2.1 The Equilibrium Constant, Keq
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Chemists have found that there is a mathematical relationship that exists between the concentration of the reactants and products once equilibrium has reached that is independent of the initial concentration of the participants. For any general reaction:
\[\ce{aA + bB \rightleftharpoons cD + dD} \tag{2.1.1}\]
an equilibrium constant expression can be written as:
\[\mathrm{K_{eq}=\dfrac{[C]^c \times [D]^d}{[A]^a \times [B]^b}} \tag{2.1.2}\]
This mathematical relationship exists for all equilibrium systems, and produces a constant ratio called the equilibrium constant, Keq .
Note
Equation 2.1.2 is sometimes called the law of mass-action.
This relationship will be very important to us for the next few units, so it is important that you understand how to set this relationship up and what it tells us about an equilibrium system.
The products of the reaction (C and D) are placed in the numerator, and their concentrations are raised to the power of the coefficients from the balanced equation. The reactants (A and B) are placed in the denominator, with their concentrations raised to the power of their coefficients.
Example
For the reaction between hydrogen and iodine gas to produce hydrogen iodide:
\[\ce{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)} \tag{2.1.3}\]
the equilibrium constant expression will be:
\[\mathrm{K_{eq}=\dfrac{[HI]^2}{[H_2] \times [I_2]}} \tag{2.1.4}\]
Using the example we examined in our last section, equilibrium concentrations for each substance were measured at equilibrium and found to be:
At equilibrium: | \(\mathrm{[H_2]=0.022\: M}\) \(\mathrm{[I_2]=0.022\: M}\) \(\mathrm{[HI]=0.156\: M}\) |
We substitute these values into our equilibrium expression and solve for Keq:
\(\mathrm{K_{eq}=\dfrac{[HI]^2}{[H_2] \times [I_2]}=\dfrac{(0.156)^2}{(0.022)(0.022)}=50.3}\)
The value of Keq, which has no units, is a constant for any particular reaction, and its value does not change unless the temperature of the system is changed. It does not depend on the initial concentrations used to reach the point of equilibrium.
For example, the following data were obtained for equilibrium concentrations of H2, I2 and HI, and the value of Keq was calculated for each trial:
Trial | [HI] | [ H2] | [ I2] | Keq |
1 | 0.156 | 0.0220 | 0.0220 | 50.3 |
2 | 0.750 | 0.106 | 0.106 | 50.1 |
3 | 1.00 | 0.820 | 0.0242 | 50.4 |
4 | 1.00 | 0.0242 | 0.820 | 50.4 |
5 | 1.56 | 0.220 | 0.220 | 50.3 |
Aside from accounting for slight experimental variation between trials, the value for Keq is the same despite differences in equilibrium concentrations for the individual participants.
There is one other important point to make at this time.
Keq relates the concentrations of products to reactants at equilibrium.
For aqueous solutions, concentration is often measured as mol · L-1. For gases, concentration is often measured as partial pressure.
The concentrations of both aqueous solutions and gases change during the progress of a reaction. For reactions involving a solid or a liquid, while the amounts of the solid or liquid will change during a reaction, their concentrations (much like their densities) will not change during the reaction.
Instead, their values will remain constant. Because they are constant, their values are not included in the equilibrium constant expression.
Example
For example, consider the reaction showing the formation of solid calcium carbonate from solid calcium oxide and carbon dioxide gas:
\[\ce{CaO(s) + CO_2(g) \leftrightarrow CaCO_3(s)}\]
The equilibrium constant for this reaction is (before modification)
\[\mathrm{K_{eq}=\dfrac{[CaCO_3]}{[CaO] \times [CO_2]}}\]
But we remove those participants whose state is either a solid or a liquid, which leaves us with the following equilibrium constant expression:
\[\mathrm{K_{eq}=\dfrac{1}{[CO_2]}}\]