10.2: Chemical Potential
- Last updated
- Save as PDF
- Page ID
- 199247
- Contributed by Patrick Fleming
- Assistant Professor (Chemistry and Biochemistry) at California State University East Bay
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 \(p_A/p_B = K_p\) 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 and Attributions
Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)