6.3: Manipulating Equilibrium Constants
- Page ID
- 5708
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We will take advantage of two useful relationships when working with equilibrium constants. First, if we reverse a reaction’s direction, the equilibrium constant for the new reaction is simply the inverse of that for the original reaction. For example, the equilibrium constant for the reaction
\[\textrm A+2\textrm B\rightleftharpoons A\textrm B_2\hspace{5mm}K_1=\mathrm{\dfrac{[AB_2]}{[A][B]^2}}\]
is the inverse of that for the reaction
\[A\textrm B_2\rightleftharpoons\textrm A + 2\textrm B\hspace{5mm}K_2=(K_1)^{-1}=\dfrac{[A][\textrm B]^2}{[\textrm{AB}_2]}\]
Second, if we add together two reactions to obtain a new reaction, the equilibrium constant for the new reaction is the product of the equilibrium constants for the original reactions.
\[\mathrm{A+C\rightleftharpoons AC}\hspace{4mm}K_3=\mathrm{\dfrac{[AC]}{[A][C]}}\]
\[\mathrm{AC+C\rightleftharpoons AC_2}\hspace{5mm}K_4=\mathrm{\dfrac{[AC_2]}{[AC][C]}}\]
\[\mathrm{A+2C\rightleftharpoons AC_2}\hspace{5mm}K_5=K_3\times K_4=\mathrm{\dfrac{[AC]}{[A][C]}\times\dfrac{[AC_2]}{[AC][C]}=\dfrac{[AC_2]}{[A][C]^2}}\]
Example 6.1
Calculate the equilibrium constant for the reaction
\[\mathrm{2A+B\rightleftharpoons C+3D}\]
given the following information
\[\begin{align}
&\textrm{Rxn 1: A + B} \rightleftharpoons \textrm D &&K_1 = 0.40\\
&\textrm{Rxn 2: A + E} \rightleftharpoons \textrm{C + D + F} &&K_2 = 0.10\\
&\textrm{Rxn 3: C + E} \rightleftharpoons \textrm B &&K_3 = 2.0\\
&\textrm{Rxn 4: F + C} \rightleftharpoons \textrm{D + B} &&K_4 = 5.0
\end{align}\]
Solution
The overall reaction is equivalent to
\[\textrm{Rxn 1 + Rxn 2} - \textrm{Rxn 3 + Rxn 4}\]
Subtracting a reaction is equivalent to adding the reverse reaction; thus, the overall equilibrium constant is
\[K=\dfrac{K_1\times K_2\times K_4}{K_3}=\dfrac{0.40\times0.10\times5.0}{2.0}=0.10\]
Practice Exercise 6.1
Calculate the equilibrium constant for the reaction
\[\mathrm{C+D+F\rightleftharpoons 2A+3B}\]
using the equilibrium constants from Example 6.1.
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