13.3 Variations in the Form of the Equilibrium Constant Expression

Learning Objectives

Skills to Develop

  • Know the relationships between the equilibrium constant and how the chemical equation is written
  • Calculate the equilibrium constant of an equation using Hess’s Law

Variations in the Form of the Equilibrium Constant Expression

[latex]a\text{A}+b\text{B}+\rightleftharpoons c\text{C}+d\text{D}[/latex]

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 shown above 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 $$O_2$$.

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:

\[ K′′=\dfrac{[N_2O_4]}{[NO_2]}^{1/2} \label{Eq14}\]

The values for K′ and K″ are related as follows:

\[ K′′=(K’)^{1/2}=\sqrt{K’} \label{Eq15}\]

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 1: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 each related reaction at 745 K?

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

 

Constant Expressions for the Sums of Reactions

Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions.

To illustrate this procedure, let’s consider the reaction of $$N_2$$ with $$O_2$$ to give $$NO_2$$. This reaction is an important source of the $$NO_2$$ that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps. In the first reaction (step 1), $$N_2$$ reacts with $$O_2$$ at the high temperatures inside an internal combustion engine to give $$NO$$. The released $$NO$$ then reacts with additional $$O_2$$ to give $$NO_2$$ (step 2). The equilibrium constant for each reaction at 100°C is also given.

$$N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}\;\; K_1=2.0 \times 10^{−25} \tag{step 1}$$

$$2NO_{(g)}+O_{2(g)} \rightleftharpoons 2NO_{2(g)}\;\;\;K_2=6.4 \times 10^9 \tag{step 2}$$

Summing reactions (step 1) and (step 2) gives the overall reaction of $$N_2$$ with $$O_2$$:

$$N_{2(g)}+2O_{2(g)} \rightleftharpoons 2NO_{2(g)} \;\;\;K_3=? \tag{overall reaction 3}$$

The equilibrium constant expressions for the reactions are as follows:

\[K_1=\dfrac{[NO]^2}{[N_2][O_2]}\;\;\; K_2=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\;\;\; K_3=\dfrac{[NO_2]^2}{[N_2][O_2]^2}\]

What is the relationship between $$K_1$$, $$K_2$$, and $$K_3$$, all at 100°C? The expression for $$K_1$$ has $$[NO]^2$$ in the numerator, the expression for $$K_2$$ has $$[NO]^2$$ in the denominator, and $$[NO]^2$$ does not appear in the expression for $$K_3$$. Multiplying $$K_1$$ by $$K_2$$ and canceling the $$[NO]^2$$ terms,

\[ K_1K_2=\dfrac{\cancel{[NO]^2}}{[N_2][O_2]} \times \dfrac{[NO_2]^2}{\cancel{[NO]^2}[O_2]}=\dfrac{[NO_2]^2}{[N_2][O_2]^2}=K_3\]

Thus the product of the equilibrium constant expressions for $$K_1$$ and $$K_2$$ is the same as the equilibrium constant expression for $$K_3$$:

\[K_3 = K_1K_2 = (2.0 \times 10^{−25})(6.4 \times 10^9) = 1.3 \times 10^{−15}\]

The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions. In contrast, recall that according to Hess’s Law, $$ΔH$$ for the sum of two or more reactions is the sum of the ΔH values for the individual reactions.

Example 2

The following reactions occur at 1200°C:

  1. $$CO_{(g)}+3H_{2(g)} \rightleftharpoons CH_{4(g)}+H_2O_{(g)} \;\;\;K_1=9.17 \times 10^{−2}$$
  2. $$CH_{4(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+4H_{2(g})\;\;\; K_2=3.3 \times 10^4$$

Calculate the equilibrium constant for the following reaction at the same temperature.

  1. $$CO_{(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}\;\;\; K_3=?$$

Key Concepts and Summary

When a reaction is written in the reverse direction, $$K$$ and the equilibrium constant expression are inverted. 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 nth power. If all the coefficients in a balanced chemical equation were subsequently divided by n, then the new equilibrium constant is the original equilibrium constant raised to the 1/nth power. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions.

Exercises

1. 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:

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

2. In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide. In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide. The reaction for each step is shown, as is the value of the corresponding equilibrium constant at 25°C. Calculate the equilibrium constant for the overall reaction at this same temperature.

  1. $$\frac{1}{8}S_{8(s)}+O_{2(g)} \rightleftharpoons SO_{2(g)}\;\;\; K_1=4.4 \times 10^{53}$$
  2. $$SO_{2(g)}+\frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_2=2.6 \times 10^{12}$$
  3. $$\frac{1}{8}S_{8(s)}+\frac{3}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_3=?$$