4.6: Coupled 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.
1. 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.
\[\begin{array}{ll}
A \rightleftharpoons B & K_{ c }=\frac{[ B ]}{[ A ]} \\[4pt]
B \rightleftharpoons A & K_{ c^{\prime}}=\frac{[ A ]}{[ B ]}
\end{array} \nonumber \]
\[K_{ c^{\prime}}=\frac{1}{ K_{ c }} \nonumber \]
2. Changing the stoichiometric coefficients in an equation by some factor x results in an exponential change in the equilibrium constant by that same factor:
\[\begin{array}{ll}
A \rightleftharpoons B & K_{ c }=\frac{[ B ]}{[ A ]} \\[4pt]
xA \rightleftharpoons xB & K_{ c }=\frac{[ B ]^{ x }}{[ A ]^{ x }}
\end{array} \nonumber \]
\[K_{ c^{\prime}}= K_{ c }{ }^{ x } \nonumber \]
3. 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:
\[\begin{array}{ll}
A \rightleftharpoons B & K_{ c 1}=\frac{[ B ]}{[ A ]} \\[4pt]
B \rightleftharpoons C & K_{ c 2}=\frac{[ C ]}{[ B ]}
\end{array} \nonumber \]
The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies:
\[\begin{aligned}
& A + B \rightleftharpoons B + C \\[4pt]
& A + B \rightleftharpoons B + C \\[4pt]
& A \rightleftharpoons C
\end{aligned} \nonumber \]
\[K_{ c^{\prime}}=\frac{[ C ]}{[ A ]} \nonumber \]
Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship:
\[K_{ c 1} K_{ c 2}=\frac{[ B ]}{[ A ]} \times \frac{[ C ]}{[ B ]}=\frac{ \cancel{[ B ]}[ C ]}{[ A ] \cancel{[ B ]}}=\frac{[ C ]}{[ A ]}= K_{ c^{\prime}} \nonumber \]
\[K_{ c^{\prime}}= K_{ c 1} K_{ c 2} \nonumber \]
Example \(\PageIndex{5}\) demonstrates the use of this strategy in describing coupled equilibrium processes.
A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C:
\[\ce{2 NH3(g) + 3 I2(g) \rightleftharpoons N2(g) + 6 HI(g)} \nonumber \]
Use the information below to calculate Kc for this reaction.
\[\begin{align*}
\ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)} & K_{ c 1}=0.50 \text { at } 400{ }^{\circ} C \\[4pt]
\ce{H2(g) + I2(g) \rightleftharpoons 2 HI(g)} & K_{ c 2}=50 \text { at } 400{ }^{\circ} C
\end{align*} \nonumber \]
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:
\[2 NH_3(g) \rightleftharpoons N_2(g) + 3 H_2(g) \quad K_{ c 1}{ }^{\prime}=\frac{1}{ K_{ c 1}}=\frac{1}{0.50}=2.0 \nonumber \]
Multiply the second coupled reaction by 3:
\[3 H_2(g) + 3 I_2(g) \rightleftharpoons 6 HI (g) \quad K_{ c 2}{ }^{\prime}= K_{ c 2}^3=50^3=1.2 \times 10^5 \nonumber \]
Finally, add the two revised equations:
\[\begin{align*}
\ce{2 NH3(g) + 3 H2(g) + 3 I2(g) &\rightleftharpoons N2(g) + 3 H2(g) + 6 HI(g)} \\[4pt]
\ce{2 NH3(g) + 3 I2(g) &\rightleftharpoons N2(g) + 6HI(g)}
\end{align*} \nonumber \]
\[K_{ c }= K_{ c 1}^\prime K_{ c 2}^3=(2.0)\left(1.2 \times 10^5\right)=2.5 \times 10^5 \nonumber \]
Use the provided information to calculate Kc for the following reaction at 550 °C:
\[\begin{align*}
\ce{H2(g) + CO2(g) &\rightleftharpoons CO(g) + H2O(g)} && K_{ c }=? \\[4pt]
\ce{CoO(s) + CO(g) &\rightleftharpoons Co(s)+ CO2(g)} && K_{ c 1}=490 \\[4pt]
\ce{CoO(s) + H2(g) &\rightleftharpoons Co(s)+ H2O(g)} && K_{ c 1}=67
\end{align*} \nonumber \]
- Answer
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K c = 0.14