20.2: Chemical Potential
- Page ID
- 238268
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Equilibrium can be understood as accruing at the composition of a reaction mixture at which the aggregate chemical potential of the products is equal to that of the reactants. Consider the simple reaction
\[A(g) \rightleftharpoons B(g)\]
The criterion for equilibrium will be
\[ \mu_A=\mu_B\]
If the gases behave ideally, the chemical potentials can be described in terms of the mole fractions of \(A\) and \(B\)
\[ \mu_A^o + RT \ln\left( \dfrac{P_A}{P_{tot}} \right) = \mu_B^o + RT \ln\left( \dfrac{P_B}{P_{tot}} \right) \label{eq2}\]
where Dalton’s Law has been used to express the mole fractions.
\[ \chi_i = \dfrac{P_i}{P_{tot}}\]
Equation \ref{eq2} can be simplified by collecting all chemical potentials terms on the left
\[ \mu_A^o - \mu_B^o = RT \ln \left( \dfrac{P_B}{P_{tot}} \right) - RT \ln\left( \dfrac{P_A}{P_{tot}} \right) \label{eq3}\]
Combining the logarithms terms and recognizing that
\[\mu_A^o - \mu_B^o =–\Delta G^o\]
for the reaction, one obtains
\[–\Delta G^o = RT \ln \left( \dfrac{P_B}{P_{A}} \right)\]
And since the equilibrium constant is \(P_A/P_B = K_P\) for this reaction (assuming perfectly ideal behavior), one can write
\[ \Delta G^o = RT \ln K_P\]
Another way to achieve this result is to consider the Gibbs function change for a reaction mixture in terms of the reaction quotient. The reaction quotient can be expressed as
\[ Q_P = \dfrac{\prod_i P_i^{\nu_i}}{\prod_j P_j^{\nu_j}} \]
where \(\nu_i\) are the stoichiometric coefficients for the products, and \(\nu_j\) are those for the reactants. Or if the stoichiometric coefficients are defined by expressing the reaction as a sum
\[ 0 =\sum_i \nu_i X_i\]
where \(X_i\) refers to one of the species in the reaction, and \(\nu_i\) is then the stoichiometric coefficient for that species, it is clear that \(\nu_i\) will be negative for a reactant (since its concentration or partial pressure will reduce as the reaction moves forward) and positive for a product (since the concentration or partial pressure will be increasing.) If the stoichiometric coefficients are expressed in this way, the expression for the reaction quotient becomes
\[Q_P = \prod_i P_i^{\nu_i}\]
Using this expression, the Gibbs function change for the system can be calculated from
\[ \Delta G =\Delta G^o + RT \ln Q_P\]
And since at equilibrium
\[\Delta G = 0\]
and
\[Q_P=K_P\]
It is evident that
\[ \Delta G_{rxn}^o = -RT \ln K_P \label{triangle}\]
It is in this simple way that \(K_P\) and \(\Delta G^o\) are related.
It is also of value to note that the criterion for a spontaneous chemical process is that \(\Delta G_{rxn}\ < 0\), rather than \(\Delta G_{rxn}^o\), as is stated in many texts! Recall that \(\Delta G_{rxn}^o\) is a function of all of the reactants and products being in their standard states of unit fugacity or activity. However, the direction of spontaneous change for a chemical reaction is dependent on the composition of the reaction mixture. Similarly, the magnitude of the equilibrium constant is insufficient to determine whether a reaction will spontaneously form reactants or products, as the direction the reaction will shift is also a function of not just the equilibrium constant, but also the composition of the reaction mixture!
Example \(\PageIndex{1}\):
Based on the data below at 298 K, calculate the value of the equilibrium constant (\(K_P\)) for the reaction
\[2 NO(g) + O_2(g) \rightleftharpoons 2 NO_2(g)\]
\(NO(g)\) | \(NO_2(g)\) | |
---|---|---|
\(G_f^o\) (kJ/mol) | 86.55 | 51.53 |
Solution:
First calculate the value of \(\Delta G_{rxn}^o\) from the \(\Delta G_{f}^o\) data.
\[ \Delta G_{rxn}^o = 2 \times (51.53 \,kJ/mol) - 2 \times (86.55 \,kJ/mol) = -70.04 \,kJ/mol\]
And now use the value to calculate \(K_p\) using Equation \ref{triangle}.
\[ -70040\, J/mol = -(8.314 J/(mol\, K) (298 \, K) \ln K_p\]
\[ K_p = 1.89 \times 10^{12}\]
Note: as expected for a reaction with a very large negative \(\Delta G_{rxn}^o\), the equilibrium constant is very large, favoring the formation of the products.
Contributors
Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)