## 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 K_{c} 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 $ | $CO\, (g) + H_2O\, (g)$ | $\mathbf{K_c=0.14}$ |