Skip to main content

# 6.3: Manipulating Equilibrium Constants

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

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 6.3.1

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 6.3.1

Calculate the equilibrium constant for the reaction

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

using the equilibrium constants from Example 6.3.1 .

Answer

The overall reaction is equivalent to

$\operatorname{Rxn} 4-2 \times \operatorname{Rxn} 1 \nonumber$

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

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

This page titled 6.3: Manipulating Equilibrium Constants is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.

• Was this article helpful?