# More on Equilibrium Constants

### 1  Pressures can express concentrations

Although reaction quotients and equilibrium constants are often expressed in terms of molar concentrations, any concentration term can be used, including mole fractions and molalities. Sometimes the symbols Kc , Kx , and Km are used to denote these forms of the equilibrium constant. The numerical values of K’s and Q’s expressed in these different ways are not generally the same.

Most commonly-encountered equilibria occur in liquid solutions and gaseous mixtures. Kc values can be expressed in terms of moles per liter for both, but when dealing with gases it is often more convenient to use partial pressures. These two measures of concentration are directly proportional:

$c = \frac{n}{V} = \frac{\frac{PV}{RT}}{V} = \frac{P}{RT}$

Therefore, for a reaction A(g) → B(g) the equilibrium constant is written as follows:

$K_p = \frac{P_B}{P_A}$

All these forms of the equilibrium constant are at best approximately correct, most accurate at low concentrations or pressures. The only equilibrium constant that is truly “constant” (neglecting the variation with temperature) is expressed in terms of activities, which are best thought of as “effective concentrations” that allow for interactions between molecules. In practice, this distinction is only important for equilibria involving gases at very high pressures (such as are often encountered in chemical engineering) and in ionic solutions more concentrated than about 0.001 M

For a reaction such as CO2(g) + OH(aq) → HCO3(aq) involving both gaseous and dissolved components, a “hybrid” equilibrium constant is commonly used:

$K = \frac{[HCO_3^-]}{P_{CO_2}[OH^-]}$

### Converting between Kp and Kc

It is sometimes necessary to convert between equilibrium constants expressed in different units. The most common case involves pressure and concentration equilibrium constants.

Note that when V is expressed in liters and P in atmospheres, R must have the value 0.08206 L atm mol​- K-. The ideal gas law relates the partial pressure of a gas to the number of moles and its volume:
$PV = nRT$

Concentrations are expressed in moles/unit volume n/V, so rearranging the above equation gives the explicit relation of pressure to concentration:

$P = \frac{n}{V}RT$

Conversely, c = (n/V) = P/RT. Thus, a concentration [A] can be expressed as PA(RT). For a reaction of the form  2 A = B + 3 C, the following can be written:

$K_p = \frac{P_C^2}{P_A P_B^3} = \frac{([C]RT)^2}{([A]RT) ([B]RT)^3} = \underbrace{\frac{[C]^2}{[A][B]^3}}_{K_c} (RT)^{\underbrace{-2}_{\Delta n_g}}$

where, assuming a homogeneous gaseous reaction,

$\Delta n_g = (moles\ of\ gas\ in\ products) - (moles\ of\ gas\ in\ reactants)$

(moles of gas in products) – (moles of gas in reactants) = Δng

The general equation, therefore, is the following:

$K_p = K_c(RT)^{\Delta n_g}$

### 2  Omit unchanging concentrations

Substances whose concentrations undergo no significant change in a chemical reaction do not appear in equilibrium constant expressions. There are two general cases for consideration.

#### The substance is also the solvent

This is a frequent occurrence in acid-base chemistry. For the hydrolysis of the cyanide ion:

$CN^- + H_2O \rightleftharpoons HCN + OH^-$

the equilibrium constant is written as follows:

$K_b = \frac{[HCN][OH^-]}{[CN^-]}$

No [H2O] term appears. The justification for this omission is that water is both the solvent and reactant, but only the small amount acting as a reactant would ordinarily be included in the equilibrium expression. The amount of water consumed in the reaction is so minute (because Kb is very small) that any change in the concentration of H2O from that of pure water (55.6 mol L–1) is negligible.

Similarly, for the "dissociation" of water H2O = H+ + OH the equilibrium constant is expressed as the "ion product" Kw = [H+][OH].

However, care must be taken not to discard H2O wherever it appears. In the esterification reaction discussed in a previous section:

$CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5+ H_2O$

a [H2O] term must be present in the equilibrium expression if the reaction is assumed to be between the two liquids acetic acid and ethanol. If, on the other hand, the reaction takes place between a dilute aqueous solution of the acid and the alcohol, then the [H2O] term is not included.

#### The substance is a solid or a pure liquid phase.

This is most frequently seen in solubility equilibria, but there are many other reactions in which solids are directly involved:

$CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$

$Fe_3O_4(s) + 4H_2(g) \rightleftharpoons 4H_2O(g) + 3Fe(s)$

These are heterogeneous reactions (meaning reactions in which some components are in different phases); concentration is only meaningful when applied to a substance within a single phase.

Thus the term [CaF2] would refer to the “concentration of calcium fluoride within the solid CaF2", which is a constant depending on the molar mass of CaF2 and the density of that solid. The concentrations of the two ions are independent of the quantity of solid CaF2 in contact with the water; in other words, the system can be in equilibrium as long as any CaF2 at all is present.

Throwing out the constant-concentration terms can lead to some rather sparse-looking equilibrium expressions. For example, the equilibrium expression for each of the processes shown in the following table consists solely of a single term involving the partial pressure of a gas:

 1) CaCO3(s) → CaO(s) + CO2(g) Thermal decomposition of limestone, a first step in the manufacture of cement. 2) Na2SO4·10 H2O(s) → Na2SO4(s) + 10 H2O(g) Sodium sulfate decahydrate is a solid in which H2O molecules (“waters of hydration") are incorporated into the crystal structure.) 3) I2(s) → I2(g) sublimation of solid iodine; this is the source of the purple vapor above solid iodine in a closed container. 4) H2O(l) → H2O(g) Vaporization of water. When the partial pressure of water vapor in the air is equal to K, the relative humidity is 100%.

Processes 3 and 4 represent changes of state (phase changes) which can be treated exactly the same as chemical reactions.

In each of the heterogeneous processes shown in the table, the reactants and products can be in equilibrium (that is, permanently coexist) only when the partial pressure of the gaseous product has the value consistent with the indicated Kp. Remember that Kp increases with temperature.

Example 1

What are the values of Kp for the equilibrium between liquid water and its vapor at 25°C, 100°C, and 120°C? The vapor pressure of water at these three temperatures is 23.8 torr, 760 torr (1 atm), and 1489 torr, respectively.

Comment: These vapor pressures are the partial pressures of water vapor in equilibrium with the liquid, so they are identical with the Kp's when expressed in units of atmospheres.

#### Solution

 25°C 100°C 120°C The partial pressure of H2O above the surface of liquid water in a closed container at 25°C will build up to this value. If the cover is removed so that this pressure cannot be maintained, the system will cease to be at equilibrium and the water will evaporate. This temperature corresponds, of course, to the boiling point of water. The normal boiling point of a liquid is the temperature at which the partial pressure of its vapor is 1 atm. The only way to heat water above its normal boiling point is to do so in a closed container that can withstand the increased vapor pressure. Thus a pressure cooker that operates at 120°C must be designed to withstand an internal pressure of at least 2 atm.

### 3  Values of equilibrium constants

The ability to interpret the numerical value of a quantity in terms of what it means in a practical sense is an essential part of developing a working understanding of chemistry. This is particularly the case for equilibrium constants, whose values span the entire range of positive numbers.

Although there is no explicit rule, for most practical purposes equilibrium constants within the range of roughly 0.01 to 100 indicate that a chemically significant amount of all components of the reaction system are present in an equilibrium mixture and that the reaction is incomplete or “reversible”.

As an equilibrium constant approaches the limits of zero or infinity, the reaction can be increasingly characterized as a one-way process; it is said to “run to completion” or to be “irreversible”. The latter term must of course not be taken literally; the Le Châtelier principle still applies (especially insofar as temperature is concerned), but addition or removal of reactants or products will have less effect.

Although it is not a general rule, it frequently happens that reactions having very large equilibrium constants are kinetically hindered, often to the extent that the reaction essentially does not take place.

The examples in the following table are intended to show that numbers (values of K), no matter how dull they may look, do have

Reaction
K
Remarks
N2(g) + O2(g) → 2 NO(g) 5×10–31 at 25°C,
0.0013 at 2100°C
These two very different values of K illustrate why reducing combustion-chamber temperatures in automobile engines is environmentally friendly.
3 H2(g) + N2(g) → 2 NH3(g) 7×105 at 25°C,
56 at 1300°C
See the discussion of this reaction in the section on the Haber process.
H2(g) → 2 H(g) 10–36 at 25°C,
6×10–5 at 5000°
Dissociation of any stable molecule into its atoms is endothermic. This means that all molecules will decompose at sufficiently high temperatures.
H2O(g) → H2(g) + ½ O2(g) 8×10–41 at 25°C Water is a poor source of oxygen gas at ordinary temperatures.
CH3COOH(l)
2 H2O(l) + 2 C(s)
Kc = 1013 at 25°C Acetic acid has a strong tendency to decompose to carbon, but nobody has ever found graphite (or diamonds) forming in a bottle of vinegar. This is good example of a kinetically-hindered reaction

### 4  Do equilibrium constants have units?

The equilibrium expression for the synthesis of ammonia, given by the equation

$3H_2(g) + N_2(g) \rightleftharpoons 2NH_3(g)$

can be expressed as

$K_p = \frac{P_{NH_3}^2}{P_{N_2}P_{H_2}^3}$

or

$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$

Kp and Qp for this process appear to have units of atm–1, and Kc and Qc are expressed in mol–2 L2. And yet these quantities are often represented as being dimensionless. Which is correct? The answer is that both forms are acceptable. There are some situations in which K must be considered dimensionless, but in simply quoting the value of an equilibrium constant it is permissible to include the units, and this may even be useful in order to remove any doubt about the units of the individual terms in equilibrium expressions containing both pressure and concentration terms.

Strictly speaking, equilibrium expressions do not have units because the concentration or pressure terms are really ratios having the forms (n mol L–1)/(1 mol L–1) or (n atm)/(1 atm). The unit quantity in the denominator refers to the standard state of the substance; thus the units always cancel out.

For substances that are liquids or solids, the standard state is just the concentration of the substance within the liquid or solid, so for something like CaF(s), the term going into the equilibrium expression is [CaF2]/[CaF2] which cancels to unity; this is the reason the terms for solid or liquid phases are excluded in equilibrium expressions. The subject of standard states is beyond the scope of this article, but the concept is made necessary by the fact that energy, which ultimately governs chemical change, is always relative to some arbitrarily defined zero value which, for chemical substances, is the standard state.

### 5   How the reaction equation affects K

It is important to remember that an equilibrium quotient or constant is always tied to a specific chemical equation, and if the equation is written in reverse or its coefficients are multiplied by a common factor, the value of Q or K will change.

The rules are very simple:

• Writing the equation in reverse inverts the equilibrium expression;
• Multiplying the coefficients by a common factor raises Q or K to the corresponding power.

Here are some of the possibilities for the reaction involving the equilibrium between gaseous water and its elements:

 2 H2 + O2→ 2 H2O 10 H2 + 5 O2 → 10 H2O H2 + ½ O2 → H2O H2O → H2 + ½ O2

#### Equilibrium constant for a sequence of reactions

Many chemical changes can be regarded as the sum or difference of two or more other reactions. If the equilibrium constants of the individual processes are known, the equilibrium constant for the overall reaction can be calculatedaccording to this rule:

The equilibrium constant for the sum of two or more reactions is the product of the equilibrium constants for each of the steps.
Problem Example 2

Given the following equilibrium constants:

 CaCO3(s) → Ca2+(aq) + CO32–(aq) K1 = 10–6.3 HCO3–(aq) → H+(aq) + CO32–(aq) K2 = 10–10.3

Calculate the value of K for the reaction CaCO3(s) + H+(aq) → Ca2+(aq) + HCO3(aq)

Solution: The net reaction is the sum of reaction 1 and the reverse of reaction 2:

 CaCO3(s) → Ca2+(aq) + CO32–(aq) K1 = 10–6.3 H+(aq) + CO32–(aq) → HCO3–(aq) K–2 = 10–(–10.3) CaCO3(s) + H+(aq) → Ca2+(aq) + HCO3–(aq) K = K1/K2 = 10(-8.4+10.3) = 10+1.9

Comment: This net reaction describes the dissolution of limestone by acid; it is responsible for the eroding effect of acid rain on buildings and statues. This an example of a reaction that has practically no tendency to take place by itself (small K1) being "driven" by a second reaction having a large equilibrium constant (K–2). From the standpoint of the LeChâtelier principle, the first reaction is "pulled to the right" by the removal of carbonate by hydrogen ion. Coupled reactions of this type are widely encountered in all areas of chemistry, and especially in biochemistry, in which a dozen or so reactions may be linked.

Example 3

The synthesis of HBr from hydrogen and liquid bromine has an equilibrium constant Kp = 4.5×​1015 at 25°C. Given that the vapor pressure of liquid bromine is 0.28 atm, find Kp for the homogeneous gas-phase reaction at the same temperature.

#### Solution

The net reaction is the sum of the heterogeneous synthesis of HBr and the reverse of the vaporization of liquid bromine:

 H2(g) + Br2(l) → 2 HBr(g) Kp = 4.5×1015 Br2(g) → Br2(l) Kp = (0.28)–1 H2(g) + Br2(g)  →​ 2 HBr(g) Kp = 1.6×1019

### 6  More on heterogeneous reactions

Heterogeneous reactions are those involving more than one phase. Some examples:

 Fe(s) + O2(g) → FeO2(s) air-oxidation of metallic iron (formation of rust) CaF2(s) → Ca(aq) + F+(aq) dissolution of calcium fluoride in water H2O(s) → H2O(g) sublimation of ice (a phase change) NaHCO3(s) + H+(aq) → CO2(g) + Na+(aq) + H2O(g) formation of carbon dioxide gas from sodium bicarbonate when water is added to baking powder (the hydrogen ions come from tartaric acid, the other component of baking powder.)

### The vapor pressure of solid hydrates

A particularly interesting type of heterogeneous reaction is one in which a solid is in equilibrium with a gas. The sublimation of ice illustrated in the above table is a very common example. The equilibrium constant for this process is simply the partial pressure of water vapor in equilibrium with the solid— the vapor pressure of the ice.

Many common inorganic salts form solids which incorporate water molecules into their crystal structures. These water molecules are usually held rather loosely and can escape as water vapor. Copper(II) sulfate, for example forms a pentahydrate in which four of the water molecules are coordinated to the Cu2+ ion while the fifth is hydrogen-bonded to SO42–. This latter water is more tightly bound, so that the pentahydrate loses water in two stages on heating:

$CuSO_4 \bullet 5H_2O \xrightarrow{140^{\circ}C} CuSO_4 \bullet 5H_2O \xrightarrow{400^{\circ}C} CuSO_4$

These dehydration steps are carried out at the temperatures indicated above, but at any temperature, some moisture can escape from a hydrate. For the complete dehydration of the pentahydrate we can define an equilibrium constant:

$CuSO_4 \bullet 5H_2O(s) \rightarrow CuSO_4(s) + 5H_2O(g)\ \ \ \ K_p = 1.14 \times 10^{10}$

The vapor pressure of the hydrate (for this reaction) is the partial pressure of water vapor at which the two solids can coexist indefinitely; its value is Kp1/5 atm. If a hydrate is exposed to air in which the partial pressure of water vapor is less than its vapor pressure, the reaction will proceed to the right and the hydrate will lose moisture. Vapor pressures always increase with temperature, so any of these compounds can be dehydrated by heating.

Loss of water usually causes a breakdown in the structure of the crystal; this is commonly seen with sodium sulfate, whose vapor pressure is sufficiently large that it can exceed the partial pressure of water vapor in the air when the relative humidity is low. What one sees is that the well-formed crystals of the decahydrate undergo deterioration into a powdery form, a phenomenon known as efflorescence.

When a solid is able to take up moisture from the air, it is described as hygroscopic. A small number of anhydrous solids that have low vapor pressures not only take up atmospheric moisture on even the driest of days, but will become wet as water molecules are adsorbed onto their surfaces; this is most commonly observed with sodium hydroxide and calcium chloride. With these solids, the concentrated solution that results continues to draw in water from the air so that the entire crystal eventually dissolves into a puddle of its own making; solids exhibiting this behavior are said to be deliquescent.

name formula vapor pressure, torr
25°C 30°C
sodium sulfate decahydrate Na2SO4·10H2O 19.2 25.3
copper(II) sulfate pentahydrate CuSO4·5H2O 7.8 12.5
calcium chloride monohydrate CaCl2·H2O 3.1 5.1
(water) H2O 23.5 31.6
Example 4

At what relative humidity will copper sulfate pentahydrate lose its waters of hydration when the air temperature is 30°C? What is Kp for this process at this temperature?
Solution: From the table above, the vapor pressure of the hydrate is 12.5 torr, which corresponds to a relative humidity of 12.5/31.6 = 0.40 or 40%. This is the humidity that will be maintained if the hydrate is placed in a closed container of dry air

For this hydrate, Kp = P(H2O)0.5, so the partial pressure of water vapor that will be in equilibrium with the hydrate and the dehydrated solid (remember that both solids must be present to have an equilibrium), expressed in atmospheres, will be (12.5/760)5 = 1.20 ×​10-9.

One of the first hydrates investigated in detail was calcium sulfate hemihydrate (CaSO4·½ H2O) which LeChâtelier showed to be the hardened form of CaSO4 known as plaster of Paris. Anhydrous CaSO4 forms compact, powdery crystals, whereas the elongated crystals of the hemihydrate bind themselves into a cement-like mass that makes this material useful for making art objects, casts for immobilizing damaged limbs, and as a construction material (fireproofing, drywall.)

### Summary

The following is a list of the essential ideas presented. The precise definitions of all italicized terms are of particular importance.

• The equilibrium quotient Q is the value of the equilibrium expression of a reaction for any arbitrary set of concentrations or partial pressures of the reaction components.
• The equilibrium constant K is the value of Q  when the reaction is at equilibrium. K has a unique value for a given reaction at a fixed temperature and pressure.
• Q and K can be expressed in terms of concentrations, partial pressures, or, when appropriate, in some combination of these.
• For a reaction in which all the components are gases, Qc and Kc will have different values except in the special case in which the total number of moles of gas does not change.
• Concentration terms for substances whose concentrations do not change in the reaction do not appear in equilibrium expressions. The most common examples are [H2O] when the reaction takes place in aqueous solution (so that [H2O] is effectively constant at 55.6 M), and in heterogeneous reactions involving solids, in which the concentration of the solid is determined by the density of the solid itself.
• A reaction whose equilibrium constant is in the range of about 0.01 to 100 is said to be incomplete or [thermodynamically] reversible.
• Q and K are conventionally treated as dimensionless quantities, and need not ordinarily have units associated with them.
• Heterogeneous reactions are those in which two or more phases are involved; homogeneous reactions take place in a single phase. A common type of heterogeneous reaction is the loss of water of crystallization by a solid hydrate such as CuSO4·5H2O.
• The equilibrium expression can be manipulated and combined in the following ways:
• If the reaction is written in reverse, Q becomes Q–1;
• If the coefficients of an equation are multiplied by n, Q becomes Qn;
• Q for the sum of two reactions (that is, for two reactions that take place in sequence) is the product (Q1)(Q2).