6.3: Manipulating Equilibrium Constants

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We will take advantage of two useful relationships when we work with equilibrium constants. First, if we reverse a reaction’s direction, the equilibrium constant for the new reaction is the inverse of that for the original reaction. For example, the equilibrium constant for the reaction

$\mathrm{A}+2 \mathrm{B}\rightleftharpoons \mathrm{AB}_{2} \quad \quad K_{1}=\frac{\left[\mathrm{AB}_{2}\right]}{[\mathrm{A}][\mathrm{B}]^{2}} \nonumber$

is the inverse of that for the reaction

$\mathrm{AB}_{2}\rightleftharpoons \mathrm{A}+2 \mathrm{B} \quad \quad K_{2}=\left(K_{1}\right)^{-1}=\frac{[\mathrm{A}][\mathrm{B}]^{2}}{\left[\mathrm{AB}_{2}\right]} \nonumber$

Second, if we add together two reactions to form a new reaction, the equilibrium constant for the new reaction is the product of the equilibrium constants for the original reactions.

$A+C\rightleftharpoons A C \quad \quad K_{3}=\frac{[A C]}{[A][C]} \nonumber$

$\mathrm{AC}+\mathrm{C}\rightleftharpoons\mathrm{AC}_{2} \quad \quad K_{4}=\frac{\left[\mathrm{AC}_{2}\right]}{[\mathrm{AC}][\mathrm{C}]} \nonumber$

$\mathrm{A}+2 \mathrm{C}\rightleftharpoons \mathrm{AC}_{2} \quad \quad K_{5}=K_{3} \times K_{4}=\frac{[\mathrm{AC}]}{[\mathrm{A}][\mathrm{C}]} \times \frac{\left[\mathrm{AC}_{2}\right]}{[\mathrm{AC}][\mathrm{C}]}=\frac{\left[\mathrm{AC}_{2}\right]}{[\mathrm{A}][\mathrm{C}]^{2}} \nonumber$

Example $$\PageIndex{1}$$

Calculate the equilibrium constant for the reaction

$2 \mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+3 \mathrm{D} \nonumber$

given the following information

$\begin{array}{ll}{\text{Rxn} \ 1 : A+B\rightleftharpoons D} & {K_{1}=0.40} \\ {\text{Rxn} \ 2 : A+E\rightleftharpoons C+D+F} & {K_{2}=0.10} \\ {\text{Rxn} \ 3 : C+E\rightleftharpoons B} & {K_{3}=2.0} \\ {\text{Rxn} \ 4 : F+C\rightleftharpoons D+B} & {K_{4}=5.0}\end{array} \nonumber$

Solution

The overall reaction is equivalent to

$\text{Rxn} \ 1+\text{Rxn} \ 2-\text{Rxn} \ 3+\text{Rxn} \ 4 \nonumber$

Subtracting a reaction is equivalent to adding the reverse reaction; thus, the overall equilibrium constant is

$K=\frac{K_{1} \times K_{2} \times K_{4}}{K_{3}}=\frac{0.40 \times 0.10 \times 5.0}{2.0}=0.10 \nonumber$

Exercise $$\PageIndex{1}$$

Calculate the equilibrium constant for the reaction

$C+D+F \rightleftharpoons 2 A+3 B \nonumber$

using the equilibrium constants from Example $$\PageIndex{1}$$.

$\operatorname{Rxn} 4-2 \times \operatorname{Rxn} 1 \nonumber$
$K=\frac{K_{4}}{\left(K_{1}\right)^{2}}=\frac{(5.0)}{(0.40)^{2}}=31.25 \approx 31 \nonumber$