# Thermodynamic Factor

The First Order Derivative of Gibbs Energy w.r.t. Molar Fraction

Let $$G$$ be the molar Gibbs energy of a $$c$$-component phase and $$\mu _k$$ the chemical potential of component $$k$$. $$G = G(x_1,x_2,\cdots,x_c)$$ The chemical potential of the component $$k$$, $$\mu _k$$, is calculated by the following equation, \begin{align} \mu_k = G-\sum_{j=1}^{c} x_j \frac{\partial G}{\partial x_j} +\frac{\partial G}{\partial x_k} \quad \quad \quad \quad (k=1,2,\cdots,c) \label {A2} \end{align} If the component $$c$$ is taken as the dependent component, \begin{align} x_c =1 - \sum_{j=1}^{c-1} x_j \end{align} then, the Gibbs energy is a function of $$c-1$$ variables only, $$G = G(x_1,x_2,\cdots,x_{c-1})$$ The chemical potential of the component $$k$$, $$\mu _k$$, is now calculated by these following equations, \begin{align} \mu_k = &G-\sum_{j=1}^{c-1} x_j \frac{\partial G}{\partial x_j} +\frac{\partial G}{\partial x_k} \quad \quad \quad \quad (k=1,2,\cdots,c-1) \label {A5} \\ \mu_c = &G-\sum_{j=1}^{c-1} x_j \frac{\partial G}{\partial x_j} \label {A6} \end{align} Please note that $$\displaystyle \frac{\partial G}{\partial x_k}$$ in Eq.\eqref{A5} and Eq.\eqref{A6} is different from that in Eq.\eqref{A2}. Therefore, the first order derivative of $$G$$ w.r.t molar fraction, $$x_j$$, is given by \begin{align} \frac{\partial G}{\partial x_k} = \mu_k - \mu_c \quad \quad \quad \quad (k=1,2,\cdots,c-1) \label {A7} \end{align}
Thermodynamic Factor and Partial Derivatives of Gibbs Free Energy

In Pandat, thermodynamic factors are available with the format $$\tt ThF(*,*@*)$$, which is defined by $${\tt ThF(x_i,x_j) }= \frac{\partial \mu_i}{\partial x_j} \quad \quad \quad \quad (i,j=1,2,\cdots,c) \label {A25}$$ where $$(x_1,x_2,\cdots,x_c)$$ are treated as the independent compositional variables. If the component $$1$$ is assumed to be the solvent component and its molar fraction $$x_1$$ is taken as the dependent variable, the independent compositional variables now are $$(X_2,X_3,\cdots,X_{c})$$. Here we use capital $$X$$ to distinguish this set of variables from $$(x_1,x_2,\cdots,x_c)$$. Then we have \begin{align} \frac{\partial \mu_j}{\partial X_k}= &\frac{\partial \mu_j}{\partial x_k} - \frac{\partial \mu_1}{\partial x_k}\quad \quad \quad \quad (j, k=2,3,\cdots,c) \\ = &{\tt ThF(x_j,x_k) } - {\tt ThF(x_1,x_k) } \quad \quad \quad \quad (j, k=2,3,\cdots,c) \end{align} Now let's see how to express the second derivatives of Gibbs free energy in terms of the thermodynamic factors. If we use this compositional variable set $$(X_2,X_3,\cdots,X_c)$$, the first and second derivatives of $$G$$ are \begin{align} \frac{\partial G}{\partial X_j}= & \sum_{i=2}^{c} \frac{\partial G}{\partial x_i} \frac{\partial x_i}{\partial x_j} + \frac{\partial G}{\partial x_1} \frac{\partial x_1}{\partial x_j} \quad \quad \quad \quad (j=2,3,\cdots,c) \\ = & \frac{\partial G}{\partial x_j} - \frac{\partial G}{\partial x_1} \quad \quad \quad \quad \quad \quad (j=2,3,\cdots,c) \\ = & \mu_j - \mu_1 \quad \quad \quad \quad \quad \quad (j=2,3,\cdots,c) \\ \frac{\partial^2 G}{\partial X_j \partial X_k} = &\sum_{i=2}^{c} \frac{\partial (\mu_j - \mu_1)}{\partial x_i} \frac{\partial x_i}{\partial x_k} + \frac{\partial (\mu_j - \mu_1)} {\partial x_1} \frac{\partial x_1}{\partial x_k} \nonumber \\ &\quad \quad\quad \quad\quad \quad\quad \quad\ (j, k=2,3,\cdots,c) \\ = & (\frac{\partial \mu_j}{\partial x_k} - \frac{\partial \mu_1}{\partial x_k}) - (\frac{\partial \mu_j}{\partial x_1} - \frac{\partial \mu_1}{\partial x_1}) \nonumber \\ &\quad \quad\quad \quad\quad \quad\quad \quad (j, k=2,3,\cdots,c) \\ = & {\tt ThF(x_j,x_k) } - {\tt ThF(x_1,x_k) } - {\tt ThF(x_j,x_1) } + {\tt ThF(x_1,x_1) } \nonumber \\ &\quad \quad\quad \quad\quad \quad\quad \quad (j, k=2,3,\cdots,c) \end{align} Traditionally, chemical potential can be expressed as a function of activity or activity coefficient, \begin{align} \mu_i =& \mu_i^{\circ} + RT \ln a_i \quad \quad \quad \quad (i=1,2,\cdots,c) \\ =& \mu_i^{\circ} + RT \ln \gamma_i x_i \quad \quad \quad \quad (i=1,2,\cdots,c) \label {A35} \end{align} Substituting Eq.(\ref{A35}) into Eq.(\ref{A25}), we have \begin{align} {\tt ThF(x_i,x_j) } =& \frac{\partial (\mu_i^{\circ} + RT \ln \gamma_i x_i)}{\partial x_j} &\quad \quad \quad \quad (i,j=1,2,\cdots,c) \nonumber \\ =& RT \frac{\partial ( \ln \gamma_i x_i)}{\partial x_j} &\quad \quad \quad \quad (i,j=1,2,\cdots,c) \end{align} When $$i=j$$, above equation becomes \begin{align} {\tt ThF(x_i,x_i) }=& \frac {RT}{x_i} \frac{\partial ( \ln \gamma_i x_i)}{\partial \ln x_i} &\quad \quad \quad \quad (i=1,2,\cdots,c) \nonumber \\ =& \frac {RT}{x_i} \left( 1+ \frac{\partial \ln \gamma_i}{\partial \ln x_i}\right) &\quad \quad \quad \quad (i=1,2,\cdots,c) \end{align} Sometimes, the term $$1+ \frac{\partial \ln \gamma_i}{\partial \ln x_i}$$ is called as thermodynamic factor. \begin{align} 1+ \frac{\partial \ln \gamma_i}{\partial \ln x_i} &= {\tt ThF(x_i,x_i) } \frac{x_i} {RT} &\quad \quad \quad \quad (i=1,2,\cdots,c) \end{align} For a dilute solution of component $$i$$, $$x_i \rightarrow 0$$ and $$\gamma_i \rightarrow constant$$ (Henry's law). Then we have \begin{align} {\tt ThF(x_i,x_i) } \frac{x_i} {RT} = 1 &\quad \quad \quad \quad (i=1,2,\cdots,c) \end{align} If component $$i$$ is the solvent and $$x_i \rightarrow 1$$, then $$\gamma_i \rightarrow 1$$ according to Raoult's law. We also have \begin{align} {\tt ThF(x_i,x_i) } \frac{1} {RT} = 1 &\quad \quad \quad \quad (i=1,2,\cdots,c) \end{align} or \begin{align} {\tt ThF(x_i,x_i) }= RT &\quad \quad \quad \quad (i=1,2,\cdots,c) \end{align}