Combining Equilibria
The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more coupled equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below.
Changing Direction of a Reaction (Reversing a Reaction)
Changing the direction of a chemical equation essentially swaps the identities of “reactants” and “products,” and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation.
| $A \rightleftharpoons B$ | $K_1=\frac{[B]}{[A]}$ |
| $B \rightleftharpoons A$ | $K_{2}=\frac{[A]}{[B]}$ |
$$K_{2}=\frac{1}{K_1}$$
Multiplying a Reaction by a Constant (Changing the Stoichiometric Coefficients)
Changing the stoichiometric coefficients in an equation by some factor $x$ results in the equilibrium constant being raised to the power of $x$:
| $A \rightleftharpoons B$ | $K_1=\frac{[B]}{[A]}$ |
| $x\, A \rightleftharpoons x\, B$ | $K_{2}=\frac{[B]^x}{[A]^x} = \left ( \frac {[B]}{[A]} \right ) ^x $ |
$$K_{2}={K_1}^x $$
Combining Reactions
Adding two or more equilibrium equations together yields an overall equation whose equilibrium constant is the mathematical product of the individual reaction’s K values:
| $1. \quad A \rightleftharpoons B$ | $K_1=\frac{[B]}{[A]}$ |
| $2. \quad B \rightleftharpoons C$ | $K_{2}=\frac{[C]}{[B]}$ |
| $3. \quad A \rightleftharpoons C$ | $K_{3}= ? $ |
Reaction 3 is obtained by summing Reactions 1 and 2 together: $$\; \; \; A \rightleftharpoons B \\ +\, B \rightleftharpoons C \\ ————————- \\ A+\require{enclose}\enclose{horizontalstrike}{B} \rightleftharpoons \require{enclose}\enclose{horizontalstrike}{B}+C \\ A \rightleftharpoons C \qquad K_{3}=\frac{[C]}{[A]} $$ Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship: $$K_{1}K_{2}=\frac{[B]}{[A]}×\frac{[C]}{[B]} \\ =\frac{\enclose{horizontalstrike}{[B]}[C]}{[A]\enclose{horizontalstrike}{[B]}} \\ =\frac{[C]}{[A]}=K_{3}$$$$K_{3}=K_{1}K_{2}$$ i.e. When adding two reactions together, multiply the two original equilibrium constants together to find the equilibrium constant for the combined reaction.
The example below demonstrates the use of this strategy in describing coupled equilibrium processes.
Equilibrium Constants for Coupled Reactions
A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C: $$2\, NH_3\, (g) + 3\, I_2\, (g) \rightleftharpoons N_2\, (g) + 6\, HI\, (g)$$ Use the information below to calculate Kc for this reaction. $$N_2\, (g) + 3\, H_2\, (g) \rightleftharpoons 2\, NH_3\, (g) \qquad K_{1}= 0.50\;\text{at}\; 400°\text{C}$$ $$H_2\, (g) + I_2\, (g) \rightleftharpoons 2\, HI\, (g) \qquad K_{2}=50.\;\text{at}\;400°\text{C}$$
Solution
The equilibrium equation of interest and its K value may be derived from the equations for the two coupled reactions as follows.
Reverse the first coupled reaction equation:
$$2NH_3\, (g) \rightleftharpoons N_2\, (g) +3\, H_2\, (g) \qquad K’_{1}=\frac{1}{K_{1}}=\frac{1}{0.50}=2.0$$Multiply the second coupled reaction by 3:
$$3\, H_2\, (g) + 3\, I_2\, (g) \rightleftharpoons 6\, HI\, (g) \qquad K’_{2}=K_{2}^3=50^3=1.2×10^5$$Finally, add the two revised equations:
$$2NH_3\, (g) +\enclose{horizontalstrike}{3\, H_2\, (g)} + 3\, I_2\, (g) \rightleftharpoons N_2\, (g) +\enclose{horizontalstrike}{3\, H_2\, (g)} + 6\, HI\, (g) \\ 2\, NH_3\, (g) + 3\, I_2\, (g) \rightleftharpoons N_2\, (g) + 6\, HI\, (g) K_c=K’_{1}\times K’_{2} \\ K_c =(2.0)\times (1.2×10^5) \\ \mathbf{K_c =2.5×10^5}$$Check Your Learning
Use the provided information to calculate Kc for the following reaction at 550 °C:
| $H_2\, (g) + CO_2\, (g)$ | $\rightleftharpoons $ | $CO\, (g) + H_2O\, (g)$ | $K_c=?$ |
| $1. \qquad CoO\, (s) + CO\, (g)$ | $\rightleftharpoons $ | $Co\, (s) + CO_2\, (g)$ | $K_{1}=490.$ |
| $2. \qquad CoO\, (s) + H_2\, (g)$ | $\rightleftharpoons $ | $Co\, (s) + H_2O\, (g)$ | $K_{2}=67$ |
Answer
In order to obtain the target reaction from Reactions 1 and 2, we must reverse Reaction 1 and sum the two resulting reactions:
| $Co\, (s) + CO_2\, (g)$ | $\rightleftharpoons $ | $CoO\, (s) + CO\, (g) $ | $K’_{1}=\frac{1}{490.}$ |
| $+\;\;CoO\, (s) + H_2\, (g)$ | $\rightleftharpoons $ | $Co\, (s) + H_2O\, (g)$ | $K_{2}=67$ |
| $\enclose{horizontalstrike}{Co\, (s)} + CO_2\, (g) + \enclose{horizontalstrike}{CoO\, (s)} + H_2\, (g) $ | $\rightleftharpoons $ | $\enclose{horizontalstrike}{CoO\, (s)} + CO\, (g) + \enclose{horizontalstrike}{Co\, (s)} + H_2O\, (g)$ | $K_c=K’_1 \times K_2 = \frac{67}{490}$ |
| $H_2\, (g) + CO_2\, (g)$ | $\rightleftharpoons $ | $CO\, (g) + H_2O\, (g)$ | $\mathbf{K_c=0.14}$ |