# 8.2 The Equilibrium Constant

- Page ID
- 220918

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

## Developing an Equilibrium Constant Expression

In 1864, the Norwegian chemists Cato Guldberg (1836–1902) and Peter Waage (1833–1900) carefully measured the compositions of many reaction systems at equilibrium. They discovered that for any reversible reaction of the general form

\[aA+bB \rightleftharpoons cC+dD \label{Eq6}\]

where A and B are reactants, C and D are products, and a, b, c, and d are the stoichiometric coefficients in the balanced chemical equation for the reaction, the ratio of the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced chemical equation) to the product of the equilibrium concentrations of the reactants (raised to their coefficients in the balanced chemical equation) is always a constant under a given set of conditions. This relationship is known as the law of mass action and can be stated as follows:

\[K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{Eq7}\]

where \(K\) is the equilibrium constant for the reaction. Equation \(\ref{Eq6}\) is called the equilibrium equation, and the right side of Equation \(\ref{Eq7}\) is called the equilibrium constant expression. The relationship shown in Equation \(\ref{Eq7}\) is true for any pair of opposing reactions regardless of the mechanism of the reaction or the number of steps in the mechanism.

The equilibrium constant can vary over a wide range of values. The values of \(K\) shown in Table \(\PageIndex{2}\), for example, vary by 60 orders of magnitude. Because products are in the numerator of the equilibrium constant expression and reactants are in the denominator, values of K greater than 1 indicate a reaction that is product-favored at equilibrium. If K is greater than \(10^3\), the reaction has a strong tendency for the reactants to form products. In this case, chemists say that equilibrium lies far to the right as written, favoring the formation of a great deal of products. An example is the reaction between \(H_2\) and \(Cl_2\) to produce \(HCl\), which has an equilibrium constant of \(1.6 \times 10^{33}\) at 300 K. Because \(H_2\) is a good reductant and \(Cl_2\) is a good oxidant, the reaction proceeds essentially to completion. In contrast, values of \(K\) less than 1 indicate a reaction that is reactant-favored at equilibrium. If K is less than \(10^{-3}\) the reaction has a very slight tendency for the reactants to form products, so that the ratio of products to reactants at equilibrium is very small. That is, reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of reactants.

Reaction |
Temperature (K) |
Equilibrium Constant (K) |
---|---|---|

*Equilibrium constants vary with temperature. The K values shown are for systems at the indicated temperatures. | ||

\(S_{(s)}+O_{2(g)} \rightleftharpoons SO_{2(g)}\) | 300 | \(4.4 \times 10^{53}\) |

\(2H_{2(g)}+O_{2(g)} \rightleftharpoons 2H2O_{(g)}\) | 500 | \(2.4 \times 10^{47}\) |

\(H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)}\) | 300 | \(1.6 \times 10^{33}\) |

\(H_{2(g)}+Br_{2(g)} \rightleftharpoons 2HBr_{(g)}\) | 300 | \(4.1 \times 10^{18}\) |

\(2NO_{(g)}+O_{2(g)} \rightleftharpoons 2NO_{2(g)}\) | 300 | \(4.2 \times 10^{13}\) |

\(3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)}\) | 300 | \(2.7 \times 10^{8}\) |

\(H_{2(g)}+D_{2(g)} \rightleftharpoons 2HD_{(g)}\) | 100 | \(1.92\) |

\(H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\) | 300 | \(2.9 \times 10^{−1}\) |

\(I_{2(g)} \rightleftharpoons 2I_{(g)}\) | 800 | \(4.6 \times^{ 10−7}\) |

\(Br_{2(g)} \rightleftharpoons 2Br_{(g)}\) | 1000 | \(4.0 \times 10^{−7}\) |

\(Cl_{2(g)} \rightleftharpoons 2Cl_{(g)}\) | 1000 | \(1.8 \times 10^{−9}\) |

\(F_{2(g)} \rightleftharpoons 2F_{(g)}\) | 500 | \(7.4 \times 10^{−13}\) |

You will also notice in Table \(\PageIndex{2}\) that equilibrium constants have no units, even though Equation \(\ref{Eq7}\) suggests that the units of concentration might not always cancel because the exponents may vary. **In fact, equilibrium constants are calculated using “effective concentrations,” or activities, of reactants and products, which are the ratios of the measured concentrations to a standard state of 1 M. **As shown in Equation \(\ref{Eq8}\), the units of concentration cancel, which makes \(K\) unitless as well:

\[ \dfrac{[A]_{measured}}{[A]_{standard\; state}}=\dfrac{\cancel{M}}{\cancel{M}} = \dfrac{\cancel{\frac{mol}{L}}}{\cancel{\frac{mol}{L}}} \label{Eq8}\]

In fact, equilibrium constants are calculated using “effective concentrations,” or activities, of reactants and products, which are the ratios of the measured concentrations to a standard state of 1 M.

Many reactions have equilibrium constants between 1000 and 0.001 (\(10^3 \ge K \ge 10^{−3}\)), neither very large nor very small. At equilibrium, these systems tend to contain significant amounts of both products and reactants, indicating that there is not a strong tendency to form either products from reactants or reactants from products. An example of this type of system is the reaction of gaseous hydrogen and deuterium, a component of high-stability fiber-optic light sources used in ocean studies, to form \(HD\):

\[H_{2(g)}+D_{2(g)} \rightleftharpoons 2HD_{(g)} \label{Eq9}\]

The equilibrium constant expression for this reaction is

\[K= \dfrac{[HD]^2}{[H_2][D_2]}\]

with \(K\) varying between 1.9 and 4 over a wide temperature range (100–1000 K). Thus an equilibrium mixture of \(H_2\), \(D_2\), and \(HD\) contains significant concentrations of both product and reactants.

Figure \(\PageIndex{3}\) summarizes the relationship between the magnitude of K and the relative concentrations of reactants and products at equilibrium for a general reaction, written as reactants \(\rightleftharpoons\) products.

**Figure \(\PageIndex{3}\):** The Relationship between the Composition of the Mixture at Equilibrium and the Magnitude of the Equilibrium Constant. The larger the K, the farther the reaction proceeds to the right before equilibrium is reached, and the greater the ratio of products to reactants at equilibrium.

A large value of the equilibrium constant \(K\) means that products predominate at equilibrium; a small value means that reactants predominate at equilibrium.

Example \(\PageIndex{1}\): equilibrium constant expressionS

Write the equilibrium constant expression for each reaction.

- \(N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}\)
- \(CO_{(g)}+\frac{1}{2}O_{2(g)} \rightleftharpoons CO_{2(g)}\)
- \(2CO_{2(g)} \rightleftharpoons 2CO_{(g)}+O_{2(g)}\)

**Given**: balanced chemical equations

**Asked for**: equilibrium constant expressions

**Strategy**:

Refer to Equation \(\ref{Eq7}\). Place the arithmetic product of the concentrations of the products (raised to their stoichiometric coefficients) in the numerator and the product of the concentrations of the reactants (raised to their stoichiometric coefficients) in the denominator.

**Solution**:

The only product is ammonia, which has a coefficient of 2. For the reactants, \(N_2\) has a coefficient of 1 and \(\ce{H2}\) has a coefficient of 3. The equilibrium constant expression is as follows:

\[\dfrac{[NH_3]^2}{[N_2][H_2]^3}\]

The only product is carbon dioxide, which has a coefficient of 1. The reactants are \(CO\), with a coefficient of 1, and \(O_2\), with a coefficient of \(\frac{1}{2}\). Thus the equilibrium constant expression is as follows:

\[\dfrac{[CO_2]}{[CO][O_2]^{1/2}}\]

This reaction is the reverse of the reaction in part b, with all coefficients multiplied by 2 to remove the fractional coefficient for \(O_2\). The equilibrium constant expression is therefore the inverse of the expression in part b, with all exponents multiplied by 2

\[\dfrac{[CO]^2[O_2]}{[CO_2]^2}\]

Exercise \(\PageIndex{1}\)

Write the equilibrium constant expression for each reaction.

- \(N_2O_{(g)} \rightleftharpoons N_{2(g)}+\frac{1}{2}O_{2(g)}\)
- \(2C_8H_{18(g)}+25O_{2(g)} \rightleftharpoons 16CO_{2(g)}+18H_2O_{(g)}\)
- \(H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\)

**Answer a**-
\(K=\dfrac{[N_2][O_2]^{1/2}}{[N_2O]}\)

**Answer b**-
\(K=\dfrac{[CO_2]^{16}[H_2O]^{18}}{[C_8H_{18}]^2[O_2]^{25}}\)

**Answer c**-
\(K=\dfrac{[HI]^2}{[H_2][I_2]}\)

Example \(\PageIndex{2}\)

Predict which systems at equilibrium will (a) contain essentially only products, (b) contain essentially only reactants, and (c) contain appreciable amounts of both products and reactants.

- \(H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\;\;\; K_{(700K)}=54\)
- \(2CO_{2(g)} \rightleftharpoons 2CO_{(g)}+O_{2(g)}\;\;\; K_{(1200K)}=3.1 \times 10^{−18}\)
- \(PCl_{5(g)} \rightleftharpoons PCl_{3(g)}+Cl_{2(g)}\;\;\; K_{(613K)}=97\)
- \(2O_{3(g)} \rightleftharpoons 3O_{2(g)} \;\;\; K_{(298 K)}=5.9 \times 10^{55}\)

**Given**: systems and values of \(K\)

**Asked for**: composition of systems at equilibrium

**Strategy**:

Use the value of the equilibrium constant to determine whether the equilibrium mixture will contain essentially only products, essentially only reactants, or significant amounts of both.

**Solution**:

- Only system 4 has \(K \gg 10^3\), so at equilibrium it will consist of essentially only products.
- System 2 has \(K \ll 10^{−3}\), so the reactants have little tendency to form products under the conditions specified; thus, at equilibrium the system will contain essentially only reactants.
- Both systems 1 and 3 have equilibrium constants in the range \(10^3 \ge K \ge 10^{−3}\), indicating that the equilibrium mixtures will contain appreciable amounts of both products and reactants.

Exercise \(\PageIndex{2}\)

Hydrogen and nitrogen react to form ammonia according to the following balanced chemical equation:

\[3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)}\]

Values of the equilibrium constant at various temperatures were reported as

- \(K_{25°C} = 3.3 \times 10^8\),
- \(K_{177°C} = 2.6 \times 10^3\), and
- \(K_{327°C} = 4.1\).

- At which temperature would you expect to find the highest proportion of \(H_2\) and \(N_2\) in the equilibrium mixture?
- Assuming that the reaction rates are fast enough so that equilibrium is reached quickly, at what temperature would you design a commercial reactor to operate to maximize the yield of ammonia?

**Answer a**-
327°C, where \(K\) is smallest

**Answer b**-
25°C

## Variations in the Form of the Equilibrium Constant Expression

Because equilibrium can be approached from either direction in a chemical reaction, the equilibrium constant expression and thus the magnitude of the equilibrium constant depend on the form in which the chemical reaction is written. For example, if we write the reaction described in Equation \(\ref{Eq6}\) in reverse, we obtain the following:

\[cC+dD \rightleftharpoons aA+bB \label{Eq10}\]

The corresponding equilibrium constant \(K′\) is as follows:

\[K'=\dfrac{[A]^a[B]^b}{[C]^c[D]^d} \label{Eq11}\]

This expression is the inverse of the expression for the original equilibrium constant, so \(K′ = 1/K\). That is, when we write a reaction in the reverse direction, the equilibrium constant expression is inverted. For instance, the equilibrium constant for the reaction \(N_2O_4 \rightleftharpoons 2NO_2\) is as follows:

\[K=\dfrac{[NO_2]^2}{[N_2O_4]} \label{Eq12}\]

but for the opposite reaction, \(2 NO_2 \rightleftharpoons N_2O_4\), the equilibrium constant K′ is given by the inverse expression:

\[K'=\dfrac{[N_2O_4]}{[NO_2]^2} \label{Eq13}\]

Consider another example, the formation of water: \(2H_{2(g)}+O_{2(g)} \rightleftharpoons 2H_2O_{(g)}\). Because \(H_2\) is a good reductant and \(O_2\) is a good oxidant, this reaction has a very large equilibrium constant (\(K = 2.4 \times 10^{47}\) at 500 K). Consequently, the equilibrium constant for the reverse reaction, the decomposition of water to form \(O_2\) and \(H_2\), is very small: \(K′ = 1/K = 1/(2.4 \times 10^{47}) = 4.2 \times 10^{−48}\). As suggested by the very small equilibrium constant, and fortunately for life as we know it, a substantial amount of energy is indeed needed to dissociate water into \(H_2\) and \(O_2\).

The equilibrium constant for a reaction written in reverse is the

inverseof the equilibrium constant for the reaction as written originally.

Example \(\PageIndex{3}\): The Haber Process

At 745 K, K is 0.118 for the following reaction:

\[N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}\]

What is the equilibrium constant for \(2NH_{3(g)} \rightleftharpoons N2(g)+3H_{2(g)}\) at 745 K?

**Given**: balanced equilibrium equation, K at a given temperature, and equations of related reactions

**Asked for**: values of \(K\) for the reverse reaction

**Strategy**:

Write the equilibrium constant expression for the given reaction and for the reverse reaction. From these expressions, calculate \(K\) for each reaction.

**Solution**:

The equilibrium constant expression for the given reaction of \(N_{2(g)}\) with \(H_{2(g)}\) to produce \(NH_{3(g)}\) at 745 K is as follows:

\[K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}=0.118\]

This reaction is the reverse of the one given, so its equilibrium constant expression is as follows:

\[K'=\dfrac{1}{K}=\dfrac{[N_2][H_2]^3}{[NH_3]^2}=\dfrac{1}{0.118}=8.47\]

Exercise \(\PageIndex{3}\)

At 527°C, the equilibrium constant for the reaction

\[2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)} \nonumber\]

is \(7.90 \times 10^4\). Calculate the equilibrium constant for the following reaction at the same temperature:

**Answer**-
\(1.27 \times 10^{−5}\)