Chemical potential and condition for equilibrium in case of binary liquid reaction mixture

Suppose there is a liquid mixture with components A and B in equilibrium along with their vapours. Then with the Gibbs-Duhem equation we know that $$μ\_A(\mathrm{liquid}) = μ\_A(\mathrm{vapor})$$ and
$$μ\_B(\mathrm{liquid}) = μ\_B(\mathrm{vapor})$$
But then imagine a case where A(liquid) is getting converted to B(liquid), so for that case $$d G = - μ\_A(\mathrm{liquid})\cdot dn+μ\_B(\mathrm{liquid})\cdot dn$$ (for this reaction), so if $μ\_A(\mathrm{liquid})$ is not equal to $μ\_B(\mathrm{liquid})$, then this reaction will take place in one of the directions.


So doesn't it mean that $μ\_A(\mathrm{liquid})$ must also be equal to $μ\_B(\mathrm{liquid})$?




Yes it does, if stoichiometric coefficients are the same.


For eventual reaction of 2 volatile miscible liquids $\ce{A}$ and $\ce{B}$


$$\ce{A <=> B}$$


all the respective chemical potentials must be equal, otherwise the system would not be at the Gibbs function minimum:


$$\mu(\ce{A(g)})=\mu(\ce{A(sol)})=\mu(\ce{B(sol)})=\mu(\ce{B(g)})$$


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> 
> Then with gibbs durhem eqn we know that μA(liquid) = μA(vapor) μB(liquid) = μB(vapor).
> 
> 
> 


That is not Gibbs/Duhem equation. Gibbs/Duhem equation states a restriction between activity coefficients at constant $P$ and $T$ must satisfy, this is, $\sum\_j x\_j d\ln(\gamma\_j) = 0$. What you have written is the condition of chemical equilibrium of species between diferent phases.



> 
> But then imagine a case where A(liquid) is getting converted to B(liquid) so for that case Delta G = - μA(liquid)\*dn+μB(liquid)\*dn. (for this reaction),
> 
> 
> 


We need to address how the chemical potentials are related between each other in this case. This is a bit more difficult than usual, because we have a reaction but two phases, i.e., a heterogeneous system. The main idea is that **we can separate** (1) the chemical reaction, and (2) the conventional chemical equilibria. This is not so straightforward to prove, but I will show what happens in this case.


First, lets apply the Gibbs phase rule for reacting systems, so we know how many relationships we need
\begin{align}
 F &= 2 - \pu{Num. of phases} + \pu{Num. of components} - \pu{Num. of chemical reactions} \\
 F &= 2 - 2 + 2 - 1 \rightarrow \boxed{F=1}
\end{align}
because we have two components ($A$ and $B$), two phases (liquid and vapor), and one chemical reaction.


The chemical reaction that you proposed, that we can generalize
$$ \ce{aA (l)} \rightarrow \ce{bB (l)} $$
gives us one condition that must be satisfied **in the liquid phase** ($a$ and $b$ are positive)
$$ \sum\_j \nu\_j \mu\_j^L = 0 \rightarrow
 \boxed{-a\mu\_A^L(P,T,x\_A) + b\mu\_B^L(P,T,x\_B) = 0} \tag{1} $$
The other two, are simply the conditions of chemical equilibrium for species in a closed system


$$ \boxed{\mu\_A^V(P,T,y\_A) = \mu\_A^L(P,T,x\_A)} \hspace{1 cm}
 \boxed{\mu\_B^V(P,T,y\_B) = \mu\_B^L(P,T,x\_B)} \tag{2,3} $$


Eqs. (1-3) are the relationships we need to solve the system. The number of intensive variables are **four**:


1. The pressure $P$ [mechanical equilibrium implies $(P^V=P^L=P)$].
2. The temperature $T$ [thermal equilibrium implies $(T^V=T^L=T)$].
3. Two molar fractions, since the other one can be calculated with $y\_B = 1 - y\_A$ and $x\_B = 1 - x\_A$.


And the number of equations are **three**, so everything is fine.


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