# Equilibria in Electrochemistry and Maximal Rates of Reaction

Tristram de Piro  Flat 3, Redesdale House, 85 The Park, Cheltenham, GL50 2RP

###### Abstract.

We consider Gibbs’ definition of chemical equilibrium and connect it with
dynamic equilibrium, in terms of no substance formed. We determine the
activity coefficient as a function of temperature and pressure, in reactions
with or without interaction of a solvent, incorporating the error terms from
Raoult’s Law and Henry’s Law, if necessary. We compute the maximal reaction
paths and apply the results to electrochemistry, using the Nernst equation.

##  1\. Introduction

This paper is divided into $12$ sections. In section 2, we give some basic
definitions, and derive the Nernst equation for the standard cell. We prove
some results about the activity coefficient $Q$, assuming an idealised law in
the behaviour of the activities and chemical potentials,
$\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})$, for $1\leq i\leq c$, with $c$
substances, which holds throughout the section. In Lemma 2.8, we use the van’t
Hoff, Gibbs-Helmholtz equations to find an expression for $\Delta
G^{\circ}(T)$ along quasi-chemical equilibrium paths. In Lemma 2.9, we use an
entropy calculation to find $({\partial G\over\partial\xi})_{T,P}$ and combine
the result with Lemma 2.8 to calculate the activity coefficient $Q$. In Lemma
2.14, we prove every straight line chemical equilibrium path is a dynamic
equilibrium path, partially confirming a speculation of Gibbs. The method of
constant $Q$ along a path implying dynamic equilibrium is repeatedly used and
generalised later in the paper. The question of the existence of feasible
paths for a reaction, given a curve in the temperature/pressure plane, is
answered in Lemma 2.16 and again later generalised.
In Section 3, we consider ideal solutions and introduce a fixed error term
from Raoult’s law. The results from section $0$ generalise and in Lemma 3.6,
we find the paths of maximal reaction, in the sense of maximising extent
$\xi$, implicitly, in terms of temperature and pressure $(T,P)$. We apply the
results to electrochemistry in Section 4. In Section 5, we consider dilute
solutions, adding substance $0$, and consider the definition of $Q$, involving
the activity $a_{0}$, obtaining the formula for the activity coefficient in
Lemma 5.2. In Section 6, we consider dilute solutions with interaction of the
solvent, in which the solvent is ideal and the solutes obey Henry’s law,
introducing a new fixed error term in Definition 6.1, and obtaining the
maximal reaction paths in Lemma 6.7. In Section 7, we introduce new fixed
error terms from fugacity, in Definition 7.1. We alter the conventional
definition of $Q$ to incorporate fugacity in the error term and obtain the
paths of maximal reaction in Lemma 7.7. We apply the results to
electrochemistry in Section 8, in particular the reaction in catalyzers, and
give a strategy for improving the efficiency of hydrogen and oxygen production
from water in Remarks 11.5. In Section 9, we consider the case when there is
no interaction of the solvent, and in Sections 9 and 10, we derive the main
results quickly by altering $Q$ to ignore the activity $a_{0}$. However, it an
interesting but difficult exercise to try and derive the results using the
definition in Section 5. We suggest the results here could be used in
maximising ethanol production. We gain apply the results to electrochemistry
in Section 11, in particularly the standard cell. In Section 12, we reconsider
the assumption that $\Delta H^{\circ}(T)$ is constant, made throughout the
paper. We show that by increasing the mass of the mixture, in particularly the
amount of solvent, we can make the error involved here disappear in the limit.
Finally, in Section 13, we consider independence and existence of paths.

##  2\. The Idealised Case

######  Definition 2.1.

For $c$ substances, we define the Gibbs energy $G(T,P,n_{1},\ldots,n_{c})$ by;

$G=U+PV-TS$

where $U$ is the internal energy, $P$ is pressure, $V$ is volume, $T$ is
temperature and $S$ is entropy. We define the enthalpy
$H(T,P,n_{1},\ldots,n_{c})$ by;

$H=U+PV=G+TS$

see [7] and [5]. We define the Gibbs energy at standard pressure
$G^{\circ}(T,n_{1},\ldots,n_{c})=G(T,P^{\circ},n_{1},\ldots,n_{c})$, where
$P^{\circ}$ is the standard pressure. We define the chemical potentials,
$1\leq i\leq c$, by;

$\mu_{i}(T,P)=({\partial G\over\partial n_{i}})_{T,P,n^{\prime}}$

where $n_{i}$ is the amount of substance $i$ measured in moles, and
$T,P,n^{\prime}$ fixes the pressure, temperature and the amount of every
substance except substance $i$.

We consider an electrolyte as a solute in a dilute solution and define the
activities $a_{i}$, $1\leq i\leq c$, by;

$a_{1}=\gamma_{1}x_{1}\simeq 1$

$a_{i}={\gamma_{i}m_{i}\over m^{\circ}}$ $(2\leq i\leq c)$

where the molality $m_{i}={n_{i}\over w_{1}}$, and $w_{1}$ is the mass of the
solvent, component $1$, $m^{\circ}=1$, $x_{i}={n_{i}\over n}$,
$n=\sum_{i=1}^{c}n_{i}$, $\gamma_{i},1\leq i\leq c$, are the activity
coefficients, $c_{i}={n_{i}\over V}$, $c^{\circ}=1$, and the activity
quotient;

$Q=\prod_{i=1}^{c}a_{i}^{\nu_{i}}$

where $\nu_{i}$, for $1\leq i\leq c$ are the stoichiometric coefficients.

We have, for a solute in a dilute solution, that
$\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})$, see [7], noting that $\mu_{i}$ is
independent of the amount of substance $n_{i}$, and $\mu_{i}^{\circ}$ in the
molality description is equal to $\mu_{i}^{(m)}$, where $m_{i}$ is equal to
$m^{\circ}=1$ in a hypothetical solution.

We define $\Delta G^{\circ}(T)$ and $\Delta H^{\circ}(T)$ to be the changes in
Gibbs energy and enthalpy at standard pressure $P^{\circ}$ and temperature
$T$, for $1$ mole of reaction, see [1]and [4].

We define the extent $\xi(T,P)$ of a reaction by;

$n_{i,0}+\nu_{i}\xi=n_{i}$

where $n_{i,0}=n_{i}(initial)$. We assume that if;

$h_{T,P,n_{1,0},\ldots,n_{c,0}}(\xi)=G(T,P,n_{1,0}+\nu_{1}\xi,\ldots,n_{c,0}+\nu_{c}\xi)$

then ${dh_{T,P,n_{1,0},\ldots,n_{c,0}}\over d\xi}(\xi)$ is independent of
$\\{n_{1,0},\ldots,n_{c,0},\\}\subset\mathcal{R}_{>0}$ and
$\xi\in\mathcal{R}_{\geq 0}$, and define this as $({\partial
G\over\partial\xi})|_{T,P}$.

We define chemical equilibrium by $({\partial G\over\partial\xi})|_{T,P}=0$,
and dynamic equilibrium by a path
$\gamma:[0,1]\rightarrow(T,P,n_{1},\ldots,n_{c})$, such that
$n_{i}^{\prime}(t)=pr_{2+i}(\gamma)^{\prime}(t)=0$, for $1\leq i\leq c$, so
that no substance is formed.

We define $E(T,P,n_{1},\ldots,n_{c})$ to be the potential in the standard cell
and $E^{\circ}(T,n_{1},\ldots,n_{c})=E(T,P^{\circ},n_{1},\ldots,n_{c})$ to be
the potential at $P^{\circ}$. We let $F$ denote Faraday’s constant, $R=N_{A}k$
the gas constant, where $N_{Av}$ is Avogadro’s constant and $k$ is Boltzmann’s
constant. We have that $N=N_{Av}n$, where $N$ is the number of particles, $n$
is the amount in moles. We define the electric potential $\phi$ by
$\overline{E}=-\bigtriangledown(\phi)$, $e$ is the charge on an electron,
$z_{i}$ is the valence of an ion.

######  Lemma 2.2.

$dG=-SdT+VdP+\sum_{i=1}^{c}\mu_{i}dn_{i}$

###### Proof.

We have, by the definition of the chemical potential and the laws of
differentials, that;

$dG={\partial G\over\partial T}_{P,n}dT+{\partial G\over\partial
P}_{T,n}dP+\sum_{i=1}^{c}{\partial G\over\partial
n_{i}}_{T,P,n^{\prime}}dn_{i}$

$={\partial G\over\partial T}_{P,n}dT+{\partial G\over\partial
P}_{T,n}dP+\sum_{i=1}^{c}\mu_{i}dn_{i}$, $(*)$

Fixing $n$, using $G=U+PV-TS$, the first law of thermodynamics, $dU=dQ-PdV$,
the definition of entropy, $dQ=TdS$, the product rule for differentials, and
$(*)$, we have that;

$dU=TdS-PdV$

$dG=dU+VdP+PdV-SdT-TdS$

$=(TdS-PdV)+VdP+PdV-SdT-TdS$

$=-SdT+VdP$

$dG={\partial G\over\partial T}_{P,n}dT+{\partial G\over\partial P}_{T,n}dP$
$(**)$

so that, from $(**)$, and equating coefficients, ${\partial G\over\partial
T}_{P,n}=-S$, ${\partial G\over\partial P}_{T,n}=V$. Substituting into $(*)$,
we obtain that;

$dG=-SdT+VdP+\sum_{i=1}^{c}\mu_{i}dn_{i}$

∎

######  Definition 2.3.

We define electrical chemical equilibrium by $({\partial
G\over\partial\xi})_{T,P}=0$ where $G$ is the Gibbs energy function for a
charged and uncharged species.

######  Lemma 2.4.

The Nernst Equation for the Standard Cell

At electrical chemical equilibrium $(T,P)$ and $(T,P^{\circ})$;

$E-E^{\circ}=-{RTln(Q)\over 2F}$

See [7].

###### Proof.

For $c$ substances, with $c^{\prime}$ the number of the charged species,using
Definition 2.1, we have that the electrostatic potential energy;

$U_{el}=\sum_{i=1}^{c^{\prime}}\phi(\overline{x}_{i})q_{i}$, where
$q_{i}=N_{i}ez_{i}=N_{A}n_{i}ez_{i}$

where $\\{\overline{x}_{i}:1\leq i\leq c^{\prime}\\}$ are the positions of the
charged species, $N_{i}$ is the number of particles at $\overline{x}_{i}$.

We have that;

$U=U_{chem}+U_{el}$, so that;

$G(T,P,n_{1},\ldots,n_{c})=U+PV-TS$

$=U_{chem}+U_{el}+PV-TS$

$=U_{el}+G_{chem}$

$=\sum_{j=1}^{c}\phi(\overline{x}_{j})q_{j}+G_{chem}$

$=\sum_{j=1}^{c}\phi(\overline{x}_{j})N_{A}n_{j}ez_{j}+G_{chem}$

so that;

$\mu_{i}=({\partial G\over\partial n_{i}})_{T,P}$

$=({\partial(\sum_{j=1}^{c}\phi(\overline{x}_{j})N_{A}n_{j}ez_{j}+G_{chem})\over\partial
n_{i}})_{T,P}$

$=\mu_{i,chem}$, $(c^{\prime}+1\leq i\leq c)$

$=\mu_{i,chem}+{\partial(\phi(\overline{x}_{i})N_{A}n_{i}ez_{i})\over\partial
n_{i}}$, $(1\leq i\leq c^{\prime})$

$=\mu_{i,chem}+\phi(\overline{x}_{i})N_{A}ez_{i}$

$=\mu_{i,chem}+\phi(\overline{x}_{i})Fz_{i}$, $(*)$

We consider the standard cell reaction $H_{2}(g)+2AgCl(s)+2e^{-}(R)\rightarrow
2HCl+2Ag(s)+2e^{-}(L)$. At chemical equilibrium, similarly to Lemma 2.7,
generalised to a collection involving charged species, using $(*)$, we have
that;

$({\partial G\over\partial\xi})_{T,P}={\sum}_{i=1}^{c}\nu_{i}\mu_{i}$

$=2\mu(HCl)+2\mu(Ag)-\mu(H_{2})-2\mu(AgCl)+2\mu(e^{-}(L))-2\mu(e^{-}(R))$

$=({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}+2\mu(e^{-}(L))-2\mu(e^{-}(R))$

$=({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}+((2\mu_{chem}(e^{-}(L))-2F\phi(L))-(2\mu_{chem}(e^{-}(L))-2F\phi(R)))$

$=({\partial G_{chem^{\prime}}\over\partial\xi})_{T,P}+2F(\phi(R)-\phi(L))$

$=({\partial G_{chem^{\prime}}\over\partial\xi})_{T,P}+2EF=0$ $({\dagger})$

where $G_{chem^{\prime}}$ is the Gibbs energy restricted to the uncharged
species. We have that;

$({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P^{\circ}}=\sum_{i=c^{\prime}+1}^{c}\nu_{i}\mu_{i}^{\circ}$

$=(\Delta G_{chem^{\prime}}^{\circ}+RTln(Q_{chem^{\prime}}(T,P^{\circ})))$

$=\Delta G_{chem^{\prime}}^{\circ}$, $({\dagger}{\dagger})$

using the definition of $Q_{chem^{\prime}}$ in Definition 2.1, the fact that
$\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})$, $\mu_{i}=\mu_{i}^{\circ}$, for
$c^{\prime}+1\leq i\leq c$, so that $Q_{chem^{\prime}}(T,P^{\circ})=1$.

From $({\dagger}),({\dagger}{\dagger})$, we obtain;

$2E^{\circ}F=-({\partial G_{chem^{\prime}}\over\partial\xi})_{T,P^{\circ}}$

$=-\Delta G_{chem^{\prime}}^{\circ}$ $({\dagger}{\dagger}{\dagger})$

Similarly, we have that;

$({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}=\sum_{i=c^{\prime}+1}^{c}\nu_{i}\mu_{i}$

$=(\Delta G_{chem^{\prime}}^{\circ}+RTln(Q_{chem^{\prime}}(T,P)))$,
$({\dagger}{\dagger}{\dagger}{\dagger})$

so from $({\dagger}),({\dagger}{\dagger}{\dagger}{\dagger}))$, we obtain that;

$2EF=-({\partial G_{chem^{\prime}}\over\partial\xi})_{T,P}$

$=-(\Delta G_{chem^{\prime}}^{\circ}+RTln(Q_{chem^{\prime}}(T,P)))$,
$(\sharp)$

Combining $(\sharp),({\dagger}{\dagger}{\dagger})$, we obtain that;

$2EF-2E^{\circ}F=-(\Delta
G_{chem^{\prime}}^{\circ}+RTln(Q_{chem^{\prime}}(T,P)))-(-\Delta
G_{chem^{\prime}}^{\circ})$

$=-RTln(Q_{chem^{\prime}}(T,P))$

so that;

$E-E^{\circ}=-{RTln(Q_{chem^{\prime}}(T,P))\over 2F}$

∎

######  Lemma 2.5.

For the energy function $G$ involving only an uncharged species;

$({\partial G\over\partial\xi})_{T,P}=\Delta G^{\circ}+RTln(Q)$

see [7].

###### Proof.

By Lemma 2.7, we have that;

$({\partial G\over\partial\xi})_{T,P}=\sum_{i=1}^{c}\nu_{i}\mu_{i}$

$\Delta G^{\circ}=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}$, $(*)$

Using $(*)$, the fact that $\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})$, and
Definition 2.1, we have that;

$({\partial G\over\partial\xi})_{T,P}-\Delta
G^{\circ}=\sum_{i=1}^{c}\nu_{i}(\mu_{i}-\mu_{i}^{\circ})$

$=\sum_{i=1}^{c}\nu_{i}(\mu_{i}^{\circ}+RTln(a_{i})-\mu_{i}^{\circ})$

$=\sum_{i=1}^{c}\nu_{i}RTln(a_{i})$

$=RTln(\prod_{i=1}^{c}a_{i}^{\nu_{i}})=RTln(Q)$

∎

######  Lemma 2.6.

At electrical chemical equilibrium $(T,P)$ and $(T,P^{\circ})$, and chemical
equilibrium $(T,P)$;

$\Delta G^{\circ}=2F(E-E^{0})$

###### Proof.

By Lemma 2.4, we have that;

$E-E^{\circ}=-{RTln(Q)\over 2F}$, $(*)$

and, by Lemma 2.5, we have that;

$0=({\partial G\over\partial\xi})_{T,P}=\Delta G^{\circ}+RTln(Q)$, $(**)$

Rearranging $(*),(**)$, we obtain the result.

∎

######  Lemma 2.7.

We have that;

$({\partial G\over\partial\xi})_{T,P}=\sum_{i=1}^{c}\nu_{i}\mu_{i}$

$\Delta G^{\circ}=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}$

At chemical equilibrium $T,P$, $({\partial G\over\partial\xi})_{T,P}=0$ and at
$T,P^{0}$, $\Delta G^{\circ}=0$.

If chemical and electrical chemical equilibrium exists at $(T,P^{\circ})$ and
$(T,P)$, $Q(T,P)=1$. and $E=E^{\circ}$. Conversely, if $Q(T,P)=1$ and chemical
equilibrium exists at $(T,P^{\circ})$ then chemical equilibrium exists at
$(T,P)$.

Chemical equilibrium exists at $(T,P)$ iff $Q(T,P)=e^{-\Delta G^{\circ}\over
RT}$

We always have that $Q(T,P^{\circ})=1$.

###### Proof.

For the first claim, we have, using the definition of $\xi$, that;

$dn_{i}=\nu_{i}d\xi$, $(1\leq i\leq c)$, $(*)$

By Lemma 2.2, fixing $T$ and $P$, and using $(*)$, we have that;

$dG=\sum_{i=1}^{c}\mu_{i}dn_{i}$

$=(\sum_{i=1}^{c}\mu_{i}\nu_{i})d\xi$, $({\dagger})$

so that;

$({\partial G\over\partial\xi})_{T,P}=\sum_{i=1}^{c}\mu_{i}\nu_{i}$,
$({\dagger}{\dagger})$

The second claim from the first, as;

$\Delta G^{\circ}(T)=\int_{0}^{1}({\partial G\over\partial\xi})_{T,P^{\circ}}$

$=\int_{0}^{1}(\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}(T))d\xi$

$=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}(T)\int_{0}^{1}d\xi$

$=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}(T)$

noting that $({\partial G\over\partial\xi})_{T,P^{\circ}}$ doesn’t vary with
$\xi$. For the third claim, at chemical equilibrium, $T,P$, noting again that
$({\partial G\over\partial\xi})_{T,P}$ doesn’t vary with $\xi$, and using
$({\dagger},{\dagger}{\dagger})$, we have that;

$dG=({\partial G\over\partial\xi})_{T,P}=0$, (independently of $\xi$)
$({\dagger}{\dagger}{\dagger})$

At chemical equilibrium $T,P^{\circ}$, using the first and second claims, and
$({\dagger}{\dagger}{\dagger})$, we have that;

$dG=({\partial G\over\partial\xi})_{T,P^{\circ}}$

$=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}$

$=\Delta G^{\circ}=0$

For the second to last claim, and the first direction, we have, by Lemma 2.5,
that $RTln(Q)=0$, so that $Q=1$, and, by Lemma 2.4, that
$E-E^{\circ}=-{RTln(Q)\over 2F}=0$. For the converse, we have by Lemma 2.5,
using the fact that $Q(T,P)=1$;

$({\partial G\over\partial\xi})_{T,P}=\Delta G^{\circ}$

and, if chemical equilibrium exists at $(T,P^{\circ})$, then, as
$Q(T,P^{\circ})=1$ we have that;

$({\partial G\over\partial\xi})_{T,P^{\circ}}=\Delta G^{\circ}+RTln(Q)=\Delta
G^{\circ}=0$

so that $({\partial G\over\partial\xi})_{T,P}=0$

For the penultimate claim, in one direction, use Lemma 2.5, together with the
fact that $({\partial G\over\partial\xi})_{T,P}=0$ and rearrange, the converse
is also clear, applying $ln$.

For the final claim, we have, by the definition of activities, that;

$\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})$

so that $a_{i}(T,P^{\circ})=1$. Now use the definition of $Q$ in Definition
2.1.

∎

######  Lemma 2.8.

van’t Hoff,Gibbs-Helmholtz

Along a chemical equilibrium path, we have that;

$ln({Q(T_{2})\over Q(T_{1})})={1\over R}\int_{T_{1}}^{T_{2}}{\Delta
H^{\circ}\over T^{2}}dT$

${\Delta G^{\circ}(T_{2})\over T_{2}}-{\Delta G^{\circ}(T_{1})\over
T_{1}}=-\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}\over T^{2}}dT$

In particularly, if $\Delta H^{\circ}$ is temperature independent;

$ln({Q(T_{2})\over Q(T_{1})})=-{\Delta H^{\circ}\over R}({1\over
T_{2}}-{1\over T_{1}})$

${\Delta G^{\circ}(T_{2})\over T_{2}}-{\Delta G^{\circ}(T_{1})\over
T_{1}}=\Delta H^{\circ}({1\over T_{2}}-{1\over T_{1}})$

$\Delta G^{\circ}(T_{1})={T_{1}\over T_{2}}\Delta
G^{\circ}(T_{2})-({T_{1}\over T_{2}}-1)\Delta H^{\circ}$

For $c\in\mathcal{R}$, Let $D_{c}$ intersect the line $P=P^{\circ}$ at
$(T_{1},P^{\circ})$, then, for $(T_{2},P)\in D_{c}$, we have that;

$Q(T_{2},P)=e^{\Delta G^{\circ}(T_{1})-\Delta G^{\circ}(T_{2})\over RT_{2}}$
$({\dagger}{\dagger}{\dagger})$

$c=\Delta G^{\circ}(T_{1})$

$ln({Q(T_{2})\over Q(T_{1})})=ln(Q(T_{2}))={1\over
R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}-c\over T^{2}}dT$

${\Delta G^{\circ}(T_{2})-\Delta G^{\circ}(T_{1})\over
T_{2}}=-\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}-c\over T^{2}}dT$

and if $\Delta H^{\circ}$ is temperature independent;

$ln({Q(T_{2})\over Q(T_{1})})=ln(Q(T_{2}))=-({\Delta H^{\circ}-c\over
R})({1\over T_{2}}-{1\over T_{1}})$ $({\dagger}{\dagger})$

${\Delta G^{\circ}(T_{2})-\Delta G^{\circ}(T_{1})\over T_{2}}=(\Delta
H^{\circ}-c)({1\over T_{2}}-{1\over T_{1}})$ $({\dagger})$

$\Delta G^{\circ}(T_{1})={T_{1}\over T_{2}}\Delta G^{\circ}(T_{2})-\Delta
H^{\circ}({T_{1}\over T_{2}}-1)$ $({\dagger}{\dagger}{\dagger}{\dagger})$

See also [7].

###### Proof.

By Lemma 2.7, we have that;

$\Delta G^{\circ}=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}$

so that differentiating with respect to $T$;

${d(\Delta G^{\circ})\over dT}=\sum_{i=1}^{c}\nu_{i}{d\mu_{i}^{\circ}\over
dT}$

$=\sum_{i=1}^{c}\nu_{i}({\partial\mu_{i}^{\circ}\over\partial T})_{P,n}$

By Euler reciprocity, we have that;

$({\partial\mu_{i}^{\circ}\over\partial T})_{P,n}=-({\partial
S^{\circ}\over\partial n_{i}})_{T,P,n^{\prime}}=-\overline{S}^{\circ}_{i}$

so that, noting $\overline{S}^{\circ}_{i}$ is independent of $n_{i}$, so we
can replace $\overline{S}^{\circ}_{i}$ by $\overline{S}^{\circ}_{m,i}$, the
absolute molar entropy of substance $i$, and using thermodynamic arguments;

${d(\Delta G^{\circ})\over
dT}=-\sum_{i=1}^{c}\nu_{i}\overline{S}^{\circ}_{i}=$

$=-\sum_{i=1}^{c}\nu_{i}\overline{S}^{\circ}_{m,i}$

$=-\Delta S^{\circ}$ $(*)$

Using the product rule, $(*)$ and the definition of enthalpy, we have that;

${d\over dT}({\Delta G^{\circ}\over T})={1\over T}{d(\Delta G^{\circ})\over
dT}-{1\over T^{2}}\Delta G^{\circ}$

$=-{\Delta S^{\circ}\over T}-{\Delta G^{\circ}\over T^{2}}$

$=-{\Delta(ST+G)^{\circ}\over T^{2}}$

$=-{\Delta H^{\circ}\over T^{2}}$ $(**)$

By Lemma 2.7, along a chemical equilibrium path, we have that $Q=e^{-\Delta
G^{\circ}\over RT}$, so that $ln(Q)={-\Delta G^{\circ}\over RT}$. It follows
from $(**)$ that;

${dln(Q)\over dT}={d\over dT}({-\Delta G^{\circ}\over RT})$

$={\Delta H^{\circ}\over RT^{2}}$

It follows, integrating between $T_{1}$ and $T_{2}$, that;

$ln({Q(T_{2})\over Q(T_{1})})=ln(Q)(T_{2})-ln(Q)(T_{1})$

$={-\Delta G^{\circ}(T_{2})\over RT_{2}}+{\Delta G^{\circ}(T_{1})\over
RT_{1}}$

$=\int_{T_{1}}^{T_{2}}{dln(Q)\over dT}dT$

$={1\over R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}\over T^{2}}$ $(P)$

so that, rearranging, we obtain the first claim. Using the fact, by Lemma 2.5,
that;

$ln(Q(T_{2}))=-{\Delta G^{\circ}(T_{2})\over RT_{2}}$

$ln(Q(T_{1}))=-{\Delta G^{\circ}(T_{1})\over RT_{1}}$

we obtain, substituting into $(P)$, canceling $R$, and performing the
integration, if $\Delta H^{\circ}$ is temperature independent, that;

${\Delta G^{\circ}(T_{2})\over T_{2}}-{\Delta G^{\circ}(T_{1})\over
T_{1}}=-\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}\over T^{2}}=\Delta
H^{\circ}({1\over T_{2}}-{1\over T_{1}})$ $(Q)$

For the fifth claim, rearrange $(Q)$. If $D_{c}$ intersects the line
$P=P^{\circ}$ at $(T_{1},P^{\circ})$, for the sixth
$({\dagger}{\dagger}{\dagger})$ and seventh claims, we have, using Lemma 2.5
and the fact from Lemma 2.7 that $Q(T_{1},P^{\circ})=1$;

$({\partial G\over\partial\xi})_{T_{2},P}=\Delta
G^{\circ}(T_{2})+RT_{2}ln(Q(T_{2},P))$

$=({\partial G\over\partial\xi})_{T_{1},P^{\circ}}$

$=\Delta G^{\circ}(T_{1})+RT_{1}ln(Q(T_{1},P^{\circ}))$

$=\Delta G^{\circ}(T_{1})=c$

so that, again rearranging, we obtain the result. Along $D_{c}$, we have,
using Lemma 2.5, that;

$ln(Q)={c-\Delta G^{\circ}\over RT}$

so that, using the first part;

${dln(Q)\over dT}={d\over dT}({c-\Delta G^{\circ}\over RT})$

$={-c\over RT^{2}}+{d\over dT}({-\Delta G^{\circ}\over RT})$

$={\Delta H^{\circ}-c\over RT^{2}}$

so that, performing the integration, using the fact that
$Q(T_{1},P^{\circ})=1$;

$ln(Q(T_{2}))-ln(Q(T_{1}))=ln(Q(T_{2}))={1\over R}\int_{T_{1}}^{T_{2}}{\Delta
H^{\circ}-c\over T^{2}}dT$

We have that, by Lemma 2.5;

$ln(Q(T_{2}))={c-\Delta G^{\circ}(T_{2})\over RT_{2}}$

$ln(Q(T_{1}))=0$

so that;

$ln(Q(T_{2}))=ln(Q(T_{2}))-ln(Q(T_{1}))$

$={c-\Delta G^{\circ}(T_{2})\over RT_{2}}$

$={\Delta G^{\circ}(T_{1})-\Delta G^{\circ}(T_{2})\over RT_{2}}$

$={1\over R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}-c\over T^{2}}dT$

$={-1\over R}(\Delta H^{\circ}-c)({1\over T_{2}}-{1\over T_{1}})$

and rearranging;

${\Delta G^{\circ}(T_{2})-\Delta G^{\circ}(T_{1})\over T_{2}}=(\Delta
H^{\circ}-c)({1\over T_{2}}-{1\over T_{1}})$

$=(\Delta H^{\circ}-\Delta G^{\circ}(T_{1}))({1\over T_{2}}-{1\over T_{1}})$

so that, rearranging again;

$\Delta G^{\circ}(T_{1})({1\over T_{1}}+{1\over T_{2}}-{1\over T_{2}})={\Delta
G^{\circ}(T_{1})\over T_{1}}$

$={\Delta G^{\circ}(T_{2})\over T_{2}}-\Delta H^{\circ}({1\over T_{2}}-{1\over
T_{1}})$

to obtain;

$\Delta G^{\circ}(T_{1})={T_{1}\over T_{2}}\Delta G^{\circ}(T_{2})-\Delta
H^{\circ}({T_{1}\over T_{2}}-1)$

∎

######  Lemma 2.9.

If there exists a component $D_{c}$, $c\in\mathcal{R}$, which projects onto a
closed bounded subinterval $I$ of the line $P=P^{\circ}$, not containing $0$,
and intersects $P=P^{\circ}$ at $(T_{1},P^{\circ})$, with $T_{1}>0$, then, for
$T_{2}\in I$, $\Delta G^{\circ}$ is linear, with;

$\Delta G^{\circ}(T_{2})=T_{2}({(\Delta G^{\circ}(T_{1})-\Delta
H^{\circ})\over T_{1}})+\Delta H^{\circ}$

for $T_{2}\in I$. If $\epsilon\neq 0$, have that;

$({dG\over d\xi})_{T,P}=\lambda+\epsilon ln(P)+\beta T$

where $\\{\lambda,\epsilon,\beta\\}\subset\mathcal{R}$ and
$\\{\beta,\epsilon\\}$ can be effectively determined, and we have that the
activity coefficient is given by;

$Q(T_{2},P^{\prime})=e^{{\epsilon ln({P^{\prime}\over P^{\prime\circ}})\over
RT_{2}}}$

and the dynamic equilibrium paths are given by;

$({P^{\prime}\over P^{\prime\circ}})^{\epsilon\over RT_{2}}=c$

for $c\in\mathcal{R}_{\geq 0}$, see Definition 2.1, while the quasi-chemical
equilibrium paths are given by;

$\lambda+\epsilon ln(P^{\prime})+\beta T_{2}=c$

for $c\in\mathcal{R}$.

If $\epsilon=0$;

$({dG\over d\xi})_{T,P}=\lambda+\beta T+\sigma ln(T)$

where $\\{\lambda,\beta,\sigma\\}\subset\mathcal{R}$, and $\\{\beta,\sigma\\}$
can be effectively determined. For every $T_{1}>0$, there exists a straight
line feasible chemical path $\gamma$ with $pr_{12}(\gamma)\subset(T=T_{1})$,
which is both a dynamic and quasi-chemical equilibrium path, and $Q(T,P)=1$.

###### Proof.

For the first claim, by Lemma 2.8, we have that;

$\Delta G^{\circ}(T_{2})={T_{2}\over T_{1}}\Delta G^{\circ}(T_{1})-\Delta
H^{\circ}({T_{2}\over T_{1}}-1)$

$=T_{2}({(\Delta G^{\circ}(T_{1})-\Delta H^{\circ})\over T_{1}})+\Delta
H^{\circ}$

For the next claim, by Lemma 2.7 and the proof of Lemma 2.8, we have that;

$({\partial({dG\over d\xi})_{T,P}\over\partial
T})_{P}=({\partial(\sum_{i=1}^{c}\nu_{i}\mu_{i})\over\partial T})_{P}$

$=\sum_{i=1}^{c}\nu_{i}({\partial\mu_{i}\over\partial T})_{P}$

$=\sum_{i=1}^{c}\nu_{i}({\partial\mu_{i}\over\partial T})_{P,n}$

$=-\sum_{i=1}^{c}\nu_{i}\overline{S}_{m,i}$ $(*)$

To compute $\overline{S}_{m,i}$, we have by the first law of thermodynamics;

$dQ=dU+dL=dU+pdV$

where $L$ is the work done by the system, see [5]. We can assume that the
liquid mixture is in thermal equilibrium with a mixture of ideal gases in the
vapour phase, and using the ideal gas law, the definition of temperature for
ideal gases, obtain the calculation of internal energy for the mixture;

$U(T,P,n_{1},\ldots,n_{c})=\sum_{i=1}^{c}({3\over 2}N_{A}n_{i}kT-
N_{A}n_{i}m_{i}\rho_{i})$

where $m_{i}$ is the molecular mass of species $i$, $\rho_{i}$ is the specific
latent heat of evaporation of species $i$, which we assume is independent of
temperature $T$. By a result in [11], using the fact that entropy difference
is independent of path, see [9], we have that $Q$ is independent of $P$. We
then have;

$dU=\sum_{i=1}^{c}{3\over 2}N_{A}kTdn_{i}+\sum_{i=1}^{c}{3\over
2}N_{A}kn_{i}dT-\sum_{i=1}^{c}N_{A}m_{i}\rho_{i}dn_{i}$

$dQ=\sum_{i=1}^{c}{3\over 2}N_{A}kTdn_{i}+\sum_{i=1}^{c}{3\over
2}N_{A}kn_{i}dT-\sum_{i=1}^{c}N_{A}m_{i}\rho_{i}dn_{i}+dL$

${dQ\over T}=\sum_{i=1}^{c}{3\over 2}N_{A}kdn_{i}+\sum_{i=1}^{c}{3\over
2}N_{A}kn_{i}{dT\over T}-\sum_{i=1}^{c}N_{A}m_{i}\rho_{i}{dn_{i}\over
T}+{g(T,\overline{n})dT\over T}$

$+\sum_{i=1}^{c}h_{i}(T,\overline{n}){dn_{i}\over T}$

$({dQ\over T})_{n^{\prime},T,P}={3\over
2}N_{A}kdn_{i}-N_{A}m_{i}\rho_{i}{dn_{i}\over
T}+h_{i}(T,\overline{n}){dn_{i}\over T}$

It follows that;

$\overline{S}_{m,i}=\int_{\Delta n_{i}=1}({dQ\over
T})_{n^{\prime},T,P}={3\over 2}N_{A}k-{N_{A}m_{i}\rho_{i}\over
T}+{k_{i}(T)\over T}$ $(**)$

So that, from $(*)$;

$({\partial({dG\over d\xi})_{T,P}\over\partial
T})_{P}=-\sum_{i=1}^{c}\nu_{i}({3\over 2}N_{A}k-{N_{A}m_{i}\rho_{i}\over
T})-\sum_{i=1}^{c}\nu_{i}{k_{i}(T)\over T}$

$=-{3\over 2}N_{A}k(\sum_{i=1}^{c}\nu_{i})+{N_{A}\over
T}\sum_{i=1}^{c}\nu_{i}\mu_{i}\rho_{i}-\sum_{i=1}^{c}\nu_{i}{k_{i}(T)\over T}$

$=-{3\over 2}N_{A}k(\sum_{i=1}^{c}\nu_{i})+{N_{A}\over
T}\sum_{i=1}^{c}\nu_{i}\mu_{i}\rho_{i}-{G(T)\over T}$ $(***)$

From $(***)$, which is uniform $P$, we see that $({dG\over d\xi})_{T,P}$ is of
the form $\alpha(P)+\beta T+\gamma ln(T)-\int{G(T)\over T}$, $(B)$, where
$\\{\beta,\gamma\\}\subset\mathcal{R}$, and, assuming that $({dG\over
d\xi})_{T,P}$ is differentiable, $\alpha\in C^{1}(\mathcal{R})$. By a similar
calculation, we have that;

$({\partial({dG\over d\xi})_{T,P}\over\partial
P})_{T}=({\partial(\sum_{i=1}^{c}\nu_{i}\mu_{i})\over\partial P})_{T}$

$=\sum_{i=1}^{c}\nu_{i}({\partial\mu_{i}\over\partial P})_{T}$

$=\sum_{i=1}^{c}\nu_{i}({\partial\mu_{i}\over\partial P})_{T,n}$

$=\sum_{i=1}^{c}\nu_{i}({\partial V\over\partial n_{i}})_{T,P,n^{\prime}}$

$=\sum_{i=1}^{c}\nu_{i}\overline{V}_{i}$

$=\sum_{i=1}^{c}\nu_{i}{N_{A}m_{i}\over\kappa_{i}(T,P)}$ $(A)$

where $\kappa_{i}$ is the density of substance $i$. We also have that;

$P(\sum_{i=1}^{c}\nu_{i}{N_{A}m_{i}\over\kappa_{i}(T,P)})=P(\sum_{i=1}^{c}\nu_{i}\overline{V}_{i})=G(T)$
$(dL=PdV)$ $(C)$

and from $(A),(B),(C)$;

$P({\partial({dG\over d\xi})_{T,P}\over\partial P})_{T}=G(T)$

$=P\alpha^{\prime}(P)$

so that $G(T)=\epsilon$, $\alpha(P)=\lambda+\epsilon ln(P)$

$({dG\over d\xi})_{T,P}$ is of the form;

$\alpha(P)+\beta T+\gamma ln(T)-\int{G(T)\over T}$

$=\lambda+\epsilon ln(P)+\beta T+\gamma ln(T)-\epsilon ln(T)$

$=\lambda+\epsilon ln(P)+\beta T+\sigma ln(T)$ $(D)$

where $\sigma=\gamma-\epsilon$,
$\\{\beta,\epsilon,\lambda,\sigma\\}\subset\mathcal{R}$.

If $\epsilon=0$, then $({dG\over d\xi})_{T,P}$ is independent of $P$, and the
components $D_{c}$ are all straight line paths. In this case, if $D_{c}$
intersects the line $P=P^{\circ}$ at $(T_{1},P^{\circ})$, then, for all $P>0$;

$c=\Delta G^{\circ}(T_{1})+RT_{1}ln(Q(T_{1},P)$

$=\Delta G^{\circ}(T_{1})$

implies that $RT_{1}ln(Q(T_{1},P)=0$, so that $Q(T_{1},P)=1$ and, by Lemmas
2.14 and 2.16, there exists a straight line feasible chemical path $\gamma$
with $pr_{12}(\gamma)\subset(T=T_{1})$, which is both a dynamic and quasi-
chemical equilibrium path. From $(D)$, we have that;

$({dG\over d\xi})_{T,P}=\lambda+\beta T+\sigma ln(T)$

We have that, for any $T_{1}>0$, we can find $c\in\mathcal{R}$ with
$\lambda+\beta T_{1}+\sigma ln(T_{1})=c$, so that $D_{c}$ defines a component
straight line path passing through $(T_{1},P^{\circ})$. Then we can apply the
previous result.

If $\epsilon\neq 0$, for any $c\in\mathcal{R}$, we can solve the equation;

$\lambda+\epsilon ln(P)+\beta T+\sigma ln(T)=c$

for any given $T>0$ and an appropriate choice of $P(T)$. In particularly,
there exists a component $D_{c}$ projecting onto the line $P=P^{0}$ and, by
the first part, $\Delta G^{\circ}$ is linear, with;

$\Delta G^{\circ}(T_{2})=T_{2}({(\Delta G^{\circ}(T_{1})-\Delta
H^{\circ})\over T_{1}})+\Delta H^{\circ}$

for an intersection at $(T_{1},P^{\circ})$. We also have, using $(D)$, that;

$\Delta G^{\circ}(T_{2})=({\partial
G\over\partial\xi})_{T,P}(T_{2},P^{\circ})$

$=\lambda+\epsilon ln(P^{\circ})+\beta T_{2}+\sigma ln(T_{2})$

so that, equating coefficients;

$\sigma=0$

$\lambda+\epsilon ln(P^{\circ})=\Delta H^{\circ}$

$\beta={(\Delta G^{\circ}(T_{1})-\Delta H^{\circ})\over T_{1}}$

$\Delta G^{\circ}(T_{1})=\beta T_{1}+\Delta H^{\circ}$

$\Delta G^{\circ}(T_{2})=\beta T_{2}+\Delta H^{\circ}$

We can then, using Lemma 2.5, obtain a formula for the activity coefficient;

$Q(T_{2},P^{\prime})=e^{{(({\partial
G\over\partial\xi})_{T,P}|_{T_{2},P^{\prime}}-\Delta G^{\circ}(T_{2}))\over
RT_{2}}}$

$e^{{(\Delta H^{\circ}-\epsilon ln(P^{\prime\circ})+\epsilon
ln(P^{\prime})+\beta T_{2}-(\beta T_{2}+\Delta H^{\circ}))\over RT_{2}}}$

$=e^{{\epsilon ln({P^{\prime}\over P^{\prime\circ}})\over RT_{2}}}$ $(W)$

as required. The claim about the coefficients being determined is clear from
the proof. The determination of the dynamical and quasi-chemical equilibrium
lines, see Definitions 2.1 and Lemma 2.17, follows from a simple rearrangement
of the formulas $Q(T_{1},P^{\prime})=c$, for some $c\in\mathcal{R}_{\geq 0}$,
using $(W)$, and $({dG\over d\xi})_{T,P}=c$, for some $c\in\mathcal{R}$, using
$(D)$, with $\sigma=0$.

∎

######  Lemma 2.10.

With notation as in Lemma 2.9, if $\epsilon=0$, then if either;

$(i)$. $\beta>0,\sigma>0$

$(ii)$. $\beta<0,\sigma<0$

$(iii)$. $\beta>0,\sigma<0,\lambda-\sigma+\sigma ln({-\sigma\over\beta})\leq
0$

$(iv)$. $\beta<0,\sigma>0,\lambda-\sigma+\sigma ln({-\sigma\over\beta})\geq 0$

we can always find $T_{0}>0$, defined as the solution to $\lambda+\beta
T+\sigma ln(T)=0$, such that $T=T_{0}$ defines a chemical equilibrium line.

###### Proof.

By the proof of Lemma 2.9, if $T_{0}$ is a solution to;

$\lambda+\beta T+\sigma ln(T)=0$ $(*)$

then $({\partial G\over\partial\xi})_{T,P}|_{T=T_{0}}=0$

so that $T=T_{0}$ defines a chemical equilibrium line. By considering limits
at $\infty$ and noting that the derivative $\beta+{\sigma\over T}$ of
$\lambda+\beta T+\sigma ln(T)$, is of a fixed sign in cases $(i),(ii)$, so
that $\lambda+\beta T+\sigma ln(T)$ is monotonic, we can see there does exist
a unique solution $T_{0}$ in cases $(i),(ii)$. In cases $(iii),(iv)$,
computing limits at $\infty$ again, and noting that $\beta+{\sigma\over T}$ is
increasing/decreasing, we have that, if the minimum/maximum $T_{1}$
respectively of $\lambda+\beta T+\sigma ln(T)$, given by the solution to;

$\beta+{\sigma\over T}=0$

so that $T_{1}={-\sigma\over\beta}$, then, in case $(iii)$, if;

$\lambda+\beta(T_{1})+\sigma ln(T_{1})<0$

iff $\lambda-\sigma+\sigma ln({-\sigma\over\beta})<0$

there exist two possible solutions $T_{0}$, with a unique solution if equality
holds. Similarly, then, in case $(iv)$, if;

$\lambda-\sigma+\sigma ln({-\sigma\over\beta})>0$

there exist at least two possible solutions $T_{0}$, with again a unique
solution if equality holds.

∎

######  Lemma 2.11.

If $\epsilon=0$, we have, for all $T_{1}>0$, that;

$({\partial G\over\partial\xi})_{T,P}|_{(T_{1},P_{1})}=({\partial
G\over\partial\xi})_{T,P}|_{(T_{1},P_{1}^{\circ})}$

iff;

$E(T_{1},P_{1})=E(T_{1},P_{1}^{\circ})=E^{\circ}(T_{1})$

where $G$ is the Gibbs energy function for the charged and uncharged species.

###### Proof.

By $({\dagger})$ of Lemma 2.4, we have that;

$({\partial G\over\partial\xi})_{T,P}=({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}+2EF$ $(*)$

By Lemma 2.9, we have that $({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}$ is independent of $P$, in
particularly, we have that;

$({\partial G_{chem^{\prime}}\over\partial\xi})_{T_{1},P_{1}}=({\partial
G_{chem^{\prime}}\over\partial\xi})_{T_{1},P_{1}^{\circ}}$ $(**)$

so that, combining $(*),(**)$, we obtain the result. ∎

######  Lemma 2.12.

We have, for all $T_{1}>0,P_{1}>0$, that;

$2F(E(T_{1},P_{1})-E^{\circ}(T_{1}))=({\partial
G\over\partial\xi})_{T,P}|_{(T_{1},P_{1})}-({\partial
G\over\partial\xi})_{T,P}|_{(T_{1},P_{1}^{\circ})}-RT_{1}ln(Q(T_{1},P_{1}))$

###### Proof.

Following the proof of Lemma 2.4, we have that;

$({\partial G\over\partial\xi})_{T,P}|_{T_{1},P_{1}}=({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}|_{T_{1},P_{1}}+2E(T_{1},P_{1})F$
$(*)$

$2E^{\circ}(T_{1})F=({\partial
G\over\partial\xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}|_{T_{1},P_{1}^{\circ}}$

$=({\partial G\over\partial\xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-\Delta
G^{\circ}_{chem^{\prime}}(T_{1})$ $(**)$

so from $(*),(**)$ and Lemma 2.5;

$2E(T_{1},P_{1})F=({\partial
G\over\partial\xi})_{T,P}|_{T_{1},P_{1}}-({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}|_{T_{1},P_{1}}$

$=({\partial G\over\partial\xi})_{T,P}|_{T_{1},P_{1}}-(\Delta
G^{\circ}_{chem^{\prime}}(T_{1})+RT_{1}ln(Q_{chem^{\prime}}(T_{1},P_{1})))$

$2E(T_{1},P_{1})F-2E^{\circ}(T_{1})F=({\partial
G\over\partial\xi})_{T,P}|_{T_{1},P_{1}}-(\Delta
G^{\circ}_{chem^{\prime}}(T_{1})+RT_{1}ln(Q_{chem^{\prime}}(T_{1},P_{1})))$

$-(({\partial G\over\partial\xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-\Delta
G^{\circ}_{chem^{\prime}}(T_{1}))$

$=({\partial G\over\partial\xi})_{T,P}|_{T_{1},P_{1}}-(({\partial
G\over\partial\xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-RT_{1}ln(Q_{chem^{\prime}}(T_{1},P_{1}))$

∎

######  Definition 2.13.

Ideal Solution

We let $pr_{1}$ is the projection onto the first factor, $pr_{12}$ be the
projection onto the first two factors, in coordinates
$(T,P,n_{1},\ldots,n_{c})$. We define a feasible chemical path
$\gamma:[0,1]\rightarrow\mathcal{R}_{>0}^{2+c}$, such that if
$n_{i}(t)=pr_{2+i}(t)$, for $1\leq i\leq c$, then;

${n_{i}^{\prime}\over\nu_{i}}={n_{j}^{\prime}\over\nu_{j}}$, for $1\leq
i<j\leq c$

where $\\{\nu_{1},\ldots,\nu_{c}\\}$ are the stoichiometric coefficients. If
$n(t)=\sum_{i=1}^{c}n_{i}(t)$, and $x_{i}(t)=a_{i}(t)={n_{i}\over n}(t)$, then
$Q(pr_{12}(\gamma(t)))=\prod_{i=1}^{c}a_{i}(t)^{\nu_{i}}$ and ${n_{i,0}\over
n_{0}}=f_{i}(pr_{12}(\gamma(0)))$, where
$f_{i}=e^{\mu_{i}-\mu_{i}^{\circ}\over RT}$, $1\leq i\leq c$, see Section 13.
Note that $n>0$ and the $x_{i}$ are well defined.

We define a chemical equilibrium path to be a feasible chemical path $\gamma$
with the additional property that chemical equilibrium exists at
$pr_{12}(\gamma(t))$, for $t\in[0,1]$. We define a quasi-chemical equilibrium
path to be a feasible chemical path $\gamma$ with the additional property that
$({\partial G\over\partial\xi})_{T,P}|_{pr_{12}([0,1])}$ is constant. We
define a dynamic equilibrium path to be a feasible chemical path $\gamma$ with
the additional property that $pr_{2+c}(\gamma)^{\prime}(t)=0$, for $1\leq
i\leq c$.

We define a straight line feasible path from $(T,P^{\circ})$ to $(T,P)$ to be
a map $\gamma:[0,1]\rightarrow\mathcal{R}_{\geq 0}^{2+c}$ such that
$pr_{12}\gamma(0)=(T,P^{\circ})$, $pr_{12}\gamma(1)=(T,P)$,
$pr_{1}(\gamma(t))=T$. We say that a point $(T,P)$ is a simple dynamic
equilibrium point, if it lies on the locus $Q(T,P)=1$.

######  Lemma 2.14.

In an ideal solution, a straight line chemical equilibrium path from
$(T,P^{\circ})$ to $(T,P)$ is a dynamic equilibrium path. Every
$(T,P^{\circ})$ is a simple dynamic equilibrium point.

###### Proof.

By Lemma 2.7, and the definition of activities for an ideal solution, we have
that;

$1=Q(T,P)=\prod_{i=1}^{c}a_{i}^{\nu_{i}}=\prod_{i=1}^{c}x_{i}^{\nu_{i}}$ $(*)$

and ${n_{i}^{\prime}\over\nu_{i}}={n_{j}^{\prime}\over\nu_{j}}$, for $1\leq
i,j\leq c$, $\sum_{i=1}^{c}n_{i}=n$

Using the relation $(*)$, differentiating and using the facts that, for $1\leq
i\leq c-1$;

$n_{i}^{\prime}={\nu_{i}n_{c}^{\prime}\over\nu_{c}}$,
$n_{i}={\nu_{i}n_{c}\over\nu_{c}}+d_{i}$ $(!)$

we obtain that;

$(\prod_{i=1}^{c}x_{i}^{\nu_{i}})^{\prime}=\sum_{i=1}^{c}\nu_{i}x_{i}^{\nu_{i}-1}x_{i}^{\prime}\prod_{j\neq
i}x_{j}^{\nu_{j}}$

$=\sum_{i=1}^{c}\nu_{i}x_{i}^{\nu_{i}-1}x_{i}^{\prime}x_{i}^{-\nu_{i}}$

$=\sum_{i=1}^{c}\nu_{i}x_{i}^{-1}x_{i}^{\prime}$

$=\sum_{i=1}^{c}\nu_{i}{n\over n_{i}}{(n_{i}^{\prime}n-n_{i}n^{\prime})\over
n^{2}}$

$=\sum_{i=1}^{c}\nu_{i}({n_{i}^{\prime}\over n_{i}}-{n^{\prime}\over n})$

$=\sum_{i=1}^{c-1}{\nu_{i}^{2}n_{c}^{\prime}\over\nu_{i}n_{c}+\nu_{c}d_{i}}+{\nu_{c}n_{c}^{\prime}\over
n_{c}}-\lambda({\sum_{i=1}^{c}n_{i}^{\prime}\over\sum_{i=1}^{c}n_{i}})$

$=\sum_{i=1}^{c-1}{\nu_{i}^{2}n_{c}^{\prime}\over\nu_{i}n_{c}+\nu_{c}d_{i}}+{\nu_{c}n_{c}^{\prime}\over
n_{c}}-\lambda({(\sum_{i=1}^{c-1}{\nu_{i}\over\nu_{c}}+1)n_{c}^{\prime}\over(\sum_{i=1}^{c-1}{\nu_{i}\over\nu_{c}}+1)n_{c}+\sum_{i=1}^{c-1}d_{i}})$

$=0$ $(B)$

where $\lambda=\sum_{i=1}^{c}\nu_{i}$. If $\nu_{c}^{\prime}\neq 0$, we obtain
that;

$=\sum_{i=1}^{c-1}{\nu_{i}^{2}\over\nu_{i}n_{c}+\nu_{c}d_{i}}+{\nu_{c}\over
n_{c}}-\lambda({(\sum_{i=1}^{c-1}{\nu_{i}\over\nu_{c}}+1)\over(\sum_{i=1}^{c-1}{\nu_{i}\over\nu_{c}}+1)n_{c}+\sum_{i=1}^{c-1}d_{i}})=0$

so that;

$\sum_{i=1}^{c-1}\nu_{i}^{2}n_{c}(\alpha n_{c}+\beta)\prod_{j\neq
i}(\nu_{j}n_{c}+\nu_{c}d_{j})+\nu_{c}(\alpha
n_{c}+\beta)\prod_{i=1}^{c-1}(\nu_{i}n_{c}+\nu_{c}d_{i})$

$-\lambda n_{c}\prod_{i=1}^{c-1}(\nu_{i}n_{c}+\nu_{c}d_{i})=0$

where $\alpha=\sum_{i=1}^{c-1}{\nu_{i}\over\nu_{c}}+1$ and
$\beta=\sum_{i=1}^{c-1}d_{i}$

which we can write in the form;

$\sum_{j=0}^{c}\gamma_{j}\nu_{c}^{j}=0$

We have that;

$\gamma_{c}=\sum_{i=1}^{c-1}\nu_{i}^{2}\alpha\prod_{j\neq
i}\nu_{j}+\alpha\nu_{c}\prod_{i=1}^{c-1}\nu_{i}-\lambda\prod_{i=1}^{c-1}\nu_{i}$

$=\alpha\delta\sum_{i=1}^{c-1}\nu_{i}+\alpha\delta\nu_{c}-\lambda\delta$

$=\delta(\alpha(\sum_{i=1}^{c}\nu_{i})-\lambda)$

$=\delta\lambda(\alpha-1)$

Noting that $\delta\neq 0$ and
$\alpha-1=\sum_{i=1}^{c-1}{\nu_{i}\over\nu_{c}}$, we have that $\gamma_{c}\neq
0$ iff $\sum_{i=1}^{c-1}\nu_{i}\neq 0$ and $\sum_{i=1}^{c}\nu_{i}\neq 0$. In
this case, we obtain a nontrivial polynomial relation $p(n_{c})=0$, so that,
by continuity and discreteness of roots, $n_{c}$ is a constant and
$n_{c}^{\prime}=0$. By the connecting relations $(!)$, we obtain that
$n_{j}^{\prime}=0$ as well, for $\leq j\leq c-1$.

If $\sum_{i=1}^{c}\nu_{i}=0$ $(G)$ then $\sum_{i=1}^{c-1}\nu_{i}\neq 0$, and,
we have, from $(*)$, that;

$\prod_{i=1}^{c}n_{i}^{\nu_{i}}=n^{\sum_{i=1}^{c}\nu_{i}}=1$ $(A)$

and, following the calculation $(B)$, we obtain;

$\sum_{i=1}^{c-1}{\nu_{i}^{2}n_{c}^{\prime}\over\nu_{i}n_{c}+\nu_{c}d_{i}}+{\nu_{c}n_{c}^{\prime}\over
n_{c}}=0$

so that, if $n_{c}^{\prime}\neq 0$;

$\sum_{i=1}^{c-1}{\nu_{i}^{2}\over\nu_{i}n_{c}+\nu_{c}d_{i}}+{\nu_{c}\over
n_{c}}=0$

and, differentiating $k$ times, for $k\geq 0$, cancelling $n_{c}^{\prime}$ if
$n_{c}^{\prime}\neq 0$, and using the chain rule, we obtain the relations;

$\sum_{i=1}^{c-1}{\nu_{i}^{k+2}\over(\nu_{i}n_{c}+\nu_{c}d_{i})^{k+1}}+{\nu_{c}\over
n_{c}^{k+1}}=0$

Let $g_{i}={1\over\nu_{i}+{\nu_{c}d_{i}\over
n_{c}}}={n_{c}\over\nu_{c}n_{i}}<0$, for $1\leq i\leq c-1$. Then, for $k\geq
0$;

$\sum_{i=1}^{c-1}\nu_{i}^{k+2}g_{i}^{k+1}+\nu_{c}=\sum_{i=1}^{c-1}\nu_{i}(\nu_{i}g_{i})^{k+1}+\nu_{c}$

$=0$ $(E)$

If $g_{i}=-1$, then;

${n_{c}\over\nu_{c}n_{i}}={n_{c}\over\nu_{c}({\nu_{i}n_{c}\over\nu_{c}}+d_{i})}={n_{c}\over\nu_{i}n_{c}+\nu_{c}d_{i}}=-1$

so that;

$n_{c}=-(\nu_{i}n_{c}+\nu_{c}d_{i})$

$(1+\nu_{i})n_{c}=\nu_{c}d_{i}$

Then if $n_{c}^{\prime}\neq 0$, we must have that $\nu_{i}=-1$, $d_{i}=0$.
Rescaling each $\nu_{i}$ by a factor of $2$, we still have the conditions
$(*)$, $(E)$,$(G)$, so we can assume that $|\nu_{i}|\geq 2$, which is a
contradiction. Hence, we can assume that $|g_{i}|\neq 1$, for $1\leq i\leq
c-1$, so that taking the limit as $k\rightarrow\infty$, with $k$ even, so that
$\nu_{i}^{k+2}>0$, $g_{i}^{k+1}<0$, for $1\leq i\leq c-1$, we obtain a
contradiction, and conclude that $n_{c}^{\prime}=0$, and $n_{i}^{\prime}=0$,
for $1\leq i\leq c-1$.

If for every $c-1$ element subset $I_{j}\subset\\{\nu_{1},\ldots\nu_{c}\\}$,
$1\leq j\leq c$, we have that $\sum_{i\in I_{j}}\nu_{i}=0$, then clearly;

$\sum_{1\leq i\leq c}\nu_{i}=\sum_{i\in I_{j}}\nu_{i}+\nu_{j}$

$=\nu_{j}$ for $1\leq j\leq c$

which we can exclude. It follows that there exists some $j_{0}$, with $1\leq
j_{0}\leq c$, such that $\sum_{i\in I_{j_{0}}}\nu_{i}\neq 0$. Using
$n_{j_{0}}$ as the pivot and following the above proof, replacing $n_{c}$ by
$n_{j_{0}}$, we can, without loss of generality, assume that
$\sum_{i=1}^{c-1}\nu_{i}\neq 0$, and the proof is complete.

The second claim follows from the fact in Lemma 2.7 that $Q(T,P^{\circ})=1$.

∎

######  Definition 2.15.

For $c\in\mathcal{R}_{>0}$, we define $C_{c}\subset\mathcal{R}^{2}$ to be the
zero locus of $Q(T,P)-c$. We define $D\subset\mathcal{R}^{2}$ to be the
condition of chemical equilibrium. We define $D_{c}$ to be the zero locus of
$({\partial G\over\partial\xi})_{T,P}-c=0$ .

######  Lemma 2.16.

For every smooth curve $W\subset\mathcal{R}^{2}$, there exists a locally
feasible path $\gamma:[0,1]\rightarrow\mathcal{R}^{2+c}$ with
$pr_{12}(\gamma)\subset W$.

###### Proof.

As $W$ is smooth, we can choose a local parametrisation
$\delta:[0,1]\rightarrow W$. Let $\epsilon(t)=Q(\delta(t))>0$ and
$w=\sum_{i=1}^{c}\nu_{i}$. Without loss of generality, we can assume that
$pr_{12}(\gamma(0))=(T_{0},P_{0})$, see Section 13. We have that;

$\prod_{i=1}^{c}x_{i}^{\nu_{i}}(t)=\epsilon(t)$ $({\dagger})$

iff
$\prod_{i=1}^{c}n_{i}^{\nu_{i}}(t)=\epsilon(t)n^{\sum_{i=1}^{c}\nu_{i}}(t)$

iff
$\prod_{i=1}^{c-1}n_{i}^{\nu_{i}}(t)n_{c}^{\nu_{c}}(t)=\epsilon(t)(\sum_{i=1}^{c}n_{i})^{\sum_{i=1}^{c}\nu_{i}}(t)$

iff
$\prod_{i=1}^{c-1}({\nu_{i}\over\nu_{c}}n_{c}+d_{i})^{\nu_{i}}(t)n_{c}^{\nu_{c}}(t)=\epsilon(t)(\sum_{i=1}^{c}n_{i})^{\sum_{i=1}^{c}\nu_{i}}(t)$

iff
$\prod_{i=1}^{c-1}({\nu_{i}\over\nu_{c}}n_{c}+d_{i})^{\nu_{i}}(t)n_{c}^{\nu_{c}}(t)=\epsilon(t)((\sum_{i=1}^{c-1}{\nu_{i}\over\nu_{c}}+1)n_{c}+\sum_{i=1}^{c-1}d_{i})^{w}(t)$

Let $d_{i}>0$, for $1\leq i\leq c-1$, then;

$\prod_{i=1}^{c}x_{i}^{\nu_{i}}(t)=\epsilon(t)$

iff
$\prod_{i=1}^{c-1}({\nu_{i}\over\nu_{c}}n_{c}+d_{i})^{\nu_{i}}(t)n_{c}^{\nu_{c}}(t)=\epsilon(t)((\sum_{i=1}^{c-1}{\nu_{i}\over\nu_{c}}+1)n_{c}+\sigma)^{w}(t)$

iff
$\prod_{i=1}^{c-1}({\nu_{i}\over\nu_{c}}n_{c}+d_{i})^{\nu_{i}}(t)n_{c}^{\nu_{c}}(t)=\epsilon(t)({w\over\nu_{c}}n_{c}+\sigma)^{w}(t)$

where $\sigma=\sum_{i=1}^{c-1}d_{i}>0$. Assume that $w>0$, then we have that;

$\prod_{i=1}^{c}x_{i}^{\nu_{i}}(t)=\epsilon(t)$

iff
$\prod_{i=1}^{p}({\nu_{i}\over\nu_{c}}n_{c}+d_{i})^{\nu_{i}}=\epsilon(t)({w\over\nu_{c}}n_{c}+\sigma)^{w}(t)n_{c}^{-\nu_{c}}\prod_{i=p+1}^{c-1}({\nu_{i}\over\nu_{c}}n_{c}+d_{i})^{-\nu_{i}}$

iff
$\prod_{i=1}^{p}({\nu_{i}\over\nu_{c}})^{\nu_{i}}n_{c}^{\kappa}+r(n_{c})=\epsilon(t)[({w\over\nu_{c}})^{w}\prod_{i=p+1}^{c-1}({\nu_{i}\over\nu_{c}})^{-\nu_{i}}n_{c}^{w-\nu_{c}-\lambda}+s(n_{c})]$

iff
$\prod_{i=1}^{p}({\nu_{i}\over\nu_{c}})^{\nu_{i}}n_{c}^{\kappa}+r(n_{c})=\epsilon(t)[({w\over\nu_{c}})^{w}\prod_{i=p+1}^{c-1}({\nu_{i}\over\nu_{c}})^{-\nu_{i}}n_{c}^{\kappa}+s(n_{c})]$

iff $\alpha n_{c}^{\kappa}+r(n_{c})=\epsilon(t)(\beta
n_{c}^{\kappa}+s(n_{c}))$ $({\dagger}{\dagger}{\dagger})$

where $\alpha=\prod_{i=1}^{p}({\nu_{i}\over\nu_{c}})^{\nu_{i}}\neq 0$,
$\beta=({w\over\nu_{c}})^{w}\prod_{i=p+1}^{c-1}({\nu_{i}\over\nu_{c}})^{-\nu_{i}}\neq
0$, $\lambda=\sum_{i=p+1}^{c-1}\nu_{i}$, $\\{r,s\\}\subset\mathcal{R}[x]$ have
degree less than $\kappa$, $r(0)=\prod_{i=1}^{p}d_{i}>0$, $s(0)=0$, as
divisible by $x^{-\nu_{c}}$. Dividing $({\dagger}{\dagger}{\dagger})$ by
$n_{c}^{\kappa}>0$, we obtain;

$\alpha+r_{1}({1\over n_{c}})=\epsilon(t)(\beta+s_{1}({1\over n_{c}}))$ $(C)$

where $\\{r_{1},s_{1}\\}\subset\mathcal{R}[x]$ have degree $\kappa$, with
$r_{1}(0)=s_{1}(0)=0$, $deg(r_{1})=\kappa$ and
$c_{\kappa}=r(0)=\prod_{i=1}^{p}d_{i}>0$ where
$r_{1}=\sum_{j=0}^{\kappa}c_{j}x^{j}$, $deg(s_{1})\leq\kappa+\nu_{c}<\kappa$,
so that $lim_{x\rightarrow\infty}{\alpha+r_{1}(x)\over\beta+s_{1}(x)}=\infty$.

Let $v(x)=\alpha x^{\kappa}+r(x)$, $w(x)=\beta x^{\kappa}+s(x)$. Then, the
roots $v_{i}$ of $r(x)$, $1\leq i\leq p$, are given by
$v_{i}=-{d_{i}\nu_{c}\over\nu_{i}}>0$, while the roots $w_{i}$ of $w(x)$ are
given by $w_{0}=0$, $w_{i}={-d_{i}\nu_{c}\over\nu_{i}}<0$, $p+1\leq i\leq c-1$
and $w_{c}={-\sigma\nu_{c}\over w}$. If $\nu_{i}=w$, for $1\leq i\leq p$, it
would follow that $w>\sum_{i=1}^{p}\nu_{i}=pw$, which is a contradiction, as
$p\geq 1$. It follows that we can choose $i_{0}$ with $1\leq i_{0}\leq p$ such
that ${-\nu_{c}\over\nu_{i_{0}}}\neq{-\nu_{c}\over w}$ and
${-\nu_{c}d_{i_{0}}\over\nu_{i_{0}}}\neq{-\nu_{c}d_{i_{0}}\over w}$,(111As
$n_{c}$ is mobile, the conditions that
$n_{i,0}=d_{i}+{\nu_{i}\over\nu_{c}}n_{c,0}$, $1\leq i\leq c-1$, $(B)$,
together with the requirement that $\overline{n}_{0}=(n_{1,0},\ldots,n_{c,0})$
lies in $Ker(M)\cap\mathcal{R}_{>0}^{c}$, see Section 13, places a $1$
dimensional restriction on the tuple $(d_{1},\ldots d_{c-1})$, defined as
$pr_{c-1}(W)$, where $W=V\cap g^{-1}(Ker(M))$, $V=\\{(d_{1},\ldots
d_{c-1},n_{c,0}(d_{1},\ldots d_{c-1})):(d_{1},\ldots
d_{c-1})\in\mathcal{R}^{c-1}\\}$, and
$g(x_{1},\ldots,x_{c})=(y_{1},\ldots,y_{c})$ is the morphism defined by;
$y_{c}=x_{c}$
$y_{i}=x_{i}+{\nu_{i}\over\nu_{c}}x_{c}$, $(1\leq i\leq c-1)$
determined by the conditions $(B)$. Moreover, we can assume that
$pr_{c-1}(W)\cap\mathcal{R}^{c-1}\neq\emptyset$. Moving the tuple
$(d_{1},\ldots d_{c-1})$ along $pr_{c-1}(W)$, we can fine $d_{i_{0}}>0$, and
$d_{j}$, $1\leq j\leq c-1$, $j\neq{i_{0}}$, with $d_{j}>0$, so that
${-\nu_{c}d_{i_{0}}\over\nu_{i_{0}}}\neq{-\nu_{c}\sigma\over w}$ and
$v_{i_{0}}$ is the highest root of $r(x)$, so that the roots $w_{i}$ of $w(x)$
do not coincide with the highest root $v_{i_{0}}$ of $r(x)$.). Dividing by
$x^{\kappa}$ doesn’t effect the positive roots $v_{i}$, $1\leq i\leq p$ of
${r(x)\over x^{\kappa}}=\alpha+r_{1}({1\over x})$, and the positive roots of
$\alpha+r_{1}(x)$, are the positive reciprocals $v_{i}^{\prime}={1\over
v_{i}}$, $1\leq i\leq p$. As
$lim_{x\rightarrow\infty}{\alpha+r_{1}(x)\over\beta+s_{1}(x)}=\infty$ and
$\epsilon(0)>0$, we can choose $v_{0}$ with $q(v_{0})=\epsilon(0)$ and
$v_{0}>v_{i}^{\prime}$, for $1\leq i\leq p$, where
$q(x)={\alpha+r_{1}(x)\over\beta+s_{i}(x)}$. Then ${1\over v_{0}}<{1\over
v_{i}^{\prime}}=v_{i}$, for $1\leq i\leq p$, and as ${\nu_{i}\over\nu_{c}}<0$,
${\nu_{i}\over\nu_{c}}{1\over v_{0}}>{\nu_{i}\over\nu_{c}}v_{i}$,
${\nu_{i}\over\nu_{c}}{1\over
v_{0}}+d_{i}>{\nu_{i}\over\nu_{c}}v_{i}+d_{i}=0$, for $1\leq i\leq p$. In
particularly, as $n_{c}(0)={1\over v_{0}}$, we have, by the linking relations,
that
$n_{i}(0)={\nu_{i}\over\nu_{c}}n_{c}(0)+d_{i}={\nu_{i}\over\nu_{c}}{1\over
v_{0}}+d_{i}>0$, for $1\leq i\leq p$. Moreover, as ${\nu_{i}\over\nu_{c}}>0$,
for $i+1\leq i\leq c-1$, and ${1\over v_{0}}>0$, we have that
$n_{i}(0)={\nu_{i}\over\nu_{c}}n_{c}(0)+d_{i}={\nu_{i}\over\nu_{c}}{1\over
v_{0}}+d_{i}>0$ as well, for $i+1\leq i\leq p-1$, $(D)$

By $(C)$, we have that $q({1\over n_{c}})=\epsilon(t)$ and, by the above
construction, it follows that $q(v_{0})=\epsilon(0)$. Consider the real
algebraic curve defined by $\theta(x,y)=q(x)-y-\epsilon(0)$, so that
$\theta(v_{0},0)=0$. Computing the differential $(q^{\prime}(x),-1)$, if
$q^{\prime}(v_{0})\neq 0$, we see that the projection $pr_{y}$ is unramified
at $(v_{0},0)$, so that we can apply the inverse function theorem, see [10],
to obtain a real branch $\gamma(y)$ with $\gamma(0)=v_{0}$ and;

$\theta(\gamma(y),y)=q(\gamma(y))-y-\epsilon(0)=0$

Replacing $y$ with $\epsilon(t)-\epsilon(0)$, and letting
$\delta(t)=\gamma(\epsilon(t)-\epsilon(0))$, we have that;

$\theta(\delta(t),\epsilon(t)-\epsilon(0))=q(\delta(t))-(\epsilon(t)-\epsilon(0))-\epsilon(0)=q(\delta(t))-\epsilon(t)=0$

Then $\delta(t)>0$, and we can set $n_{c}={1\over\delta(t)}$, with
$n_{c}(0)={1\over v_{0}}$ Using the linkage relations, we can define
$n_{i}(t)$, for $1\leq i\leq c-1$ from $n_{c}$. As, by $(D)$, we have that
$n_{i}(0)>0$, for $1\leq i\leq c$, by continuity, for sufficiently small $t$,
we have that $n_{i}(t)>0$, for $1\leq i\leq c$ as well.

If $w<0$, we can take the reciprocal of the relation $({\dagger})$, replace
$\nu_{i}$ by $-\nu_{i}$ and $\epsilon(t)$ by ${1\over\epsilon(t)}>0$, to get
$w>0$. Reordering so that the pivot $\nu_{c}<0$, $\nu_{i}>0$, for $1\leq i\leq
p^{\prime}$, $\nu_{i}<0$, for $p^{\prime}+1\leq i\leq c-1$, we can carry out
the above proof to get the result.

If $w=0$, we can carry out the first calculation with $\beta$ replaced by
$\prod_{i=p+1}^{c-1}({\nu_{i}\over\nu_{c}})^{-\nu_{i}}\neq 0$.

∎

######  Lemma 2.17.

A feasible path $\gamma$ is a dynamic equilibrium path iff
$pr(\gamma_{12})\subset C_{f}$, for some $f\in\mathcal{R}_{>0}$, iff ${dQ\over
dt}=0$. In particular, for any feasible path $\gamma$ in which
$pr_{12}(\gamma)$ is fixed, we have dynamic equilibrium and ${dQ\over dt}=0$.

###### Proof.

For the first claim, we have that $f>0$ and if $pr(\gamma_{12})\subset C_{f}$,
we have that;

$\prod_{i=1}^{c}x_{i}^{\nu_{i}}=f$

with the same linkage relations as Lemma 2.14. Now follow through Lemma 2.14,
noting that differentiating reduces the constant $f$ to $0$, as in the proof.
Conversely, if $pr(\gamma_{12})\not\subset C_{f}$, then we have that;

$(\prod_{i=1}^{c}x_{i}^{\nu_{i}})^{\prime}|_{0}=Q(\gamma(t))^{\prime}|_{0}$

$=(grad(Q)|_{\gamma(0)}\centerdot\gamma^{\prime}(0))\neq 0$

but, if $\gamma$ is a dynamic equilibrium path, then clearly each $n_{i}$ is
constant, $1\leq i\leq c$, $n$ is constant and $x_{i}$ is constant, so that
$x_{i}^{\prime}=0$, for $1\leq i\leq c$ and
$(\prod_{i=1}^{c}x_{i}^{\nu_{i}})^{\prime}|_{0}=0$, which is a contradiction.
For the second claim, if $pr(\gamma_{12})\subset C_{c}$, for some
$c\in\mathcal{R}$, it follows from Definition 2.15 and the proof of Lemma
2.14, that $Q$ is constant and ${dQ\over dt}=0$. Conversely, if ${dQ\over
dt}=0$, then $Q$ is constant along $pr(\gamma_{12})$, so that
$pr(\gamma_{12})\subset C_{c}$, for some $c\in\mathcal{R}$. The final claim
follows from the fact that $Q$ depends only on the coordinates $(T,P)$, so
that ${dQ\over dt}=0$, and the first claim.

∎

######  Lemma 2.18.

We have that the condition of chemical equilibrium defines a $1$-dimensional
curve $D$ in the state space $(T,P)$. Similarly, the conditions that
$Q(T,P)=c$ define $1$-dimensional curves $C_{c}$ in $(T,P)$, and if
$\gamma:[0,1]\rightarrow(T,P,n_{1},\ldots n_{c})$ is a path, such that
$pr_{12}(\gamma)$ lies in $C_{c}$, then it must be a dynamic equilibrium path.
Let $D^{\prime}$ be a component of $D$, then $Q$ is constant along
$D^{\prime}$ iff ${\Delta G^{\circ}\over T}$ is constant along $D^{\prime}$.
Let $C_{c}^{\prime}$ be a component of $C_{c}$, then $({\partial
G\over\partial\xi})|_{T,P}=0$ along $C_{c}^{\prime}$ iff ${\Delta
G^{\circ}\over T}=-Rln(c)$. Assuming that ${\Delta G^{\circ}\over T}$ is non
constant, we have that $Q$ is constant along $D^{\prime}$ iff
$pr_{1}(D^{\prime})$ is a fixed temperature $T$, and $({\partial
G\over\partial\xi})|_{T,P}=0$ along $C_{c}^{\prime}$ iff
$pr_{1}(C_{c}^{\prime})$ is a fixed temperature $T$. The only feasible paths
which are both chemical and dynamic equilibrium paths are straight line
chemical equilibrium paths. There exists a feasible dynamic equilibrium path,
with $pr_{12}(\gamma)\subset P=P^{\circ}$.

###### Proof.

For the first part, either use the fact that $({\partial
G\over\partial\xi})|_{T,P}$ only depends on $(T,P)$ and differentiability
properties, or the result from Lemma 2.7 that chemical equilibrium is defined
by $Q(T,P)-e^{-\Delta G^{\circ}\over RT}=0$, and the fact that $\Delta
G^{\circ}$ depends on $T$. For the second part, either use differentiability
properties of $Q(T,P)$ or the fact from Lemma 2.7 that $Q=1$ iff $({\partial
G\over\partial\xi})|_{T,P}-\Delta G^{\circ}=0$. The second claim is clear from
Lemma 2.17. For the third claim, we have, by Lemma 2.7, that along
$D^{\prime}$, $Q=e^{\Delta G^{\circ}\over RT}$, so that clearly $Q$ is
constant along $D^{\prime}$ iff ${\Delta G^{\circ}\over T}$ is constant. The
fourth claim is clear by the fact that $({\partial
G\over\partial\xi})|_{T,P}-\Delta G^{\circ}=RTln(c)$ along $C_{c}^{\prime}$.
The fifth and sixth claims follow immediately from the fact that ${\Delta
G^{\circ}\over T}$ is a function of $T$ and is non constant. For the seventh
claim, if $\gamma$ is a chemical and dynamic equilibrium path, then
$pr_{12}(\gamma)\subset D$, and, by Lemma 2.17, $pr_{12}(\gamma)\subset
C_{c}$, for some $c\in\mathcal{R}$. It follows that $Q$ is constant along the
$pr_{12}(\gamma)\subset D^{\prime}$ for some component $D^{\prime}$, and then,
by the fifth claim, $pr_{1}(\gamma)\subset pr_{1}(D^{\prime})$ is a fixed
temperature $T$, so that $\gamma$ is a straight line chemical equilibrium
path. For the final claim, we have that $Q(T,P^{\circ})=1$, by Lemma 2.14, so
that $P=P^{\circ}$ lies in $C_{1}$. It follows, by Lemma 2.17, that a feasible
path $\gamma$ with $pr_{12}(\gamma)\subset P=P^{\circ}$ is a dynamic
equilibrium path. ∎

##  3\. Ideal Solutions

######  Remarks 3.1.

We have, using the phase rule for an ideal solution in equilibrium with its
vapour, and using the ideal gas law, see [8], that;

$\mu_{i}^{(g)}=\mu_{i}^{\circ(g)}+RTln({P_{i}\over P^{\circ}})$

$\mu_{i}^{(sol)}=\mu_{i}^{\circ(sol)}+RTln({P_{i}\over P^{\circ}})$ $(*)$

By the definition of an ideal solution, we have that;

$\mu_{i}=\mu_{i}^{*}+RTln(x_{i})$ $(**)$

where, by $\mu_{i}^{*}(T,P)$, we mean the chemical potential of substance $i$
on its own, at temperature and pressure $(T,P)$. By Raoult’s law
$P_{i}=x_{i}P_{i}^{*}$, see [7], combined with $(**)$, we obtain;

$\mu_{i}=\mu_{i}^{*}+RTln(x_{i})$

$=\mu_{i}^{*}+RTln({P_{i}\over P_{i}^{*}})$

$=\mu_{i}^{*}+RTln({P_{i}\over P^{\circ}})-RTln({P_{i}^{*}\over P^{\circ}})$
$(***)$

Combining $(*),(***)$, we obtain that;

$\mu_{i}^{*}=\mu_{i}-RTln({P_{i}\over P^{\circ}})+RTln({P_{i}^{*}\over
P^{\circ}})$

$=(\mu_{i}^{\circ}+RTln({P_{i}\over P^{\circ}}))-RTln({P_{i}\over
P^{\circ}})+RTln({P_{i}^{*}\over P^{\circ}})$

$=\mu_{i}^{\circ}+RTln({P_{i}^{*}\over P^{\circ}})$ $({\dagger})$

Letting $P_{i}^{*}=P^{\circ}$, we obtain that
$\mu_{i}^{*}(T,P^{\prime})=\mu_{i}^{\circ}$, $(****)$, where $(T,P^{\prime})$
is the temperature and pressure at which the equilibrium pressure
$P_{i}^{\prime*}=P^{\circ}$. From $(**)$, the fact that
$\mu_{i}^{*}(T,P)\simeq\mu_{i}^{*}(T,P^{\prime})$ and $(****)$, we obtain
that;

$\mu_{i}\simeq\mu_{i}^{\circ}+RTln(x_{i})$, $(*****)$

as a very good approximation. This avoids the contradiction that $x_{i}=1$ for
a solution involving more than one component, at $P=P^{\circ}$. To make the
results here more precise, we need to compute the error term, but the proof is
still consistent if we allow that $n_{i}(T,P)\rightarrow 0$ as $P\rightarrow
P^{\circ}$, so that $x_{i}={n_{i}\over n}\rightarrow 1$, and $x_{i}$ is not
defined at $P=P^{\circ}$.

More specifically, we have that;

$\mu_{i}^{*}(T,P)=\mu_{i}^{*}(T,P^{\prime})+\delta$

where $\delta=\mu_{i}^{*}(T,P)-\mu_{i}^{*}(T,P^{\prime})$, so that;

$\mu_{i}=\mu_{i}^{\circ}+RTln(x_{i})+\delta$

For Raoult’s law, see [7], we also need an approximation. We have that, by the
definition of an ideal solution, the phase rule, Dalton’s law that each gas in
a mixture of ideal gases behaves as if it were alone in the container at the
equilibrium pressures $\\{P_{i},P_{i}^{*}\\}$, see [7], that;

$\mu_{i}=\mu_{i}^{*}+RTln(x_{i})$

$=\mu_{i}^{*}(T,P_{i}^{*})+RTln(x_{i})+\epsilon$

$=\mu_{i}^{\circ(g)}+RTln({P_{i}^{*}\over P^{\circ}})+RTln(x_{i})+\epsilon$

$=\mu_{i}^{\circ(g)}+RTln({P_{i}\over P^{\circ}})$

so that;

$RTln(x_{i})=RTln({P_{i}\over P^{\circ}})-RTln({P_{i}^{*}\over
P^{\circ}})-\epsilon$

where $\epsilon=\mu_{i}^{*}(T,P)-\mu_{i}^{*}(T,P_{i}^{*})$, so that $(***)$
becomes;

$\mu_{i}=\mu_{i}^{*}+RTln(x_{i})$

$=\mu_{i}^{*}+RTln({P_{i}\over P^{\circ}})-RTln({P_{i}^{*}\over
P^{\circ}})-\epsilon$ $(***)^{\prime}$

Combining $(*),(***)^{\prime}$, we obtain that;

$\mu_{i}^{*}=\mu_{i}-RTln({P_{i}\over P^{\circ}})+RTln({P_{i}^{*}\over
P^{\circ}})+\epsilon$

$=(\mu_{i}^{\circ}+RTln({P_{i}\over P^{\circ}}))-RTln({P_{i}\over
P^{\circ}})+RTln({P_{i}^{*}\over P^{\circ}})+\epsilon$

$=\mu_{i}^{\circ}+RTln({P_{i}^{*}\over P^{\circ}})+\epsilon$
$({\dagger})^{\prime}$

Letting $P_{i}^{*}=P^{\circ}$ again, we obtain that
$\mu_{i}^{*}(T,P^{\prime})=\mu_{i}^{\circ}+\epsilon$, $(****)^{\prime}$

From $(**)$, $(****)^{\prime}$, we obtain that;

$\mu_{i}=\mu_{i}^{*}+RTln(x_{i})$

$=\mu_{i}^{*}(T,P^{\prime})+\delta+RTln(x_{i})$

$=\mu_{i}^{\circ}+\epsilon+\delta+RTln(x_{i})$

$=\mu_{i}^{\circ}+RTln(x_{i})+\gamma_{i}$

where
$\gamma_{i}=\epsilon+\delta=\mu_{i}^{*}(T,P)-\mu_{i}^{*}(T,P_{i}^{*})+\mu_{i}^{*}(T,P)-\mu_{i}^{*}(T,P^{\prime})$

$=2\mu_{i}^{*}(T,P)-\mu_{i}^{*}(T,P_{i}^{*})-\mu_{i}^{*}(T,P^{\prime})\simeq
0$

Using Lemma 2.2, we have that $dG=-SdT+VdP$, so that, if temperature is fixed,
$dG=VdP$, then, for the Gibbs energy function of substance $i$ on it own, in
the liquid phase;

$\mu_{i}^{*}(T,P)-\mu_{i}^{*}(T,P_{i}^{*})={G(T,P,n)-G(T,P_{i}^{*},n)\over n}$

$={1\over n}\int_{P_{i}^{*}}^{P}dG$

$={1\over n}\int_{P_{i}^{*}}^{P}VdP$

$={1\over n}\int_{P_{i}^{*}}^{P}{nN_{A}m_{i}\over\kappa_{i}(T,P)}dP$

$\simeq{N_{A}m_{i}(P-P_{i}^{*})\over\kappa}$

$=V_{m,i}(P-P_{i}^{*})$

where $\kappa(T,P)$ is the density of substance $i$ in the liquid phase, and
which we assume to be approximately constant, and $V_{m.i}$ is the molar
volume. Similarly;

$\mu_{i}^{*}(T,P)-\mu_{i}^{*}(T,P^{\prime})\simeq{N_{A}m_{i}(P-P^{\prime})\over\kappa}$

$=V_{m,i}(P-P^{\prime})$

so that;

$\gamma_{i}(P)\simeq V_{m,i}(2P-P_{i}^{*}-P^{\prime})\simeq 0$

We reformulate Lemmas 2.5, 2.7, 2.8 and 2.9 with this error term;

######  Lemma 3.2.

In the ideal solution case, for the energy function $G$ involving only $c$
uncharged species;

$({\partial G\over\partial\xi})_{T,P}=\Delta G^{\circ}+RTln(Q)+\epsilon$

where $\epsilon(P)=\sum_{i=1}^{c}\nu_{i}\gamma_{i}(P)\simeq 0$ and
$\gamma_{i}(P)\simeq 0$ is the error term for the $i$’th uncharged species in
Remark 3.1.

###### Proof.

By Lemma 2.7, we have that;

$({\partial G\over\partial\xi})_{T,P}=\sum_{i=1}^{c}\nu_{i}\mu_{i}$

$\Delta G^{\circ}=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}$, $(*)$

Using $(*)$, the fact that $\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})+\gamma_{i}$,
and Definition 2.1, we have that;

$({\partial G\over\partial\xi})_{T,P}-\Delta
G^{\circ}=\sum_{i=1}^{c}\nu_{i}(\mu_{i}-\mu_{i}^{\circ})$

$=\sum_{i=1}^{c}\nu_{i}(\mu_{i}^{\circ}+RTln(a_{i})+\gamma_{i}-\mu_{i}^{\circ})$

$=\sum_{i=1}^{c}\nu_{i}RTln(a_{i})+\sum_{i=1}^{c}\nu_{i}\gamma_{i}$

$=RTln(\prod_{i=1}^{c}a_{i}^{\nu_{i}})+\sum_{i=1}^{c}\nu_{i}\gamma_{i}=RTln(Q)+\epsilon$

∎

######  Lemma 3.3.

For an ideal solution, we have, using the definition of $\epsilon(P)$ in Lemma
2.5, and the error terms $\gamma_{i}(P)$, $1\leq i\leq c$ in Remark 3.1, that;

$({\partial G\over\partial\xi})_{T,P}=\sum_{i=1}^{c}\nu_{i}\mu_{i}$

$\Delta G^{\circ}=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}$

At chemical equilibrium $T,P$, $({\partial G\over\partial\xi})_{T,P}=0$ and at
$T,P^{0}$, $\Delta G^{\circ}=0$.

If chemical and electrical chemical equilibrium exists at $(T,P^{\circ})$ and
$(T,P)$, $Q(T,P)=e^{-\epsilon(P)\over RT}\simeq 1$ and $E=E^{\circ}$.
Conversely, if $Q(T,P)=e^{-\epsilon(P)\over RT}\simeq 1$ and chemical
equilibrium exists at $(T,P^{\circ})$ then chemical equilibrium exists at
$(T,P)$.

Chemical equilibrium exists at $(T,P)$ iff $Q(T,P)=e^{-\Delta
G^{\circ}-\epsilon(P)\over RT}$

We always have that $Q(T,P^{\circ})=e^{-\delta\over RT}\simeq 1$, where;

$\delta=\epsilon(P^{\circ})=\sum_{i=1}^{c}\nu_{i}\gamma_{i}(P^{\circ})$

###### Proof.

For the first claim, we have, using the definition of $\xi$, that;

$dn_{i}=\nu_{i}d\xi$, $(1\leq i\leq c)$, $(*)$

By Lemma 2.2, fixing $T$ and $P$, and using $(*)$, we have that;

$dG=\sum_{i=1}^{c}\mu_{i}dn_{i}$

$=(\sum_{i=1}^{c}\mu_{i}\nu_{i})d\xi$, $({\dagger})$

so that;

$({\partial G\over\partial\xi})_{T,P}=\sum_{i=1}^{c}\mu_{i}\nu_{i}$,
$({\dagger}{\dagger})$

The second claim from the first, as;

$\Delta G^{\circ}(T)=\int_{0}^{1}({\partial G\over\partial\xi})_{T,P^{\circ}}$

$=\int_{0}^{1}(\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}(T))d\xi$

$=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}(T)\int_{0}^{1}d\xi$

$=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}(T)$

noting that $({\partial G\over\partial\xi})_{T,P^{\circ}}$ doesn’t vary with
$\xi$. For the third claim, at chemical equilibrium, $T,P$, noting again that
$({\partial G\over\partial\xi})_{T,P}$ doesn’t vary with $\xi$, and using
$({\dagger},{\dagger}{\dagger})$, we have that;

$dG=({\partial G\over\partial\xi})_{T,P}=0$, (independently of $\xi$)
$({\dagger}{\dagger}{\dagger})$

At chemical equilibrium $T,P^{\circ}$, using the first and second claims, and
$({\dagger}{\dagger}{\dagger})$, we have that;

$dG=({\partial G\over\partial\xi})_{T,P^{\circ}}$

$=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}$

$=\Delta G^{\circ}=0$

For the second to last claim, and the first direction, we have, by Lemma 2.5,
that $RTln(Q)=-\epsilon\simeq 0$, so that $Q(T,P)=e^{-\epsilon(P)\over
RT}\simeq 1$, and, by Lemma 4.1, that $E-E^{\circ}=-{RTln(Q)\over
2F}-{\epsilon(P)\over 2F}={\epsilon(P)\over 2F}-{\epsilon(P)\over 2F}=0$. For
the converse, we have by Lemma 2.5, using the fact that
$Q(T,P)=e^{-\epsilon(P)\over RT}\simeq 1$;

$({\partial G\over\partial\xi})_{T,P}=\Delta
G^{\circ}-\epsilon(P)+\epsilon(P)=\Delta G^{\circ}$

and, if chemical equilibrium exists at $(T,P^{\circ})$, then, as
$Q(T,P^{\circ})=e^{-\epsilon(P^{\circ})\over RT}$ we have that;

$({\partial G\over\partial\xi})_{T,P^{\circ}}=\Delta
G^{\circ}+RTln(Q(T,P^{\circ})+\epsilon(P^{\circ})=\Delta G^{\circ}=0$

so that $({\partial G\over\partial\xi})_{T,P}=0$

For the penultimate claim, in one direction, use Lemma 2.5, together with the
fact that $({\partial G\over\partial\xi})_{T,P}=0$ and rearrange, the converse
is also clear, applying $ln$.

For the final claim, we have, by the definition of activities, that;

$\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})+\gamma_{i}(P)$

so that $RTln(a_{i}(T,P^{\circ}))=-\gamma_{i}(P^{\circ})$. Now use the
definition of $Q$ in Definition 2.1.

∎

######  Lemma 3.4.

Along a chemical equilibrium path, we have that;

$ln({Q(T_{2})\over Q(T_{1})})={1\over R}\int_{T_{1}}^{T_{2}}{\Delta
H^{\circ}\over T^{2}}dT++{1\over R}({\epsilon(P(T_{1}))\over
T_{1}}-{\epsilon(P(T_{2}))\over T_{2}})$

${\Delta G^{\circ}(T_{2})\over T_{2}}-{\Delta G^{\circ}(T_{1})\over
T_{1}}=-\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}\over T^{2}}dT$

In particularly, if $\Delta H^{\circ}$ is temperature independent;

$ln({Q(T_{2})\over Q(T_{1})})=-{\Delta H^{\circ}\over R}({1\over
T_{2}}-{1\over T_{1}})+{1\over R}({\epsilon(P(T_{1}))\over
T_{1}}-{\epsilon(P(T_{2}))\over T_{2}})$

${\Delta G^{\circ}(T_{2})\over T_{2}}-{\Delta G^{\circ}(T_{1})\over
T_{1}}=\Delta H^{\circ}({1\over T_{2}}-{1\over T_{1}})$

$\Delta G^{\circ}(T_{1})={T_{1}\over T_{2}}\Delta
G^{\circ}(T_{2})-({T_{1}\over T_{2}}-1)\Delta H^{\circ}$

For $c\in\mathcal{R}$, Let $D_{c}$ intersect the line $P=P^{\circ}$ at
$(T_{1},P^{\circ})$, then, for $(T_{2},P)\in D_{c}$, we have that;

$Q(T_{2},P)=e^{\Delta G^{\circ}(T_{1})-\Delta
G^{\circ}(T_{2})-\epsilon(P(T_{2}))\over RT_{2}}$
$({\dagger}{\dagger}{\dagger})$

$c=\Delta G^{\circ}(T_{1})$

$ln({Q(T_{2})\over Q(T_{1})})=ln(Q(T_{2}))={1\over
R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}-c\over
T^{2}}dT-({\epsilon(P(T_{2}))\over RT_{2}}-{\epsilon(P(T_{1}))\over RT_{1}})$

${\Delta G^{\circ}(T_{2})-\Delta G^{\circ}(T_{1})\over
T_{2}}=-\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}-c\over T^{2}}dT$

and if $\Delta H^{\circ}$ is temperature independent;

$ln({Q(T_{2})\over Q(T_{1})})=ln(Q(T_{2}))=-({\Delta H^{\circ}-c\over
R})({1\over T_{2}}-{1\over T_{1}})-({\epsilon(P(T_{2}))\over
RT_{2}}-{\epsilon(P(T_{1}))\over RT_{1}})$ $({\dagger}{\dagger})$

${\Delta G^{\circ}(T_{2})-\Delta G^{\circ}(T_{1})\over T_{2}}=(\Delta
H^{\circ}-c)({1\over T_{2}}-{1\over T_{1}})$ $({\dagger})$

$\Delta G^{\circ}(T_{1})={T_{1}\over T_{2}}\Delta G^{\circ}(T_{2})-\Delta
H^{\circ}({T_{1}\over T_{2}}-1)$ $({\dagger}{\dagger}{\dagger}{\dagger})$

###### Proof.

By Lemma 2.7, we have that;

$\Delta G^{\circ}=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}$

so that differentiating with respect to $T$;

${d(\Delta G^{\circ})\over dT}=\sum_{i=1}^{c}\nu_{i}{d\mu_{i}^{\circ}\over
dT}$

$=\sum_{i=1}^{c}\nu_{i}({\partial\mu_{i}^{\circ}\over\partial T})_{P,n}$

By Euler reciprocity, we have that;

$({\partial\mu_{i}^{\circ}\over\partial T})_{P,n}=-({\partial
S^{\circ}\over\partial n_{i}})_{T,P,n^{\prime}}=-\overline{S}^{\circ}_{i}$

so that, noting $\overline{S}^{\circ}_{i}$ is independent of $n_{i}$, so we
can replace $\overline{S}^{\circ}_{i}$ by $\overline{S}^{\circ}_{m,i}$, the
absolute molar entropy of substance $i$, and using thermodynamic arguments;

${d(\Delta G^{\circ})\over
dT}=-\sum_{i=1}^{c}\nu_{i}\overline{S}^{\circ}_{i}=$

$=-\sum_{i=1}^{c}\nu_{i}\overline{S}^{\circ}_{m,i}$

$=-\Delta S^{\circ}$ $(*)$

Using the product rule, $(*)$ and the definition of enthalpy, we have that;

${d\over dT}({\Delta G^{\circ}\over T})={1\over T}{d(\Delta G^{\circ})\over
dT}-{1\over T^{2}}\Delta G^{\circ}$

$=-{\Delta S^{\circ}\over T}-{\Delta G^{\circ}\over T^{2}}$

$=-{\Delta(ST+G)^{\circ}\over T^{2}}$

$=-{\Delta H^{\circ}\over T^{2}}$ $(**)$

By Lemma 3.3, along a chemical equilibrium path, we have that $Q=e^{-\Delta
G^{\circ}-\epsilon(P)\over RT}$, so that $ln(Q)={-\Delta
G^{\circ}-\epsilon(P)\over RT}$. It follows from $(**)$ that;

${dln(Q)\over dT}={d\over dT}({-\Delta G^{\circ}\over RT})-{d\over
dT}({\epsilon(P)\over RT})$

$={\Delta H^{\circ}\over RT^{2}}-{d\over dT}({\epsilon(P)\over RT})$

It follows, integrating between $T_{1}$ and $T_{2}$, and using the fundamental
theorem of calculus, that;

$ln({Q(T_{2})\over Q(T_{1})})=ln(Q)(T_{2})-ln(Q)(T_{1})$

$=\int_{T_{1}}^{T_{2}}{dln(Q)\over dT}dT$

$={1\over R}\int_{T_{1}}^{T_{2}}[{\Delta H^{\circ}\over T^{2}}-{d\over
dT}({\epsilon(P)\over RT})]dT$

$={-\Delta H^{\circ}\over RT_{2}}+{\Delta H^{\circ}\over
RT_{1}}-({\epsilon(P(T_{2}))\over RT_{2}}-{\epsilon(P(T_{1}))\over RT_{1}})$
$(P)$

so that, rearranging, we obtain the first claim. Using the fact, by Lemma 3.2,
that;

$ln(Q(T_{2}))={-\Delta G^{\circ}(T_{2})-\epsilon(P(T_{2}))\over RT_{2}}$

$ln(Q(T_{1}))={-\Delta G^{\circ}(T_{1})-\epsilon(P(T_{1}))\over RT_{1}}$

we obtain, substituting into $(P)$, canceling $R$, and performing the
integration, if $\Delta H^{\circ}$ is temperature independent, that;

${-\Delta G^{\circ}(T_{2})-\epsilon(P(T_{2}))\over RT_{2}}-{-\Delta
G^{\circ}(T_{1})-\epsilon(P(T_{1}))\over RT_{1}}={-\Delta H^{\circ}\over
RT_{2}}+{\Delta H^{\circ}\over RT_{1}}-({\epsilon(P(T_{2}))\over
RT_{2}}-{\epsilon(P(T_{1}))\over RT_{1}})$

so that;

${\Delta G^{\circ}(T_{2})\over T_{2}}-{\Delta G^{\circ}(T_{1})\over
T_{1}}=\Delta H^{\circ}({1\over T_{2}}-{1\over T_{1}})$ $(Q)$

For the fifth claim, rearrange $(Q)$. If $D_{c}$ intersects the line
$P=P^{\circ}$ at $(T_{1},P^{\circ})$, for the sixth
$({\dagger}{\dagger}{\dagger})$ and seventh claims, we have, using Lemma 2.5
and the fact from Lemma 3.3 that $Q(T_{1},P^{\circ})=e^{-\delta\over RT_{1}}$;

$({\partial G\over\partial\xi})_{T_{2},P}=\Delta
G^{\circ}(T_{2})+RT_{2}ln(Q(T_{2},P))+\epsilon(P(T_{2}))$

$=({\partial G\over\partial\xi})_{T_{1},P^{\circ}}$

$=\Delta G^{\circ}(T_{1})+RT_{1}ln(Q(T_{1},P^{\circ}))+\epsilon(P(T_{1}))$

$=\Delta G^{\circ}(T_{1})+RT_{1}ln(Q(T_{1},P^{\circ}))+\epsilon(P^{\circ})$

$=\Delta G^{\circ}(T_{1})-\delta+\epsilon(P^{\circ})$

$=\Delta G^{\circ}(T_{1})=c$

so that, again rearranging, we obtain the result. Along $D_{c}$, we have,
using Lemma 2.5, that;

$ln(Q(T))={c-\Delta G^{\circ}(T)-\epsilon(P(T))\over RT}$

so that, using the first part;

${dln(Q)\over dT}={d\over dT}({c-\Delta G^{\circ}(T)-\epsilon(P(T))\over RT})$

$={-c\over RT^{2}}+{d\over dT}({-\Delta G^{\circ}(T)\over RT})-{d\over
dT}({\epsilon(P(T))\over RT})$

$={\Delta H^{\circ}-c\over RT^{2}}-{d\over dT}({\epsilon(P(T))\over RT})$

so that, performing the integration, using the fact that
$Q(T_{1},P^{\circ})=e^{-\delta\over RT_{1}}$;

$ln(Q(T_{2}))-ln(Q(T_{1}))=ln(Q(T_{2}))+{\delta\over RT_{1}}={1\over
R}\int_{T_{1}}^{T_{2}}[{\Delta H^{\circ}-c\over T^{2}}-{d\over
dT}({\epsilon(P(T))\over RT})]dT$

$={1\over R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}-c\over
T^{2}}dT-({\epsilon(P(T_{2}))\over RT_{2}}-{\epsilon(P(T_{1}))\over RT_{1}})$

We have that, by Lemma 3.2;

$ln(Q(T_{2}))={c-\Delta G^{\circ}(T_{2})-\epsilon(P(T_{2}))\over RT_{2}}$

$ln(Q(T_{1}))=-{\delta\over RT_{1}}$

so that;

$ln(Q(T_{2}))-ln(Q(T_{1}))={c-\Delta G^{\circ}(T_{2})-\epsilon(P(T_{2}))\over
RT_{2}}+{\delta\over RT_{1}}$

$={\Delta G^{\circ}(T_{1})-\Delta G^{\circ}(T_{2})-\epsilon(P(T_{2}))\over
RT_{2}}+{\delta\over RT_{1}}$

$={1\over R}\int_{T_{1}}^{T_{2}}[{\Delta H^{\circ}-c\over
T^{2}}-({\epsilon(P(T_{2}))\over RT_{2}}-{\epsilon(P(T_{1}))\over RT_{1}})$

$={-1\over R}(\Delta H^{\circ}-c)({1\over T_{2}}-{1\over
T_{1}})-({\epsilon(P(T_{2}))\over RT_{2}}-{\delta\over RT_{1}})$

so that, cancelling $R$ and the the error terms;

${\Delta G^{\circ}(T_{2})-\Delta G^{\circ}(T_{1})\over T_{2}}=(\Delta
H^{\circ}-c)({1\over T_{2}}-{1\over T_{1}})$

$=(\Delta H^{\circ}-\Delta G^{\circ}(T_{1}))({1\over T_{2}}-{1\over T_{1}})$

so that, rearranging again;

$\Delta G^{\circ}(T_{1})({1\over T_{1}}+{1\over T_{2}}-{1\over T_{2}})={\Delta
G^{\circ}(T_{1})\over T_{1}}$

$={\Delta G^{\circ}(T_{2})\over T_{2}}-\Delta H^{\circ}({1\over T_{2}}-{1\over
T_{1}})$

to obtain;

$\Delta G^{\circ}(T_{1})={T_{1}\over T_{2}}\Delta G^{\circ}(T_{2})-\Delta
H^{\circ}({T_{1}\over T_{2}}-1)$

∎

######  Lemma 3.5.

If there exists a component $D_{c}$, $c\in\mathcal{R}$, which projects onto a
closed bounded subinterval $I$ of the line $P=P^{\circ}$, not containing $0$,
and intersects $P=P^{\circ}$ at $(T_{1},P^{\circ})$, with $T_{1}>0$, then, for
$T_{2}\in I$, $\Delta G^{\circ}$ is linear, with;

$\Delta G^{\circ}(T_{2})=T_{2}({(\Delta G^{\circ}(T_{1})-\Delta
H^{\circ})\over T_{1}})+\Delta H^{\circ}$

for $T_{2}\in I$. If $\epsilon\neq 0$, have that;

$({dG\over d\xi})_{T,P}=\lambda+\epsilon ln(P)+\beta T$

where $\\{\lambda,\epsilon,\beta\\}\subset\mathcal{R}$ and
$\\{\beta,\epsilon\\}$ can be effectively determined, and we have that the
activity coefficient is given by;

$Q(T_{1},P^{\prime})=e^{{\epsilon ln({P^{\prime}\over
P^{\prime\circ}})-\epsilon(P^{\prime})\over RT_{1}}}$

and the dynamic equilibrium paths are given by;

$({P^{\prime}\over P^{\prime\circ}})^{\epsilon\over
RT_{1}}=ce^{\epsilon(P^{\prime})}$

for $c\in\mathcal{R}_{\geq 0}$, see Definition 2.1, while the quasi-chemical
equilibrium paths are given by;

$\lambda+\epsilon ln(P^{\prime})+\beta T_{1}=c$

for $c\in\mathcal{R}$.

If $\epsilon=0$;

$({dG\over d\xi})_{T,P}=\lambda+\beta T+\sigma ln(T)$

where $\\{\lambda,\beta,\sigma\\}\subset\mathcal{R}$, and $\\{\beta,\sigma\\}$
can be effectively determined. The activity coefficient $Q$ is given by;

$Q(T_{1},P^{\prime})=e^{-\epsilon(P^{\prime})\over RT_{1}}$

The dynamic equilibrium paths are given by;

$e^{-\epsilon(P^{\prime})\over RT_{1}}=c$

for $c\in\mathcal{R}_{\geq 0}$, see Definition 2.1, while the quasi-chemical
equilibrium paths are given by;

$\lambda+\epsilon ln(P^{\prime})+\beta T_{1}=c$

for $c\in\mathcal{R}$.

###### Proof.

For the first claim, by Lemma 3.4, we have that;

$\Delta G^{\circ}(T_{2})={T_{2}\over T_{1}}\Delta G^{\circ}(T_{1})-\Delta
H^{\circ}({T_{2}\over T_{1}}-1)$

$=T_{2}({(\Delta G^{\circ}(T_{1})-\Delta H^{\circ})\over T_{1}})+\Delta
H^{\circ}$

For the next claim, by Lemma 3.3 and the proof of Lemma 2.9, we have that;

$({\partial({dG\over d\xi})_{T,P}\over\partial
T})_{P}=({\partial(\sum\nu_{i}\mu_{i})\over\partial T})_{P}$

$=\sum\nu_{i}({\partial\mu_{i}\over\partial T})_{P}$

$=\sum\nu_{i}({\partial\mu_{i}\over\partial T})_{P,n}$

$=-\sum\nu_{i}\overline{S}_{m,i}$ $(*)$

Again, to compute $\overline{S}_{m,i}$, we have by the first law of
thermodynamics;

$dQ=dU+dL=dU+pdV$

where $L$ is the work done by the system. We can assume that the liquid
mixture is in thermal equilibrium with a mixture of ideal gases in the vapour
phase, and using the ideal gas law, the definition of temperature for ideal
gases, obtain the calculation of internal energy for the mixture;

$U(T,P,n_{1},\ldots,n_{c})=\sum_{i=1}^{c}({3\over 2}N_{A}n_{i}kT-
N_{A}n_{i}m_{i}\rho_{i})$

where $m_{i}$ is the molecular mass of species $i$, $\rho_{i}$ is the specific
latent heat of evaporation of species $i$, which we assume is independent of
temperature $T$. By a result in [11], using the fact that entropy difference
is independent of path, we have that $Q$ is independent of $P$. We then have;

$dU=\sum_{i=1}^{c}{3\over 2}N_{A}kTdn_{i}+\sum_{i=1}^{c}{3\over
2}N_{A}kn_{i}dT-\sum_{i=1}^{c}N_{A}m_{i}\rho_{i}dn_{i}$

$dQ=\sum_{i=1}^{c}{3\over 2}N_{A}kTdn_{i}+\sum_{i=1}^{c}{3\over
2}N_{A}kn_{i}dT-\sum_{i=1}^{c}N_{A}m_{i}\rho_{i}dn_{i}+dL$

${dQ\over T}=\sum_{i=1}^{c}{3\over 2}N_{A}kdn_{i}+\sum_{i=1}^{c}{3\over
2}N_{A}kn_{i}{dT\over T}-\sum_{i=1}^{c}N_{A}m_{i}\rho_{i}{dn_{i}\over
T}+{g(T,\overline{n})dT\over T}$

$+\sum_{i=1}^{c}h_{i}(T,\overline{n}){dn_{i}\over T}$

$({dQ\over T})_{n^{\prime},T,P}={3\over
2}N_{A}kdn_{i}-N_{A}m_{i}\rho_{i}{dn_{i}\over
T}+h_{i}(T,\overline{n}){dn_{i}\over T}$

It follows that;

$\overline{S}_{m,i}=\int_{\Delta n_{i}=1}({dQ\over
T})_{n^{\prime},T,P}={3\over 2}N_{A}k-{N_{A}m_{i}\rho_{i}\over
T}+{k_{i}(T)\over T}$ $(**)$

So that, from $(*)$;

$({\partial({dG\over d\xi})_{T,P}\over\partial
T})_{P}=-\sum_{i=1}^{c}\nu_{i}({3\over 2}N_{A}k-{N_{A}m_{i}\rho_{i}\over
T})-\sum_{i=1}^{c}\nu_{i}{k_{i}(T)\over T}$

$=-{3\over 2}N_{A}k(\sum_{i=1}^{c}\nu_{i})+{N_{A}\over
T}\sum_{i=1}^{c}\nu_{i}\mu_{i}\rho_{i}-\sum_{i=1}^{c}\nu_{i}{k_{i}(T)\over T}$

$=-{3\over 2}N_{A}k(\sum_{i=1}^{c}\nu_{i})+{N_{A}\over
T}\sum_{i=1}^{c}\nu_{i}\mu_{i}\rho_{i}-{G(T)\over T}$ $(***)$

From $(***)$, which is uniform $P$, we see that $({dG\over d\xi})_{T,P}$ is of
the form $\alpha(P)+\beta T+\gamma ln(T)-\int{G(T)\over T}$, $(B)$, where
$\\{\beta,\gamma\\}\subset\mathcal{R}$, and, assuming that $({dG\over
d\xi})_{T,P}$ is differentiable, $\alpha\in C^{1}(\mathcal{R})$. By a similar
calculation, we have that;

$({\partial({dG\over d\xi})_{T,P}\over\partial
P})_{T}=({\partial(\sum\nu_{i}\mu_{i})\over\partial P})_{T}$

$=\sum_{i=1}^{c}\nu_{i}({\partial\mu_{i}\over\partial P})_{T}$

$=\sum_{i=1}^{c}\nu_{i}({\partial\mu_{i}\over\partial P})_{T,n}$

$=\sum_{i=1}^{c}\nu_{i}({\partial V\over\partial n_{i}})_{T,P,n^{\prime}}$

$=\sum_{i=1}^{c}\nu_{i}\overline{V}_{i}$

$=\sum_{i=1}^{c}\nu_{i}{N_{A}m_{i}\over\kappa_{i}(T,P)}$ $(A)$

where $\kappa_{i}$ is the density of substance $i$. We also have that;

$P(\sum_{i=1}^{c}\nu_{i}{N_{A}m_{i}\over\kappa_{i}(T,P)})=P(\sum_{i=1}^{c}\nu_{i}\overline{V}_{i})=G(T)$
$(dL=PdV)$ $(C)$

and from $(A),(B),(C)$;

$P({\partial({dG\over d\xi})_{T,P}\over\partial P})_{T}=G(T)$

$=P\alpha^{\prime}(P)$

so that $G(T)=\epsilon$, $\alpha(P)=\lambda+\epsilon ln(P)$

$({dG\over d\xi})_{T,P}$ is of the form;

$\alpha(P)+\beta T+\gamma ln(T)-\int{G(T)\over T}$

$=\lambda+\epsilon ln(P)+\beta T+\gamma ln(T)-\epsilon ln(T)$

$=\lambda+\epsilon ln(P)+\beta T+\sigma ln(T)$ $(D)$

where $\sigma=\gamma-\epsilon$,
$\\{\beta,\epsilon,\lambda,\sigma\\}\subset\mathcal{R}$.

If $\epsilon=0$, then $({dG\over d\xi})_{T,P}$ is independent of $P$, and the
components $D_{c}$ are all straight line paths. In this case, if $D_{c}$
intersects the line $P=P^{\circ}$ at $(T_{1},P^{\circ})$, then, for all $P>0$;

$c=\Delta G^{\circ}(T_{1})+RT_{1}ln(Q(T_{1},P)+\epsilon(P)$

$=\Delta G^{\circ}(T_{1})$

implies that $RT_{1}ln(Q(T_{1},P)=-\epsilon(P)$, so that
$Q(T_{1},P)=e^{-\epsilon(P)\over RT_{1}}$, $(X)$. From $(D)$, we have that;

$({dG\over d\xi})_{T,P}=\lambda+\beta T+\sigma ln(T)$, $(Y)$

The calculation of the dynamical and chemical equilibrium paths follows
easily, from the equations $Q=c$, for $c\in\mathcal{R}_{\geq 0}$ and
$({dG\over d\xi})_{T,P}=c$, for $c\in\mathcal{R}$, using $(X),(Y)$.

If $\epsilon\neq 0$, for any $c\in\mathcal{R}$, we can solve the equation;

$\lambda+\epsilon ln(P)+\beta T+\sigma ln(T)=c$

for any given $T>0$ and an appropriate choice of $P(T)$. In particularly,
there exists a component $D_{c}$ projecting onto the line $P=P^{0}$.
Calculating limits at $\\{+\infty,-\infty\\}$, we have that for
$\beta>0,\sigma>0$ or $\beta<0,\sigma<0$, we can solve the equation;

$\lambda+\epsilon ln(P^{\circ})+\beta T+\sigma ln(T)=c$ $(Z)$

for $T$. If $\beta>0,\sigma<0$ or $\beta<0,\sigma>0$, observing that $(\beta
T+\sigma ln(T))^{\prime}=\beta+{\sigma\over T}$, $(\beta T+\sigma
ln(T))^{\prime\prime}=-{\sigma\over T^{2}}$, so there exists a min/max at
$T={-\sigma\over\beta}$, we have that, if;

$-\sigma+\sigma ln({-\sigma\over\beta})\leq c-\lambda-\epsilon
ln(ln(P^{\circ}))$

$-\sigma+\sigma ln({-\sigma\over\beta})\geq c-\lambda-\epsilon
ln(ln(P^{\circ}))$

we can again solve the equation $(Z)$ for $T$, so that, for an appropriate
choice of $c$, there exists an intersection of the component $D_{c}$ with the
line $P=P^{\circ}$.

By the first part, $\Delta G^{\circ}$ is linear, with;

$\Delta G^{\circ}(T_{2})=T_{2}({(\Delta G^{\circ}(T_{1})-\Delta
H^{\circ})\over T_{1}})+\Delta H^{\circ}$

for an intersection at $(T_{1},P^{\circ})$. We also have, using $(D)$, that;

$\Delta G^{\circ}(T_{2})=({\partial
G\over\partial\xi})_{T,P}(T_{2},P^{\circ})$

$=\lambda+\epsilon ln(P^{\circ})+\beta T_{2}+\sigma ln(T_{2})$

so that, equating coefficients;

$\sigma=0$

$\lambda+\epsilon ln(P^{\circ})=\Delta H^{\circ}$

$\beta={(\Delta G^{\circ}(T_{1})-\Delta H^{\circ})\over T_{1}}$

$\Delta G^{\circ}(T_{1})=\beta T_{1}+\Delta H^{\circ}$

We can then, using Lemma 3.2, obtain a formula for the activity coefficient;

$Q(T_{1},P^{\prime})=e^{{(({\partial
G\over\partial\xi})_{T,P}|_{T_{1},P^{\prime}}-\Delta
G^{\circ}(T_{1}))-\epsilon(P^{\prime})\over RT_{1}}}$

$=e^{{(\Delta H^{\circ}-\epsilon ln(P^{\prime\circ})+\epsilon
ln(P^{\prime})+\beta T_{1}-(\beta T_{1}+\Delta
H^{\circ}))-\epsilon(P^{\prime})\over RT_{1}}}$

$=e^{{\epsilon ln({P^{\prime}\over P^{\prime\circ}})-\epsilon(P^{\prime})\over
RT_{1}}}$ $(W)$

as required. The claim about the coefficients being determined is clear from
the proof. The determination of the dynamical and quasi-chemical equilibrium
lines, see Definitions 2.1 and Lemma 2.17, follows from a simple rearrangement
of the formulas $Q(T_{1},P^{\prime})=c$, for some $c\in\mathcal{R}_{\geq 0}$,
using $(W)$, and $({dG\over d\xi})_{T,P}=c$, for some $c\in\mathcal{R}$, using
$(D)$, with $\sigma=0$.

∎

######  Lemma 3.6.

Let notation be as in Lemma 2.9, then if $\epsilon\neq 0$, with
$Q(T,P)=({P\over P^{\circ}})^{\epsilon\over RT}e^{-\epsilon(P)\over RT}$,
then, using the definition of $grad$ in [2];

$grad(Q)(T,P)=({-\epsilon ln({P\over P^{\circ}})+\epsilon(P)\over
RT^{2}}({P\over P^{\circ}})^{\epsilon\over RT}e^{-\epsilon(P)\over
RT},{{\epsilon\over P}-\epsilon^{\prime}(P)\over RT}({P\over
P^{\circ}})^{\epsilon\over RT}e^{-\epsilon(P)\over RT})$

In particular the paths of maximal reaction, for the region
$|grad(Q)(T,P)|>1$, $Q(T,P)>0$, are given by;

$\int{\epsilon Pln({P\over P^{\circ}})-P\epsilon(P)\over
P\epsilon^{\prime}(P)-\epsilon}dP=-{T^{2}\over 2}+c$

for $c\in\mathcal{R}$

If $\epsilon(P)=0$;

$grad(Q)(T,P)=({-\epsilon ln({P\over P^{\circ}})\over RT^{2}}({P\over
P^{\circ}})^{\epsilon\over RT},{\epsilon\over RTP}({P\over
P^{\circ}})^{\epsilon\over RT})$

and the paths of maximal reaction, for the region $|grad(Q)(T,P)|>1$,
$Q(T,P)>0$, are given by;

$P^{2}({ln(P)\over 2}-{ln(P^{\circ})\over 2}-{1\over 4})+{T^{2}\over 2}=c$

for $c\in\mathcal{R}$. If $\epsilon=0$, with with $Q(T,P)=e^{-\epsilon(P)\over
RT}$, then;

$grad(Q)(T,P)=({\epsilon(P)\over RT^{2}}e^{-\epsilon(P)\over
RT},-{\epsilon^{\prime}(P)\over RT}e^{-\epsilon(P)\over RT})$

and the paths of maximal reaction, for the region $|grad(Q)(T,P)|>1$,
$Q(T,P)>0$, are given by;

$\int{\epsilon(P)\over\epsilon^{\prime}(P)}dP={-T^{2}\over 2}+c$

for $c\in\mathcal{R}$.

###### Proof.

The determination of $grad(Q)(T,P)=({\partial Q\over\partial T},{\partial
Q\over\partial P})$ is a simple application of the chain rule and the formula
for $Q$. By the definition of the extent $\xi$ of a reaction, see Definition
2.1, we have that, for $1\leq i\leq c$;

$n_{i}(t)=\nu_{i}\xi(t)+n_{i,0}$

$n(t)=\sum_{j=1}^{c}n_{i}(t)$

$=\sum_{i=1}^{c}(\nu_{i}\xi(t)+n_{i,0})$

$=\alpha\xi(t)+\beta$

where $\alpha=\sum_{i=1}^{c}\nu_{i}$ and $\beta=\sum_{i=1}^{c}n_{i,0}$

so that;

$x_{i}(t)={n_{i}(t)\over n(t)}={\nu_{i}\xi(t)+n_{i,0}\over\alpha\xi(t)+\beta}$

It follows that for a feasible path $\gamma$;

$\prod_{i=1}^{c}x_{i}^{\nu_{i}}(t)=Q(\gamma_{12}(t))$

$=\prod_{i=1}^{c}({\nu_{i}\xi(t)+n_{i,0}\over\alpha\xi(t)+\beta})^{\nu_{i}}$

$={\prod_{i=1}^{c}(\nu_{i}\xi(t)+n_{i,0})^{\nu_{i}}\over(\alpha\xi(t)+\beta)^{c}}$

$=G_{\gamma}(\xi(t))$

where $G_{\gamma}(x)={\prod_{i=1}^{c}(\nu_{i}x+n_{i,0})^{\nu_{i}}\over(\alpha
x+\beta)^{c}}$

We have that $\xi(0)=0$, and, as we can assume that $\beta>0$, we have that;

$G_{\gamma}^{\prime}(0)={\sum_{i=1}^{c}\nu_{i}(\prod_{j\neq
i}n_{j,0})\over\beta^{c}}-{c\prod_{j=1}^{c}n_{j,0}\over\beta^{c+1}}$

$={1\over\beta^{c+1}}(\beta(\sum_{i=1}^{c}\nu_{i}(\prod_{j\neq
i}n_{j,0})-c\prod_{j=1}^{c}n_{j,0}))$

$={\prod_{j=1}^{c}n_{j,0}\over\beta^{c+1}}(\beta{\sum_{i=1}^{c}\nu_{i}\over
n_{i,0}}-c)$

$={\prod_{j=1}^{c}n_{j,0}\over\beta^{c+1}}(n_{init}{\sum_{i=1}^{c}\nu_{i}\over
n_{i,0}}-c)$

$={\prod_{j=1}^{c}n_{j,0}\over\beta^{c+1}}(\sum_{i=1}^{c}{\nu_{i}\over
x_{i,init}}-c)$

$={\prod_{j=1}^{c}n_{j,0}\over\beta^{c+1}}((\sum_{i=1}^{c}\nu_{i}log(x_{i}))^{\prime}_{init}-c)$

$={\prod_{j=1}^{c}n_{j,0}\over\beta^{c+1}}(log(\prod_{i=1}^{c}x_{i}^{\nu_{i}})^{\prime}_{init}-c)$

$={\prod_{j=1}^{c}n_{j,0}\over\beta^{c+1}}(log(Q(\gamma_{12}(t)))^{\prime}|_{0}-c)$

$={\prod_{j=1}^{c}n_{j,0}\over\beta^{c+1}}({grad(Q)\centerdot\gamma_{12}^{\prime}(0)\over
Q(\gamma_{12}(0))}-c)$

so that;

$G_{\gamma}^{\prime}(0)=0$ iff ${grad(Q)\centerdot\gamma_{12}^{\prime}(0)\over
Q(\gamma_{12}(0))}=c$

which we can exclude by an appropriate parametrisation of the feasible path
$\gamma$, without altering the direction of $\gamma_{12}^{\prime}(0)$. By the
inverse function theorem, we can invert $G_{\gamma}$ locally, to obtain that
$\xi(t)=(G_{\gamma}^{-1}\circ Q)(\gamma_{12}(t))$. Then;

$\xi^{\prime}(0)=(G_{\gamma}^{-1})^{\prime}|_{Q(T_{0},P_{0})}grad(Q)(T_{0},P_{0})\centerdot\gamma_{12}^{\prime}(0)$

$={grad(Q)(T_{0},P_{0})\centerdot\gamma_{12}^{\prime}(0)\over
G_{\gamma}^{\prime}(0)}$

$=grad(Q)(T_{0},P_{0})\centerdot\gamma_{12}^{\prime}(0){\beta^{c+1}\over\prod_{j=1}^{c}n_{j,0}}{Q(\gamma_{12}(0))\over
grad(Q)\centerdot\gamma_{12}^{\prime}(0)-cQ(\gamma_{12}(0))}$

$={\alpha_{1}\beta_{1}[grad(Q)(T_{0},P_{0})\centerdot\gamma_{12}^{\prime}(0)]\over
grad(Q)(T_{0},P_{0})\centerdot\gamma_{12}^{\prime}(0)-c\beta_{1}}$

where $\gamma_{12}(0)=(T_{0},P_{0})$,
$\alpha_{1}(T_{0},P_{0})={\beta^{c+1}\over\prod_{j=1}^{c}n_{j,0}}$,
$\beta_{1}(T_{0},P_{0})=Q(\gamma_{12}(0))$.

Writing $\gamma_{12}^{\prime}(0)=\lambda(cos(\theta),sin(\theta))$, we have
that;

$\xi^{\prime}(0)={\lambda\alpha_{1}\beta_{1}[{\partial Q\over\partial
T}|_{(T_{0},P_{0})}cos(\theta)+{\partial Q\over\partial
T}|_{(T_{0},P_{0})}sin(\theta)]\over\lambda[{\partial Q\over\partial
T}|_{(T_{0},P_{0})}cos(\theta)+{\partial Q\over\partial
T}|_{(T_{0},P_{0})}sin(\theta)]-c\beta_{1}}=h(\lambda,\theta)={\lambda\alpha_{1}\beta_{1}r(\theta)\over\lambda
r(\theta)-c\beta_{1}}$

where $r(\theta)={\partial Q\over\partial
T}|_{(T_{0},P_{0})}cos(\theta)+{\partial Q\over\partial
T}|_{(T_{0},P_{0})}sin(\theta)$. We have that;

${\partial h\over\partial\lambda}={\alpha_{1}\beta_{1}r(\theta)\over\lambda
r(\theta)-c\beta_{1}}-{\lambda\alpha_{1}\beta_{1}r^{2}(\theta)\over(\lambda
r(\theta)-c\beta_{1})^{2}}$

so that;

${\partial h\over\partial\lambda}=0$ iff $\alpha_{1}\beta_{1}r(\theta)(\lambda
r(\theta)-c\beta_{1})-\lambda\alpha_{1}\beta_{1}r^{2}(\theta)=0$

iff $-c\alpha_{1}\beta_{1}^{2}r(\theta)=0$

so that, as $\alpha_{1}\neq 0$, $Q(T_{0},P_{0})\neq 0$, and
$(cos(\theta),sin(\theta))$ is not tangent to the dynamic equilibrium path at
$(T_{0},P_{0})$, then $h(\lambda,\theta)$ is monotonic in $\lambda$.

${\partial
h\over\partial\theta}={\lambda\alpha_{1}\beta_{1}r^{\prime}(\theta)\over\lambda
r(\theta)-c\beta_{1}}-{\lambda^{2}\alpha_{1}\beta_{1}r^{\prime}(\theta)\over(\lambda
r(\theta)-c\beta_{1})^{2}}$

so that;

${\partial h\over\partial\theta}=0$ iff
$\lambda\alpha_{1}\beta_{1}r^{\prime}(\theta)(\lambda
r(\theta)-c\beta_{1})-\lambda^{2}\alpha_{1}\beta_{1}r^{\prime}(\theta)=0$

iff $\lambda\alpha_{1}\beta_{1}(\lambda
r(\theta)-c\beta_{1})-\lambda^{2}\alpha_{1}\beta_{1}=0$

iff
$r(\theta)={\lambda^{2}\alpha_{1}\beta_{1}+c\lambda\alpha_{1}\beta_{1}^{2}\over\lambda^{2}\alpha_{1}\beta_{1}}$

$=1+{c\beta_{1}\over\lambda}$

If $|1+{c\beta_{1}\over\lambda}|\leq|grad(Q)(T_{0},P_{0})|$, and
$|grad(Q)(T_{0},P_{0})|>1$, we can solve
$r(\theta)=1+{c\beta_{1}\over\lambda}$, for $\lambda>0$, so that, when
$r(\theta(\lambda))=1+{c\beta_{1}\over\lambda}$;

$h(\lambda,\theta(\lambda))={\lambda\alpha_{1}\beta_{1}(1+{c\beta_{1}\over\lambda})\over\lambda(1+{c\beta_{1}\over\lambda})-c\beta_{1}}$

$={\lambda\alpha_{1}\beta_{1}+c\alpha_{1}\beta_{1}^{2}\over\lambda}$

$=\alpha_{1}\beta_{1}+{c\alpha_{1}\beta_{1}^{2}\over\lambda}$

$=\alpha_{1}\beta_{1}(1+{c\beta_{1}\over\lambda})$

so a maximum/minimum occurs when
$|1+{c\beta_{1}\over\lambda}|=|grad(Q)(T_{0},P_{0})|$, in which case
$(cos(\theta),sin(\theta))$ is parallel to $grad(Q)(T_{0},P_{0})$, and
perpendicular to the tangent of the level curve of $Q$, through
$(T_{0},P_{0})$. We, therefore, have to solve the paired differential
equation;

${dT\over dt}={-\epsilon ln({P(t)\over P^{\circ}})+\epsilon(P(t))\over
RT(t)^{2}}({P(t)\over P^{\circ}})^{\epsilon\over RT(t)}e^{-\epsilon(P(t))\over
RT(t)}$

${dP\over dt}={{\epsilon\over P(t)}-\epsilon^{\prime}(P(t))\over
RT(t)}({P(t)\over P^{\circ}})^{\epsilon\over RT(t)}e^{-\epsilon(P(t))\over
RT(t)}$

so that;

${dP\over dT}={{dP\over dt}\over{dT\over dt}}$

$={{{\epsilon\over P}-\epsilon^{\prime}(P)\over RT}({P\over
P^{\circ}})^{\epsilon\over RT}e^{-\epsilon(P)\over RT}\over{-\epsilon
ln({P\over P^{\circ}})+\epsilon(P)\over RT^{2}}({P\over
P^{\circ}})^{\epsilon\over RT}e^{-\epsilon(P)\over RT}}$

$={T({\epsilon\over P}-\epsilon^{\prime}(P))\over-\epsilon ln({P\over
P^{\circ}})+\epsilon(P)}$

Separating variables, we obtain that;

${\epsilon ln({P\over
P^{\circ}})-\epsilon(P)\over\epsilon^{\prime}(P)-{\epsilon\over P}}dP=-TdT$

which has an implicit solution given by;

$\int{\epsilon Pln({P\over P^{\circ}})-P\epsilon(P)\over
P\epsilon^{\prime}(P)-\epsilon}dP=-{T^{2}\over 2}+c$

for $c\in\mathcal{R}$. By a result due to [3], we have that these implicit
solutions are integral curves for $grad(Q)$ as required. If $\epsilon(P)=0$,
then $\epsilon^{\prime}(P)=0$ and the implicit solutions are given by;

$\int Pln({P\over P^{\circ}})dP=-{T^{2}\over 2}+c$

for $c\in\mathcal{R}$. We have, integrating by parts, that;

$\int Pln({P\over P^{\circ}})=\int Pln(P)-Pln(P^{\circ})$

$={P^{2}ln(P)\over 2}-\int{P\over 2}-{P^{2}ln(P^{\circ})\over 2}$

$={P^{2}ln(P)\over 2}-{P^{2}\over 4}-{P^{2}ln(P^{\circ})\over 2}$

$=P^{2}({ln(P)\over 2}-{ln(P^{\circ})\over 2}-{1\over 4})$

so that the implicit solutions are given by;

$P^{2}({ln(P)\over 2}-{ln(P^{\circ})\over 2}-{1\over 4})+{T^{2}\over 2}=c$

for $c\in\mathcal{R}$ as required. The determination of $grad(Q)(T,P)$ when
$\epsilon=0$ is again a simple application of the chain rule. As before, we
compute;

${dP\over dT}={{-\epsilon^{\prime}(P)\over RT}e^{-\epsilon(P)\over
RT}\over{\epsilon(P)\over RT^{2}}e^{-\epsilon(P)\over RT}}$

$={-\epsilon^{\prime}(P)T\over\epsilon(P)}$

so that, separating variables;

${\epsilon(P)\over\epsilon^{\prime}(P)}dP=-TdT$

and the implicit solutions are given by;

$\int{\epsilon(P)\over\epsilon^{\prime}(P)}dP={-T^{2}\over 2}+c$

∎

######  Remarks 3.7.

The fact that, in the case $\epsilon(P)=0$, the paths of maximal reaction
depend on an arbitrary choice of $P^{\circ}$ suggest that some approximation
is needed in the formula;

$\mu_{i}(T,P)=\mu_{i}^{\circ}(T)+RTlog(x_{i}(T,P))$

for ideal solutions. Of course, once $P^{\circ}$ is fixed, $\epsilon(P)$
depends on this choice of $P^{\circ}$ as well.

##  4\. Electrochemistry with error terms and ideal solution

We consider the reaction $H_{2}(g)+2AgCl(s)+2e^{-}(R)\rightarrow
2HCl+2Ag(s)+2e^{-}(L)$, for the standard cell, even though the uncharged
species probably don’t form an ideal solution. The reader can easily
reformulate the results in the context of an ideal solution, by just changing
the electron count, see Section 8.

######  Lemma 4.1.

The Nernst Equation for the Standard Cell

At electrical chemical equilibrium $(T,P)$ and $(T,P^{\circ})$;

$(E-E^{\circ})(T,P)=-{RTln(Q(T,P))\over 2F}-{\epsilon(P)\over 2F}$

###### Proof.

For $c$ substances, with $c^{\prime}$ the number of the charged species,using
Definition 2.1, we have that the electrostatic potential energy;

$U_{el}=\sum_{i=1}^{c^{\prime}}\phi(\overline{x}_{i})q_{i}$, where
$q_{i}=N_{i}ez_{i}=N_{A}n_{i}ez_{i}$

where $\\{\overline{x}_{i}:1\leq i\leq c^{\prime}\\}$ are the positions of the
charged species, $N_{i}$ is the number of particles at $\overline{x}_{i}$.

We have that;

$U=U_{chem}+U_{el}$, so that;

$G(T,P,n_{1},\ldots,n_{c})=U+PV-TS$

$=U_{chem}+U_{el}+PV-TS$

$=U_{el}+G_{chem}$

$=\sum_{j=1}^{c}\phi(\overline{x}_{j})q_{j}+G_{chem}$

$=\sum_{j=1}^{c}\phi(\overline{x}_{j})N_{A}n_{j}ez_{j}+G_{chem}$

so that;

$\mu_{i}=({\partial G\over\partial n_{i}})_{T,P}$

$=({\partial(\sum_{j=1}^{c}\phi(\overline{x}_{j})N_{A}n_{j}ez_{j}+G_{chem})\over\partial
n_{i}})_{T,P}$

$=\mu_{i,chem}$, $(c^{\prime}+1\leq i\leq c)$

$=\mu_{i,chem}+{\partial(\phi(\overline{x}_{i})N_{A}n_{i}ez_{i})\over\partial
n_{i}}$, $(1\leq i\leq c^{\prime})$

$=\mu_{i,chem}+\phi(\overline{x}_{i})N_{A}ez_{i}$

$=\mu_{i,chem}+\phi(\overline{x}_{i})Fz_{i}$, $(*)$

We consider the standard cell reaction $H_{2}(g)+2AgCl(s)+2e^{-}(R)\rightarrow
2HCl+2Ag(s)+2e^{-}(L)$. At electrical chemical equilibrium, similarly to Lemma
2.7, generalised to a collection involving charged species, using $(*)$, we
have that;

$({\partial G\over\partial\xi})_{T,P}={\sum}_{i=1}^{c}\nu_{i}\mu_{i}$

$=2\mu(HCl)+2\mu(Ag)-\mu(H_{2})-2\mu(AgCl)+2\mu(e^{-}(L))-2\mu(e^{-}(R))$

$=({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}+2\mu(e^{-}(L))-2\mu(e^{-}(R))$

$=({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}+((2\mu_{chem}(e^{-}(L))-2F\phi(L))-(2\mu_{chem}(e^{-}(L))-2F\phi(R)))$

$=({\partial G_{chem^{\prime}}\over\partial\xi})_{T,P}+2F(\phi(R)-\phi(L))$

$=({\partial G_{chem^{\prime}}\over\partial\xi})_{T,P}+2EF=0$ $({\dagger})$

where $G_{chem^{\prime}}$ is the Gibbs energy restricted to the uncharged
species. By Lemmas 3.2 and 3.3, we have that;

$({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P^{\circ}}=\sum_{i=c^{\prime}+1}^{c}\nu_{i}\mu_{i}^{\circ}$

$=(\Delta
G_{chem^{\prime}}^{\circ}+RTln(Q_{chem^{\prime}}(T,P^{\circ}))+\epsilon(P^{\circ}))$

$=(\Delta G_{chem^{\prime}}^{\circ}-\epsilon(P^{\circ}))+\epsilon(P^{\circ}))$

$=\Delta G_{chem^{\prime}}^{\circ}$, $({\dagger}{\dagger})$

From $({\dagger}),({\dagger}{\dagger})$, we obtain;

$2E^{\circ}F=-({\partial G_{chem^{\prime}}\over\partial\xi})_{T,P^{\circ}}$

$=-\Delta G_{chem^{\prime}}^{\circ}$ $({\dagger}{\dagger}{\dagger})$

Similarly, we have that;

$({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}=\sum_{i=c^{\prime}+1}^{c}\nu_{i}\mu_{i}$

$=(\Delta
G_{chem^{\prime}}^{\circ}+RTln(Q_{chem^{\prime}}(T,P))+\epsilon(P))$,
$({\dagger}{\dagger}{\dagger}{\dagger})$

so from $({\dagger}),({\dagger}{\dagger}{\dagger}{\dagger}))$, we obtain that;

$2EF=-({\partial G_{chem^{\prime}}\over\partial\xi})_{T,P}$

$=-(\Delta
G_{chem^{\prime}}^{\circ}+RTln(Q_{chem^{\prime}}(T,P))+\epsilon(P))$,
$(\sharp)$

Combining $(\sharp),({\dagger}{\dagger}{\dagger})$, we obtain that;

$2EF-2E^{\circ}F=-(\Delta
G_{chem^{\prime}}^{\circ}+RTln(Q_{chem^{\prime}}(T,P))+\epsilon(P))-(-\Delta
G_{chem^{\prime}}^{\circ})$

$=-RTln(Q_{chem^{\prime}}(T,P))-\epsilon(P)$

so that;

$E-E^{\circ}=-{RTln(Q_{chem^{\prime}}(T,P))\over 2F}-{\epsilon(P)\over 2F}$

∎

######  Lemma 4.2.

At electrical chemical equilibrium $(T,P)$ and $(T,P^{\circ})$, and chemical
equilibrium $(T,P)$;

$\Delta G^{\circ}=2F(E-E^{0})$

###### Proof.

By Lemma 4.1, we have that;

$E-E^{\circ}=-{RTln(Q)\over 2F}-{\epsilon(P)\over 2F}$, $(*)$

and, by Lemma 3.2, we have that;

$0=({\partial G\over\partial\xi})_{T,P}=\Delta G^{\circ}+RTln(Q)+\epsilon(P)$,
$(**)$

Rearranging $(*),(**)$, we obtain the result.

∎

######  Lemma 4.3.

If $\epsilon=0$, we have, for all $T_{1}>0$, that;

$({\partial G\over\partial\xi})_{T,P}|_{(T_{1},P_{1})}=({\partial
G\over\partial\xi})_{T,P}|_{(T_{1},P_{1}^{\circ})}$

iff;

$E(T_{1},P_{1})=E(T_{1},P_{1}^{\circ})=E^{\circ}(T_{1})$

where $G$ is the Gibbs energy function for the charged and uncharged species.

###### Proof.

By $({\dagger})$ of Lemma 4.1, we have that;

$({\partial G\over\partial\xi})_{T,P}=({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}+2EF$ $(*)$

By Lemma 3.5, we have that $({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}$ is independent of $P$, in
particularly, we have that;

$({\partial G_{chem^{\prime}}\over\partial\xi})_{T_{1},P_{1}}=({\partial
G_{chem^{\prime}}\over\partial\xi})_{T_{1},P_{1}^{\circ}}$ $(**)$

so that, combining $(*),(**)$, we obtain the result. ∎

######  Lemma 4.4.

We have, for all $T_{1}>0,P_{1}>0$, that;

$2F(E(T_{1},P_{1})-E^{\circ}(T_{1}))=({\partial
G\over\partial\xi})_{T,P}|_{(T_{1},P_{1})}-({\partial
G\over\partial\xi})_{T,P}|_{(T_{1},P_{1}^{\circ})}-RT_{1}ln(Q(T_{1},P_{1}))-\epsilon(P_{1})$

###### Proof.

Following the proof of Lemma 4.1, we have that;

$({\partial G\over\partial\xi})_{T,P}|_{T_{1},P_{1}}=({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}|_{T_{1},P_{1}}+2E(T_{1},P_{1})F$
$(*)$

$2E^{\circ}(T_{1})F=({\partial
G\over\partial\xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}|_{T_{1},P_{1}^{\circ}}$

$=({\partial G\over\partial\xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-\Delta
G^{\circ}_{chem^{\prime}}(T_{1})$ $(**)$

so from $(*),(**)$ and Lemma 3.2;

$2E(T_{1},P_{1})F=({\partial
G\over\partial\xi})_{T,P}|_{T_{1},P_{1}}-({\partial
G_{chem^{\prime}}\over\partial\xi})_{T,P}|_{T_{1},P_{1}}$

$=({\partial G\over\partial\xi})_{T,P}|_{T_{1},P_{1}}-(\Delta
G^{\circ}_{chem^{\prime}}(T_{1})+RT_{1}ln(Q_{chem^{\prime}}(T_{1},P_{1}))+\epsilon(P_{1}))$

$2E(T_{1},P_{1})F-2E^{\circ}(T_{1})F=({\partial
G\over\partial\xi})_{T,P}|_{T_{1},P_{1}}-(\Delta
G^{\circ}_{chem^{\prime}}(T_{1})+RT_{1}ln(Q_{chem^{\prime}}(T_{1},P_{1}))+\epsilon(P_{1}))$

$-(({\partial G\over\partial\xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-\Delta
G^{\circ}_{chem^{\prime}}(T_{1}))$

$=({\partial G\over\partial\xi})_{T,P}|_{T_{1},P_{1}}-(({\partial
G\over\partial\xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-RT_{1}ln(Q_{chem^{\prime}}(T_{1},P_{1}))-\epsilon(P_{1})$

∎

##  5\. dilute solutions

######  Definition 5.1.

As mentioned in Definition 2.1, we can consider an electrolyte as a solute in
a dilute solution. Sometimes the solvent is involved in an electrolytic
reaction, for example;

$2H_{2}O+4e^{-}(R)\rightarrow O_{2}+2H_{2}+4e^{-}(L)$ $(*)$

and sometimes not, as in the standard cell, where we can consider $H_{2}O$ as
the solvent not involved in the reaction. In Lemmas 0.4-0.12, for the standard
cell, we can replace $Q$ defined as $\prod_{i=1}^{c}a_{i}^{\nu_{i}}$ by
$a_{0}(\prod_{i=1}^{c}a_{i}^{\nu_{i}})$, considering $H_{2}O$ as substance
$0$.

We ideally have that $\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})$, $0\leq i\leq c$,
$({\dagger})$, when we define the activities $a_{i}$, for $0\leq i\leq c$,
which, when $({\dagger})$ holds, involves a contradiction.

######  Lemma 5.2.

In Lemmas 0.4-0.12, for the standard cell, and considering a dilute solution
with no interaction of the solvent, replacing $Q$ defined as
$\prod_{i=1}^{c}a_{i}^{\nu_{i}}$ by;

$a_{0}(\prod_{i=1}^{c}a_{i}^{\nu_{i}})$

If we assume without approximation that $\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})$,
$0\leq i\leq c$, then;

Lemma 2.4; $E-E^{\circ}=-{RTln(Q)\over 2F}+{RTln({a_{0}(T,P)\over
a_{0}(T,P^{\circ})})\over 2F}$

Lemma 2.5; $({\partial G\over\partial\xi})_{T,P}=\Delta G^{\circ}+RTln({Q\over
a_{0}(T,P)})$

where $\Delta G^{\circ}$ is the Gibbs energy change for $1$ mole of reaction
without the solvent.

Lemma 2.6; $\Delta G^{\circ}=2F(E-E^{\circ})+RTln(a_{0}(T,P^{\circ}))$

Lemma 2.7; The same with the modification that if chemical and electrical
equilibrium exist at $(T,P^{\circ})$ and $(T,P)$, $Q(T,P)=a_{0}(T,P)$ and
$E-E^{\circ}=-{RTln(a_{0}(T,P^{\circ}))\over 2F}$. Conversely, if
$Q(T,P)={a_{0}(T,P)\over a_{0}(T,P^{\circ})}$ and chemical equilibrium exists
at $(T,P^{\circ})$ then chemical equilibrium exists at $(T,P)$.

Chemical equilibrium exists at $(T,P)$ iff $Q(T,P)=a_{0}e^{-\Delta
G^{\circ}\over RT}$

We always have that $Q(T,P^{\circ})=1$

Lemma 2.8; The same, with the modification that along a chemical equilibrium
path, we have that;

$ln({Q(T_{2})\over Q(T_{1})})={1\over R}\int_{T_{1}}^{T_{2}}{\Delta
H^{\circ}\over T^{2}}dT+ln({a_{0}(T_{2},P_{2})\over a_{0}(T_{1},P_{1})})$

and, if $\Delta H^{\circ}$ is temperature independent;

$ln({Q(T_{2})\over Q(T_{1})})=-{\Delta H^{\circ}\over R}({1\over
T_{2}}-{1\over T_{1}})+ln({a_{0}(T_{2},P_{2})\over a_{0}(T_{1},P_{1})})$

For $c\in\mathcal{R}$, if $D_{c}$ intersect the line $P=P^{\circ}$ at
$(T_{1},P^{\circ})$, then, for $(T_{2},P)\in D_{c}$, we have that;

$Q(T_{2},P)=e^{\Delta G^{\circ}(T_{1})-\Delta G^{\circ}(T_{2})\over
RT_{2}}a_{0}(T_{1},P^{\circ})^{-T_{1}\over T_{2}}a_{0}(T_{2},P)$
$({\dagger}{\dagger}{\dagger})^{\prime}$

$c=\Delta G^{\circ}(T_{1})-RT_{1}ln(a_{0}(T_{1},P^{\circ})$

$ln({Q(T_{2})\over Q(T_{1})})=ln(Q(T_{2}))={1\over
R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}-c\over
T^{2}}dT+ln({a_{0}(T_{2},P)\over a_{0}(T_{1},P^{\circ})})$

${\Delta G^{\circ}(T_{2})-\Delta G^{\circ}(T_{1})\over
T_{2}}=-\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}-c\over T^{2}}dT+R(1-{T_{1}\over
T_{2}})ln(a_{0}(T_{1},P^{\circ}))$

and if $\Delta H^{\circ}$ is temperature independent;

$ln({Q(T_{2})\over Q(T_{1})})=ln(Q(T_{2}))=-({\Delta H^{\circ}-c\over
R})({1\over T_{2}}-{1\over T_{1}})+ln({a_{0}(T_{2},P)\over
a_{0}(T_{1},P^{\circ})})$ $({\dagger}{\dagger})^{\prime}$

${\Delta G^{\circ}(T_{2})-\Delta G^{\circ}(T_{1})\over T_{2}}=(\Delta
H^{\circ}-c)({1\over T_{2}}-{1\over T_{1}})+R(1-{T_{1}\over
T_{2}})ln(a_{0}(T_{1},P^{\circ}))$ $({\dagger})^{\prime}$

to obtain;

$\Delta G^{\circ}(T_{1})={T_{1}\over T_{2}}\Delta G^{\circ}(T_{2})-\Delta
H^{\circ}({T_{1}\over T_{2}}-1)$

again.

Lemma 2.9; The same, with the modification that if $\epsilon\neq 0$;

$Q(T_{2},P^{\prime})=e^{\epsilon ln({P^{\prime}\over P^{\prime\circ}})\over
RT_{2}}a_{0}(T_{2},P^{\prime})$

and the dynamic equilibrium paths are given by;

$a_{0}(T_{2},P^{\prime})({P^{\prime}\over P^{\prime\circ}})^{\epsilon\over
RT_{2}}=c$

for $c\in\mathcal{R}_{\geq 0}$, while if $\epsilon=0$,
$Q(T_{2},P^{\prime})={a_{0}(T_{2},P^{\prime})\over
a_{0}(T_{2},P^{\prime\circ})}$, the dynamic equilibrium lines are given by;

${a_{0}(T_{2},P^{\prime})\over a_{0}(T_{2},P^{\prime\circ})}=c$

for $c\in\mathcal{R}_{\geq 0}$ and the quasi-chemical equilibrium lines are
given by;

$\lambda+\beta T_{2}+\sigma ln(T_{2})=c$

for $c\in\mathcal{R}$.

###### Proof.

Following the proof of Lemma 2.7, we note that for Gibbs function $G$ with
$c+1$ species, including the solvent, substance $0$, as $dn_{0}=0$, that;

$dG=\sum_{i=0}^{c}\mu_{i}dn_{i}$

$=\sum_{i=1}^{c}\mu_{i}dn_{i}$

so the first three claims in Lemma 2.7 go through as before. Going through the
proof of Lemma 2.5, we then obtain that;

$({\partial G\over\partial\xi})_{T,P}=\Delta
G^{\circ}+RTln(\prod_{i=1}^{c}a_{i}^{\nu_{i}})$

$=\Delta G^{\circ}+RTln({\prod_{i=0}^{c}a_{i}^{\nu_{i}}\over a_{0}(T,P)})$

$=\Delta G^{\circ}+RTln({Q\over a_{0}(T,P)})$

Going back through the proof of Lemma 2.7, we obtain that $RTln({Q\over
a_{0}(T,P)})=0$, so that $Q(T,P)=a_{0}(T,P)$. Going through the proof of Lemma
2.4, using the fact that $\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})$, for $0\leq
i\leq c$, so that $Q(T,P^{\circ})=1$, we have that;

$({\partial G_{chem^{\prime}}\over\partial\xi})_{T,P^{\circ}}=\Delta
G_{chem^{\prime}}^{\circ}+RTln(Q_{chem^{\prime}}(T,P^{\circ})$

$=\Delta G_{chem^{\prime}}^{\circ}+RTln({Q\over a_{0}(T,P^{\circ})})$

$=\Delta G_{chem^{\prime}}^{\circ}-RTln(a_{0}(T,P^{\circ}))$
$({\dagger}{\dagger})^{\prime}$

where $G_{chem^{\prime}}$ is the Gibbs energy restricted to the uncharged
species without the solvent. Using $({\dagger})$ from Lemma 2.4 and
$({\dagger}{\dagger})^{\prime}$, we obtain;

$2E^{\circ}F=-\Delta G_{chem^{\prime}}^{\circ}+RTln(a_{0}(T,P^{\circ}))$
$({\dagger}{\dagger}{\dagger})^{\prime}$

Similarly, we obtain;

$({\partial G_{chem^{\prime}}\over\partial\xi})_{T,P}=\Delta
G_{chem^{\prime}}^{\circ}+RTln(Q_{chem^{\prime}}(T,P)$

$=\Delta G_{chem^{\prime}}^{\circ}+RTln({Q\over a_{0}(T,P)})$

so that, from $({\dagger})$ from Lemma 2.4;

$2EF=-(\Delta G_{chem^{\prime}}^{\circ}+RTln({Q\over a_{0}(T,P)}))$
$(\sharp)^{\prime}$

Combining $(\sharp)^{\prime}$, $({\dagger}{\dagger}{\dagger})^{\prime}$, we
obtain;

$2EF-2E^{\circ}F=-(\Delta G_{chem^{\prime}}^{\circ}+RTln({Q\over
a_{0}(T,P)}))-(-\Delta G_{chem^{\prime}}^{\circ}+RTln(a_{0}(T,P^{\circ})))$

$=-RTln({Q\over a_{0}(T,P)})-RTln(a_{0}(T,P^{\circ}))$

$=-RTln(Q)+RTln({a_{0}(T,P)\over a_{0}(T,P^{\circ})})$

which gives the result.of Lemma 2.4. Going back through the proof of Lemma 2.7
again, we then have that;

$E-E^{\circ}=-{RTln(Q)\over 2F}+{RTln({a_{0}(T,P)\over
a_{0}(T,P^{\circ})})\over 2F}$

$=-{RTln(a_{0}(T,P))\over 2F}+{RTln({a_{0}(T,P)\over a_{0}(T,P^{\circ})})\over
2F}$

$=-{RTln(a_{0}(T,P^{\circ}))\over 2F}$

For the converse claim, we have by the modification of Lemma 2.5, and the
facts that $Q(T,P)={a_{0}(T,P)\over a_{0}(T,P^{\circ})}$,$Q(T,P^{\circ})=1$,
$({\partial G\over\partial\xi})_{T,P^{\circ}}=0$, that;

$({\partial G\over\partial\xi})_{T,P}=\Delta G^{\circ}+RTln({Q\over
a_{0}(T,P)})$

$=\Delta G^{\circ}-RTln(a_{0}(T,P^{\circ}))$

$({\partial G\over\partial\xi})_{T,P^{\circ}}=\Delta G^{\circ}+RTln({Q\over
a_{0}(T,P^{\circ})})$

$=0$

so that we have chemical equilibrium at $(T,P)$. For the penultimate claim of
Lemma 2.7, rearrange the formula from the modification of Lemma 2.5, with the
definition of chemical equilibrium;

$({\partial G\over\partial\xi})_{T,P}=\Delta G^{\circ}+RTln({Q\over
a_{0}(T,P)})=0$

The last claim is clear from;

$\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})$, for $0\leq i\leq c$

For Lemma 2.6, we have by the modification of Lemma 2.4, that;

$E-E^{\circ}=-{RTln(Q)\over 2F}+{RTln({a_{0}(T,P)\over
a_{0}(T,P^{\circ})})\over 2F}$

and, by the modification of Lemma 2.5, that;

$0=({\partial G\over\partial\xi})_{T,P}=\Delta
G^{\circ}+RTln(Q)-RTln(a_{0}(T,P))$

so that;

$\Delta G^{\circ}=RTln(a_{0}(T,P))-RTln(Q)$

$=RTln(a_{0}(T,P))+2F(E-E^{\circ})-RTln({a_{0}(T,P)\over a_{0}(T,P^{\circ})})$

$=2F(E-E^{\circ})+RTln(a_{0}(T,P^{\circ}))$

For the modification of Lemma 2.8, the first part of the proof goes through
with the chemical potentials $\mu_{i},1\leq i\leq c$, defined relative to the
Gibbs energy including the solvent. By the modification of Lemma 2.7, we have
that $Q=a_{0}e^{-\Delta G^{\circ}\over RT}$ along a chemical equilibrium path,
so that $ln(Q)={-\Delta G^{\circ}\over RT}+ln(a_{0}(T,P))$, $(K)^{\prime}$. It
follows that;

${dln(Q)\over dT}={d\over dT}({-\Delta G^{\circ}\over RT})+{d\over
dT}(ln(a_{0}(T,P)))$

$={\Delta H^{\circ}\over RT^{2}}+{d\over dT}(ln(a_{0}(T,P)))$

It follows, integrating between $T_{1}$ and $T_{2}$, using $(K)^{\prime}$ and
the fundamental theorem of calculus, that;

$ln({Q(T_{2})\over Q(T_{1})})=ln(Q)(T_{2})-ln(Q)(T_{1})$

$={-\Delta G^{\circ}(T_{2})\over RT_{2}}+{\Delta G^{\circ}(T_{1})\over
RT_{1}}+ln(a_{0}(T_{2},P_{2}))-ln(a_{0}(T_{1},P_{1}))$

$=\int_{T_{1}}^{T_{2}}{dln(Q)\over dT}dT$

$={1\over R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}\over
T^{2}}+\int_{T_{1}}^{T_{2}}{d\over dT}(ln(a_{0}(T,P)))dT$

$={1\over R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}\over
T^{2}}+ln(a_{0}(T_{2},P_{2}))-ln(a_{0}(T_{1},P_{1}))$ $(P)^{\prime}$

so that, rearranging, we obtain the first claim. Using the fact, by the
modification of Lemma 2.5, that;

$ln(Q(T_{2}))=-{\Delta G^{\circ}(T_{2})\over RT_{2}}+ln(a_{0}(T_{2},P_{2}))$

$ln(Q(T_{1}))=-{\Delta G^{\circ}(T_{1})\over RT_{1}}+ln(a_{0}(T_{1},P_{1}))$

we obtain, substituting into $(P)^{\prime}$, canceling $R$, and performing the
integration, if $\Delta H^{\circ}$ is temperature independent, that;

${\Delta G^{\circ}(T_{2})\over T_{2}}-{\Delta G^{\circ}(T_{1})\over
T_{1}}=-\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}\over T^{2}}=\Delta
H^{\circ}({1\over T_{2}}-{1\over T_{1}})$ $(Q)^{\prime}$

For the fifth claim, rearrange $(Q)^{\prime}$. If $D_{c}$ intersects the line
$P=P^{\circ}$ at $(T_{1},P^{\circ})$, for the sixth
$({\dagger}{\dagger}{\dagger})$ and seventh claims, we have, using the
modification of Lemma 2.5 and the fact from Lemma 2.7 that
$Q(T_{1},P^{\circ})=1$;

$({\partial G\over\partial\xi})_{T_{2},P}=\Delta
G^{\circ}(T_{2})+RT_{2}ln(Q(T_{2},P))-RT_{2}ln(a_{0}(T_{2},P))$

$=({\partial G\over\partial\xi})_{T_{1},P^{\circ}}$

$=\Delta
G^{\circ}(T_{1})+RT_{1}ln(Q(T_{1},P^{\circ}))-RT_{1}ln(a_{0}(T_{1},P^{\circ}))$

$=\Delta G^{\circ}(T_{1})-RT_{1}ln(a_{0}(T_{1},P^{\circ}))=c$

so that, again rearranging, we obtain the result. Along $D_{c}$, we have,
using Lemma 2.5, that;

$ln(Q)={c-\Delta G^{\circ}\over RT}+ln(a_{0}(T,P))$

so that, using the first part;

${dln(Q)\over dT}={d\over dT}({c-\Delta G^{\circ}\over RT})+{d\over
dT}(ln(a_{0}(T,P)))$

$={-c\over RT^{2}}+{d\over dT}({-\Delta G^{\circ}\over RT})+{d\over
dT}(ln(a_{0}(T,P)))$

$={\Delta H^{\circ}-c\over RT^{2}}+{d\over dT}(ln(a_{0}(T,P)))$

so that, performing the integration, using the fact that
$Q(T_{1},P^{\circ})=1$;

$ln(Q(T_{2}))-ln(Q(T_{1}))=ln(Q(T_{2}))={1\over R}\int_{T_{1}}^{T_{2}}{\Delta
H^{\circ}-c\over T^{2}}dT+ln(a_{0}(T_{2},P_{2}))-ln(a_{0}(T_{1},P^{\circ}))$

We have that, by the modification of Lemma 2.5;

$ln(Q(T_{2}))={c-\Delta G^{\circ}(T_{2})\over RT_{2}}+ln(a_{0}(T_{2},P))$

$ln(Q(T_{1}))=0$

so that, using the formula for $c$;

$ln(Q(T_{2}))=ln(Q(T_{2}))-ln(Q(T_{1}))$

$={c-\Delta G^{\circ}(T_{2})\over RT_{2}}+ln(a_{0}(T_{2},P))$

$={\Delta G^{\circ}(T_{1})-RT_{1}ln(a_{0}(T_{1},P^{\circ}))-\Delta
G^{\circ}(T_{2})\over RT_{2}}+ln(a_{0}(T_{2},P))$

$={1\over R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}-c\over
T^{2}}dT+ln(a_{0}(T_{2},P_{2}))-ln(a_{0}(T_{1},P^{\circ}))$

$={-1\over R}(\Delta H^{\circ}-c)({1\over T_{2}}-{1\over
T_{1}})+ln(a_{0}(T_{2},P_{2}))-ln(a_{0}(T_{1},P^{\circ}))$

and rearranging;

${\Delta G^{\circ}(T_{1})-RT_{1}ln(a_{0}(T_{1},P^{\circ}))-\Delta
G^{\circ}(T_{2})\over RT_{2}}={-1\over R}(\Delta H^{\circ}-c)({1\over
T_{2}}-{1\over T_{1}})-ln(a_{0}(T_{1},P^{\circ}))$

${\Delta G^{\circ}(T_{2})-\Delta G^{\circ}(T_{1})\over
T_{2}}+{RT_{1}ln(a_{0}(T_{1},P^{\circ}))\over T_{2}}$

$=(\Delta H^{\circ}-c)({1\over T_{2}}-{1\over
T_{1}})+Rln(a_{0}(T_{1},P^{\circ}))$

${\Delta G^{\circ}(T_{2})-\Delta G^{\circ}(T_{1})\over T_{2}}$

$=(\Delta H^{\circ}-c)({1\over T_{2}}-{1\over
T_{1}})+Rln(a_{0}(T_{1},P^{\circ}))-{RT_{1}ln(a_{0}(T_{1},P^{\circ}))\over
T_{2}}$

$=(\Delta H^{\circ}-(\Delta
G^{\circ}(T_{1})-RT_{1}ln(a_{0}(T_{1},P^{\circ}))))({1\over T_{2}}-{1\over
T_{1}})+Rln(a_{0}(T_{1},P^{\circ}))-{RT_{1}ln(a_{0}(T_{1},P^{\circ}))\over
T_{2}}$

$=(\Delta H^{\circ}-\Delta G^{\circ}(T_{1}))({1\over T_{2}}-{1\over
T_{1}})+RT_{1}ln(a_{0}(T_{1},P^{\circ}))({1\over T_{2}}-{1\over
T_{1}})+(R-{RT_{1}\over T_{2}})ln(a_{0}(T_{1},P^{\circ}))$

$=(\Delta H^{\circ}-\Delta G^{\circ}(T_{1}))({1\over T_{2}}-{1\over
T_{1}})+ln(a_{0}(T_{1},P^{\circ}))(RT_{1}({1\over T_{2}}-{1\over
T_{1}})+(R-{RT_{1}\over T_{2}}))$

$=(\Delta H^{\circ}-\Delta G^{\circ}(T_{1}))({1\over T_{2}}-{1\over T_{1}})$

so that, rearranging again;

$\Delta G^{\circ}(T_{1})({1\over T_{1}}+{1\over T_{2}}-{1\over T_{2}})={\Delta
G^{\circ}(T_{1})\over T_{1}}$

$={\Delta G^{\circ}(T_{2})\over T_{2}}-\Delta H^{\circ}({1\over T_{2}}-{1\over
T_{1}})$

to obtain;

$\Delta G^{\circ}(T_{1})={T_{1}\over T_{2}}\Delta G^{\circ}(T_{2})-\Delta
H^{\circ}({T_{1}\over T_{2}}-1)$

For the modification of Lemma 2.9, be careful to use the restricted summation
for $c$ substances, in the calculation of $({\partial({{\partial
G\over\partial\xi}})_{T,P}\over\partial T})_{P}$. while the calculation for
$dU$ involves $c+1$ substances, including the solvent. If $\epsilon=0$, we
have that;

$c=\Delta
G^{\circ}(T_{2})+RT_{2}ln(Q(T_{2},P^{\prime}))-RT_{2}ln(a_{0}(T_{2},P^{\prime}))$

$=\Delta G^{\circ}(T_{2})-RT_{2}ln(a_{0}(T_{2},P^{\prime\circ}))$

so that rearranging, $Q(T_{2},P^{\prime})={a_{0}(T_{2},P^{\prime})\over
a_{0}(T_{2},P^{\prime\circ})}$

and the claims about dynamic and quasi-chemical equilibrium lines follows from
$Q=c$ and $({\partial G\over\partial\xi})_{T,P}=c$, for $c\in\mathcal{R}_{\geq
0}$ and $c\in\mathcal{R}$ respectively. If $\epsilon\neq 0$, we use the
modification of Lemma 2.5, to obtain the formula for the activity coefficient;

$Q(T_{2},P^{\prime})=e^{({\partial
G\over\partial\xi})_{T,P}|{T_{2},P^{\prime}}-\Delta
G^{\circ}(T_{2})+RT_{2}ln(a_{0}(T_{2},P^{\prime}))\over RT_{2}}$

$=e^{\epsilon ln({P^{\prime}\over P^{\prime\circ}})\over
RT_{2}}a_{0}(T_{2},P^{\prime})$

Again, the determination of the dynamical and quasi-chemical equilibrium lines
follows again from rearrangement of $Q=c$ and $({\partial
G\over\partial\xi})_{T,P}=c$, for $c\in\mathcal{R}_{\geq 0}$ and
$c\in\mathcal{R}$ respectively.

∎

######  Lemma 5.3.

In Lemmas 0.4-0.12, for the case of a reaction like $(*)$ in Definition 5.1,
and considering a dilute solution with interaction of the solvent, replacing
$Q$ defined as $\prod_{i=1}^{c}a_{i}^{\nu_{i}}$ by;

$\prod_{i=0}^{c}a_{i}^{\nu_{i}}$

If we assume without approximation that $\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})$,
$0\leq i\leq c$, then the proofs go through as before, with the modification
that we have $c+1$ rather than $c$ substances.

######  Remarks 5.4.

In the case of a solvent with no interaction, if we define the activities by
$a_{i}=x_{i}$, $0\leq i\leq c$, with the definition of $Q$ as
$a_{0}\prod_{1\leq i\leq c}a_{i}^{\nu_{i}}$, then we need to modify the proofs
of Lemmas 2.14 and 2.17. This is done in Lemma 9.4. For the existence of a
feasible path, where we require that the $n_{0}$ term is fixed, see Remark
9.1, we need to change Lemma 2.16. Letting $d_{0}>0$ denote the fixed molar
amount of the solvent, we obtain the relation, modifying the proof of Lemma
2.16;

$d_{0}\prod_{i=1}^{c-1}({\nu_{i}\over\nu_{c}}n_{c}+d_{i})^{\nu_{i}}(t)n_{c}^{\nu_{c}}(t)=\epsilon(t)((\sum_{i=1}^{c-1}{\nu_{i}\over\nu_{c}}+1)n_{c}+\sum_{i=0}^{c-1}d_{i})^{w}(t)$

where $w=1+\sum_{i=1}^{c}\nu_{i}$.

In the $w>0$ case, absorb the constant $d_{0}$ into $\epsilon(t)$, by setting
$\epsilon_{1}(t)={\epsilon(t)\over d_{0}}>0$, and redefine
$\sigma=\sum_{i=0}^{c-1}d_{i}$. Then use the proof of Lemma 2.16, noting that
if $\nu_{i}=w$, for $1\leq i\leq p$, then
$w=1+\sum_{i=1}^{c}\nu_{i}>\sum_{i=1}^{p}\nu_{i}=\sum_{i=1}^{p}w=pw$ which is
a contradiction again. If $w<0$, then use reciprocality again to reduce to
$w>0$, replacing $d_{0}$ with ${1\over d_{0}}>0$. The $w=0$ case is again
similar, using the $w>0$ calculation.

##  6\. Dilute solutions with Henry’s Law for Solutes, Raoult’s Law for the
Solvent and Interaction of the Solvent

######  Definition 6.1.

As mentioned in Definition 2.1, we can consider an electrolyte as a solute in
a dilute solution and define the activities $a_{i}$, $0\leq i\leq c$, by;

$a_{0}=x_{0}\simeq 1$

$a_{i}=x_{i}$ $(1\leq i\leq c)$

and define;

$Q=a_{0}\prod_{i=1}^{c}a_{i}^{\nu_{i}}\simeq\prod_{i=1}^{c}a_{i}^{\nu_{i}}$
(no interaction of the solvent)

$Q=\prod_{i=0}^{c}a_{i}^{\nu_{i}}$ (interaction of the solvent)

We ideally have that $\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})$, $0\leq i\leq c$,
which involves a contradiction. By Henry’s Law, we have that
$P_{i}=k_{i}x_{i}$, $1\leq i\leq c$, so that, by Henry’s Law, phase
equilibrium and the ideal gas law;

$\mu_{i}^{(sol)}=\mu_{i}^{(g)\circ}+RTln({P_{i}\over P^{\circ}})$

$=\mu_{i}^{(g)\circ}+RTln({k_{i}x_{i}\over P^{\circ}})$

$=\mu_{i}^{(H)\circ}+RTln(x_{i})$ $(*)$

where $\mu_{i}^{(H)\circ}=\mu_{i}^{(g)\circ}+RTln({k_{i}\over P^{\circ}})$
$({\dagger})$. From $(*)$, we obtain that;

$\mu_{i}^{\circ}=\mu_{i}^{(H)\circ}+RTln(x_{i}(T,P^{\circ}))$
$({\dagger}{\dagger})$

and, for $1\leq i\leq c$;

$\mu_{i}=\mu_{i}^{\circ}-RTln(x_{i}(T,P^{\circ}))+RTln(x_{i})$

$=\mu_{i}^{\circ}+RTln(x_{i})+\kappa(T)$ $({\dagger}{\dagger}{\dagger})$

where, from phase equilibrium and $({\dagger}),({\dagger}{\dagger})$;

$\kappa(T)=-RTln(x_{i}(T,P^{\circ}))$

$=\mu_{i}^{(H)\circ}-\mu_{i}^{\circ}$

$=\mu_{i}^{(g)\circ}+RTln({k_{i}\over P^{\circ}})-\mu_{i}^{(sol)\circ}$

$=RTln({k_{i}\over P^{\circ}})$

We can use Raoult’s law for the solvent as $P_{0}=P_{0}^{*}x_{0}$, see [7].
Then, using equilibrium with an ideal gas mixture;

$\mu_{0}^{(sol)}=\mu_{0}^{\circ(g)}+RTln({P_{0}\over P^{\circ}})$

$=\mu_{0}^{\circ(g)}+RTln({P_{0}^{*}x_{0}\over P^{\circ}})$

$\mu_{0}^{\circ(g)}+RTln(x_{0})+RTln({P_{0}^{*}\over P^{\circ}})$

$=\mu_{0}^{*}(T,P_{0}^{*})+RTln(x_{0})$

$=\mu_{0}^{*}(T,P)+RTln(x_{0})+\theta(T,P)$ $(**)$

where $\theta(T,P)=\mu_{0}^{*}(T,P_{0}^{*})-\mu_{0}^{*}(T,P)\simeq 0$

so that;

$RTln(x_{0})=RTln({P_{i}\over P^{\circ}})-RTln({P_{i}^{*}\over P^{\circ}})$

and;

$\mu_{0}=\mu_{0}^{*}+RTln(x_{0})+\theta$

$=\mu_{0}^{*}+RTln({P_{0}\over P^{\circ}})-RTln({P_{0}^{*}\over
P^{\circ}})+\theta$ $(***)^{\prime}$

We also have, using the phase rule for the solvent in equilibrium with an
ideal gas mixture, that;

$\mu_{0}^{(g)}=\mu_{0}^{\circ(g)}+RTln({P_{0}\over P^{\circ}})$

$\mu_{0}^{(sol)}=\mu_{0}^{\circ(sol)}+RTln({P_{0}\over P^{\circ}})$ $(*)$

Combining $(*),(***)^{\prime}$, we obtain that;

$\mu_{0}^{*}=\mu_{0}-RTln({P_{0}\over P^{\circ}})+RTln({P_{0}^{*}\over
P^{\circ}})-\theta$

$=(\mu_{0}^{\circ}+RTln({P_{0}\over P^{\circ}}))-RTln({P_{0}\over
P^{\circ}})+RTln({P_{0}^{*}\over P^{\circ}})-\theta$

$=\mu_{0}^{\circ}+RTln({P_{0}^{*}\over P^{\circ}})-\theta$
$({\dagger})^{\prime}$

Letting $P_{0}^{*}=P^{\circ}$, we obtain that
$\mu_{0}^{*}(T,P^{\prime})=\mu_{0}^{\circ}-\theta$, for the corresponding
$P^{\prime}$, $(****)^{\prime}$

From $(**)$, $(****)^{\prime}$, we obtain that;

$\mu_{0}=\mu_{0}^{*}+RTln(x_{0})+\theta$

$=\mu_{0}^{*}(T,P^{\prime})+\delta+RTln(x_{0})+\theta$

$=\mu_{0}^{\circ}-\theta+\delta+RTln(x_{0})+\theta$

$=\mu_{0}^{\circ}+RTln(x_{0})+\delta$

where $\delta=\mu_{0}^{*}(T,P)-\mu_{0}^{*}(T,P^{\prime})\simeq 0$

Using the same calculation as before, we have that;

$\mu_{0}=\mu_{0}^{\circ}+RTln(x_{0})+\gamma(P)$

######  Lemma 6.2.

In the case of dilute solutions, with interaction of the solvent, a feasible
path $\gamma$ is a dynamic equilibrium path iff $pr(\gamma_{12})\subset
C_{c}$, for some $c\in\mathcal{R}_{>0}$ iff ${dQ\over dt}=0$.

###### Proof.

We have that;

$\prod_{i=0}^{c}a_{i}^{\nu_{i}}=\prod_{i=0}^{c}x_{i}^{\nu_{i}}$

$=Q=c=f$

where $x_{i}={n_{i}\over n}$. Now follow through the proof of Lemma 2.14, as
we are differentiating, the proof works with a constant $f>0$. ∎

We reformulate Lemmas 3.2 to 3.6 in this context, assuming Henry’s law for the
solutes and the solvent an ideal solution;

######  Lemma 6.3.

In the dilute solution case, with interaction of the solvent, for the energy
function $G$ involving $c+1$ uncharged species;

$({\partial G\over\partial\xi})_{T,P}=\Delta G^{\circ}+RTln(Q)+\epsilon$

where
$\epsilon(T,P)=\nu_{0}\gamma_{0}(P)+\sum_{i=1}^{c}\nu_{i}\kappa_{i}(T)\simeq
0$, $\gamma_{0}(P)\simeq 0$ and $\kappa_{i}(T)\simeq 0$ are the error term for
the $i$’th uncharged species in Definition 6.1, $1\leq i\leq c$.

###### Proof.

The proof is clear from Lemma 3.2.

∎

######  Lemma 6.4.

For a dilute solution, with interaction of the solvent, we have, using the
definition of $\epsilon(T,P)$ in Lemma 6.3, the error term $\gamma_{0}(P)$ and
the error terms $\kappa_{i}(T)$, $1\leq i\leq c$ in Definition 5.1, that the
same results as Lemma 3.3 hold, replacing $\epsilon(P)$ with $\epsilon(T,P)$
and $\delta$ with
$\epsilon(T,P^{\circ})=\nu_{0}\gamma_{0}(P^{\circ})+\sum_{i=1}^{c}\nu_{i}\kappa_{i}(T)$

###### Proof.

The proof is clear from the proof of Lemma 3.3.

∎

######  Lemma 6.5.

For a dilute solution, with interaction of the solvent, we have the same
result as Lemma 3.4 hold, replacing $\epsilon(P(T))$ by $\epsilon(T,P(T))$
along the quasi-chemical equilibrium lines.

###### Proof.

The proof is clear from the proof of Lemma 3.4

∎

######  Lemma 6.6.

For a dilute solution, with interaction of the solvent, we have the same
results as Lemma 3.5 hold, replacing $\epsilon(P^{\prime})$ by
$\epsilon(T_{1},P^{\prime})$. In particularly, if $\epsilon\neq 0$, we have
that;

$Q(T,P)=e^{\epsilon ln({P\over P^{\circ}})-\epsilon(T,P)\over RT}$

and, if $\epsilon=0$, we have that;

$Q(T,P)=e^{-\epsilon(T,P)\over RT}$

###### Proof.

The proof is clear from the proof of Lemma 3.5.

∎

######  Lemma 6.7.

For a dilute solution, with interaction of the solvent, we have that, if
$\epsilon\neq 0$;

$grad(Q)(T,P)=(({-\epsilon ln({P\over P^{\circ}})\over
RT^{2}}+{\epsilon(T,P)\over RT^{2}}-{{\partial\epsilon\over\partial
T}(T,P)\over RT})({P\over P^{\circ}})^{\epsilon\over RT}e^{-\epsilon(T,P)\over
RT},$

$({\epsilon P^{\circ}\over RTP}-{{\partial\epsilon\over\partial P}(T,P)\over
RT})({P\over P^{\circ}})^{\epsilon\over RT}e^{-\epsilon(T,P)\over RT})$

and, if $\epsilon=0$;

$grad(Q)(T,P)=(({\epsilon(T,P)\over RT^{2}}-{{\partial\epsilon\over\partial
T}(T,P)\over RT})e^{-\epsilon(T,P)\over RT},{-{\partial\epsilon\over\partial
P}(T,P)\over RT}e^{-\epsilon(T,P)\over RT})$

The paths of maximal reaction in the region $|grad(Q)(T,P)|>1$, $Q(T,P)>0$,
are given by implicit solutions to the differential equations;

${dP\over dT}={\epsilon TP^{\circ}-PT{\partial\epsilon\over\partial
P}(T,P)\over(-\epsilon Pln({P\over
P^{\circ}})+P\epsilon(T,P)-PT{\partial\epsilon\over\partial T}(T,P))}$

${dP\over dT}={-T{\partial\epsilon\over\partial
P}(T,P)\over\epsilon(T,P)-T{\partial\epsilon\over\partial T}(T,P)}$

respectively.

###### Proof.

The computation of $grad(Q)$ in both cases is a simple application of the
chain and product rules. Following the method of Lemma 3.6, noting the claim
about maximal reaction is still valid with the same definition of $Q$, if
$\epsilon\neq 0$, we compute;

${dP\over dT}={({\epsilon P^{\circ}\over RTP}-{{\partial\epsilon\over\partial
P}(T,P)\over RT})({P\over P^{\circ}})^{\epsilon\over RT}e^{-\epsilon(T,P)\over
RT}\over({-\epsilon ln({P\over P^{\circ}})\over RT^{2}}+{\epsilon(T,P)\over
RT^{2}}-{{\partial\epsilon\over\partial T}(T,P)\over RT})({P\over
P^{\circ}})^{\epsilon\over RT}e^{-\epsilon(T,P)\over RT}}$

$={\epsilon TP^{\circ}-PT{\partial\epsilon\over\partial P}(T,P)\over(-\epsilon
Pln({P\over P^{\circ}})+P\epsilon(T,P)-PT{\partial\epsilon\over\partial
T}(T,P))}$

and, if $\epsilon=0$;

${dP\over dT}={{{-{\partial\epsilon\over\partial P}(T,P)\over
RT}e^{-\epsilon(T,P)\over RT}}\over({\epsilon(T,P)\over
RT^{2}}-{{\partial\epsilon\over\partial T}(T,P)\over
RT})e^{-\epsilon(T,P)\over RT}}$

$={-T{\partial\epsilon\over\partial
P}(T,P)\over\epsilon(T,P)-T{\partial\epsilon\over\partial T}(T,P)}$

∎

##  7\. Dilute solutions with Fugacity and Interaction of the Solvent

######  Definition 7.1.

As mentioned in Definition 2.1, we can consider an electrolyte as a solute in
a dilute solution and define the activities $a_{i}$, $0\leq i\leq c$, by;

$a_{0}=\gamma_{0}x_{0}\simeq 1$

$a_{i}=\gamma_{i}x_{i}$ $(1\leq i\leq c)$

We can define the activity coefficient
$Q=a_{0}\prod_{i=1}^{c}a_{i}^{\nu_{i}}\simeq\prod_{i=1}^{c}a_{i}^{\nu_{i}}$,
but we will adopt a new convention, see below.

We have that $\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})$, $0\leq i\leq c$, which
involves the contradiction with the definition of activity for ideal
solutions. By the approximation of Henry’s Law for the solutes, we have that
$P_{i}=k_{i}x_{i}\gamma_{i}$, $1\leq i\leq c$, (convention (II)), see [7], so
that, by the approximation of Henry’s Law, phase equilibrium and the gas law
with fugacity $\delta_{i}$;

$\mu_{i}^{(sol)}=\mu_{i}^{(g)\circ}+RTln({\delta_{i}P_{i}\over P^{\circ}})$

$=\mu_{i}^{(g)\circ}+RTln({k_{i}x_{i}\gamma_{i}\over
P^{\circ}})+RTln(\delta_{i}(T,P))$

$=\mu_{i}^{(H)\circ}+RTln(\gamma_{i}x_{i})+RTln(\delta_{i}(T,P))$ $(*)$

where $\mu_{i}^{(H)\circ}=\mu_{i}^{(g)\circ}+RTln({k_{i}\over P^{\circ}})$
$({\dagger})$. From $(*)$, we obtain that;

$\mu_{i}^{\circ}=\mu_{i}^{(H)\circ}+RTln(\gamma_{i}x_{i}(T,P^{\circ}))+RTln(\delta_{i}(T,P^{\circ})$
$({\dagger}{\dagger})$

and, for $1\leq i\leq c$;

$\mu_{i}=\mu_{i}^{\circ}-RTln(\gamma_{i}x_{i}(T,P^{\circ}))-RTln(\delta_{i}(T,P^{\circ})+RTln(\gamma_{i}x_{i})+RTln(\delta_{i}(T,P))$

$=\mu_{i}^{\circ}+RTln(\gamma_{i}x_{i})+\kappa(T,P)$
$({\dagger}{\dagger}{\dagger})$

where, from phase equilibrium and $({\dagger}),({\dagger}{\dagger})$;

$\kappa(T,P)=-RTln(\gamma_{i}x_{i}(T,P^{\circ}))+RT(ln(\delta_{i}(T,P))-ln(\delta_{i}(T,P^{\circ}))$

$=\mu_{i}^{(H)\circ}-\mu_{i}^{\circ}+RTln(\delta_{i}(T,P^{\circ}))+RTln({\delta_{i}(T,P)\over\delta_{i}(T,P^{\circ})})$

$=\mu_{i}^{(g)\circ}+RTln({k_{i}\over
P^{\circ}})-\mu_{i}^{(sol)\circ}+RTln(\delta_{i}(T,P))$

$=RTln({k_{i}\over P^{\circ}})+RTln(\delta_{i}(T,P))$

$=RTln({k_{i}\delta_{i}(T,P)\over P^{\circ}})$

We can measure the correction in Raoult’s law for the solvent by
$P_{0}=\gamma_{0}P_{0}^{*}x_{0}$, (convention I), see [7]. Then, using the gas
law with fugacity $\delta_{0}$, and the correction $\sigma$ for the difference
of the chemical potential between a gas in a non ideal mixture and on its own,
we have, at equilibrium, that;

$\mu_{0}^{(sol)}=\mu_{0}^{\circ(g)}+RTln({\delta_{0}P_{0}\over P^{\circ}})$

$=\mu_{0}^{\circ(g)}+RTln({\delta_{0}\gamma_{0}P_{0}^{*}x_{0}\over
P^{\circ}})$

$\mu_{0}^{\circ(g)}+RTln(\gamma_{0}x_{0})+RTln({\delta_{0}P_{0}^{*}\over
P^{\circ}})$

$=\mu_{0}(T,P_{0}^{*})+RTln(\gamma_{0}x_{0})$

$=\mu_{0}^{*}(T,P_{0}^{*})+\sigma(T,P_{0}^{*})+RTln(\gamma_{0}x_{0})$

$=\mu_{0}^{*}(T,P)+RTln(\gamma_{0}x_{0})+\theta(T,P)$ $(**)$

where
$\theta(T,P)=\mu_{0}^{*}(T,P_{0}^{*})-\mu_{0}^{*}(T,P)+\sigma(T,P_{0}^{*})$

so that;

$RTln(\gamma_{(}0)x_{0})=RTln({\delta_{0}P_{0}\over
P^{\circ}})-RTln({\delta_{0}P_{0}^{*}\over P^{\circ}})$

and;

$\mu_{0}=\mu_{0}^{*}+RTln(\gamma_{0}x_{0})+\theta$

$=\mu_{0}^{*}+RTln({\delta_{0}P_{0}\over
P^{\circ}})-RTln({\delta_{0}P_{0}^{*}\over P^{\circ}})+\theta$
$(***)^{\prime}$

We also have, using the phase rule for the solvent in equilibrium, that;

$\mu_{0}^{(g)}=\mu_{0}^{\circ(g)}+RTln({\delta_{0}P_{0}\over P^{\circ}})$

$\mu_{0}^{(sol)}=\mu_{0}^{\circ(sol)}+RTln({\delta_{0}P_{0}\over P^{\circ}})$
$(*)$

Combining $(*),(***)^{\prime}$, we obtain that;

$\mu_{0}^{*}=\mu_{0}-RTln({\delta_{0}P_{0}\over
P^{\circ}})+RTln({\delta_{0}P_{0}^{*}\over P^{\circ}})-\theta$

$=(\mu_{0}^{\circ}+RTln({\delta_{0}P_{0}\over
P^{\circ}}))-RTln({\delta_{0}P_{0}\over
P^{\circ}})+RTln({\delta_{0}P_{0}^{*}\over P^{\circ}})-\theta$

$=\mu_{0}^{\circ}+RTln({\delta_{0}P_{0}^{*}\over P^{\circ}})-\theta$
$({\dagger})^{\prime}$

Letting $\delta_{0}P_{0}^{*}=P^{\circ}$, we obtain that
$\mu_{0}^{*}(T,P^{\prime})=\mu_{0}^{\circ}-\theta$, for the corresponding
$P^{\prime}$, $(****)^{\prime}$

From $(**)$, $(****)^{\prime}$, we obtain that;

$\mu_{0}=\mu_{0}^{*}+RTln(\gamma_{0}x_{0})+\theta$

$=\mu_{0}^{*}(T,P^{\prime})+\delta+RTln(\gamma_{0}x_{0})+\theta$

$=\mu_{0}^{\circ}-\theta+\delta+RTln(\gamma_{0}x_{0})+\theta$

$=\mu_{0}^{\circ}+RTln(\gamma_{0}x_{0})+\delta$

where $\delta=\mu_{0}^{*}(T,P)-\mu_{0}^{*}(T,P^{\prime})\simeq 0$

Using the same calculation as before, we have that;

$\mu_{0}=\mu_{0}^{\circ}+RTln(\gamma_{0}x_{0})+\lambda(P)$

We can define a new activity coefficient by
$Z=\prod_{i=0}^{c}b_{i}^{\nu_{i}}$, where;

$b_{0}=x_{0}$

$b_{i}=x_{i}$, $1\leq i\leq c$

From the above, we have that;

$\mu_{0}(T,P)=\mu_{0}^{\circ}+RTln(x_{0})+RTln(\gamma_{0}(T,P))+\delta_{0}(P)$

$=\mu_{0}^{\circ}+RTln(b_{0})+\phi_{0}(T,P)$

and, for $1\leq i\leq c$;

$\mu_{i}(T,P)=\mu_{i}^{\circ}+RTln(x_{i})+RTln(\gamma_{i}(T,P))+\kappa_{i}(T)$

$=\mu_{i}^{\circ}+RTln(b_{i})+\psi_{i}(T,P)$

######  Lemma 7.2.

In the case of dilute solutions, with interaction of the solvent, a feasible
path $\gamma$ is a dynamic equilibrium path iff $pr(\gamma_{12})\subset
C_{f}$, for some $f\in\mathcal{R}_{>0}$ iff ${dZ\over dt}=0$.

###### Proof.

We have that;

$\prod_{i=0}^{c}a_{i}^{\nu_{i}}=\prod_{i=0}^{c}x_{i}^{\nu_{i}}$

$=Z=f$

Now copy the proof of Lemma 6.2.

∎

We reformulate Lemmas 3.2 to 3.6 in this context, assuming the approximation
to Henry’s law for the solutes and the approximation to Raoult’s law for the
solvent;

######  Lemma 7.3.