15.2: The Equilibrium Constant Expression - Chemistry LibreTexts

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 15.2.6 in reverse, we obtain the following:

\[cC+dD \rightleftharpoons aA+bB \label{15.2.10}\]

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

\[K'=\dfrac{[A]^a[B]^b}{[C]^c[D]^d} \label{15.2.11}\]

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:

\[K=\dfrac{[NO_2]^2}{[N_2O_4]} \label{15.2.12}\]

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

\[K'=\dfrac{[N_2O_4]}{[NO_2]^2} \label{15.2.13}\]

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 inverse of the equilibrium constant for the reaction as written originally.

Writing an equation in different but chemically equivalent forms also causes both the equilibrium constant expression and the magnitude of the equilibrium constant to be different. For example, we could write the equation for the reaction

\[2NO_2 \rightleftharpoons N_2O_4\]

as

\[NO_2 \rightleftharpoons \frac{1}{2}N_2O_4\]

with the equilibrium constant K″ is as follows:

\[ K′′=\dfrac{[N_2O_4]^{1/2}}{[NO_2]} \label{15.2.14}\]

The values for K′ (Equation \ref{15.2.13}) and K″ are related as follows:

\[ K′′=(K')^{1/2}=\sqrt{K'} \label{15.2.15}\]

In general, if all the coefficients in a balanced chemical equation were subsequently multiplied by \(n\), then the new equilibrium constant is the original equilibrium constant raised to the \(n^{th}\) power.

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

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

\[\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \nonumber\]

What is the equilibrium constant for each related reaction at 745 K?

1. \(2NH_{3(g)} \rightleftharpoons N2(g)+3H_{2(g)}\)
2. \(\frac{1}{2}N_{2(g)}+\frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}\)

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

Asked for: values of \(K\) for related reactions

Strategy:

Write the equilibrium constant expression for the given reaction and for each related 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:

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

In this reaction, the stoichiometric coefficients of the given reaction are divided by 2, so the equilibrium constant is calculated as follows:

\[K′′=\dfrac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}=K^{1/2}=\sqrt{K}=\sqrt{0.118} = 0.344\]

Exercise \(\PageIndex{3}\)

At 527°C, the equilibrium constant for the reaction

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

is \(7.9 \times 10^4\). Calculate the equilibrium constant for the following reaction at the same temperature:

\[SO_{3(g)} \rightleftharpoons SO_{2(g)}+\frac{1}{2}O_{2(g)}\]

Answer: \(3.6 \times 10^{−3}\)

Determining the Equilibrium Expression: https://youtu.be/ZK9cMIWFerY