7.3: Calculating the Equilibrium Constant From Measured Equilibrium Concentrations, Part 1
- Page ID
- 341919
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Learning Objectives
- To solve quantitative problems involving chemical equilibriums.
There are two fundamental kinds of equilibrium problems:
- those in which we are given the concentrations of the reactants and the products at equilibrium (or, more often, information that allows us to calculate these concentrations), and we are asked to calculate the equilibrium constant for the reaction; and
- those in which we are given the equilibrium constant and the initial concentrations of reactants, and we are asked to calculate the concentration of one or more substances at equilibrium. In this section, we describe methods for solving both kinds of problems.
Calculating an Equilibrium Constant from Equilibrium Concentrations
We saw in the exercise in Example 6 in Section 15.2 that the equilibrium constant for the decomposition of \(CaCO_{3(s)}\) to \(CaO_{(s)}\) and \(CO_{2(g)}\) is \(K = [CO_2]\). At 800°C, the concentration of \(CO_2\) in equilibrium with solid \(CaCO_3\) and \(CaO\) is \(2.5 \times 10^{-3}\; M\). Thus K at 800°C is \(2.5 \times 10^{-3}\). (Remember that equilibrium constants are unitless.)
A more complex example of this type of problem is the conversion of n-butane, an additive used to increase the volatility of gasoline, into isobutane (2-methylpropane).
This reaction can be written as follows:
\[\ce{n-butane_{(g)} \rightleftharpoons isobutane_{(g)}} \label{Eq1}\]
and the equilibrium constant \(K = [\text{isobutane}]/[\text{n-butane}]\). At equilibrium, a mixture of n-butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n-butane. Substituting these concentrations into the equilibrium constant expression,
\[K=\dfrac{[\textit{isobutane}]}{[\textit{n-butane}]}=0.041\; M = 2.6 \label{Eq2}\]
Thus the equilibrium constant for the reaction as written is 2.6.
Example \(\PageIndex{1}\)
The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid:
\[2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}\nonumber\]
A mixture of \(SO_2\) and \(O_2\) was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained
- \(5.0 \times 10^{-2}\; M\; SO_3\),
- \(3.5 \times 10^{-3}\; M\; O_2\), and
- \(3.0 \times 10^{-3}\; M\; SO_2\).
Calculate \(K\) and \(K_p\) at this temperature.
Given: balanced equilibrium equation and composition of equilibrium mixture
Asked for: equilibrium constant
Strategy
Write the equilibrium constant expression for the reaction. Then substitute the appropriate equilibrium concentrations into this equation to obtain \(K\).
Solution
Substituting the appropriate equilibrium concentrations into the equilibrium constant expression,
\[K=\dfrac{[SO_3]^2}{[SO_2]^2[O_2]}=\dfrac{(5.0 \times 10^{-2})^2}{(3.0 \times 10^{-3})^2(3.5 \times 10^{-3})}=7.9 \times 10^4 \nonumber\]
To solve for \(K_p\), we use the relationship derived previously
\[K_p = K(RT)^{Δn} \]
where \(Δn = 2 − 3 = −1\):
\[K_p=K(RT)^{Δn}\nonumber\]
\[K_p=7.9 \times 10^4 [(0.08206\; L⋅atm/mol⋅K)(800 K)]^{−1}\nonumber\]
\[K_p=1.2 \times 10^3\nonumber\]
Exercise \(\PageIndex{1}\)
Hydrogen gas and iodine react to form hydrogen iodide via the reaction
\[H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\nonumber\]
A mixture of \(H_2\) and \(I_2\) was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained
- \(1.37\times 10^{−2}\; M\; HI\),
- \(6.47 \times 10^{−3}\; M\; H_2\), and
- \(5.94 \times 10^{-4}\; M\; I_2\).
Calculate \(K\) and \(K_p\) for this reaction.
- Answer
-
\(K = 48.8\) and \(K_p = 48.8\)
Chemists are not often given the concentrations of all the substances, and they are not likely to measure the equilibrium concentrations of all the relevant substances for a particular system. In such cases, we can obtain the equilibrium concentrations from the initial concentrations of the reactants and the balanced chemical equation for the reaction, as long as the equilibrium concentration of one of the substances is known. Example \(\PageIndex{2}\) shows one way to do this.
Example \(\PageIndex{2}\)
A 1.00 mol sample of \(NOCl\) was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of \(Cl_2\). Calculate \(K\) at this temperature. The equation for the decomposition of \(NOCl\) to \(NO\) and \(Cl_2\) is as follows:
\[2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber\]
Given: balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium
Asked for: \(K\)
Strategy:
- Write the equilibrium constant expression for the reaction. Construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations (as initial concentrations plus changes in concentrations).
- Calculate all possible initial concentrations from the data given and insert them in the table.
- Use the coefficients in the balanced chemical equation to obtain the changes in concentration of all other substances in the reaction. Insert those concentration changes in the table.
- Obtain the final concentrations by summing the columns. Calculate the equilibrium constant for the reaction.
Solution
A The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows:
\[K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\nonumber\]
To obtain the concentrations of \(NOCl\), \(NO\), and \(Cl_2\) at equilibrium, we construct a table showing what is known and what needs to be calculated. We begin by writing the balanced chemical equation at the top of the table, followed by three lines corresponding to the initial concentrations, the changes in concentrations required to get from the initial to the final state, and the final concentrations.
\[2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber\]
| ICE | \([NOCl]\) | \([NO]\) | \([Cl_2]\) |
|---|---|---|---|
| Initial | |||
| Change | |||
| Final |
B Initially, the system contains 1.00 mol of \(NOCl\) in a 2.00 L container. Thus \([NOCl]_i = 1.00\; mol/2.00\; L = 0.500\; M\). The initial concentrations of \(NO\) and \(Cl_2\) are \(0\; M\) because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of \(Cl_2\) in a 2.00 L container, so \([Cl_2]_f = 0.056 \;mol/2.00 \;L = 0.028\; M\). We insert these values into the following table:
\[2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber\]
| ICE | \([NOCl]\) | \([NO]\) | \([Cl_2]\) |
|---|---|---|---|
| Initial | 0.500 | 0 | 0 |
| Change | |||
| Final | 0.028 |
C We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of \(Cl_2\), the substance for which initial and final concentrations are known:
\[Δ[Cl_2] = 0.028 \;M_{(final)} − 0.00\; M_{(initial)}] = +0.028\; M\nonumber\]
According to the coefficients in the balanced chemical equation, 2 mol of \(NO\) are produced for every 1 mol of \(Cl_2\), so the change in the \(NO\) concentration is as follows:
\[Δ[NO]=\left(\dfrac{0.028\; \cancel{mol \;Cl_2}}{ L}\right)\left(\dfrac{2\; mol\; NO}{1 \cancel{\;mol \;Cl_2}}\right)=0.056\; M\nonumber\]
Similarly, 2 mol of \(NOCl\) are consumed for every 1 mol of \(Cl_2\) produced, so the change in the \(NOCl\) concentration is as follows:
\[Δ[NOCl]= \left(\dfrac{0.028\; \cancel{mol\; Cl_2}}{L}\right) \left(\dfrac{−2\; mol \;NOCl}{1\; \cancel{mol\; Cl_2}} \right) = -0.056 \;M\nonumber\]
We insert these values into our table:
\[2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber\]
| ICE | \([NOCl]\) | \([NO]\) | \([Cl_2]\) |
|---|---|---|---|
| Initial | 0.500 | 0 | 0 |
| Change | −0.056 | +0.056 | +0.028 |
| Final | 0.028 |
D We sum the numbers in the \([NOCl]\) and \([NO]\) columns to obtain the final concentrations of \(NO\) and \(NOCl\):
\[[NO]_f = 0.000\; M + 0.056 \;M = 0.056\; M\nonumber\]
\[[NOCl]_f = 0.500\; M + (−0.056\; M) = 0.444 M\nonumber\]
We can now complete the table:
\[2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber\]
| ICE | \([NOCl] | \([NO]\) | \([Cl_2]\) |
|---|---|---|---|
| initial | 0.500 | 0 | 0 |
| change | −0.056 | +0.056 | +0.028 |
| final | 0.444 | 0.056 | 0.028 |
We can now calculate the equilibrium constant for the reaction:
\[K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}=\dfrac{(0.056)^2(0.028)}{(0.444)^2}=4.5 \times 10^{−4}\nonumber\]
Exercise \(\PageIndex{2}\)
The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia (\(NH_3\)) by reacting \(0.1248\; M \;H_2\) and \(0.0416\; M \;N_2\) at about 500°C. At equilibrium, the mixture contained 0.00272 M \(NH_3\). What is \(K\) for the reaction
\[N_2+3H_2 \rightleftharpoons 2NH_3\nonumber\]
at this temperature? What is \(K_p\)?
- Answer
-
\(K = 0.105\) and \(K_p = 2.61 \times 10^{-5}\)