

\[\ln K=-\dfrac{\Delta H^\circ}{RT}+\dfrac{\Delta S^\circ}{R} \label{18.40}\]

Assuming ΔH° and ΔS° are temperature independent, for an exothermic reaction (ΔH° < 0), the magnitude of K decreases with increasing temperature, whereas for an endothermic reaction (ΔH° > 0), the magnitude of K increases with increasing temperature. The quantitative relationship expressed in Equation \(\ref{18.40}\) agrees with the qualitative predictions made by applying Le Chatelier’s principle. Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of K. Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of K. Equation \(\ref{18.40}\) also shows that the magnitude of ΔH° dictates how rapidly K changes as a function of temperature. In contrast, the magnitude and sign of ΔS° affect the magnitude of K but not its temperature dependence.

If we know the value of K at a given temperature and the value of ΔH° for a reaction, we can estimate the value of K at any other temperature, even in the absence of information on ΔS°. Suppose, for example, that K_{1} and K_{2} are the equilibrium constants for a reaction at temperatures T_{1} and T_{2}, respectively. Applying Equation \(\ref{18.40}\) gives the following relationship at each temperature:

\[\begin{align}\ln K_1&=\dfrac{-\Delta H^\circ}{RT_1}+\dfrac{\Delta S^\circ}{R}

\\ \ln K_2 &=\dfrac{-\Delta H^\circ}{RT_2}+\dfrac{\Delta S^\circ}{R}\end{align}\]

Subtracting \(\ln K_1\) from \(\ln K_2\),

\[\ln K_2-\ln K_1=\ln\dfrac{K_2}{K_1}=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \label{18.41}\]

Thus calculating ΔH° from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature (K_{1}) allow us to calculate the value of the equilibrium constant at any other temperature (K_{2}), assuming that ΔH° and ΔS° are independent of temperature.

## Summary

For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature. If we assume ideal gas behavior, the ideal gas law allows us to express ΔG in terms of the partial pressures of the reactants and products, which gives us a relationship between ΔG and K_{p}, the equilibrium constant of a reaction involving gases, or K, the equilibrium constant expressed in terms of concentrations. If ΔG° < 0, then K > 1, and products are favored over reactants at equilibrium. Conversely, if ΔG° > 0, then K < 1, and reactants are favored over products at equilibrium. If ΔG° = 0, then K=1, and neither reactants nor products are favored at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature.

## Contributors

- Modified by Tom Neils (Grand Rapids Community College)