20.8: Calculations of Free Energy and \(K_\text{eq}\)
- Page ID
- 53928
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Formation of stalactites (pointing down) and stalagmites (pointing up) is a complex process. Solutions of minerals drip down and absorb carbon dioxide as water flows through the cave. Calcium carbonate dissolves in this liquid and re-deposits on the rock as the carbon dioxide is dissipated into the environment.
Equilibrium Constant and \(\Delta G\)
At equilibrium, the \(\Delta G\) for a reversible reaction is equal to zero. \(K_\text{eq}\) relates the concentrations of all substances in the reaction at equilibrium. Through a more advanced treatment of thermodynamics, we can write the following equation:
\[\Delta G^\text{o} = -RT \: \text{ln} \: K_\text{eq}\nonumber \]
The variable \(R\) is the ideal gas constant \(\left( 8.314 \: \text{J/K} \cdot \text{mol} \right)\), \(T\) is the Kelvin temperature, and \(\text{ln} \: K_\text{eq}\) is the natural logarithm of the equilibrium constant.
When \(K_\text{eq}\) is large, the products of the reaction are favored and the negative sign in the equation means that the \(\Delta G^\text{o}\) is negative. When \(K_\text{eq}\) is small, the reactants of the reaction are favored. The natural logarithm of a number less than one is negative, and so the sign of \(\Delta G^\text{o}\) is positive. The table below summarizes the relationship of \(\Delta G^\text{o}\) to \(K_\text{eq}\):
| Table \(\PageIndex{1}\): Relationship of \(\Delta G^\text{o}\) and \(K_\text{eq}\) | |||
|---|---|---|---|
| \(K_\text{eq}\) | \(\text{ln} \: K_\text{eq}\) | \(\Delta G^\text{o}\) | Description |
| >1 | positive | negative | Products are favored at equilibrium. |
| 1 | 0 | 0 | Reactants and products are equally favored. |
| <1 | negative | positive | Reactants are favored at equilibrium. |
Knowledge of either the standard free energy change or the equilibrium constant for a reaction allows for the calculation of the other. The following two sample problems illustrate each case.
Example \(\PageIndex{1}\)
The formation of nitrogen monoxide from nitrogen and oxygen gases is a reaction that strongly favors the reactants at \(25^\text{o} \text{C}\).
\[\ce{N_2} \left( g \right) + \ce{O_2} \left( g \right) \rightleftharpoons 2 \ce{NO} \left( g \right)\nonumber \]
The actual concentrations of each gas would be difficult to measure, and so the \(K_\text{eq}\) for the reaction can more easily be calculated from the \(\Delta G^\text{o}\), which is equal to \(173.4 \: \text{kJ/mol}\). Find the \(K_\text{eq}\).
Solution:
Step 1: List the known quantities and plan the problem.
Known
- \(\Delta G^\text{o} = +173.4 \: \text{kJ/mol}\)
- \(R = 8.314 \: \text{J/K} \cdot \text{mol}\)
- \(T = 25^\text{o} \text{C} = 298 \: \text{K}\)
Unknown
In order to make the units agree, the value of \(\Delta G^\text{o}\) will need to be converted to \(\text{J/mol}\) \(\left( 173,400 \: \text{J/mol} \right)\). To solve for \(K_\text{eq}\), the inverse of the natural logarithm, \(e^x\), will be used.
Step 2: Solve.
\[\begin{align*} \Delta G^\text{o} &= -RT \: \text{ln} \: K_\text{eq} \\ \text{ln} \: K_\text{eq} &= \frac{-\Delta G^\text{o}}{RT} \\ K_\text{eq} &= e^{\frac{-\Delta G^\text{o}}{RT}} = e^{\frac{-173,400 \: \text{J/mol}}{8.314 \: \text{J/K} \cdot \text{mol} \left( 298 \: \text{K} \right)}} = 4.0 \times 10^{-31} \end{align*}\nonumber \]
Step 3: Think about your result.
The large positive free energy change leads to a \(K_\text{eq}\) that is extremely small. Both lead to the conclusion that the reactants are highly favored and very few product molecules are present at equilibrium.
Example \(\PageIndex{2}\)
The solubility product constant \(\left( K_\text{sp} \right)\) of lead (II) iodide is \(1.4 \times 10^{-8}\) at \(25^\text{o} \text{C}\). Calculate \(\Delta G^\text{o}\) for the dissociation of lead (II) iodide in water.
\[\ce{PbI_2} \left( s \right) \rightleftharpoons \ce{Pb^{2+}} \left( aq \right) + 2 \ce{I^-} \left( aq \right)\nonumber \]
Solution:
Step 1: List the known values and plan the problem.
Known
- \(K_\text{eq} = K_\text{sp} = 1.4 \times 10^{-8}\)
- \(R = 8.314 \: \text{J/K} \cdot \text{mol}\)
- \(T = 25^\text{o} \text{C} = 298 \: \text{K}\)
Unknown
The equation relating \(\Delta G^\text{o}\) to \(K_\text{eq}\) can be solved directly.
Step 2: Solve.
\[\begin{align*} \Delta G^\text{o} &= -RT \: \text{ln} \: K_\text{eq} \\ &= -8.314 \: \text{J/K} \cdot \text{mol} \left( 298 \: \text{K} \right) \: \text{ln} \left( 1.4 \times 10^{-8} \right) \\ &= 45,000 \: \text{J/mol} \\ &= 45 \: \text{kJ/mol} \end{align*}\nonumber \]
Step 3: Think about your result.
The large, positive \(\Delta G^\text{o}\) indicates that the solid lead (II) iodide is nearly insoluble and so very little of the solid is dissociated at equilibrium.
Summary
- The relationship between \(\Delta G\) and \(K_\text{eq}\) is described.
- Calculations involving these two parameters are shown.