Episode 05. Solid - Liquid Equilibrium
Summary
In this episode, we cover the equilibrium between solids and liquids. Like before, we discuss the equilibrium using fugacity and predict the composition of the mixture based on thermodynamic properties. Finally, we compare predicted values with experimental data using naphthalene as an example.
1. Introduction
We can easily observe the phenomenon of solids dissolving in solvents, and the maximum quantity of solute that can be dissolved in the solvent is termed solubility. Let’s assume the amount of solute is gradually increased in the solvent. Initially, the solid dissolves in the solvent. However, at a certain point, it stops dissolving and remains solid. We can consider the point as reaching the equilibrium between the solid and the liquid phase. In this episode, we discuss how to predict the composition of a mixture in solid-liquid equilibrium with no reaction.
(Figure 5.1)
2. Solid - Liquid equilibrium in closed system
In Episode 03, we learned to utilize fugacity when dealing with phase equilibrium. Likewise, when dealing with the equilibrium between solids and liquids in a closed system, we use the fact that the fugacities of each phase are equal. Now, let’s discuss calculating the composition ratios in the solid-liquid equilibrium.
Pure solid – liquid
Imagine that the part of a pure solid is dissolved in a liquid. The component i in the solid phase exists in its pure form, while i in the liquid phase exists within the mixture. Since component i in each phase is in equilibrium, we can set that the fugacities of component i are equal in both the liquid mixture and pure solid phase. By (Eq 2.18), we can replace the fugacity in the liquid mixture with \({\small\displaystyle x_i\gamma_if_i^l}\)
\( {\small\displaystyle\begin{align} \hat{f}_i^l \quad&=\quad f_i^s \tag{Eq 5.1} \\[15pt] x_i\gamma_if_i^l \quad&=\quad f_i^s \tag{Eq 5.2} \end{align} } \)
Divide both sides of (Eq 5.2) by the fugacity of the pure liquid and define the ratio of the fugacity of the pure solid to that of the pure liquid as \({\small\displaystyle \psi_i}\). At the melting point, the fugacities of the pure component in each phase (liquid and solid) are equal. Therefore, if the mixture temperature is the same as the melting point temperature, we can summarize (Eq 5.3) as (Eq 5.4).
\( {\small\displaystyle\begin{align} x_i\gamma_i \quad&=\quad \psi_i,\qquad \psi_i \,\equiv\, \frac{f_i^s}{f_i^l} \tag{Eq 5.3} \\[15pt] x_i\gamma_i \quad&=\quad 1 \tag{Eq 5.4} \end{align} } \)
What happens if the solid-liquid system is in equilibrium and the temperature is below the melting point? The Gibbs free energy of each phase at temperature \({\small\displaystyle T}\) is expressed by (Eq 2.10).
\( {\small\displaystyle\begin{align} G_i^l(T,P) \quad&=\quad g(T)\,+\,RTlnf_i^l \\[10pt] G_i^s(T,P) \quad&=\quad g(T)\,+\,RTlnf_i^s \tag*{} \end{align} } \)
We can represent the difference in Gibbs free energy between the two phases, which is the Gibbs free energy of fusion, as shown in (Eq 5.5).
\( {\small\displaystyle\begin{align} \frac{\Delta G_i^{fus}}{RT} \quad=\quad \frac{G_i^l\,-\,G_i^s}{RT} \quad=\quad ln\frac{f_i^l}{f_i^s} \quad=\quad -ln\psi_i \tag{Eq 5.5} \end{align} } \)
Based on (Eq 1.9), the Gibbs free energy of fusion is:
\( {\small\displaystyle\begin{align} \Delta G_i^{fus} \quad=\quad \Delta H_i^{fus} \,-\,T\Delta S_i^{fus} \tag*{} \end{align} } \)
Now assume that the following processes occur when the solid melts at the temperature \({\small\displaystyle T}\).
- Step 1 : The temperature of the solid component rises to the melting point.
- Step 2 : The solid becomes a liquid.
- Step 3 : The liquid temperature decreases from the melting point temperature to the original temperature.
According to the steps, the enthalpy and entropy changes as follows.
\( {\small\displaystyle\begin{align} \Delta H_i^{fus}(T) \quad&=\quad \int_T^{T_m}\,C_{P,i}^s\,dT \,+\, \Delta H_i^{fus}(T_m) \,+\, \int_{T_m}^{T}\,C_{P,i}^l\,dT \\[10pt] &=\quad \Delta H_i^{fus}(T_m) \,+\, \int_{T_m}^{T}\,\Delta C_{P,i}^{sl}\,dT \tag{Eq 5.6} \\[15pt] \Delta S_i^{fus}(T) \quad&=\quad \int_T^{T_m}\,\frac{C_{P,i}^s}{T}\,dT \,+\, \Delta S_i^{fus}(T_m) \,+\, \int_{T_m}^{T}\,\frac{C_{P,i}^l}{T}\,dT \\[10pt] &=\quad \Delta S_i^{fus}(T_m) \,+\, \int_{T_m}^{T}\,\frac{\Delta C_{P,i}^{sl}}{T}\,dT \tag{Eq 5.7} \end{align} } \)
Since Gibbs free energy of fusion becomes 0 at the melting point by (Eq 5.4), the entropy change at the melting point temperature becomes:
\( {\small\displaystyle\begin{align} ln\left(x_i\gamma_i\right) \,=\,0 \quad&\longrightarrow\quad \Delta G_i^{fus}(T_m) \,=\,0 \\[15pt] \Delta S_i^{fus}(T_m) \quad&=\quad \frac{\Delta H_i^{fus}(T_m)}{T_m} \tag{Eq 5.8} \end{align} } \)
Then, we express the Gibbs free energy of fusion at temperature \({\small\displaystyle T}\) using (Eq 5.6) to (Eq 5.8).
\( {\small\displaystyle\begin{align} \Delta G_i^{fus}(T) \quad&=\quad \Delta H_i^{fus}(T_m)\,+\,\int_{T_m}^{T}\,\Delta C_{P,i}^{sl}\,dT \,-\, \frac{T\Delta H_i^{fus}(T_m)}{T_m}\,-\,T\int_{T_m}^{T}\,\frac{\Delta C_{P,i}^{sl}}{T}\,dT \\[10pt] &=\quad \Delta H_i^{fus}(T_m)\left(1\,-\,\frac{T}{T_m}\right)\,+\,\int_{T_m}^{T}\,\Delta C_{P,i}^{sl}\,dT\,-\,T\int_{T_m}^T\,\frac{\Delta C_{P,i}^{sl}}{T}\,dT \tag*{} \end{align} } \)
Dividing both sides by \({\small\displaystyle RT}\), we get (Eq 5.9). According to (Eq 5.5), the left side of (Eq 5.9) is the same as \({\small\displaystyle -\,ln\psi_i}\). Hence, (Eq 5.9) can be rearranged as (Eq 5.10).
\( {\small\displaystyle\begin{align} \frac{\Delta G_i^{fus}(T)}{RT} \quad&=\quad \frac{\Delta H_i^{fus}(T_m)}{RT}\left(1\,-\,\frac{T}{T_m}\right)\,+\,\frac{1}{RT}\int_{T_m}^{T}\,\Delta C_{P,i}^{sl}\,dT\,-\,\frac{1}{R}\int_{T_m}^T\,\frac{\Delta C_{P,i}^{sl}}{T}\,dT \tag{Eq 5.9} \\[15pt] ln\psi_i \quad&=\quad \frac{\Delta H_i^{fus}(T_m)}{RT}\left(\frac{T}{T_m}\,-\,1\right)\,-\,\frac{1}{RT}\int_{T_m}^{T}\,\Delta C_{P,i}^{sl}\,dT\,+\,\frac{1}{R}\int_{T_m}^T\,\frac{\Delta C_{P,i}^{sl}}{T}\,dT \tag{Eq 5.10} \end{align} } \)
If \({\small\displaystyle \Delta C_{P,i}^{sl}}\) is independent of temperature, we obtain (Eq 5.11). Assume the liquid mixture is an ideal solution. Then, the activity coefficient is 1, and (Eq 5.11) becomes (Eq 5.12).
\( {\small\displaystyle\begin{align} ln\psi_i \quad&=\quad \frac{\Delta H_i^{fus}(T_m)}{RT}\left(\frac{T}{T_m}\,-\,1\right)\,-\,\frac{\Delta C_{P,i}^{sl}}{R}\left(1\,-\,\frac{T_m}{T}\,+\,ln\frac{T_m}{T}\right) \tag{Eq 5.11} \\[15pt] lnx_i \quad&=\quad \frac{\Delta H_i^{fus}(T_m)}{RT}\left(\frac{T}{T_m}\,-\,1\right)\,-\,\frac{\Delta C_{P,i}^{sl}}{R}\left(1\,-\,\frac{T_m}{T}\,+\,ln\frac{T_m}{T}\right) \tag{Eq 5.12} \end{align} } \)
When it’s difficult to obtain information about the melting point temperature of a solid, sometimes (Eq 5.12) can be calculated using a triple point temperature as a substitute for the melting point temperature.
Solid mixtures – liquid
Let’s discuss the equilibrium between a solid mixture and a liquid. Since the fugacities of component i are the same in each phase, we can establish the equation as (Eq 5.13).
\( {\small\displaystyle\begin{align} \hat{f}_i^l \quad&=\quad \hat{f}_i^s \tag{Eq 5.13} \\[15pt] x_i\gamma_i^lf_i^l \quad&=\quad z_i\gamma_i^sf_i^s \tag{Eq 5.14} \end{align} } \)
Using the previously defined \({\small\displaystyle \psi_i}\), we can express (Eq 5.14) as (Eq 5.15). Additionally, if the activity coefficients in the solid phase have a negligible effect, we obtain (Eq 5.16).
\( {\small\displaystyle\begin{align} x_i\gamma_i^l \quad&=\quad z_i\gamma_i^s\psi_i \tag{Eq 5.15} \\[15pt] x_i\gamma_i^l \quad&=\quad z_i\psi_i \tag{Eq 5.16} \end{align} } \)
3. How to predict solubility of solids
Now, we can predict the composition of a mixture when it reaches solid-liquid equilibrium with the thermodynamic properties and activity coefficients. Let’s apply this to predict the solubility of a solid and compare it with experimental data. There are 100 moles of benzene and n-hexane solvents. We dissolve 100 moles of solid naphthalene in each solvent at 25ºC and 1 atm. (Table 5.1) shows the thermodynamic properties of naphthalene.
| Naphthalene | ||
|---|---|---|
| \( {\small\displaystyle MW } \) | \( {\small\displaystyle 128.19 } \) | \( {\small\displaystyle g/mol } \) |
| \( {\small\displaystyle C_P^s } \) | \( {\small\displaystyle 189.4 } \) | \( {\small\displaystyle J/mol\cdot K } \) |
| \( {\small\displaystyle C_P^l } \) | \( {\small\displaystyle 196.06 } \) | \( {\small\displaystyle J/mol\cdot K } \) |
| \( {\small\displaystyle \Delta C_P^{sl} } \) | \( {\small\displaystyle 6.66 } \) | \( {\small\displaystyle J/mol\cdot K } \) |
| \( {\small\displaystyle \Delta H^{fus} } \) | \( {\small\displaystyle 18.9 } \) | \( {\small\displaystyle kJ/mol } \) |
| \( {\small\displaystyle T_m } \) | \( {\small\displaystyle 353.2 } \) | \( {\small\displaystyle K } \) |
Case 1 : Naphthalene in benzene
Assuming that the heat capacity of naphthalene is independent of temperature and that the mixture is the ideal solution, we can use (Eq 5.12) to calculate the mole fraction of naphthalene in the liquid phase upon reaching equilibrium.
\( {\small\displaystyle\begin{align} lnx_i \,&=\, \frac{\Delta H_i^{fus}(T_m)}{RT}\left(\frac{T}{T_m}\,-\,1\right)\,-\,\frac{\Delta C_{P,i}^{sl}}{R}\left(1\,-\,\frac{T_m}{T}\,+\,ln\frac{T_m}{T}\right) \\[10pt] &=\, \frac{18.9\times 10^3}{8.314\cdot 298.15}\cdot\left(\frac{298.15}{353.2}\,-\,1\right)\,-\,\frac{6.66}{8.314}\cdot\left(1\,-\,\frac{353.2}{298.15}\,+\,ln\frac{353.2}{298.15}\right) \,=\, -\,1.176 \\[15pt] x_i \,&=\, \exp(-\,1.176) \,=\,0.3084 \end{align} } \)
The mole fraction calculated using (Eq 5.12) is 0.3084. We can confirm that this value is not significantly different from the experimental value of 0.2946 under the same conditions.
| Initial | \( {\longrightarrow} \) | Final | ||
|---|---|---|---|---|
| Solid phase | Solid phase | |||
| Naphthalene | 100 mol | Naphthalene | 55.397 mol | |
| Benzene | 0 mol | Benzene | 0 mol | |
| Total | 100 mol | Total | 55.397 mol | |
| Liquid phase | Liquid phase | |||
| Naphthalene | 0 mol | Naphthalene | 44.603 mol | |
| Benzene | 100 mol | Benzene | 100 mol | |
| Total | 100 mol | Total | 144.603 mol | |
Case 2 : Naphthalene in n-hexane
In this case, since it is expected that the mixture is not an ideal solution, we use (Eq 5.11). The activity coefficient varies with mole fraction. Therefore, we determine the final value through the trial and error method. Let’s assume 50 moles of naphthalene dissolve when the mixture is in equilibrium.
\( {\small\displaystyle\begin{align} ln\psi_i \,&=\, -\,1.176 \\[10pt] \psi_i \,&=\, 0.3084 \tag*{} \end{align} } \)
| Trial | 1 | 2 | 3 | \( {\small\displaystyle \cdots } \) | Final |
|---|---|---|---|---|---|
|
Moles of naphthalene
(Liquid phase) |
50 | 25.45 | 16.65 | 12.18 | |
|
Moles of
Liquid mixture |
150 | 125.45 | 116.65 | 112.18 | |
|
Mol fraction
(Assume) |
0.333 | 0.203 | 0.143 | 0.109 | |
|
Activity coefficient
of naphthalene |
1.817 | 2.323 | 2.635 | 0.109 | |
|
Mol fraction
(calc.) |
0.17 | 0.133 | 0.117 | 0.109 | |
| Error | 0.163 | 0.07 | 0.026 | 0 |
The predicted mole fraction of naphthalene in the liquid mixture is 0.109. The experimental value under the same conditions is approximately 0.117, indicating a difference of around 7%.
Nomenclature
\({\small\displaystyle C_P}\) : heat capacity
\({\small\displaystyle f_i}\) : fugacity of pure component
\({\small\displaystyle \hat{f}_i}\) : fugacity of component i in mixture
\({\small\displaystyle G}\) : Gibbs free energy
\({\small\displaystyle H}\) : enthalpy
\({\small\displaystyle R}\) : gas constant
\({\small\displaystyle S}\) : entropy
\({\small\displaystyle T}\) : temperature
\({\small\displaystyle T_m}\) : melting point temperature
\({\small\displaystyle x_i,\,z_i}\) : mole fraction of component i
\({\small\displaystyle \gamma}\) : activity coefficient
\({\small\displaystyle \psi}\) : the ratio of the fugacity of the pure solid to that of the pure liquid