9.2: Chemical Potential
- Page ID
- 84345
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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
\[A(g) \rightleftharpoons B(g) \nonumber \]
The criterion for equilibrium will be
\[ \mu_A=\mu_B \nonumber \]
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.
\[ \chi_i = \dfrac{p_i}{p_{tot}} \nonumber \]
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
\[\mu_A^o - \mu_B^o –\Delta G^o \nonumber \]
for the reaction, one obtains
\[–\Delta G^o = RT \ln \left( \dfrac{p_B}{p_{A}} \right) \nonumber \]
And since \(p_A/p_B = K_p\) for this reaction (assuming perfectly ideal behavior), one can write
\[ \Delta G^o = RT \ln K_p \nonumber \]
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
\[ Q_p = \dfrac{\prod_i p_i^{\nu_i}}{\prod_j p_j^{\nu_j}} \nonumber \]
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
\[ 0 =\sum_i \nu_i X_i \nonumber \]
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
\[Q_p = \prod_i p_i^{\nu_i} \nonumber \]
Using this expression, the Gibbs function change for the system can be calculated from
\[ \Delta G =\Delta G^o + RT \ln Q_p \nonumber \]
And since at equilibrium
\[\Delta G = 0 \nonumber \]
and
\[Q_p=K_p \nonumber \]
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!
Based on the data below at 298 K, calculate the value of the equilibrium constant (\(K_p\)) for the reaction
\[2 NO(g) + O_2(g) \rightleftharpoons 2 NO_2(g) \nonumber \]
\(NO(g)\) | \(NO_2(g)\) | |
---|---|---|
\(G_f^o\) (kJ/mol) | 86.55 | 51.53 |
Solution
First calculate the value of \(\Delta G_{rxn}^o\) from the \(\Delta G_{f}^o\) data.
\[ \Delta G_{rxn}^o = 2 \times (51.53 \,kJ/mol) - 2 \times (86.55 \,kJ/mol) = -70.04 \,kJ/mol \nonumber \]
And now use the value to calculate \(K_p\) using Equation \ref{triangle}.
\[ -70040\, J/mol = -(8.314 J/(mol\, K) (298 \, K) \ln K_p \nonumber \]
\[ K_p = 1.89 \times 10^{12} \nonumber \]
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.