Equilibria in electrochemistry and maximal rates of reaction

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

\[\begin{aligned} \label{2*} dG&={\partial G\over \partial T}_{P,n}dT+{\partial G\over \partial P}_{T,n}dP+\sum_{i=1}^{c}{\partial G\over \partial n_{i}}_{T,P,n'}dn_{i}\notag\\ &={\partial G\over \partial T}_{P,n}dT+{\partial G\over \partial P}_{T,n}dP+\sum_{i=1}^{c}\mu_{i}dn_{i}. \end{aligned} \tag{1}\] 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 (1), we have that;

\[\begin{aligned} \label{2**} dU&=TdS-PdV,\notag\\ dG&=dU+VdP+PdV-SdT-TdS\notag\\ &=(TdS-PdV)+VdP+PdV-SdT-TdS\notag\\ &=-SdT+VdP\notag\\ &={\partial G\over \partial T}_{P,n}dT+{\partial G\over \partial P}_{T,n}dP, \end{aligned} \tag{2}\] so that, from (2), and equating coefficients, \({\partial G\over \partial T}_{P,n}=-S\), \({\partial G\over \partial P}_{T,n}=V\). Substituting into (1), we obtain that; \[dG=-SdT+VdP+\sum_{i=1}^{c}\mu_{i}dn_{i}.\] ◻

Proof. For \(c\) substances with \(c'\) the number of the charged species, using Definition 1, we have that the electrostatic potential energy;

\[U_{el}= \sum_{i=1}^{c'}\phi(\overline{x}_{i})q_{i},\] where \(q_{i}=N_{i}ez_{i}=N_{A}n_{i}ez_{i}\),\(\{\overline{x}_{i}:1\leq i\leq c'\}\) 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;

\[\begin{aligned} G(T,P,n_{1},\ldots,n_{c})&=U+PV-TS\\ &=U_{chem}+U_{el}+PV-TS\\ &=U_{el}+G_{chem}\\ &=\sum_{j=1}^{c}\phi(\overline{x}_{j})q_{j}+G_{chem}&\\ &=\sum_{j=1}^{c}\phi(\overline{x}_{j})N_{A}n_{j}ez_{j}+G_{chem}, \end{aligned}\] so that; \[\begin{aligned} \label{21*} \mu_{i}&=({\partial G\over \partial n_{i}})_{T,P}\notag\\ &=({\partial (\sum_{j=1}^{c}\phi(\overline{x}_{j})N_{A}n_{j}ez_{j}+G_{chem})\over \partial n_{i}})_{T,P}\notag\\ &=\mu_{i,chem}, \qquad(c'+1\leq i\leq c)\notag\\ &=\mu_{i,chem}+{\partial (\phi(\overline{x}_{i})N_{A}n_{i}ez_{i})\over \partial n_{i}}, \quad(1\leq i\leq c')\notag\\ &=\mu_{i,chem}+\phi(\overline{x}_{i})N_{A}ez_{i}\notag\\ &=\mu_{i,chem}+\phi(\overline{x}_{i})F z_{i}. \end{aligned} \tag{3}\] 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 5, generalized to a collection involving charged species, using (3), we have that; \[\begin{aligned} \label{2dag} ({\partial G\over \partial \xi})_{T,P}&={\sum}_{i=1}^{c}\nu_{i}\mu_{i}\notag\\ &=2\mu(HCl)+2\mu(Ag)-\mu(H_{2})-2\mu(AgCl)+2\mu(e^{-}(L))-2\mu(e^{-}(R))\notag\\ &=({\partial G_{chem'}\over \partial \xi})_{T,P}+2\mu(e^{-}(L))-2\mu(e^{-}(R))\notag\\ &=({\partial G_{chem'}\over \partial \xi})_{T,P}+((2\mu_{chem}(e^{-}(L))-2F\phi(L))-(2\mu_{chem}(e^{-}(L))-2F\phi(R)))\notag\\ &=({\partial G_{chem'}\over \partial \xi})_{T,P}+2F(\phi(R)-\phi(L))\notag\\ &=({\partial G_{chem'}\over \partial \xi})_{T,P}+2EF=0, \end{aligned} \tag{4}\] where \(G_{chem'}\) is the Gibbs energy restricted to the uncharged species. We have that; \[\begin{aligned} \label{2dagdag} ({\partial G_{chem'}\over \partial \xi})_{T,P^{\circ}}&=\sum_{i=c'+1}^{c}\nu_{i}\mu_{i}^{\circ}\notag\\ &=(\Delta G_{chem'}^{\circ}+RTln(Q_{chem'}(T,P^{\circ})))\notag\\ &=\Delta G_{chem'}^{\circ}, \end{aligned} \tag{5}\] using the definition of \(Q_{chem'}\) in Definition 1, the fact that \(\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})\), \(\mu_{i}=\mu_{i}^{\circ}\), for \(c'+1\leq i\leq c\), so that \(Q_{chem'}(T,P^{\circ})=1\).

From (4),(5), we obtain; \[\begin{aligned} \label{2dagdagdag} 2E^{\circ}F&=-({\partial G_{chem'}\over \partial \xi})_{T,P^{\circ}}\notag\\ &=-\Delta G_{chem'}^{\circ}. \end{aligned}\ tag{6}\] Similarly, we have that; \[\begin{aligned} \label{2dagdagdagdag} ({\partial G_{chem'}\over \partial \xi})_{T,P}&=\sum_{i=c'+1}^{c}\nu_{i}\mu_{i}=(\Delta G_{chem'}^{\circ}+RTln(Q_{chem'}(T,P))), \end{aligned} \tag{7}\] so from (4),(7), we obtain that; \[\begin{aligned} \label{2sharp} 2EF&=-({\partial G_{chem'}\over \partial \xi})_{T,P}\notag\\ &=-(\Delta G_{chem'}^{\circ}+RTln(Q_{chem'}(T,P))). \end{aligned} \tag{8}\] Combining (8), (6), we obtain that; \[\begin{aligned} 2EF-2E^{\circ}F&=-(\Delta G_{chem'}^{\circ}+RTln(Q_{chem'}(T,P)))-(-\Delta G_{chem'}^{\circ})\notag\\ &=-RTln(Q_{chem'}(T,P)),\notag \end{aligned}\] so that; \[E-E^{\circ}=-{RTln(Q_{chem'}(T,P))\over 2F}.\] ◻

Proof. By Lemma 5, we have that; \[\begin{aligned} ({\partial G\over \partial \xi})_{T,P}&=\sum_{i=1}^{c}\nu_{i}\mu_{i}\notag\\ \Delta G^{\circ}&=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}.\label{22*} \end{aligned} \tag{9}\] Using (9), the fact that \(\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})\), and Definition 1, we have that; \[\begin{aligned} ({\partial G\over \partial \xi})_{T,P}-\Delta G^{\circ}&=\sum_{i=1}^{c}\nu_{i}(\mu_{i}-\mu_{i}^{\circ})\\ &=\sum_{i=1}^{c}\nu_{i}(\mu_{i}^{\circ}+RTln(a_{i})-\mu_{i}^{\circ})\\ &=\sum_{i=1}^{c}\nu_{i}RTln(a_{i})\\ &=RTln(\prod_{i=1}^{c}a_{i}^{\nu_{i}})=RTln(Q). \end{aligned}\] ◻

Proof. By Lemma 2, we have that

\[\begin{aligned} \label{23*}E-E^{\circ}=-{RTln(Q)\over 2F}, \end{aligned} \tag{10}\] and, by Lemma 3, we have that \[\begin{aligned} 0=({\partial G\over \partial \xi})_{T,P}=\Delta G^{\circ}+RTln(Q).\label{23**} \end{aligned} \tag{11}\] Rearranging (10) and (11), we obtain the result. ◻

Proof. For the first claim, using the definition of \(\xi\), we have \[\begin{aligned} \label{lem5*} dn_{i}&=\nu_{i}d\xi, &(1\leq i\leq c). \end{aligned} \tag{12}\] By Lemma 1, fixing \(T\) and \(P\), and using (12), we have \[\begin{aligned} dG=\sum_{i=1}^{c}\mu_{i}d n_{i}=(\sum_{i=1}^{c}\mu_{i}\nu_{i})d\xi,\label{lem5dag} \end{aligned} \tag{13}\] so that; \[\begin{aligned} ({\partial G\over \partial \xi})_{T,P}=\sum_{i=1}^{c}\mu_{i}\nu_{i}.\label{lem5dagdag} \end{aligned} \tag{14}\] The second claim from the first, as \[\begin{aligned} \Delta G^{\circ}(T)&=\int_{0}^{1}({\partial G\over \partial \xi})_{T,P^{\circ}}\\ &=\int_{0}^{1}(\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}(T))d\xi\\ &=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}(T)\int_{0}^{1}d\xi\\ &=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}(T). \end{aligned}\] 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 (13) and (14), we have \[\begin{aligned} dG=\left({\partial G\over \partial \xi}\right)_{T,P}=0,\;\;\;\;\;\;\; \text{(independently of $\xi$)}. \label{lem5dagdagdag} \end{aligned} \tag{15}\] At chemical equilibrium \(T,P^{\circ}\), using the first and second claims, and (15), we have

\[dG=\left({\partial G\over \partial \xi}\right)_{T,P^{\circ}}=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}=\Delta G^{\circ}=0.\] For the second to last claim, and the first direction, we have, by Lemma 3, that \(RTln(Q)=0\), so that \(Q=1\), and, by Lemma 2, that \(E-E^{\circ}=-{RTln(Q)\over 2F}=0\). For the converse, we have by Lemma 3, using the fact that \(Q(T,P)=1\);

\[\left({\partial G\over \partial \xi}\right)_{T,P}=\Delta G^{\circ},\] and, if chemical equilibrium exists at \((T,P^{\circ})\), then, as \(Q(T,P^{\circ})=1\) we have

\[({\partial G\over \partial \xi})_{T,P^{\circ}}=\Delta G^{\circ}+RTln(Q)=\Delta G^{\circ}=0,\] so that \(({\partial G\over \partial \xi})_{T,P}=0\).

For the penultimate claim, in one direction, use Lemma 3, together with the fact that \(({\partial G\over \partial \xi})_{T,P}=0\) and rearrange, the converse is also clear, by 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 1. ◻

Proof. By Lemma 5, we have \[\Delta G^{\circ}=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ},\] 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 \[({\partial \mu_{i}^{\circ}\over \partial T})_{P,n}=-({\partial S^{\circ}\over \partial n_{i}})_{T,P,n'}=-\overline{S}^{\circ}_{i},\] 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; \[\begin{aligned} {d(\Delta G^{\circ})\over dT}=-\sum_{i=1}^{c}\nu_{i}\overline{S}^{\circ}_{i}=-\sum_{i=1}^{c}\nu_{i}\overline{S}^{\circ}_{m,i}=-\Delta S^{\circ}.\label{lem6*} \end{aligned} \tag{20}\] Using the product rule, (20) and the definition of enthalpy, we have \[\begin{aligned} {d\over dT}({\Delta G^{\circ}\over T})={1\over T}{d(\Delta G^{\circ})\over dT}-{1\over T^{2}}\Delta G^{\circ} =-{\Delta S^{\circ}\over T}-{\Delta G^{\circ}\over T^{2}} =-{\Delta (ST+G)^{\circ}\over T^{2}} =-{\Delta H^{\circ}\over T^{2}}.\label{lem6**} \end{aligned} \tag{21}\] By Lemma 5, 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 (21) that; \[{d ln(Q)\over dT}={d\over dT}({-\Delta G^{\circ}\over RT}) ={\Delta H^{\circ}\over RT^{2}}.\] It follows, integrating between \(T_{1}\) and \(T_{2}\), that;
\[\begin{aligned} ln({Q(T_{2})\over Q(T_{1})})&=ln(Q)(T_{2})-ln(Q)(T_{1})\notag\\ &={-\Delta G^{\circ}(T_{2})\over RT_{2}}+{\Delta G^{\circ}(T_{1})\over RT_{1}}\notag\\ &=\int_{T_{1}}^{T_{2}}{d ln(Q)\over dT}dT\notag\\ &={1\over R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}\over T^{2}}.\label{lem6P} \end{aligned} \tag{22}\] So that, rearranging, we obtain the first claim. Using the fact, by Lemma 3, that; \[\begin{aligned} &ln(Q(T_{2}))=-{\Delta G^{\circ}(T_{2})\over RT_{2}},\\ &ln(Q(T_{1}))=-{\Delta G^{\circ}(T_{1})\over RT_{1}}, \end{aligned}\] we obtain, substituting into (22), canceling \(R\), and performing the integration, if \(\Delta H^{\circ}\) is temperature independent, that; \[\begin{aligned} {\Delta G^{\circ}(T_{2})\over T_{2}}-{\Delta G^{\circ}(T_{1})\over T_{1}}=-\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}\over T^{2}}=\Delta H^{\circ}({1\over T_{2}}-{1\over T_{1}}).\label{lem6Q} \end{aligned} \tag{23}\] For the fifth claim, rearrange (23). If \(D_{c}\) intersects the line \(P=P^{\circ}\) at \((T_{1},P^{\circ})\), for the sixth (16) and seventh claims, we have, using Lemma 3 and the fact from Lemma 5 that \(Q(T_{1},P^{\circ})=1\); \[\begin{aligned} ({\partial G\over \partial \xi})_{T_{2},P}&=\Delta G^{\circ}(T_{2})+RT_{2}ln(Q(T_{2},P))\\ &=({\partial G\over \partial \xi})_{T_{1},P^{\circ}}\\ &=\Delta G^{\circ}(T_{1})+RT_{1}ln(Q(T_{1},P^{\circ}))\\ &=\Delta G^{\circ}(T_{1})=c, \end{aligned}\] so that, again rearranging, we obtain the result. Along \(D_{c}\), we have, using Lemma 3, that; \[ln(Q)={c-\Delta G^{\circ}\over RT},\] so that, using the first part; \[\begin{aligned} {d ln(Q)\over dT}&={d\over dT}({c-\Delta G^{\circ}\over RT})\\ &={-c\over RT^{2}}+{d\over dT}({-\Delta G^{\circ}\over RT})\\ &={\Delta H^{\circ}-c\over RT^{2}}, \end{aligned}\] 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 3; \[\begin{aligned} &ln (Q(T_{2}))={c-\Delta G^{\circ}(T_{2})\over RT_{2}},\\ &ln(Q(T_{1}))=0, \end{aligned}\] so that; \[\begin{aligned} ln (Q(T_{2}))&=ln (Q(T_{2}))-ln(Q(T_{1}))\\ &={c-\Delta G^{\circ}(T_{2})\over RT_{2}}\\ &={\Delta G^{\circ}(T_{1})-\Delta G^{\circ}(T_{2})\over RT_{2}}\\ &={1\over R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}-c\over T^{2}}dT\\ &={-1\over R}(\Delta H^{\circ}-c)({1\over T_{2}}-{1\over T_{1}}), \end{aligned}\] 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; \[\Delta G^{\circ}(T_{1})={T_{1}\over T_{2}}\Delta G^{\circ}(T_{2})-\Delta H^{\circ}({T_{1}\over T_{2}}-1).\] ◻

Proof. For the first claim, by Lemma 6, 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 5 and the proof of Lemma 6, we have that \[\begin{aligned} ({\partial ({dG\over d\xi})_{T,P}\over \partial T})_{P}&=({\partial (\sum_{i=1}^{c} \nu_{i}\mu_{i})\over \partial T})_{P}\notag\\ &=\sum_{i=1}^{c} \nu_{i}({\partial \mu_{i}\over \partial T})_{P}\notag\\ &=\sum_{i=1}^{c} \nu_{i}({\partial \mu_{i}\over \partial T})_{P,n}\notag\\ &=-\sum_{i=1}^{c} \nu_{i}\overline{S}_{m,i}.\label{lem7*} \end{aligned} \tag{24}\] 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; \[\begin{aligned} dU&=\sum_{i=1}^{c}{3\over 2}N_{A}kTdn_{i}+\sum_{i=1}^{c}{3\over 2}N_{A}kn_{i}dT-\sum_{i=1}^{c}N_{A}m_{i}\rho_{i}dn_{i},\\ dQ&=\sum_{i=1}^{c}{3\over 2}N_{A}kTdn_{i}+\sum_{i=1}^{c}{3\over 2}N_{A}kn_{i}dT-\sum_{i=1}^{c}N_{A}m_{i}\rho_{i}dn_{i}+dL,\\ {dQ\over T}&=\sum_{i=1}^{c}{3\over 2}N_{A}kdn_{i}+\sum_{i=1}^{c}{3\over 2}N_{A}kn_{i}{dT\over T}-\sum_{i=1}^{c}N_{A}m_{i}\rho_{i}{dn_{i}\over T}+{g(T,\overline{n})dT\over T} +\sum_{i=1}^{c}h_{i}(T,\overline{n}){dn_{i}\over T},\\ ({dQ\over T})_{n',T,P}&={3\over 2}N_{A}kdn_{i}-N_{A}m_{i}\rho_{i}{dn_{i}\over T}+h_{i}(T,\overline{n}){dn_{i}\over T}. \end{aligned}\] It follows that \[\begin{aligned} \overline{S}_{m,i}=\int_{\Delta n_{i}=1}({dQ\over T})_{n',T,P}={3\over 2}N_{A}k-{N_{A}m_{i}\rho_{i}\over T}+{k_{i}(T)\over T}.\label{lem7**} \end{aligned} \tag{25}\] So that, from (24); \[\begin{aligned} ({\partial ({dG\over d\xi})_{T,P}\over \partial T})_{P}&=-\sum_{i=1}^{c} \nu_{i}({3\over 2}N_{A}k-{N_{A}m_{i}\rho_{i}\over T})-\sum_{i=1}^{c}\nu_{i}{k_{i}(T)\over T}\notag\\ &=-{3\over 2}N_{A}k(\sum_{i=1}^{c} \nu_{i})+{N_{A}\over T}\sum_{i=1}^{c}\nu_{i}\mu_{i}\rho_{i}-\sum_{i=1}^{c}\nu_{i}{k_{i}(T)\over T}\notag\\ &=-{3\over 2}N_{A}k(\sum_{i=1}^{c} \nu_{i})+{N_{A}\over T}\sum_{i=1}^{c}\nu_{i}\mu_{i}\rho_{i}-{G(T)\over T}.\label{lem7***} \end{aligned} \tag{26}\] From (26), 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 \[\begin{aligned} ({\partial ({dG\over d\xi})_{T,P}\over \partial P})_{T}&=({\partial (\sum_{i=1}^{c} \nu_{i}\mu_{i})\over \partial P})_{T}\notag\\ &=\sum_{i=1}^{c} \nu_{i}({\partial \mu_{i}\over \partial P})_{T}\notag\\ &=\sum_{i=1}^{c} \nu_{i}({\partial \mu_{i}\over \partial P})_{T,n}\notag\\ &=\sum_{i=1}^{c} \nu_{i}({\partial V\over \partial n_{i}})_{T,P,n'}\notag\\ &=\sum_{i=1}^{c} \nu_{i}\overline{V}_{i}\notag\\ &=\sum_{i=1}^{c}\nu_{i}{N_{A}m_{i}\over \kappa_{i}(T,P)},\label{lem7A} \end{aligned} \tag{27}\] where \(\kappa_{i}\) is the density of substance \(i\). We also have that

\[\begin{aligned} P(\sum_{i=1}^{c}\nu_{i}{N_{A}m_{i}\over \kappa_{i}(T,P)})&=P(\sum_{i=1}^{c} \nu_{i}\overline{V}_{i})=G(T),&(dL=PdV),\label{lem7C} \end{aligned} \tag{28}\] and from (27) and (28); \[P({\partial ({dG\over d\xi})_{T,P}\over \partial P})_{T}=G(T) =P\alpha'(P),\] so that \(G(T)=\epsilon\), \(\alpha(P)=\lambda+\epsilon ln(P)\), \(({dG\over d\xi})_{T,P}\) is of the form \[\begin{aligned} \alpha(P)+\beta T+\gamma ln(T)-\int {G(T)\over T} &=\lambda+\epsilon ln(P)+\beta T+\gamma ln(T)-\epsilon ln(T)\notag\\ &=\lambda+\epsilon ln(P)+\beta T+\sigma ln(T),\label{lem7D} \end{aligned} \tag{29}\] 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 11 and 12, 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 (29), 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; \[\begin{aligned} &\sigma=0,\\ &\lambda+\epsilon ln(P^{\circ})=\Delta H^{\circ},\\ &\beta={(\Delta G^{\circ}(T_{1})-\Delta H^{\circ})\over T_{1}},\\ &\Delta G^{\circ}(T_{1})=\beta T_{1}+\Delta H^{\circ},\\ &\Delta G^{\circ}(T_{2})=\beta T_{2}+\Delta H^{\circ}. \end{aligned}\] We can then, using Lemma 3, obtain a formula for the activity coefficient;
\[\begin{aligned} Q(T_{2},P')=e^{{(({\partial G\over \partial \xi})_{T,P}|_{T_{2},P'}-\Delta G^{\circ}(T_{2}))\over RT_{2}}} e^{{(\Delta H^{\circ}-\epsilon ln(P'^{\circ})+\epsilon ln(P')+\beta T_{2}-(\beta T_{2}+\Delta H^{\circ}))\over RT_{2}}} =e^{{\epsilon ln({P'\over P'^{\circ}})\over RT_{2}}},\label{lem7W} \end{aligned} \tag{30}\] 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 Definition 1 and Lemma 13, follows from a simple rearrangement of the formulas \(Q(T_{1},P')=c\), for some \(c\in\mathcal{R}_{\geq 0}\), using (30), and \(({dG\over d\xi})_{T,P}=c\), for some \(c\in \mathcal{R}\), using (29), with \(\sigma=0\). ◻

Proof. By the proof of Lemma 7, if \(T_{0}\) is a solution to; \[\begin{aligned} \lambda+\beta T+\sigma ln(T)=0,\label{lem8*} \end{aligned} \tag{31}\] 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. ◻

Proof. By (4) of Lemma 2, we have that \[\begin{aligned} ({\partial G\over \partial \xi})_{T,P}=({\partial G_{chem'}\over \partial \xi})_{T,P}+2EF.\label{lem9*} \end{aligned} \tag{32}\] By Lemma 7, we have that \(({\partial G_{chem'}\over \partial \xi})_{T,P}\) is independent of \(P\), in particularly, we have that \[\begin{aligned} ({\partial G_{chem'}\over \partial \xi})_{T_{1},P_{1}}=({\partial G_{chem'}\over \partial \xi})_{T_{1},P_{1}^{\circ}},\label{lem9**} \end{aligned} \tag{33}\] so that, combining (32) and (33), we obtain the result. ◻

Proof. Following the proof of Lemma 2, we have that \[\begin{aligned} ({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}}=({\partial G_{chem'}\over \partial \xi})_{T,P}|_{T_{1},P_{1}}+2E(T_{1},P_{1})F,\label{lem10*} \end{aligned} \tag{34}\] \[\begin{aligned} 2E^{\circ}(T_{1})F&=({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-({\partial G_{chem'}\over \partial \xi})_{T,P}|_{T_{1},P_{1}^{\circ}}\notag\\ &=({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-\Delta G^{\circ}_{chem'}(T_{1}).\label{lem10**} \end{aligned} \tag{35}\] So from (34) and (35) and Lemma 3; \[\begin{aligned} 2E(T_{1},P_{1})F&=({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}}-({\partial G_{chem'}\over \partial \xi})_{T,P}|_{T_{1},P_{1}}\\ &=({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}}-(\Delta G^{\circ}_{chem'}(T_{1})+RT_{1}ln(Q_{chem'}(T_{1},P_{1}))). \end{aligned}\] \[\begin{aligned} 2E(T_{1},P_{1})F-2E^{\circ}(T_{1})F&=({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}}-(\Delta G^{\circ}_{chem'}(T_{1})+RT_{1}ln(Q_{chem'}(T_{1},P_{1}))) -(({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-\Delta G^{\circ}_{chem'}(T_{1}))\\ &=({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}}-(({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-RT_{1}ln(Q_{chem'}(T_{1},P_{1})). \end{aligned}\] ◻

Proof. By Lemma 5, and the definition of activities for an ideal solution, we have that \[\begin{aligned} 1=Q(T,P)=\prod_{i=1}^{c}a_{i}^{\nu_{i}}=\prod_{i=1}^{c}x_{i}^{\nu_{i}},\label{lem11*} \end{aligned} \tag{36}\] and \({n_{i}'\over \nu_{i}}={n_{j}'\over \nu_{j}}\), for \(1\leq i,j\leq c\), \(\sum_{i=1}^{c}n_{i}=n\).

Using the relation (36), differentiating and using the facts that, for \(1\leq i\leq c-1\); \[\begin{aligned} n_{i}'={\nu_{i}n_{c}'\over \nu_{c}}, \quad n_{i}={\nu_{i}n_{c}\over \nu_{c}}+d_{i},\label{lem11!} \end{aligned} \tag{37}\] we obtain that \[\begin{aligned} (\prod_{i=1}^{c}x_{i}^{\nu_{i}})'&=\sum_{i=1}^{c}\nu_{i}x_{i}^{\nu_{i}-1}x_{i}'\prod_{j\neq i}x_{j}^{\nu_{j}} =\sum_{i=1}^{c}\nu_{i}x_{i}^{\nu_{i}-1}x_{i}'x_{i}^{-\nu_{i}}=\sum_{i=1}^{c}\nu_{i}x_{i}^{-1}x_{i}'\notag\\ &=\sum_{i=1}^{c}\nu_{i}{n\over n_{i}}{(n_{i}'n-n_{i}n')\over n^{2}} =\sum_{i=1}^{c}\nu_{i}({n_{i}'\over n_{i}}-{n'\over n})\notag\\ &=\sum_{i=1}^{c-1}{\nu_{i}^{2}n_{c}'\over \nu_{i}n_{c}+\nu_{c}d_{i}}+{\nu_{c}n_{c}'\over n_{c}}- \lambda({\sum_{i=1}^{c}n_{i}'\over \sum_{i=1}^{c}n_{i}})\notag\\ &=\sum_{i=1}^{c-1}{\nu_{i}^{2}n_{c}'\over \nu_{i}n_{c}+\nu_{c}d_{i}}+{\nu_{c}n_{c}'\over n_{c}}- \lambda({(\sum_{i=1}^{c-1}{\nu_{i}\over \nu_{c}}+1)n_{c}'\over (\sum_{i=1}^{c-1}{\nu_{i}\over \nu_{c}}+1)n_{c}+\sum_{i=1}^{c-1}d_{i}}) =0,\label{lem11B} \end{aligned} \tag{38}\] where \(\lambda=\sum_{i=1}^{c}\nu_{i}\). If \(\nu_{c}'\neq 0\), 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 \[\begin{aligned} \gamma_{c}&=\sum_{i=1}^{c-1}\nu_{i}^{2}\alpha\prod_{j\neq i}\nu_{j}+\alpha \nu_{c}\prod_{i=1}^{c-1}\nu_{i}-\lambda\prod_{i=1}^{c-1}\nu_{i}\\ &=\alpha\delta\sum_{i=1}^{c-1}\nu_{i}+\alpha\delta\nu_{c}-\lambda\delta\\ &=\delta(\alpha(\sum_{i=1}^{c}\nu_{i})-\lambda)\\ &=\delta\lambda(\alpha-1). \end{aligned}\]

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 \(n_{c}'=0\). By the connecting relations (37), we obtain that \(n_{j}'=0\) 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 (36), that \[\begin{aligned} \prod_{i=1}^{c}n_{i}^{\nu_{i}}=n^{\sum_{i=1}^{c}\nu_{i}}=1,\label{lem11A} \end{aligned} \tag{39}\] and, following the calculation (38), we obtain \[\sum_{i=1}^{c-1}{\nu_{i}^{2}n_{c}'\over \nu_{i}n_{c}+\nu_{c}d_{i}}+{\nu_{c}n_{c}'\over n_{c}}=0,\] so that, if \(n_{c}'\neq 0\); \[\sum_{i=1}^{c-1}{\nu_{i}^{2}\over \nu_{i}n_{c}+\nu_{c}d_{i}}+{\nu_{c}\over n_{c}}=0,\] and, differentiating \(k\) times, for \(k\geq 0\), canceling \(n_{c}'\) if \(n_{c}'\neq 0\), 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 \(k\geq 0\); \[\begin{aligned} \sum_{i=1}^{c-1}\nu_{i}^{k+2}g_{i}^{k+1}+\nu_{c}=\sum_{i=1}^{c-1}\nu_{i}(\nu_{i}g_{i})^{k+1}+\nu_{c} =0.\label{lem11E} \end{aligned} \tag{40}\] 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 \[n_{c}=-(\nu_{i}n_{c}+\nu_{c}d_{i}),\] implies \[(1+\nu_{i})n_{c}=\nu_{c}d_{i}.\] Then if \(n_{c}'\neq 0\), we must have that \(\nu_{i}=-1\), \(d_{i}=0\). Re-scaling each \(\nu_{i}\) by a factor of \(2\), we still have the conditions (36), \((E)\), \((G)\), 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 \(n_{c}'=0\), and \(n_{i}'=0\), 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; \[\sum_{1\leq i\leq c}\nu_{i}=\sum_{i\in I_{j}}\nu_{i}+\nu_{j} =\nu_{j},\;\;\;\text{ for}\;\;\;1\leq j\leq c,\] 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 5 that \(Q(T,P^{\circ})=1\). ◻

Proof. As \(W\) is smooth, we can choose a local parametrization \(\delta:[0,1]\rightarrow W\). 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 §13. We have that \[\begin{aligned} \prod_{i=1}^{c}x_{i}^{\nu_{i}}(t)=\epsilon(t),\label{lem12dag} \end{aligned} \tag{41}\] 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 \[\begin{aligned} \alpha n_{c}^{\kappa}+r(n_{c})=\epsilon(t)(\beta n_{c}^{\kappa}+s(n_{c})),\label{lem12dagdagdag} \end{aligned} \tag{42}\] 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 (42) by \(n_{c}^{\kappa}>0\), we obtain \[\begin{aligned} \alpha+r_{1}({1\over n_{c}})=\epsilon(t)(\beta+s_{1}({1\over n_{c}})),\label{lem12C} \end{aligned} \tag{43}\] 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 \({-\nu_{c}d_{i_{0}}\over \nu_{i_{0}}}\neq {-\nu_{c}d_{i_{0}}\over w}\),(¹). 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 \(v_{i}'={1\over v_{i}}\), \(1\leq i\leq p\). 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 \(v_{0}>v_{i}'\), for \(1\leq i\leq p\), where \(q(x)={\alpha+r_{1}(x)\over \beta+s_{i}(x)}\). Then \({1\over v_{0}}<{1\over v_{i}'}=v_{i}\), 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 \[\begin{aligned} n_{i}(0)={\nu_{i}\over \nu_{c}}n_{c}(0)+d_{i}={\nu_{i}\over \nu_{c}}{1\over v_{0}}+d_{i}>0,\quad \text{as well, for}\quad i+1\leq i\leq p-1,\label{lem12D} \end{aligned} \tag{44}\]

By (43), 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 \((q'(x),-1)\), if \(q'(v_{0})\neq 0\), we see that the projection \(pr_{y}\) is un-ramified 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 (44), 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 (41), 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 \(1\leq i\leq p'\), \(\nu_{i}<0\), for \(p'+1\leq i\leq c-1\), 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 \(\prod_{i=p+1}^{c-1}({\nu_{i}\over \nu_{c}})^{-\nu_{i}}\neq 0\). ◻

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 11. Now follow through Lemma 11, 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 \[(\prod_{i=1}^{c}x_{i}^{\nu_{i}})'|_{0}=Q(\gamma(t))'|_{0} =(grad(Q)|_{\gamma(0)}\centerdot \gamma'(0))\neq 0,\] 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 \(x_{i}'=0\), for \(1\leq i\leq c\) and \((\prod_{i=1}^{c}x_{i}^{\nu_{i}})'|_{0}=0\), 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 4 and the proof of Lemma 11, 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. ◻

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 5 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 5 that \(Q=1\) iff \(({\partial G\over \partial \xi})|_{T,P}-\Delta G^{\circ}=0\). The second claim is clear from Lemma 13.

For the third claim, we have, by Lemma 5, that along \(D'\), \(Q=e^{\Delta G^{\circ}\over RT}\), so that clearly \(Q\) is constant along \(D'\) 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 \(C_{c}'\).

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 13, \(pr_{12}(\gamma)\subset C_{c}\), for some \(c\in\mathcal{R}\). It follows that \(Q\) is constant along the \(pr_{12}(\gamma)\subset D'\) for some component \(D'\), and then, by the fifth claim, \(pr_{1}(\gamma)\subset pr_{1}(D')\) 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 11, so that \(P=P^{\circ}\) lies in \(C_{1}\). It follows, by Lemma 13, that a feasible path \(\gamma\) with \(pr_{12}(\gamma)\subset P=P^{\circ}\) is a dynamic equilibrium path. ◻

Proof. By Lemma 5, we have that \[\begin{aligned} &({\partial G\over \partial \xi})_{T,P}=\sum_{i=1}^{c}\nu_{i}\mu_{i}, &\Delta G^{\circ}=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}.\label{lem15*} \end{aligned} \tag{53}\] Using (53), the fact that \(\mu_{i}=\mu_{i}^{\circ}+RTln(a_{i})+\gamma_{i}\), and Definition 1, we have that \[\begin{aligned} ({\partial G\over \partial \xi})_{T,P}-\Delta G^{\circ}&=\sum_{i=1}^{c}\nu_{i}(\mu_{i}-\mu_{i}^{\circ}) =\sum_{i=1}^{c}\nu_{i}(\mu_{i}^{\circ}+RTln(a_{i})+\gamma_{i}-\mu_{i}^{\circ})\\ &=\sum_{i=1}^{c}\nu_{i}RTln(a_{i})+\sum_{i=1}^{c}\nu_{i}\gamma_{i} =RTln(\prod_{i=1}^{c}a_{i}^{\nu_{i}})+\sum_{i=1}^{c}\nu_{i}\gamma_{i}=RTln(Q)+\epsilon. \end{aligned}\] ◻

Proof. For the first claim, we have, using the definition of \(\xi\), that \[\begin{aligned} dn_{i}&=\nu_{i}d\xi,& (1\leq i\leq c).\label{lem16*} \end{aligned} \tag{54}\] By Lemma 1, fixing \(T\) and \(P\), and using (54), we have that \[\begin{aligned} dG=\sum_{i=1}^{c}\mu_{i}d n_{i}=(\sum_{i=1}^{c}\mu_{i}\nu_{i})d\xi,\label{lem16dag} \end{aligned} \tag{55}\] so that \[\begin{aligned} ({\partial G\over \partial \xi})_{T,P}=\sum_{i=1}^{c}\mu_{i}\nu_{i}.\label{lem16dagdag} \end{aligned} \tag{56}\]

The second claim follows 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 (55) and (56), we have that \[\begin{aligned} dG&=({\partial G\over \partial \xi})_{T,P}=0, &\text{(independently of $\xi$)}.\label{lem16dagdagdag} \end{aligned} \tag{57}\] At chemical equilibrium \((T,P^{\circ})\), using the first and second claims, and (57), 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 3, that \(RTln(Q)=-\epsilon\simeq 0\), so that \(Q(T,P)=e^{-\epsilon(P)\over RT}\simeq 1\), and, by Lemma 20, 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 3, 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 3, 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 1. ◻

Proof. By Lemma 5, we have that \[\begin{aligned} \Delta G^{\circ}=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}, \end{aligned}\] 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 \[({\partial \mu_{i}^{\circ}\over \partial T})_{P,n}=-({\partial S^{\circ}\over \partial n_{i}})_{T,P,n'}=-\overline{S}^{\circ}_{i},\] 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; \[\begin{aligned} {d(\Delta G^{\circ})\over dT}=-\sum_{i=1}^{c}\nu_{i}\overline{S}^{\circ}_{i}= =-\sum_{i=1}^{c}\nu_{i}\overline{S}^{\circ}_{m,i} =-\Delta S^{\circ}.\label{lem17*} \end{aligned} \tag{62}\] Using the product rule, (62) and the definition of enthalpy, we have that \[\begin{aligned} {d\over dT}({\Delta G^{\circ}\over T})={1\over T}{d(\Delta G^{\circ})\over dT}-{1\over T^{2}}\Delta G^{\circ} =-{\Delta S^{\circ}\over T}-{\Delta G^{\circ}\over T^{2}} =-{\Delta (ST+G)^{\circ}\over T^{2}} =-{\Delta H^{\circ}\over T^{2}}.\label{lem17**} \end{aligned} \tag{63}\] By Lemma 16, 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 (63) that \[{d ln(Q)\over dT}={d\over dT}({-\Delta G^{\circ}\over RT})-{d\over dT}({\epsilon(P)\over RT}) ={\Delta H^{\circ}\over RT^{2}}-{d\over dT}({\epsilon(P)\over RT}).\] It follows, integrating between \(T_{1}\) and \(T_{2}\), and using the fundamental theorem of calculus, that \[\begin{aligned} ln({Q(T_{2})\over Q(T_{1})})&=ln(Q)(T_{2})-ln(Q)(T_{1})\notag\\ &=\int_{T_{1}}^{T_{2}}{d ln(Q)\over dT}dT\notag\\ &={1\over R}\int_{T_{1}}^{T_{2}}[{\Delta H^{\circ}\over T^{2}}-{d\over dT}({\epsilon(P)\over RT})]dT\notag\\ &={-\Delta H^{\circ}\over RT_{2}}+{\Delta H^{\circ}\over RT_{1}}-({\epsilon(P(T_{2}))\over RT_{2}}-{\epsilon(P(T_{1}))\over RT_{1}}),\label{lem17P} \end{aligned} \tag{64}\] so that, rearranging, we obtain the first claim. Using the fact, by Lemma 15, that \[\begin{aligned} ln(Q(T_{2}))&={-\Delta G^{\circ}(T_{2})-\epsilon(P(T_{2}))\over RT_{2}},\\ ln(Q(T_{1}))&={-\Delta G^{\circ}(T_{1})-\epsilon(P(T_{1}))\over RT_{1}}, \end{aligned}\] we obtain, substituting into (64), 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 \[\begin{aligned} {\Delta G^{\circ}(T_{2})\over T_{2}}-{\Delta G^{\circ}(T_{1})\over T_{1}}=\Delta H^{\circ}({1\over T_{2}}-{1\over T_{1}}).\label{lem17Q} \end{aligned} \tag{65}\] For the fifth claim, rearrange (65). If \(D_{c}\) intersects the line \(P=P^{\circ}\) at \((T_{1},P^{\circ})\), for the sixth (58) and seventh claims, we have, using Lemma 3 and the fact from Lemma 16 that \(Q(T_{1},P^{\circ})=e^{-\delta\over RT_{1}}\); \[\begin{aligned} ({\partial G\over \partial \xi})_{T_{2},P}&=\Delta G^{\circ}(T_{2})+RT_{2}ln(Q(T_{2},P))+\epsilon(P(T_{2})) =({\partial G\over \partial \xi})_{T_{1},P^{\circ}}\\ &=\Delta G^{\circ}(T_{1})+RT_{1}ln(Q(T_{1},P^{\circ}))+\epsilon(P(T_{1}))\\ &=\Delta G^{\circ}(T_{1})+RT_{1}ln(Q(T_{1},P^{\circ}))+\epsilon(P^{\circ})\\ &=\Delta G^{\circ}(T_{1})-\delta+\epsilon(P^{\circ})\\ &=\Delta G^{\circ}(T_{1})=c, \end{aligned}\] so that, again rearranging, we obtain the result. Along \(D_{c}\), we have, using Lemma 3, that \[ln(Q(T))={c-\Delta G^{\circ}(T)-\epsilon(P(T))\over RT},\] so that, using the first part; \[{d ln(Q)\over dT}={d\over dT}({c-\Delta G^{\circ}(T)-\epsilon(P(T))\over RT}) ={-c\over RT^{2}}+{d\over dT}({-\Delta G^{\circ}(T)\over RT})-{d\over dT}({\epsilon(P(T))\over RT}) ={\Delta H^{\circ}-c\over RT^{2}}-{d\over dT}({\epsilon(P(T))\over RT}),\] so that, performing the integration, using the fact that \(Q(T_{1},P^{\circ})=e^{-\delta\over RT_{1}}\); \[\begin{aligned} ln(Q(T_{2}))-ln(Q(T_{1}))&=ln(Q(T_{2}))+{\delta\over RT_{1}}\\ &={1\over R}\int_{T_{1}}^{T_{2}}[{\Delta H^{\circ}-c\over T^{2}}-{d\over dT}({\epsilon(P(T))\over RT})]dT\\ &={1\over R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}-c\over T^{2}}dT-({\epsilon(P(T_{2}))\over RT_{2}}-{\epsilon(P(T_{1}))\over RT_{1}}). \end{aligned}\] We have that, by Lemma 15; \[\begin{aligned} ln (Q(T_{2}))&={c-\Delta G^{\circ}(T_{2})-\epsilon(P(T_{2}))\over RT_{2}},\\ ln(Q(T_{1}))&=-{\delta\over RT_{1}}, \end{aligned}\] so that \[\begin{aligned} ln (Q(T_{2}))-ln(Q(T_{1}))&={c-\Delta G^{\circ}(T_{2})-\epsilon(P(T_{2}))\over RT_{2}}+{\delta\over RT_{1}}\\ &={\Delta G^{\circ}(T_{1})-\Delta G^{\circ}(T_{2})-\epsilon(P(T_{2}))\over RT_{2}}+{\delta\over RT_{1}}\\ &={1\over R}\int_{T_{1}}^{T_{2}}[{\Delta H^{\circ}-c\over T^{2}}-({\epsilon(P(T_{2}))\over RT_{2}}-{\epsilon(P(T_{1}))\over RT_{1}})\\ &={-1\over R}(\Delta H^{\circ}-c)({1\over T_{2}}-{1\over T_{1}})-({\epsilon(P(T_{2}))\over RT_{2}}-{\delta\over RT_{1}}), \end{aligned}\] so that, canceling \(R\) 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; \[\Delta G^{\circ}(T_{1})={T_{1}\over T_{2}}\Delta G^{\circ}(T_{2})-\Delta H^{\circ}({T_{1}\over T_{2}}-1).\] ◻

Proof. For the first claim, by Lemma 17, 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 16 and the proof of Lemma 7, we have that \[\begin{aligned} ({\partial ({dG\over d\xi})_{T,P}\over \partial T})_{P}=({\partial (\sum \nu_{i}\mu_{i})\over \partial T})_{P} =\sum \nu_{i}({\partial \mu_{i}\over \partial T})_{P} =\sum \nu_{i}({\partial \mu_{i}\over \partial T})_{P,n} =-\sum \nu_{i}\overline{S}_{m,i}.\label{lem18*} \end{aligned} \tag{66}\] 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 \[\begin{aligned} dU&=\sum_{i=1}^{c}{3\over 2}N_{A}kTdn_{i}+\sum_{i=1}^{c}{3\over 2}N_{A}kn_{i}dT-\sum_{i=1}^{c}N_{A}m_{i}\rho_{i}dn_{i},\\ dQ&=\sum_{i=1}^{c}{3\over 2}N_{A}kTdn_{i}+\sum_{i=1}^{c}{3\over 2}N_{A}kn_{i}dT-\sum_{i=1}^{c}N_{A}m_{i}\rho_{i}dn_{i}+dL,\\ {dQ\over T}&=\sum_{i=1}^{c}{3\over 2}N_{A}kdn_{i}+\sum_{i=1}^{c}{3\over 2}N_{A}kn_{i}{dT\over T}-\sum_{i=1}^{c}N_{A}m_{i}\rho_{i}{dn_{i}\over T}+{g(T,\overline{n})dT\over T} +\sum_{i=1}^{c}h_{i}(T,\overline{n}){dn_{i}\over T},\\ ({dQ\over T})_{n',T,P}&={3\over 2}N_{A}kdn_{i}-N_{A}m_{i}\rho_{i}{dn_{i}\over T}+h_{i}(T,\overline{n}){dn_{i}\over T}. \end{aligned}\] It follows that \[\begin{aligned} \overline{S}_{m,i}=\int_{\Delta n_{i}=1}({dQ\over T})_{n',T,P}={3\over 2}N_{A}k-{N_{A}m_{i}\rho_{i}\over T}+{k_{i}(T)\over T}.\label{lem18**} \end{aligned} \tag{67}\] So that, from (66) \[\begin{aligned} ({\partial ({dG\over d\xi})_{T,P}\over \partial T})_{P}&=-\sum_{i=1}^{c} \nu_{i}({3\over 2}N_{A}k-{N_{A}m_{i}\rho_{i}\over T})-\sum_{i=1}^{c}\nu_{i}{k_{i}(T)\over T}\notag\\ &=-{3\over 2}N_{A}k(\sum_{i=1}^{c} \nu_{i})+{N_{A}\over T}\sum_{i=1}^{c}\nu_{i}\mu_{i}\rho_{i}-\sum_{i=1}^{c}\nu_{i}{k_{i}(T)\over T}\notag\\ &=-{3\over 2}N_{A}k(\sum_{i=1}^{c} \nu_{i})+{N_{A}\over T}\sum_{i=1}^{c}\nu_{i}\mu_{i}\rho_{i}-{G(T)\over T}.\label{lem18***} \end{aligned} \tag{68}\] From (68), which is uniform \(P\), we see that \(({dG\over d\xi})_{T,P}\) is of the form \[\begin{aligned} \alpha(P)+\beta T+\gamma ln(T)-\int {G(T)\over T}, \label{lem18B} \end{aligned} \tag{69}\] 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 \[\begin{aligned} ({\partial ({dG\over d\xi})_{T,P}\over \partial P})_{T}=({\partial (\sum \nu_{i}\mu_{i})\over \partial P})_{T} =\sum_{i=1}^{c} \nu_{i}({\partial \mu_{i}\over \partial P})_{T} =\sum_{i=1}^{c} \nu_{i}({\partial \mu_{i}\over \partial P})_{T,n} =\sum_{i=1}^{c} \nu_{i}({\partial V\over \partial n_{i}})_{T,P,n'} =\sum_{i=1}^{c} \nu_{i}\overline{V}_{i} =\sum_{i=1}^{c}\nu_{i}{N_{A}m_{i}\over \kappa_{i}(T,P)},\label{lem18A} \end{aligned} \tag{70}\] where \(\kappa_{i}\) is the density of substance \(i\). We also have that \[\begin{aligned} P(\sum_{i=1}^{c}\nu_{i}{N_{A}m_{i}\over \kappa_{i}(T,P)})&=P(\sum_{i=1}^{c} \nu_{i}\overline{V}_{i})=G(T),&(dL=PdV),\label{lem18C} \end{aligned} \tag{71}\] and from (70),(69) and (71), we have that \[P({\partial ({dG\over d\xi})_{T,P}\over \partial P})_{T}=G(T) =P\alpha'(P),\] so that \(G(T)=\epsilon\), \(\alpha(P)=\lambda+\epsilon ln(P)\), \(({dG\over d\xi})_{T,P}\) is of the form; \[\begin{aligned} \alpha(P)+\beta T+\gamma ln(T)-\int {G(T)\over T} =\lambda+\epsilon ln(P)+\beta T+\gamma ln(T)-\epsilon ln(T) =\lambda+\epsilon ln(P)+\beta T+\sigma ln(T),\label{lem18D} \end{aligned} \tag{72}\] 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 \[\begin{aligned} Q(T_{1},P)=e^{-\epsilon(P)\over RT_{1}}. \label{lem18X} \end{aligned} \tag{73}\] From (72), we have that \[\begin{aligned} ({dG\over d\xi})_{T,P}=\lambda+\beta T+\sigma ln(T).\label{lem18Y} \end{aligned} \tag{74}\] 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 (73) and (74).

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; \[\begin{aligned} \lambda+\epsilon ln(P^{\circ})+\beta T+\sigma ln(T)=c,\label{lem18Z} \end{aligned} \tag{75}\] for \(T\). If \(\beta>0,\sigma<0\) or \(\beta<0,\sigma>0\), observing that \((\beta T+\sigma ln(T))'=\beta+{\sigma\over T}\), \((\beta T+\sigma ln(T))''=-{\sigma\over T^{2}}\), so there exists a min/max at \(T={-\sigma\over \beta}\), we have that, if \[\begin{aligned} &-\sigma+\sigma ln({-\sigma\over \beta})\leq c-\lambda-\epsilon ln(ln(P^{\circ})),\\ &-\sigma+\sigma ln({-\sigma\over \beta})\geq c-\lambda-\epsilon ln(ln(P^{\circ})), \end{aligned}\] we can again solve the Eq. (75) 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 (72), 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; \[\begin{aligned} \sigma&=0,\\ \lambda+\epsilon ln(P^{\circ})&=\Delta H^{\circ},\\ \beta&={(\Delta G^{\circ}(T_{1})-\Delta H^{\circ})\over T_{1}},\\ \Delta G^{\circ}(T_{1})&=\beta T_{1}+\Delta H^{\circ}. \end{aligned}\] We can then, using Lemma 15, obtain a formula for the activity coefficient; \[\begin{aligned} Q(T_{1},P')=e^{{(({\partial G\over \partial \xi})_{T,P}|_{T_{1},P'}-\Delta G^{\circ}(T_{1}))-\epsilon(P')\over RT_{1}}} =e^{{(\Delta H^{\circ}-\epsilon ln(P'^{\circ})+\epsilon ln(P')+\beta T_{1}-(\beta T_{1}+\Delta H^{\circ}))-\epsilon(P')\over RT_{1}}} =e^{{\epsilon ln({P'\over P'^{\circ}})-\epsilon(P')\over RT_{1}}},\label{lem18W} \end{aligned} \tag{76}\] 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 1 and Lemma 13, follows from a simple rearrangement of the formulas \(Q(T_{1},P')=c\), for some \(c\in\mathcal{R}_{\geq 0}\), using (76), and \(({dG\over d\xi})_{T,P}=c\), for some \(c\in \mathcal{R}\), using (72), with \(\sigma=0\). ◻

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 1, we have that, for \(1\leq i\leq c\); \[\begin{aligned} &n_{i}(t)=\nu_{i}\xi(t)+n_{i,0},\\ n(t)=\sum_{j=1}^{c}n_{i}(t) =\sum_{i=1}^{c}(\nu_{i}\xi(t)+n_{i,0}) =\alpha\xi(t)+\beta, \end{aligned}\] 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 \[\begin{aligned} G_{\gamma}'(0)&={\sum_{i=1}^{c}\nu_{i}(\prod_{j\neq i}n_{j,0})\over \beta^{c}}-{c\prod_{j=1}^{c}n_{j,0}\over \beta^{c+1}}\\ &={1\over \beta^{c+1}}(\beta(\sum_{i=1}^{c}\nu_{i}(\prod_{j\neq i}n_{j,0})-c\prod_{j=1}^{c}n_{j,0}))\\ &={\prod_{j=1}^{c}n_{j,0}\over \beta^{c+1}}(\beta{\sum_{i=1}^{c}\nu_{i}\over n_{i,0}}-c)\\ &={\prod_{j=1}^{c}n_{j,0}\over \beta^{c+1}}(n_{init}{\sum_{i=1}^{c}\nu_{i}\over n_{i,0}}-c)\\ &={\prod_{j=1}^{c}n_{j,0}\over \beta^{c+1}}(\sum_{i=1}^{c}{\nu_{i}\over x_{i,init}}-c)\\ &={\prod_{j=1}^{c}n_{j,0}\over \beta^{c+1}}((\sum_{i=1}^{c}\nu_{i}log(x_{i}))'_{init}-c)\\ &={\prod_{j=1}^{c}n_{j,0}\over \beta^{c+1}}(log(\prod_{i=1}^{c}x_{i}^{\nu_{i}})'_{init}-c)\\ &={\prod_{j=1}^{c}n_{j,0}\over \beta^{c+1}}(log(Q(\gamma_{12}(t)))'|_{0}-c)\\ &={\prod_{j=1}^{c}n_{j,0}\over \beta^{c+1}}({grad(Q)\centerdot \gamma_{12}'(0)\over Q(\gamma_{12}(0))}-c), \end{aligned}\] so that; \[\begin{aligned} G_{\gamma}'(0)=0,\quad \text{iff}\quad {grad(Q)\centerdot \gamma_{12}'(0)\over Q(\gamma_{12}(0))}=c, \end{aligned}\] which we can exclude by an appropriate parametrization of the feasible path \(\gamma\), without altering the direction of \(\gamma_{12}'(0)\). 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 \[\begin{aligned} \xi'(0)&=(G_{\gamma}^{-1})'|_{Q(T_{0},P_{0})}grad(Q)(T_{0},P_{0})\centerdot \gamma_{12}'(0)\\ &={grad(Q)(T_{0},P_{0})\centerdot \gamma_{12}'(0)\over G_{\gamma}'(0)}\\ &=grad(Q)(T_{0},P_{0})\centerdot \gamma_{12}'(0){\beta^{c+1}\over \prod_{j=1}^{c}n_{j,0}}{Q(\gamma_{12}(0))\over grad(Q)\centerdot \gamma_{12}'(0)-cQ(\gamma_{12}(0))}\\ &={\alpha_{1}\beta_{1}[grad(Q)(T_{0},P_{0})\centerdot \gamma_{12}'(0)]\over grad(Q)(T_{0},P_{0})\centerdot \gamma_{12}'(0)-c\beta_{1}}, \end{aligned}\] 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 \(\gamma_{12}'(0)=\lambda(cos(\theta),sin(\theta))\), we have that \[\xi'(0)={\lambda\alpha_{1}\beta_{1}[{\partial Q\over \partial T}|_{(T_{0},P_{0})}cos(\theta)+{\partial Q\over \partial P}|_{(T_{0},P_{0})}sin(\theta)]\over \lambda[{\partial Q\over \partial T}|_{(T_{0},P_{0})}cos(\theta)+{\partial Q\over \partial P}|_{(T_{0},P_{0})}sin(\theta)]-c\beta_{1}}=h(\lambda,\theta)={\lambda\alpha_{1}\beta_{1}r(\theta)\over \lambda r(\theta)-c\beta_{1}},\] where \(r(\theta)={\partial Q\over \partial T}|_{(T_{0},P_{0})}cos(\theta)+{\partial Q\over \partial P}|_{(T_{0},P_{0})}sin(\theta)\). 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 \(\lambda\). \[{\partial h\over \partial \theta}={\lambda \alpha_{1}\beta_{1}r'(\theta)\over \lambda r(\theta)-c\beta_{1}}-{\lambda^{2}\alpha_{1}\beta_{1}r'(\theta)\over (\lambda r(\theta)-c\beta_{1})^{2}},\] so that \({\partial h\over \partial \theta}=0\) iff \(\lambda \alpha_{1}\beta_{1}r'(\theta)(\lambda r(\theta)-c\beta_{1})-\lambda^{2}\alpha_{1}\beta_{1}r'(\theta)=0\) 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; \[\begin{aligned} {dT\over dt}&={-\epsilon ln({P(t)\over P^{\circ}})+\epsilon(P(t))\over RT(t)^{2}}({P(t)\over P^{\circ}})^{\epsilon\over RT(t)}e^{-\epsilon(P(t))\over RT(t)},\\ {dP\over dt}&={{\epsilon\over P(t)}-\epsilon'(P(t))\over RT(t)}({P(t)\over P^{\circ}})^{\epsilon\over RT(t)}e^{-\epsilon(P(t))\over RT(t)}, \end{aligned}\] so that \[{dP\over dT}={{dP\over dt}\over {dT\over dt}} ={{{\epsilon\over P}-\epsilon'(P)\over RT}({P\over P^{\circ}})^{\epsilon\over RT}e^{-\epsilon(P)\over RT}\over {-\epsilon ln({P\over P^{\circ}})+\epsilon(P)\over RT^{2}}({P\over P^{\circ}})^{\epsilon\over RT}e^{-\epsilon(P)\over RT}} ={T({\epsilon\over P}-\epsilon'(P))\over -\epsilon ln({P\over P^{\circ}})+\epsilon(P)}.\] Separating variables, we obtain that \[{\epsilon ln({P\over P^{\circ}})-\epsilon(P)\over \epsilon'(P)-{\epsilon\over P}}dP=-TdT,\] which has an implicit solution given by \[\int {\epsilon P ln({P\over P^{\circ}})-P\epsilon(P)\over P\epsilon'(P)-\epsilon}dP=-{T^{2}\over 2}+c,\] 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 \(\epsilon'(P)=0\) and the implicit solutions are given by \[\int P ln({P\over P^{\circ}})dP=-{T^{2}\over 2}+c,\] for \(c\in\mathcal{R}\). We have, integrating by parts, that \[\begin{aligned} \int P ln({P\over P^{\circ}})&=\int Pln(P)-Pln(P^{\circ})\\ &={P^{2}ln(P)\over 2}-\int {P\over 2}-{P^{2}ln(P^{\circ})\over 2}\\ &={P^{2}ln(P)\over 2}-{P^{2}\over 4}-{P^{2}ln(P^{\circ})\over 2}\\ &=P^{2}({ln(P)\over 2}-{ln(P^{\circ})\over 2}-{1\over 4}), \end{aligned}\] 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 \[{dP\over dT}={{-\epsilon'(P)\over RT}e^{-\epsilon(P)\over RT}\over {\epsilon(P)\over RT^{2}}e^{-\epsilon(P)\over RT}} ={-\epsilon'(P)T\over \epsilon(P)},\] so that, separating variables; \[{\epsilon(P)\over \epsilon'(P)}dP=-TdT,\] and the implicit solutions are given by \[\int {\epsilon(P)\over \epsilon'(P)}dP={-T^{2}\over 2}+c.\] ◻

Proof. For \(c\) substances, with \(c'\) the number of the charged species, using Definition 1, we have that the electrostatic potential energy; \[U_{el}= \sum_{i=1}^{c'}\phi(\overline{x}_{i})q_{i},\] where \(q_{i}=N_{i}ez_{i}=N_{A}n_{i}ez_{i}\) and where \(\{\overline{x}_{i}:1\leq i\leq c'\}\) 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 \[\begin{aligned} G(T,P,n_{1},\ldots,n_{c})&=U+PV-TS\\ &=U_{chem}+U_{el}+PV-TS\\ &=U_{el}+G_{chem}\\ &=\sum_{j=1}^{c}\phi(\overline{x}_{j})q_{j}+G_{chem}\\ &=\sum_{j=1}^{c}\phi(\overline{x}_{j})N_{A}n_{j}ez_{j}+G_{chem}, \end{aligned}\] so that \[\begin{aligned} \mu_{i}&=({\partial G\over \partial n_{i}})_{T,P}&\notag\\ &=({\partial (\sum_{j=1}^{c}\phi(\overline{x}_{j})N_{A}n_{j}ez_{j}+G_{chem})\over \partial n_{i}})_{T,P}&\notag\\ &=\mu_{i,chem}, &(c'+1\leq i\leq c)\notag\\ &=\mu_{i,chem}+{\partial (\phi(\overline{x}_{i})N_{A}n_{i}ez_{i})\over \partial n_{i}}, &(1\leq i\leq c')\notag\\ &=\mu_{i,chem}+\phi(\overline{x}_{i})N_{A}ez_{i}\notag\\ &=\mu_{i,chem}+\phi(\overline{x}_{i})F z_{i}.\label{lem20*} \end{aligned} \tag{77}\] 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 5, generalized to a collection involving charged species, using (77), we have that \[\begin{aligned} ({\partial G\over \partial \xi})_{T,P}&={\sum}_{i=1}^{c}\nu_{i}\mu_{i}\notag\\ &=2\mu(HCl)+2\mu(Ag)-\mu(H_{2})-2\mu(AgCl)+2\mu(e^{-}(L))-2\mu(e^{-}(R))\notag\\ &=({\partial G_{chem'}\over \partial \xi})_{T,P}+2\mu(e^{-}(L))-2\mu(e^{-}(R))\notag\\ &=({\partial G_{chem'}\over \partial \xi})_{T,P}+((2\mu_{chem}(e^{-}(L))-2F\phi(L))-(2\mu_{chem}(e^{-}(L))-2F\phi(R)))\notag\\ &=({\partial G_{chem'}\over \partial \xi})_{T,P}+2F(\phi(R)-\phi(L))\notag\\ &=({\partial G_{chem'}\over \partial \xi})_{T,P}+2EF=0,\label{lem20dag} \end{aligned} \tag{78}\] where \(G_{chem'}\) is the Gibbs energy restricted to the uncharged species. By Lemmas 15 and 16, we have that \[\begin{aligned} ({\partial G_{chem'}\over \partial \xi})_{T,P^{\circ}}&=\sum_{i=c'+1}^{c}\nu_{i}\mu_{i}^{\circ}\notag\\ &=(\Delta G_{chem'}^{\circ}+RTln(Q_{chem'}(T,P^{\circ}))+\epsilon(P^{\circ}))\notag\\ &=(\Delta G_{chem'}^{\circ}-\epsilon(P^{\circ}))+\epsilon(P^{\circ}))\notag\\ &=\Delta G_{chem'}^{\circ}.\label{lem20dagdag} \end{aligned} \tag{79}\] From (78) and (79), we obtain \[\begin{aligned} 2E^{\circ}F=-({\partial G_{chem'}\over \partial \xi})_{T,P^{\circ}} =-\Delta G_{chem'}^{\circ}.\label{lem20dagdagdag} \end{aligned} \tag{80}\] Similarly, we have that \[\begin{aligned} ({\partial G_{chem'}\over \partial \xi})_{T,P}=\sum_{i=c'+1}^{c}\nu_{i}\mu_{i} =(\Delta G_{chem'}^{\circ}+RTln(Q_{chem'}(T,P))+\epsilon(P)),\label{lem20dagdagdagdag} \end{aligned} \tag{81}\] so from (78) and (81), we obtain that \[\begin{aligned} 2EF=-({\partial G_{chem'}\over \partial \xi})_{T,P} =-(\Delta G_{chem'}^{\circ}+RTln(Q_{chem'}(T,P))+\epsilon(P)).\label{lem20sharp} \end{aligned} \tag{82}\] Combining (82) and (80), we obtain that \[2EF-2E^{\circ}F=-(\Delta G_{chem'}^{\circ}+RTln(Q_{chem'}(T,P))+\epsilon(P))-(-\Delta G_{chem'}^{\circ}) =-RTln(Q_{chem'}(T,P))-\epsilon(P),\] so that \[E-E^{\circ}=-{RTln(Q_{chem'}(T,P))\over 2F}-{\epsilon(P)\over 2F}.\] ◻

Proof. By Lemma 20, we have that \[\begin{aligned} E-E^{\circ}=-{RTln(Q)\over 2F}-{\epsilon(P)\over 2F},\label{lem21*} \end{aligned} \tag{83}\] and, by Lemma 15, we have that \[\begin{aligned} 0=({\partial G\over \partial \xi})_{T,P}=\Delta G^{\circ}+RTln(Q)+\epsilon(P).\label{lem21**} \end{aligned} \tag{84}\] Rearranging (83) and (84), we obtain the result. ◻

Proof. By (78) of Lemma 20, we have that \[\begin{aligned} ({\partial G\over \partial \xi})_{T,P}=({\partial G_{chem'}\over \partial \xi})_{T,P}+2EF.\label{lem22*} \end{aligned} \tag{85}\] By Lemma 18, we have that \(({\partial G_{chem'}\over \partial \xi})_{T,P}\) is independent of \(P\), in particularly, we have that \[\begin{aligned} ({\partial G_{chem'}\over \partial \xi})_{T_{1},P_{1}}=({\partial G_{chem'}\over \partial \xi})_{T_{1},P_{1}^{\circ}},\label{lem22**} \end{aligned} \tag{86}\] so that, combining (85) and (86), we obtain the result. ◻

Proof. Following the proof of Lemma 20, we have that \[\ ({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}}=({\partial G_{chem'}\over \partial \xi})_{T,P}|_{T_{1},P_{1}}+2E(T_{1},P_{1})F,\label{lem23*}\ \tag{87}\] \[\ 2E^{\circ}(T_{1})F=({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-({\partial G_{chem'}\over \partial \xi})_{T,P}|_{T_{1},P_{1}^{\circ}} =({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-\Delta G^{\circ}_{chem'}(T_{1}),\label{lem23**} \tag{88}\] so from (87), (88) and Lemma 15, we have \[\begin{aligned} 2E(T_{1},P_{1})F&=({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}}-({\partial G_{chem'}\over \partial \xi})_{T,P}|_{T_{1},P_{1}}\\ &=({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}}-(\Delta G^{\circ}_{chem'}(T_{1})+RT_{1}ln(Q_{chem'}(T_{1},P_{1}))+\epsilon(P_{1})), \end{aligned}\] \[\begin{aligned} 2E(T_{1},P_{1})F-2E^{\circ}(T_{1})F&=({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}}-(\Delta G^{\circ}_{chem'}(T_{1})+RT_{1}ln(Q_{chem'}(T_{1},P_{1}))+\epsilon(P_{1}))\\ &\;\;\;-(({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-\Delta G^{\circ}_{chem'}(T_{1}))\\ &=({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}}-(({\partial G\over \partial \xi})_{T,P}|_{T_{1},P_{1}^{\circ}}-RT_{1}ln(Q_{chem'}(T_{1},P_{1}))-\epsilon(P_{1}). \end{aligned}\] ◻

Proof. Following the proof of Lemma 5, 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 5 go through as before. Going through the proof of Lemma 3, 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 5, 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, 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 \[\begin{aligned} ({\partial G_{chem'}\over \partial \xi})_{T,P^{\circ}}&=\Delta G_{chem'}^{\circ}+RTln(Q_{chem'}(T,P^{\circ})\notag\\ &=\Delta G_{chem'}^{\circ}+RTln({Q\over a_{0}(T,P^{\circ})})\\ \Delta G_{chem'}^{\circ}-RTln(a_{0}(T,P^{\circ})), \end{aligned} \tag{94}\] where \(G_{chem'}\) is the Gibbs energy restricted to the uncharged species without the solvent. Using (4), from Lemma 2 and (94), we obtain \[\begin{aligned} 2E^{\circ}F=-\Delta G_{chem'}^{\circ}+RTln(a_{0}(T,P^{\circ})).\label{lem24dagdagdag)'} \end{aligned} \tag{95}\] Similarly, we obtain \[({\partial G_{chem'}\over \partial \xi})_{T,P}=\Delta G_{chem'}^{\circ}+RTln(Q_{chem'}(T,P) =\Delta G_{chem'}^{\circ}+RTln({Q\over a_{0}(T,P)}),\] so that, from (4), from Lemma 2; \[\begin{aligned} 2EF=-(\Delta G_{chem'}^{\circ}+RTln({Q\over a_{0}(T,P)})).\label{lem24sharp'} \end{aligned} \tag{96}\] Combining (96) and (91), we obtain \[\begin{aligned} 2EF-2E^{\circ}F&=-(\Delta G_{chem'}^{\circ}+RTln({Q\over a_{0}(T,P)}))-(-\Delta G_{chem'}^{\circ}+RTln(a_{0}(T,P^{\circ})))\\ &=-RTln({Q\over a_{0}(T,P)})-RTln(a_{0}(T,P^{\circ})) =-RTln(Q)+RTln({a_{0}(T,P)\over a_{0}(T,P^{\circ})}), \end{aligned}\] which gives the result.of Lemma 2. Going back through the proof of Lemma 5 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 3, 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 \[({\partial G\over \partial \xi})_{T,P}=\Delta G^{\circ}+RTln({Q\over a_{0}(T,P)}) =\Delta G^{\circ}-RTln(a_{0}(T,P^{\circ})),\] and \[({\partial G\over \partial \xi})_{T,P^{\circ}}=\Delta G^{\circ}+RTln({Q\over a_{0}(T,P^{\circ})})=0.\] So that we have chemical equilibrium at \((T,P)\). For the penultimate claim of Lemma 5, rearrange the formula from the modification of Lemma 3, 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 4, we have by the modification of Lemma 2, 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 3, that \[0=({\partial G\over \partial \xi})_{T,P}=\Delta G^{\circ}+RTln(Q)-RTln(a_{0}(T,P)),\] so that \[\begin{aligned} \Delta G^{\circ}&=RTln(a_{0}(T,P))-RTln(Q)\\ &=RTln(a_{0}(T,P))+2F(E-E^{\circ})-RTln({a_{0}(T,P)\over a_{0}(T,P^{\circ})})\\ &=2F(E-E^{\circ})+RTln(a_{0}(T,P^{\circ})). \end{aligned}\]

For the modification of Lemma 6, 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 5, we have that \(Q=a_{0}e^{-\Delta G^{\circ}\over RT}\) along a chemical equilibrium path, so that \[\begin{aligned} ln(Q)&={-\Delta G^{\circ}\over RT}+ln(a_{0}(T,P)). \label{lem24K'} \end{aligned} \tag{97}\] It follows that \[{d ln(Q)\over dT}={d\over dT}({-\Delta G^{\circ}\over RT})+{d\over dT}(ln(a_{0}(T,P))) ={\Delta H^{\circ}\over RT^{2}}+{d\over dT}(ln(a_{0}(T,P))).\] It follows, integrating between \(T_{1}\) and \(T_{2}\), using (97) and the fundamental theorem of calculus, that \[\begin{aligned} ln({Q(T_{2})\over Q(T_{1})})&=ln(Q)(T_{2})-ln(Q)(T_{1})\notag\\ &={-\Delta G^{\circ}(T_{2})\over RT_{2}}+{\Delta G^{\circ}(T_{1})\over RT_{1}}+ln(a_{0}(T_{2},P_{2}))-ln(a_{0}(T_{1},P_{1}))\notag\\ &=\int_{T_{1}}^{T_{2}}{d ln(Q)\over dT}dT\notag\\ &={1\over R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}\over T^{2}}+\int_{T_{1}}^{T_{2}}{d\over dT}(ln(a_{0}(T,P)))dT\notag\\ &={1\over R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}\over T^{2}}+ln(a_{0}(T_{2},P_{2}))-ln(a_{0}(T_{1},P_{1})).\label{lem24P'} \end{aligned} \tag{98}\] So that, rearranging, we obtain the first claim. Using the fact, by the modification of Lemma 3, that \[\begin{aligned} ln(Q(T_{2}))&=-{\Delta G^{\circ}(T_{2})\over RT_{2}}+ln(a_{0}(T_{2},P_{2})),\\ ln(Q(T_{1}))&=-{\Delta G^{\circ}(T_{1})\over RT_{1}}+ln(a_{0}(T_{1},P_{1})). \end{aligned}\] We obtain, substituting into (98), canceling \(R\), and performing the integration, if \(\Delta H^{\circ}\) is temperature independent, that \[\begin{aligned} {\Delta G^{\circ}(T_{2})\over T_{2}}-{\Delta G^{\circ}(T_{1})\over T_{1}}=-\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}\over T^{2}}=\Delta H^{\circ}({1\over T_{2}}-{1\over T_{1}}).\label{lem24Q'} \end{aligned} \tag{99}\] For the fifth claim, rearrange (99). If \(D_{c}\) intersects the line \(P=P^{\circ}\) at \((T_{1},P^{\circ})\), for the sixth and seventh claims, we have, using the modification of Lemma 3 and the fact from Lemma 5 that \(Q(T_{1},P^{\circ})=1\); \[\begin{aligned} ({\partial G\over \partial \xi})_{T_{2},P}&=\Delta G^{\circ}(T_{2})+RT_{2}ln(Q(T_{2},P))-RT_{2}ln(a_{0}(T_{2},P))\\ &=({\partial G\over \partial \xi})_{T_{1},P^{\circ}}\\ &=\Delta G^{\circ}(T_{1})+RT_{1}ln(Q(T_{1},P^{\circ}))-RT_{1}ln(a_{0}(T_{1},P^{\circ}))\\ &=\Delta G^{\circ}(T_{1})-RT_{1}ln(a_{0}(T_{1},P^{\circ}))=c. \end{aligned}\] So that, again rearranging, we obtain the result. Along \(D_{c}\), we have, using Lemma 3, that \[ln(Q)={c-\Delta G^{\circ}\over RT}+ln(a_{0}(T,P)),\] so that, using the first part; \[\begin{aligned} {d ln(Q)\over dT}&={d\over dT}({c-\Delta G^{\circ}\over RT})+{d\over dT}(ln(a_{0}(T,P)))\\ &={-c\over RT^{2}}+{d\over dT}({-\Delta G^{\circ}\over RT})+{d\over dT}(ln(a_{0}(T,P)))\\ &={\Delta H^{\circ}-c\over RT^{2}}+{d\over dT}(ln(a_{0}(T,P))), \end{aligned}\] 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 3; \[\begin{aligned} ln (Q(T_{2}))&={c-\Delta G^{\circ}(T_{2})\over RT_{2}}+ln(a_{0}(T_{2},P)),\\ ln(Q(T_{1}))&=0, \end{aligned}\] so that, using the formula for \(c\); \[\begin{aligned} ln (Q(T_{2}))&=ln (Q(T_{2}))-ln(Q(T_{1}))\\ &={c-\Delta G^{\circ}(T_{2})\over RT_{2}}+ln(a_{0}(T_{2},P))\\ &={\Delta G^{\circ}(T_{1})-RT_{1}ln(a_{0}(T_{1},P^{\circ}))-\Delta G^{\circ}(T_{2})\over RT_{2}}+ln(a_{0}(T_{2},P))\\ &={1\over R}\int_{T_{1}}^{T_{2}}{\Delta H^{\circ}-c\over T^{2}}dT+ln(a_{0}(T_{2},P_{2}))-ln(a_{0}(T_{1},P^{\circ}))\\ &={-1\over R}(\Delta H^{\circ}-c)({1\over T_{2}}-{1\over T_{1}})+ln(a_{0}(T_{2},P_{2}))-ln(a_{0}(T_{1},P^{\circ})), \end{aligned}\] and rearranging; \[{\Delta G^{\circ}(T_{1})-RT_{1}ln(a_{0}(T_{1},P^{\circ}))-\Delta G^{\circ}(T_{2})\over RT_{2}}={-1\over R}(\Delta H^{\circ}-c)({1\over T_{2}}-{1\over T_{1}})-ln(a_{0}(T_{1},P^{\circ})),\] \[{\Delta G^{\circ}(T_{2})-\Delta G^{\circ}(T_{1})\over T_{2}}+{RT_{1}ln(a_{0}(T_{1},P^{\circ}))\over T_{2}} =(\Delta H^{\circ}-c)({1\over T_{2}}-{1\over T_{1}})+Rln(a_{0}(T_{1},P^{\circ})),\] \[\begin{aligned} {\Delta G^{\circ}(T_{2})-\Delta G^{\circ}(T_{1})\over T_{2}}&=(\Delta H^{\circ}-c)({1\over T_{2}}-{1\over T_{1}})+Rln(a_{0}(T_{1},P^{\circ}))-{RT_{1}ln(a_{0}(T_{1},P^{\circ}))\over T_{2}}\\ &=(\Delta H^{\circ}-(\Delta G^{\circ}(T_{1})-RT_{1}ln(a_{0}(T_{1},P^{\circ}))))({1\over T_{2}}-{1\over T_{1}})+Rln(a_{0}(T_{1},P^{\circ}))-{RT_{1}ln(a_{0}(T_{1},P^{\circ}))\over T_{2}}\\ &=(\Delta H^{\circ}-\Delta G^{\circ}(T_{1}))({1\over T_{2}}-{1\over T_{1}})+RT_{1}ln(a_{0}(T_{1},P^{\circ}))({1\over T_{2}}-{1\over T_{1}})+(R-{RT_{1}\over T_{2}})ln(a_{0}(T_{1},P^{\circ}))\\ &=(\Delta H^{\circ}-\Delta G^{\circ}(T_{1}))({1\over T_{2}}-{1\over T_{1}})+ln(a_{0}(T_{1},P^{\circ}))(RT_{1}({1\over T_{2}}-{1\over T_{1}})+(R-{RT_{1}\over T_{2}}))\\ &=(\Delta H^{\circ}-\Delta G^{\circ}(T_{1}))({1\over T_{2}}-{1\over T_{1}}), \end{aligned}\] 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 7, 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 \[c=\Delta G^{\circ}(T_{2})+RT_{2}ln(Q(T_{2},P'))-RT_{2}ln(a_{0}(T_{2},P')) =\Delta G^{\circ}(T_{2})-RT_{2}ln(a_{0}(T_{2},P'^{\circ})),\] so that rearranging, \[Q(T_{2},P')={a_{0}(T_{2},P')\over a_{0}(T_{2},P'^{\circ})},\] 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 3, to obtain the formula for the activity coefficient; \[Q(T_{2},P')=e^{({\partial G\over \partial \xi})_{T,P}|{T_{2},P'}-\Delta G^{\circ}(T_{2})+RT_{2}ln(a_{0}(T_{2},P'))\over RT_{2}} =e^{\epsilon ln({P'\over P'^{\circ}})\over RT_{2}}a_{0}(T_{2},P').\] 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. ◻

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 11, as we are differentiating, the proof works with a constant \(f>0\). ◻

Proof. The proof is clear from Lemma 15. ◻

Proof. The proof is clear from the proof of Lemma 16. ◻

Proof. The proof is clear from the proof of Lemma 17. ◻

Proof. The proof is clear from the proof of Lemma 18. ◻

Proof. The computation of \(grad(Q)\) in both cases is a simple application of the chain and product rules. Following the method of Lemma 19, 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 \(\epsilon=0\); \[{dP\over dT}={{{-{\partial \epsilon\over \partial P}(T,P)\over RT}e^{-\epsilon(T,P)\over RT}}\over ({\epsilon(T,P)\over RT^{2}}-{{\partial \epsilon\over \partial T}(T,P)\over RT})e^{-\epsilon(T,P)\over RT}} ={-T{\partial \epsilon\over \partial P}(T,P)\over \epsilon(T,P)-T{\partial \epsilon\over \partial T}(T,P)}.\] ◻

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 26. ◻

Proof. The proof is clear from Lemma 15. ◻

Proof. The proof is clear from the proof of Lemma 16. ◻

Proof. The proof is again clear from the proof of Lemma 17 ◻

Proof. The proof is again clear from the proof of Lemma 18. ◻

Proof. The proof is the same as Lemma 31, replacing \(Q\) by \(Z\), noting that \(Z\) is defined the same way in terms of activities. ◻

Proof. Just follow the proof of Lemma 20, replacing \(\epsilon(P)\) with \(\epsilon(T,P)\) and use the fact that the catalyzer reaction \(2H_{2}O+4 e^{-}(R)\rightarrow O_{2}+2H_{2}+4e^{-}(L)\) occurs with \(4\) electrons rather than \(2\). ◻

Proof. Follow the proof of Lemma 21, replacing \(\epsilon(P)\) with \(\epsilon(T,P)\), noting the remark in Lemma 38. ◻

Proof. Follow the proof of Lemma 22, replacing the result there that \(({\partial G_{chem'}\over \partial \xi})_{T,P}\) is independent of \(P\), with the corresponding same result in Lemma 36. ◻

Proof. Follow the proof of Lemma 23, replacing \(\epsilon(P)\) with \(\epsilon(T,P)\). ◻

Proof. We have that \[\begin{aligned} W=\prod_{i=1}^{c}a_{i}^{\nu_{i}}=\prod_{i=1}^{c}x_{i}^{\nu_{i}} =f.\label{lem42*} \end{aligned} \tag{117}\] If \(\gamma\) is a feasible path, then \(pr_{12}(\gamma)\subset W_{>0}\), otherwise, we could find \((T,P)\), with \(x_{i}(T,P)\leq 0\), contradicting the fact that \(n_{i}>0\), \(n>0\). It follows that \(f>0\). With \(f>0\), follow through the proof of Lemma 11, replacing \(\beta\) with \(\sum_{i=0}^{c}d_{i}\), where \(d_{0}=n_{0}\). Clearly \(n_{0}'=0\) so we obtain the first direction. The rest of the proof follows from Lemma 13, using the additional fact that \(n_{0}'=0\). ◻

Proof. We have that \[\prod_{i=1}^{c}a_{i}^{\nu_{i}}={Q\over a_{0}} =\prod_{i=1}^{c}x_{i}^{\nu_{i}} ={c\over a_{0}}=d.\] If \(\gamma\) is a feasible path, then \(pr_{12}(\gamma)\subset Q_{>0}\), otherwise, we could find \((T,P)\), with \(x_{i}(T,P)\leq 0\), contradicting the fact that \(n_{i}>0\), for \(0\leq i\leq c\), \(n>0\). It follows that \(c>0\), \(d>0\). With \(d>0\), follow through the proof of Lemma 11 again, getting the other directions from Lemma 13. ◻

Proof. With the same caveat as in Lemma 43, we have that \[\begin{aligned} a_{0}\prod_{i=1}^{c}a_{i}^{\nu_{i}}=x_{0}\prod_{i=1}^{c}x_{i}^{\nu_{i}} =\prod_{i=0}^{c}x_{i}^{\mu_{i}} =Q=f,\label{lem44*} \end{aligned} \tag{118}\] where \(x_{i}={n_{i}\over n}\), \(\mu_{0}=1\), \(\mu_{i}=\nu_{i}\), \(1\leq i\leq c\). Now, using the relation (118), differentiating and using the facts that, for \(1\leq i\leq c-1\); \(n_{i}'={\mu_{i}n_{c}'\over \mu_{c}}\), \(n_{i}={\mu_{i}n_{c}\over \mu_{c}}+d_{i}\), \(n_{0}'=0\) and \(n_{0}=d_{0}\), we obtain that \[(\prod_{i=0}^{c}x_{i}^{\mu_{i}})'=\sum_{i=0}^{c}\mu_{i}x_{i}^{\mu_{i}-1}x_{i}'\prod_{j\neq i}x_{j}^{\mu_{j}} =f\sum_{i=0}^{c}\mu_{i}x_{i}^{\mu_{i}-1}x_{i}'x_{i}^{-\mu_{i}} =f\sum_{i=0}^{c}\mu_{i}x_{i}^{-1}x_{i}' =0,\] so that \[\begin{aligned} \sum_{i=0}^{c}\mu_{i}{n\over n_{i}}{(n_{i}'n-n_{i}n')\over n^{2}}&=\sum_{i=0}^{c}\mu_{i}({n_{i}'\over n_{i}}-{n'\over n})\notag\\ &=\sum_{i=1}^{c-1}{\mu_{i}^{2}n_{c}'\over \nu_{i}n_{c}+\mu_{c}d_{i}}+{\mu_{c}n_{c}'\over n_{c}}- \lambda({\sum_{i=0}^{c}n_{i}'\over \sum_{i=0}^{c}n_{i}})\notag\\ &=\sum_{i=1}^{c-1}{\mu_{i}^{2}n_{c}'\over \mu_{i}n_{c}+\mu_{c}d_{i}}+{\mu_{c}n_{c}'\over n_{c}}-\lambda ({(\sum_{i=1}^{c-1}{\mu_{i}\over \mu_{c}}+1)n_{c}'\over (\sum_{i=1}^{c-1}{\nu_{i}\over \nu_{c}}+1)n_{c}+\sum_{i=0}^{c-1}d_{i}})\notag\\ &=0,\label{lem44B} \end{aligned} \tag{119}\] where \(\lambda=\sum_{i=0}^{c}\mu_{i}=1+\sum_{i=1}^{c}\nu_{i}\).

Following the proof of Lemma 11, replacing \(\beta\) with \(\sum_{i=0}^{c-1}d_{i}\), we have, if \(\sum_{i=1}^{c-1}\mu_{i}=\sum_{i=1}^{c-1}\nu_{i}\neq 0\) and \(\lambda=1+\sum_{i=1}^{c}\mu_{i}=1+\sum_{i=1}^{c}\nu_{i}\neq 0\), then \(n_{i}'=0\), for \(1\leq i\leq c\), and clearly we have that \(n_{0}'=0\). Similarly, if \(\lambda=\sum_{i=0}^{c}\mu_{i}=1+\sum_{i=1}^{c}\nu_{i}=0\), we obtain the relation \(\prod_{i=0}^{c}n_{i}^{\mu_{i}}=f\).

Again, following the proof of Lemma 11, if \(n_{c}'\neq 0\), we obtain the relation; \[\sum_{i=1}^{c-1}{\mu_{i}^{2}\over \mu_{i}n_{c}+\mu_{c}d_{i}}+{\mu_{c}\over n_{c}}=\sum_{i=1}^{c-1}{\nu_{i}^{2}\over \nu_{i}n_{c}+\nu_{c}d_{i}}+{\nu_{c}\over n_{c}} =0,\] and, by the proof there, we obtain that \(n_{i}'=0\), for \(1\leq i\leq c\). As \(n_{0}'=0\), we obtain the result. We are left with the case \(\sum_{i=1}^{c-1}\mu_{i}=\sum_{i=1}^{c-1}\nu_{i}=0\). As in the proof of Lemma 11, we can assume this choosing the appropriate pivot. The other directions in the Lemma follow from a simple modification of Lemma 13. ◻

Proof. The proof is clear. ◻

Proof. The proof is clear. ◻

Proof. The proof is clear from Lemma 15 and the fact that, as \(dn_{0}=0\); \[dG=\sum_{i=0}^{c}\mu_{i}dn_{i}=\sum_{i=1}^{c}\mu_{i}dn_{i},\] so that \(({\partial G\over \partial \xi})_{T,P}=\sum_{i=1}^{c}\nu_{i}\mu_{i}\) and \(\Delta G^{\circ}=\sum_{i=1}^{c}\nu_{i}\mu_{i}^{\circ}\). ◻

Proof. The proof is clear from the proof of Lemma 16 and using the observation from Lemma 47. ◻

Proof. The proof is again clear from the proof of Lemma 17. ◻

Proof. The proof is clear from the proof of Lemma 18. ◻

Proof. The computation follows easily from the proof of Lemma 31, replacing \(\epsilon(P,T)\) by \(\epsilon(T)\). We also note that in the proof of maximal reaction, see Lemma 19, we have to change \(\beta\) to \(\sum_{i=0}^{c+1}n_{i,0}\), where \(n_{i,0}\) is the fixed molar amount of the solvent. This effects \(\alpha_{1}\) but we still have that \(\alpha_{1}\neq 0\) and the rest of the proof remains unchanged. ◻

Proof. The proof is similar to the proof of Lemma 42. ◻

Proof. We have that \[\prod_{i=1}^{c}b_{i}^{\nu_{i}}={Z\over b_{0}} =\prod_{i=1}^{c}x_{i}^{\nu_{i}} ={f\over b_{0}}=d.\] The remaining proof is similar to the proof of Lemma 43. ◻

Proof. We have that \[b_{0}\prod_{i=1}^{c}b_{i}^{\nu_{i}}=x_{0}\prod_{i=1}^{c}x_{i}^{\nu_{i}} =Z=f.\] The remaining proof is similar to the proof of Lemma 44. ◻

Proof. The proof follows from the proof of the Lemma 45. ◻

Proof. The proof follows from the proof of the Lemma 46. ◻

Proof. The proof is clear from Lemma 15, with the same observation as in Lemma 47. ◻

Proof. The proof is clear from the proof of Lemma 16 and Lemma 57. ◻

Proof. The proof is clear from the proof of Lemma 17 and Lemma 58. ◻

Proof. The proof is clear from the proof of Lemma 18 and Lemma 59. ◻

Proof. The proof is the same as Lemma 31, using \(W\) instead of \(Z\), noting that the error term \(\epsilon(T,P)\) depends on \(P\), unlike Lemma 51. ◻

Proof. Using the proof of Lemma 20, replacing \(\epsilon(P)\) with the error term \(\epsilon(T,P)\) from §10, noting that for the Gibbs function involving \(c+1\) species, including the solvent, \(({\partial G\over \partial \xi})_{T,P}=\sum_{i=1}^{c}\nu_{i}\mu_{i}\), we get the desired result. ◻

Proof. Following the proof of Lemma 21, replacing \(\epsilon(P)\) with \(\epsilon(T,P)\), we get our desired result. ◻

Proof. The result can be obtained by following the proof of Lemma 22, replacing the result there that \(({\partial G_{chem'}\over \partial \xi})_{T,P}\) is independent of \(P\), with the corresponding result in Lemma 60. ◻

Proof. The result follows by following the proof of Lemma 23 and replacing \(\epsilon(P)\) with \(\epsilon(T,P)\). ◻

Proof. We have, by the definition of enthalpy, the laws of differentials, the first law of thermodynamics, with \(dP=0\), that \[dH=dU+PdV+VdP =dU+PdV =(dQ-PdV)+PdV =dQ,\] where \(Q\) is the internal energy including work, or heat. From this calculation, as any internal energy change \(dU\) and change in volume \(dV\) is unaffected by adding a solvent not involved in the reaction, and by re-establishing the pressure of the original reaction, it is clear that \(\Delta H^{\circ}(T)=\Delta H^{\circ,solv}(T)\), where \(\Delta H^{\circ,solv}(T)\) is calculated by adding a solvent not involved in the reaction at the same pressure \(P^{\circ}\). As in Kirchoff’s Law of Thermodynamics, we have that;
\[\begin{aligned} \Delta H^{\circ,solv}(T)&=\Delta H^{\circ,solv}(T_{0})+\int_{T_{0}}^{T}{d(\Delta H^{\circ,solv}(T))\over dT}dT\notag\\ &=\Delta H^{\circ,solv}(T_{0})+\int_{T_{0}}^{T}{d(\Delta Q_{solv}(T))\over dT}dT\notag\\ &=\Delta H^{\circ,solv}(T_{0})+\int_{T_{0}}^{T}C_{solv}(T)dT\notag\\ &=\Delta H^{\circ}(T_{0})+\int_{T_{0}}^{T}C_{solv}(T)dT,\label{lem66M} \end{aligned} \tag{121}\] where \(C_{solv}(T)=\Delta({dQ_{solv}\over dT})(T)\) is the change in the heat capacity of the mixture, after one mole of reaction. If we confine ourselves to a small temperature range, we can, therefore assume that \(\Delta H^{\circ}\) is approximately temperature independent. However, we can also add solvent to the reaction, to lower the magnitude of \(C_{solv}(T)\), as the heat capacity before and after the reaction would approach that of the solvent, and extend the temperature range of the reaction. More precisely, we have, by the law of mixtures for heat capacities, the fact that \(m_{mix,solv}\) is conserved during \(1\) mole of reaction, that \[\begin{aligned} C_{solv}(T)&=C_{fin,solv}(T)-C_{in,solv}(T)\notag\\ &={1\over m_{mix,solv}}(\sum_{i=0}^{c}m_{i,fin}C_{i,fin}(T)-\sum_{i=1}^{c}m_{i,in}C_{i,in}(T))\notag\\ &={N_{A}\over m_{mix,solv}}(\sum_{i=0}^{c}n_{i,fin}m_{i,molec}C_{i,fin}(T)-\sum_{i=0}^{c}n_{i,in}m_{i,molec}C_{i,in}(T))\notag\\ &={N_{A}\over m_{mix,solv}}(\sum_{i=0}^{c}n_{i,fin}m_{i,molec}m_{i,fin}SC_{i}(T)-\sum_{i=0}^{c}n_{i,in}m_{i,molec}m_{i,in}SC_{i}(T))\notag\\ &={N_{A}^{2}\over m_{mix,solv}}(\sum_{i=0}^{c}n_{i,fin}^{2}m_{i,molec}^{2}SC_{i}(T) -\sum_{i=0}^{c}n_{i,in}^{2}m_{i,molec}^{2}SC_{i}(T))\notag\\ &={N_{A}^{2}\over m_{mix,solv}}(\sum_{i=0}^{c}(n_{i,in}-\nu_{i})^{2}m_{i,molec}^{2}SC_{i}(T)-\sum_{i=0}^{c}n_{i,in}^{2}m_{i,molec}^{2}SC_{i}(T))\notag\\ &=-{2\over m_{mix,solv}}(\sum_{i=1}^{c}\nu_{i}m_{i,in}m_{i,mol}SC_{i}(T))+{1\over m_{mix,solv}}(\sum_{i=1}^{c}\nu_{i}^{2}m_{i,mol}^{2}SC_{i}(T))\notag\\ &=\theta(T)+v(T),\label{lem66N} \end{aligned} \tag{122}\] where \(\nu_{0}=0\) and \(\nu_{i}\), for \(1\leq i \leq c\) are the stochiometric coefficients for the original reaction, (²).

Using the fact that \({m_{i,in}\over m_{mix,solv}}\rightarrow 0\), for \(1\leq i\leq c\), by increasing the mass of the solvent not involved in the reaction, we have that \(\theta(T)\rightarrow 0\), and, similarly, \(v(T)\rightarrow 0\) as \(m_{mix}\rightarrow \infty\) with \(\{m_{i,in},m_{i,fin},m_{i,molec},m_{i,mol}\}\) denoting the initial, final, molecular and molar masses of substance \(i\) respectively, \(\{C_{i},SC_{i},C_{in},C_{fin}\}\) denoting the heat capacities of substance \(i\), the specific heat capacities of substance \(i\), the initial heat capacity of the mixture and the final heat capacity of the mixture respectively. In particular, we see that, as \(m_{mix,solv}\rightarrow \infty\), which we can achieve, by increasing the solvent not involved in the reaction, \(C_{solv}(T)\rightarrow 0\). We then have, by \((M),(N)\), that \[\Delta H^{\circ}(T)=\Delta H^{\circ}(T_{0})+\int_{T_{0}}^{T}(\theta(S)+v(S))dS \rightarrow \Delta H^{\circ}(T_{0}),\] as we can make the convergence uniform on the interval \((T_{0},T)\), given that the specific heat capacities \(SC_{i}(S)\) are bounded on \((T_{0},T)\), for \(1\leq i \leq c\). ◻

Proof. Suppose that an amount of substance \(\xi\) is formed in a loop. We have, by Lemma 1, that \[\begin{aligned} dG=-SdT+VdP+\sum_{i=1}^{c}\mu_{i}dn_{i},\label{lem67A} \end{aligned} \tag{123}\] and, by the definition of enthalpy in Definition 1, that \[\begin{aligned} dH=d(U+PV) =dU+PdV+VdP,\label{lem67dag} \end{aligned}\tag{124}\] \[\begin{aligned} d(G+TS)=dG+TdS+SdT.\label{lem67dagdag} \end{aligned} \tag{125}\] We then have that, for a closed path \(\gamma\), using (123), (124), (125), the definition of entropy as \(dS={dQ\over T}\) and the first law of thermodynamics, \(dQ=dU+pdV\), the calculation of \(({\partial G\over \partial \xi})_{(T,P)}\) as follows; \[\begin{aligned} \int_{\gamma}(dG+SdT-VdP)&=\int_{\gamma}((dH-TdS-SdT)+SdT-VdP)\\ &=\int_{\gamma}(dH-TdS-VdP)\\ &=\int_{\gamma}(dH-dQ-VdP)\\ &=\int_{\gamma}((dU+PdV+VdP)-dQ-VdP)\\ &=\int_{\gamma}(dU+PdV-dQ)\\ &=\int_{\gamma}((dQ-PdV)+PdV-dQ)\\ &=0, \end{aligned}\] \[\begin{aligned} \int_{\gamma}(\sum_{i=1}^{c}\mu_{i}dn_{i})& =\int_{\gamma}(\sum_{i=1}^{c}\mu_{i}\nu_{i}d\xi\notag\\ &=\int_{\gamma}({\partial G\over \partial \xi})_{(T,P)}d\xi\notag\\ &=\int_{\gamma}(\lambda+\epsilon ln(P)+\beta T+\sigma ln(T))d\xi\notag\\ &=\int_{\gamma}(\lambda+\epsilon ln(P)+\beta T+\sigma ln(T))d\xi\notag\\ &=\lambda \xi+\int_{\gamma}(\epsilon ln(P)\theta(P)dP+(\beta T+\sigma ln(T))\phi(T)dT),\label{lem67C} \end{aligned} \tag{126}\] where \(\theta(P)=\theta_{1}(P)\) along \(\gamma_{1}\), \(\theta_{1}(P)dP=d\xi|_{\gamma_{1}}\), \(\theta(P)=\theta_{2}(P)\) along \(\gamma_{2}\), \(\theta_{2}(P)dP=d\xi|_{\gamma_{2}}\), \(\phi(T)=\phi_{1}(T)\) along \(\gamma_{1}\), \(\phi_{1}(T)dT=d\xi|_{\gamma_{1}}\), \(\phi(T)=\phi_{2}(T)\) along \(\gamma_{2}\), \(\phi_{2}(T)dT=d\xi|_{\gamma_{2}}\).

Now, by Stokes Theorem; \[\begin{aligned} \int_{\gamma}(\epsilon ln(P)\theta(P)dP+\beta T\phi(T)dT) =\int\int_{R}({\partial((\beta T+\sigma ln(T))\phi(T))\over \partial P}-{\partial (\epsilon ln(P)\theta(P))\over \partial T})dTdP =0,\label{lem67D} \end{aligned} \tag{127}\] and, by (126), \(\xi=0\), if \(\lambda\neq 0\). The second claim, with the assumption that \(\Delta H^{\circ}\) is constant, follows from the proof of Lemma 7, and in later sections, when we introduce error terms. The next claim follows easily as \(w(T_{1},T_{2})\simeq 0\), see §12, and the computation of \(\lambda\) without the error term \(w(T_{1},T_{2})\). The computation of \(\epsilon(T_{1},T_{2})\) is given in Lemma 7. ◻