Skip to main content
Published in:
Cover of the book

Open Access 2024 | OriginalPaper | Chapter

5. Introduction to Equilibrium

Activate our intelligent search to find suitable subject content or patents.

loading …


This chapter defines equilibrium in mixtures and ideal solutions and introduces the equilibrium constant, its relationship with the Gibbs free energy and its dependence on temperature. Homogenous and heterogenous equilibria are distinguished and solid–gas, liquid–gas and mineral-solution equilibria are presented, including Henry’s law and the solubility product.
Our thermodynamic classes have provided us the tools to determine whether a reaction at Earth surface conditions is spontaneous or not, i.e., whether it has a negative or positive Gibbs free energy. For instance, the dissolution of sodium chloride (kitchen salt) is an endothermic process (consuming heat and cooling the water; \(\Delta_{r} H^{o} > 0\)) but occurs spontaneously because it has a negative Gibbs free energy \((\Delta_{r} G^{o}\)) of −9 kJ mol−1. Spontaneity of a reaction does not imply that the reaction goes to completion, i.e., equilibrium is not where all reactants are converted into products. While addition of kitchen salt to water initially indeed results in the dissolution of NaCl, no further dissolution is observed after that about 357 gr has been added to 1 kg of water. If more than 357 gr of NaCl were dissolved in water, it would precipitate (a backward reaction). The system water-NaCl has reached equilibrium, i.e., the concentrations of reactants and products remain constant over time. An equilibrium state is attained as the rate of forward (from reactants to products) and backward (from products to reactants) are equal and there are no further changes in concentrations. In System Earth many reactions can be fully described, or be approximated, by an equilibrium approach: the exchange of gases between water and air, proton transfers (acid-base chemistry) and electron transfers (redox reactions) and the precipitation and dissolution of minerals. For instance, the dissolution of calcium carbonate leads to karst processes and the precipitation of calcium carbonate causes formation of stalactites and stalagmites in caves.
It is therefore useful to deepen our understanding and further develop our quantitative tools for predicting equilibria. In this introduction, we will focus on ideal solution chemistry and equilibria at Earth-surface conditions and leave non-ideal solutions and high-temperature and high-pressure equilibria during formation of magmatic and metamorphic rocks to subsequent courses. Following the introduction of the equilibrium constant and its relation to the Gibbs free energy, we will focus on equilibria in ideal solutions involving minerals, acids and bases, and redox reactions.

5.1 The Equilibrium Constant and Its Relation to the Gibbs Free Energy

Nitrogen dioxide (NO2) is an atmospheric gas resulting from combustion engines (cars, buses, trucks). It is a deep brown gas that contributes to smog above cities. In the atmosphere, it is in equilibrium with dinitrogen tetroxide (a colorless gas):
$$ 2 {\text{NO}}_{2} \left( {\text{g}} \right) \leftrightarrow {\text{N}}_{2} {\text{O}}_{4} \left( {\text{g}} \right) $$
The transformation of NO2(g) to N2O4(g) causes the color to fade until an equilibrium is established, and this reaction has therefore been well studied in the laboratory. Irrespective of the starting conditions (only NO2(g), only N2O4(g), high NO2(g) with low N2O4 (g), or the other way around), the final composition of the reaction mixture is always the same when expressed as the ratio of product to reactant concentrations (the law of mass action). Specifically, in this case the calculated concentration ratio \(\frac{{{\text{N}}_{{2}} {\text{O}}_{{4}} { }\left( {\text{g}} \right)}}{{ ({\text{NO}}_{2} )^{2} \left( {\text{g}} \right)}}\) is about 216 at 25 °C and 1 bar, irrespective of starting conditions.
To generalize this phenomenon, we introduce the reaction of α moles A and β moles B to form γ moles C and δ moles D
$$ \alpha A + \beta B \to \gamma C + \delta D $$
and define a reaction quotient Q:
$$ Q = \frac{{C^{\gamma } D^{\delta } }}{{A^{\alpha } B^{\beta } }} $$
This reaction quotient is dimensionless and general, i.e., it can always be calculated irrespective whether there is an equilibrium or not. When the system is at equilibrium, the value of Q is called the equilibrium constant K:
$$ K = \frac{{C_{eq}^{\gamma } D_{eq}^{\delta } }}{{A_{eq}^{\alpha } B_{eq}^{\beta } }} $$
The value of K expresses the composition of the system, i.e., concentrations or partial pressures of the reactants and products at equilibrium (taking reaction stoichiometry into account) and has a constant value at a certain temperature and pressure. The equilibrium constant has no unit because they cancel out. Reversing a reaction causes an inversion of K, and multiplying the coefficients (α, β, γ, δ) by a common factor raises the equilibrium constant to the corresponding factor. If the K value is very high, the equilibrium mixture will consist mainly of products C and D. If  \({K \ll {1}}\), then reactants will dominate the mixture, while similar quantities of reactant and products are expected if K is around 1, depending on the reaction stoichiometry of course.
$$ \begin{array}{*{20}l} {K > > { 1}} \hfill & {{\text{Products }}\,{\text{dominant}}} \hfill \\ {K \approx {1}} \hfill & {{\text{Similar }}\,{\text{quantities}}} \hfill \\ {K < < { 1} } \hfill & {{\text{Reactants }}\,{\text{dominant}}} \hfill \\ \end{array} $$
Comparing the reaction quotient Q and the equilibrium constant K provides information on the direction of change, i.e., whether a process is spontaneous. If Q > K, then the net reaction goes from right to left (products to reactant, backward reaction is spontaneous) and if Q < K then the reaction goes from left to right (reactants are transformed into products and the forward reaction is spontaneous). If Q = K then the reaction is in equilibrium.
$$ \begin{array}{*{20}l} {Q\, > \,K} \hfill & {{\text{backward }}\,{\text{reaction}}\,{\text{ is }}\,{\text{spontaneous}}} \hfill \\ { Q\, < \,K} \hfill & {{\text{forward }}\,{\text{reaction}}\,{\text{ is }}\,{\text{spontaneous}}} \hfill \\ { Q = K} \hfill & {{\text{reaction}}\,{\text{ is }}\,{\text{in }}\,{\text{equilibrium}}} \hfill \\ \end{array} $$
In thermodynamics lectures we have seen that spontaneity is related to the Gibbs free energy change at constant pressure and temperature; there should thus be a relation between the reaction quotient Q, the equilibrium constant K, and the sign of the Gibbs free energy change (\(\Delta_{r} G\)).
Returning to reaction (5.1), the Gibbs free energy change for the gas reaction (\(\Delta_{r} G) \) of NO2(g) to N2O4(g)
$$ \Delta_{r} G = \Delta G\left( {{\text{N}}_{2} {\text{O}}_{4} \left( {\text{g}} \right)} \right) - 2 \times \Delta G\left( {{\text{NO}}_{2 } \left( {\text{g}} \right)} \right) $$
We use \(\Delta_{r} G \) rather than \(\Delta_{r} G^{o}\) because the reaction is not occurring under standard conditions (i.e., for a pure gas). During the thermodynamic classes (Eq. 4.​12) we have derived that
$$ \Delta G = \Delta G^{o} + RT\ln \left( {\frac{{P_{g} }}{{P_{0} }}} \right) $$
where R is the universal gas constant, T is absolute temperature, \(\Delta G^{o} \) is the standard Gibbs free energy change for a pure gas, Pg is the partial pressure of gas g, and P0 is the total pressure of 1 bar, respectively. Combining Eqs. (5.5 and 5.6), we then arrive at
$$ \Delta_{r} G = \Delta G_{f}^{o} \left( {{\text{N}}_{{2}} {\text{O}}_{{4}} \left( {\text{g}} \right)} \right) + RT\ln \left( {\frac{{P_{{{\text{N}}_{{2}} {\text{O}}_{{4}} }} }}{{P_{0} }}} \right) - 2\left[ {\Delta G_{f}^{o} \left( {{\text{NO}}_{2 } \left( {\text{g}} \right)} \right) + RT\ln \left( {\frac{{P_{{{\text{NO}}_{{2}} }} }}{{P_{0} }}} \right)} \right] $$
Using \(\Delta_{r} G^{o} = \Delta G_{f}^{o} \left( {{\text{N}}_{{2}} {\text{O}}_{{4}} \left( {\text{g}} \right)} \right) - 2 \times \Delta G_{f}^{o} \left( {{\text{NO}}_{2 } \left( {\text{g}} \right)} \right)\), we can rewrite Eq. (5.7) as follows,
$$ \Delta_{r} G = \Delta_{r} G^{o} + RT\ln \left( {\frac{{P_{{{\text{N}}_{{2}} {\text{O}}_{{4}} }} }}{{P_{0} }}} \right) - 2 RT\ln \left( {\frac{{P_{{{\text{NO}}_{{2}} }} }}{{P_{0} }}} \right) = \Delta_{r} G^{o} + RT\ln \frac{{\left( {\frac{{P_{{{\text{N}}_{{2}} {\text{O}}_{{4}} }} }}{{P_{0} }}} \right)}}{{\left( {\frac{{P_{{{\text{NO}}_{{2}} }} }}{{P_{0} }}} \right)^{2} }} $$
Accordingly, the Gibbs free energy of reaction \(\Delta_{r} G \) contains the standard Gibbs free energy change of reaction \((\Delta_{r} G^{o} )\) that is independent of the partial pressures of the two gases and a term \(RT\ln \frac{{\left( {\frac{{P_{{{\text{N}}_{{2}} {\text{O}}_{{4}} }} }}{{P_{0} }}} \right)}}{{\left( {\frac{{P_{{{\text{NO}}_{{2}} }} }}{{P_{0} }}} \right)^{2} }}\) that depends on the partial pressure of the reactant and product.
Moreover, this ratio \(\frac{{\left( {\frac{{P_{{{\text{N}}_{{2}} {\text{O}}_{{4}} }} }}{{P_{0} }}} \right)}}{{\left( {\frac{{P_{{{\text{NO}}_{{2}} }} }}{{P_{0} }}} \right)^{2} }}\) is the reaction quotient Q for reaction 5.1, if expressed in partial pressures and a total pressure P0 of 1 bar.
Generalizing this to any reaction, we obtain
$$ \Delta_{r} G = \Delta_{r} G^{o} + RT\ln Q $$
stating that the Gibbs free energy change of a reaction is the sum of the standard Gibbs free energy change of the reaction \((\Delta_{r} G^{o} )\), calculated from the Gibbs free energy of formations, and the term \(RTlnQ\), where the latter provides a correction term for the actual composition of the mixture.
As an example, consider reaction 5.1 and a reaction vessel at 298 K with 0.350 bar NO2(g) and 0.65 bar N2O4(g). The standard Gibbs free energy of formations of the two gases are \(\Delta G_{f}^{0}\) 51.3 and 99.8 kJ mol−1, respectively.
$$ \begin{aligned} \Delta_{r} G & = \Delta_{r} G^{o} + RT\ln Q \\ & = 99{,}800 - 2 \times 51{,}300 + 8.314 \times 298\ln \frac{0.65}{{0.35^{2} }} = 1330 \;{\text{J}} \\ \end{aligned} $$
The \(\Delta_{r} G \ne 0,\) the system is thus not at equilibrium. Specifically, the net reaction proceeds towards reactants, i.e., NO2(g), because \(\Delta_{r} G\) > 0. (The same conclusion could be reached using the reaction quotient Q and the equilibrium constant K).
At equilibrium, the reaction quotient is equal to the equilibrium constant (Q = K), and the Gibbs free energy of the reaction should be zero (\(\Delta_{r} G = 0; \) see thermodynamic lectures, Eq. (4.​4)). Consequently,
$$ 0 = \Delta_{r} G = \Delta_{r} G^{o} + RT\ln K $$
$$ \Delta_{r} G^{o} = - RT\ln K $$
or alternatively written,
$$ K = \exp \left( {\frac{{ - \Delta_{r} G^{o} }}{RT}} \right) $$
Equations (5.10a and 5.10b) are completely general and refer to all types of reactions. They provide a direct link between thermodynamic data and theory on the one hand and the composition of equilibrium mixtures on the other. However, Eqs. (5.10a and 5.10b) implies that the mixture composition is expressed in activities rather than concentrations (see Eq. 4.​13). In this course, we restrict ourselves to ideal solutions, i.e., ion pairing and interactions between ions and the solvent, are ignored and we express the equilibrium constants and reaction quotients in concentrations [c], rather than thermodynamic activities a, where \(\left[ c \right] = \frac{{P_{g} }}{{P_{0} }}\) for a gas \((P_{0} = 1 \,{\text{bar}}), \) \(\left[ c \right] = \frac{{C_{a} }}{{C_{0} }}\) for a solute (\(C_{0}\) = 1 mol L−1) and \(\left[ c \right] = 1 \) for solids or the liquid medium. Activities (or effective concentrations) and concentrations are related via activity coefficients. However, the use of concentrations rather than activities implies the need of units for equilibrium constants.
Equilibrium mixture compositions are temperature dependent and there are two ways to correct equilibrium constants for lower or higher temperatures. The first method uses Eq. (5.10a and 5.10b) to link the equilibrium constant K with the \(\Delta_{r} G^{o}\) at the temperature of interest calculated from the entropy (\(\Delta_{r} S^{o}\)) and enthalpy (\(\Delta_{r} H^{o}\)) of reactions (Eq. 4.​3 in thermodynamics). Alternatively, the van ‘t Hoff equation is used.
Combining, Eq. (5.10a): \(\Delta_{r} G^{o} = - RT\ln K\).
and Eq. (4.​3): \(\Delta_{r} G^{o} = \Delta_{r} H^{o} - T\Delta_{r} S^{o}\).
we obtain the van ‘t Hoff equation after rearrangement,
$$ \ln K = \frac{{\Delta_{r} S^{o} }}{R} - \frac{{\Delta_{r} H^{o} }}{RT} $$
Or in its more useful form going from T1 to T2,
$$ \ln K_{2} = \ln K_{1} + \frac{{\Delta_{r} H^{o} }}{R}\left( {\frac{1}{{T_{1} }} - \frac{1}{{T_{2} }}} \right) $$
where T1 is usually 298 K (SATP). If the reaction is exothermic (\(\Delta_{r} H^{o} < 0),\) then K decreases with increasing temperature and equilibrium shifts towards more reactants, i.e., to the left. Conversely, equilibria shift to the right and K increases with temperature for an endothermic reaction \((\Delta_{r} H^{o} > 0).\)

5.2 Heterogenous Equilibria, Including Mineral Solubility

Equilibrium can be established when all reactants and products are in one single phase (homogenous equilibrium), or multiple phases are involved (heterogenous equilibrium). Acid-base reactions in solution are an example of the former, while mineral dissolution/precipitation reactions involving a solid and a liquid phase are heterogenous. This section will present three types of heterogenous equilibria (solid–gas, liquid–gas and mineral-solution). Homogenous acid-base equilibria will be covered in the next chapter.
Solid–gas equilibria: The production of lime (CaO, s) from limestone (CaCO3, s)
$$ {\text{CaCO}}_{{3 }} \left( {\text{s}} \right) \leftrightarrow {\text{CaO}}\left( {\text{s}} \right) + {\text{CO}}_{{2}} \left( {\text{g}} \right) $$
is a heterogenous reaction with an equilibrium constant K defined as
$$ K = \frac{{\left[ {{\text{CaO}}} \right]\left[ {{\text{CO}}_{{2}} } \right]}}{{\left[ {{\text{CaCO}}_{{3}} } \right]}} = \left[ {{\text{CO}}_{{2}} } \right] $$
Note that […] are used to indicate concentrations rather than activities and that solid phase and liquid can be excluded from heterogenous equilibria mass action laws since they have an (activity) value of one. Consequently, the equilibrium concentration of carbon dioxide gas in a system with both solid phase CaO and CaCO3 is independent of the amount of these solid phases, as long as both are present.
Liquid–gas equilibria: The solubility of gases in water is another example of heterogenous equilibrium. The solubility of a gas in water is described using Henry’s law:
$$ C_{gas} = K_{H} \cdot P_{gas} $$
where Cgas is the equilibrium concentration of a gas in water (mol kg−1), Pgas is the partial pressure (atm) and KH is the Henry’s law constant (mol kg−1 atm−1). Henry’s law constant is a function of temperature and salinity of the water. Consider the solubility of oxygen in water:
$$ {\text{O}}_{{2 }} \left( {\text{g}} \right) + {\text{H}}_{{2}} {\text{O}}\left( {\text{l}} \right) \leftrightarrow {\text{O}}_{{2}} \left( {{\text{aq}}} \right) $$
The corresponding equilibrium constant is \(K_{{{\text{H,O}}_{{2}} }} = \frac{{{\text{O}}_{{2}} \left( {{\text{aq}}} \right)}}{{P_{{{\text{O}}_{{2}} }} }}\).
Figure 5.1 shows the concentrations of dissolved oxygen as a function of temperature for freshwater and ocean water. Oxygen and most other gases dissolve better in freshwater than seawater because of a salting out effect. Dissolved oxygen concentrations in water decline with warming of the water. Cold waters in equilibrium with the atmosphere therefore have higher oxygen contents than warm waters.
Mineral-solution equilibria: The availability of certain minerals constrains the composition of natural waters and vice versa the composition of solutions can determine whether minerals dissolve or precipitate. Mineral equilibria are written as dissolution reactions (i.e., mineral on the left-hand side and ions on the right-hand side) and characterized by a solubility product. The solubility product of the mineral fluorite (CaF2(s)) is based on the following reaction:
$$ {\text{CaF}}_{2} \left( {\text{s}} \right) \leftrightarrow {\text{Ca}}^{2 + } \left( {{\text{aq}}} \right) + 2 {\text{F}}^{ - } \left( {{\text{aq}}} \right) $$
with the following equilibrium constant
$$ K = \frac{{\left[ {{\text{Ca}}^{2 + } } \right]_{eq} \left[ {{\text{F}}^{ - } } \right]^{2}_{eq} }}{{\left[ {{\text{CaF}}_{{2}} } \right]}} $$
where the subscript eq indicates that these are equilibrium concentrations. Recalling that solid phases have a (molar fraction) value of one, we can then define the solubility product for fluorite as
$$ K_{sp} = \left[ {{\text{Ca}}^{2 + } } \right]_{eq} \left[ {{\text{F}}^{ - } } \right]^{2}_{eq} $$
The solubility product \(K_{sp} \) is the equilibrium constant (K) for the special case that a solid phase (mineral) is dissolving. This can be generalized to
$$ C_{x} A_{y} \leftrightarrow xC^{a + }_{eq} + yA^{b - }_{eq} $$
$$ K_{sp} = \left[ {C^{a + } } \right]^{x}_{eq} \left[ {A^{b - } } \right]^{y}_{eq} $$
Returning to the solubility of fluorite (reaction 5.14), if we know the equilibrium concentrations of dissolved calcium (2 × 10–4 M) and fluoride (1.6 × 10–4 M), we can use Eq. (5.15) to calculate the solubility product as 5.1 × 10–12. A solubility product is often reported as a pKsp value, i.e. the −log10(Ksp), in this case pKsp = 11.29.
The same result would be obtained if calculated using Eqs. (5.10a and 5.10b) and thermochemical data with \(\Delta G_{f}^{o}\) values of −1175.6, −553.6, −278.8 kJ mol−1 for fluorite, dissolved calcium, and fluoride, respectively. First, one calculates the Gibbs free energy of reaction (\(\Delta_{r} G^{o} ): \)\({\text{Type equation here}}.\)
$$\begin{aligned}\Delta _{r} G^{o} &= \left[ {~1 \times \Delta ~G_{f}^{o} ~\left( {{\text{Ca}}^{{2 + }} \left( {{\text{aq}}} \right)} \right) + ~2 \times \Delta ~G_{f}^{o} ~\left( {{\text{F}}^{ - } \left( {{\text{aq}}} \right)} \right)} \right]\\ & \;\;\;\;- \left[ {\Delta ~G_{f}^{o} \left( {{\text{CaF}}_{2} \left( s \right)} \right)} \right] = {\text{64}}.{\text{4}}\;{\text{kJ}}\end{aligned}$$
and then uses Eqs. (5.10a and 5.10b):
$$ K = \exp \left( {\frac{{ - \Delta_{r} G^{o} }}{RT}} \right) = \exp \left( {\frac{ - 64400}{{8.31 \cdot 298}}} \right) = 5 \times 10^{ - 12} \to pKsp = 11.29 $$
Accordingly, solubility products link thermochemistry with solution chemistry for solutions in equilibrium.
Solubility products can also be used to constrain the composition of natural waters. Consider a pool in equilibrium with the minerals fluorite (CaF2, pKsp = 11.29), calcite (CaCO3, pKsp = 8.48) and dolomite (CaMg(CO3)2, pKsp = 19), knowing that the dissolved fluoride concentration is 10–4 M. From the fluorite equilibrium, we then obtain a dissolved calcium concentration of 5.13 × 10–4 M, next the calcite equilibrium can be used to derive the carbonate ion concentration 6.46 × 10–6 M and finally the dissolved magnesium concentration is calculated making use of the solubility product of dolomite (4.68 × 10–6 M). This example clearly shows that sharing common ions has major implications for the solubility of minerals.
The precipitation and dissolution of minerals depends on the saturation state of that mineral. For the dissolution of calcium carbonate in the form of calcite:
$$ {\text{CaC}}O_{3 } \left( {\text{s}} \right) \leftrightarrow {\text{Ca}}^{2 + } \left( {{\text{aq}}} \right) + {\text{CO}}_{3}^{2 - } \left( {{\text{aq}}} \right) $$
we first define the ion product (IP), which is a special form of the reaction quotient Q for mineral equilibria.
$$ {\text{IP}} = \left[ {{\text{Ca}}^{2 + } } \right]\left[ {{\text{CO}}_{3}^{2 - } } \right] $$
with, in this case, a unit of mol2 kg−2, and then the degree of saturation or saturation state (Ω)
$$ {\Omega } = \frac{{{ }\left[ {{\text{Ca}}^{2 + } } \right] \left[ {{\text{CO}}_{3}^{2 - } } \right]}}{{\left[ {{\text{Ca}}^{2 + } } \right]_{eq} \left[ {{\text{CO}}_{3}^{2 - } } \right]_{eq} }} = \frac{{{ }\left[ {{\text{Ca}}^{2 + } } \right]\left[ {{\text{CO}}_{3}^{2 - } } \right]}}{{K_{sp} }} = \frac{{{ }IP}}{{K_{sp} }} $$
Calcite will dissolve if Ω < 1, i.e., in undersaturated water, and precipitate from solution if Ω > 1, i.e., if waters are supersaturated with respect to the mineral.
Open Access This chapter is licensed under the terms of the Creative Commons Attribution 4.0 International License (http://​creativecommons.​org/​licenses/​by/​4.​0/​), which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license and indicate if changes were made.
The images or other third party material in this chapter are included in the chapter's Creative Commons license, unless indicated otherwise in a credit line to the material. If material is not included in the chapter's Creative Commons license and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder.
Introduction to Equilibrium
Jack J. Middelburg
Copyright Year

Premium Partners