9.2: Chemical Potential
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- Jul 19, 2021
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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)⇌B(g)
The criterion for equilibrium will be
μA=μB
If the gases behave ideally, the chemical potentials can be described in terms of the mole fractions of A and B
μoA+RTln(pAptot)=μoB+RTln(pBptot)
where Dalton’s Law has been used to express the mole fractions.
χi=piptot
Equation ??? can be simplified by collecting all chemical potentials terms on the left
μoA−μoB=RTln(pBptot)−RTln(pAptot)
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!
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) \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.