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Erstellen von binären Schmelzdiagrammen (flüssig-fest) aus Abkühlungskurven

Die deutsche Version sollte bald zugänlich sein. Zur Zeit existiert leider nur die englische Version (kopiert auch unten).

Few practical comments

  1. incongruent melting of sodium sulfate decahydrate
    • introduce magnetic stirrers to all samples
    • start heating at room temperature and record the heating curves
    • stop recording when the temperature gets over approx. 45 ˚C
    • remove the stirrers and clean the vials
  2. binary diagram of the mixture biphenyl - naphthalene
    • stirrers are already present in the vials
    • heat the samples to approx. 90-95 ˚C, do not go over 98 ˚C in order to avoid thermometer sensor damage
    • start the measurement, record the cooling curves
    • DO NOT clean the vials after the experiment; in order to remove the sensors from the samples it’s usually necessary to heat the vials again and melt the solid phase
  3. binary diagram of the mixture water - zinc nitrate
    • get ice from the generator
    • introduce magnetic stirrer to all samples
    • heat the samples to approx. 75 ˚C
    • start the measurement, record the cooling curves while cooling the samples with ice
    • remove the stirrers and clean the vials (heating up may be necessary)

Theoretical remarks

This part of the page should summarize, extend, or explain the introductory text.

Gibbs phase rule

A phase is a state of matter that is uniform throughout (in chemical composition, physical state - solid, liquid, gas). The number of phases is denoted P.

A constituent is any substance that is present. Chemically independent constituent is called a component. Number of components (C) gives the number of independent species to define all phases. When no reaction takes place in the mixture, numbers of components and constituents are equal. In a reaction mixture, C is usually lower than the number of constituents (it’s defined by stoichiometry).

A variance gives the number of intensive variables that can be changed without disturbing the number of phases. It’s also called degrees of freedom (F).

The Gibbs phase rule reads F = C - P + 2. It can be easily derived. A system is defined by 2 intensive variables corresponding to the physical state (pressure p, temperature T) and further by P(C - 1) variables showing the composition of every phase P. Note here, that only C - 1 mole fractions are independent in every phase. Therefore, the total number of intensive variables that we can change is P(C-1) + 2.

In equilibrium, the chemical potential of a component \mu_j has to be the same throughout all P phases, i.e. \mu_{j,\alpha}=\mu_{j,\beta}=\mu_{j,\gamma}=\dots
We have P-1 equations for each component j, decreasing the number of degrees of freedom by C(P-1).

Putting these two parts together, we obtain the Gibbs phase rule: F = P(C-1) + 2 - C(P-1) = C - P + 2

Chemical potential in equilibrium (pure substances)

As already stated before, the chemical potential of a substance in a system in equilibrium is the same throughout the sample. Let’s now assume that chemical potentials of a pure substance at points 1 and 2 are not the same, and that we are moving from 1 to 2:

\Delta G = \mu_2\,dn - \mu_1\,dn = (\mu_2 - \mu_1)\,dn

For system in equilibrium \Delta G = 0 and therefore \mu_2 - \mu_1 = 0

Temperature and pressure dependence of phase stability (pure substances)

Some useful mathematical expressions (just a repetitorium of what you should already know) will follow:

dG = dH - T\,dS - S\,dT = dU + p\,dV +V\,dp - T\,dS - S\,dT = V\,dp - S\,dT

Then, we can write

\displaystyle dG(p,T) = \left(\frac{\partial G}{\partial p}\right)_T + \left(\frac{\partial G}{\partial T}\right)_p

From these two expressions, it follows directly

\displaystyle \left(\frac{\partial G}{\partial p}\right)_T = V; \left(\frac{\partial \mu}{\partial p}\right)_T = V_m

and

\displaystyle \left(\frac{\partial G}{\partial T}\right)_p = -S; \left(\frac{\partial \mu}{\partial T}\right)_p = -S_m

Therefore, with increasing temperature, the chemical potential decreases, because S_m is always positive.

The effect of pressure on melting depends on the fact, whether the change in a molar volume is positive or negative during this phase transition. If V_m > 0 (normal case), then increasing the pressure increases \mu (l) more than \mu (s) and the melting point increases (the increased pressure prevents the formation of less dense liquid phase). In opposite case (for example for water), the sitution is reversed, and the melting point decreases with increasing pressure.

At a given temperature and pressure, the phase with the lowest chemical potential is the most stable.

From the fact that the chemical potential of a substance is the same throughout all phases in equilibrium, we can derive two important equations.

Clapeyron equation describes any phase equilibrium of a pure substance. The equality of the chemical potentials \mu_{\alpha}(p,T) = \mu_{\beta}(p,T) can be expanded to involve molar volumes and entropies.

-S_{m,\alpha}\,dT + V_{m,\alpha}\,dp = -S_{m,\beta}\,dT + V_{m,\beta}\,dp.

(V_{m,\beta} - V_{m,\alpha})\,dp = (S_{m,\beta}\,dT - S_{m,\alpha}\,dT)

The Clapeyron equation then reads \displaystyle\frac{dp}{dT} = \frac{\Delta_{\mathrm{trs}}S_m}{\Delta_{\mathrm{trs}}V_m}

For the solid-liquid phase boundary, we write \Delta_\mathrm{fus}S_m = \Delta_\mathrm{fus}H_m/T. After substituting to the Clapeyron equation, we obtain the mathematical description of the boundary

\displaystyle \frac{dp}{dT} = \frac{\Delta_\mathrm{fus}H_m}{T \Delta_\mathrm{fus}V_m}

Usually, both \Delta_\mathrm{fus}H_m and \Delta_\mathrm{fus}V_m are positive, and the phase boundary is a steep curve (when displayed in p-T diagram). When we integrate this expression, we arrive at

\displaystyle \int_{p^*}^{p} dp = \frac{\Delta_\mathrm{fus}H_m}{\Delta_\mathrm{fus}V_m} \int_{T^*}^{T} \frac{dT}{T}

The expression for the pressure then reads

\displaystyle p = p^* + \frac{\Delta_\mathrm{fus}H_m}{\Delta_\mathrm{fus}V_m} \ln\left(\frac{T}{T^*}\right)

For the temperature close to T^*, we can expand the logarithm

\displaystyle \ln \left(\frac{T}{T^*}\right) = \ln \left(1 + \frac{T - T^*}{T^*}\right) \approx \frac{T - T^*}{T^*}

and write the final expression as

\displaystyle p = p^* + \frac{T - T^*}{T^*} \frac{\Delta_\mathrm{fus}H_m}{\Delta_\mathrm{fus}V_m}

For the liquid-vapor phase boundary, it follows that

\displaystyle \frac{dp}{dT} = \frac{\Delta_\mathrm{vap}H_m}{T \Delta_\mathrm{vap}V_m}

Usually, the \Delta_\mathrm{vap}H_m is positive and \Delta_\mathrm{vap}V_m is positive and very large. The phase boundary is then not so steep as in the case of the solid-liquid boundary. Since the molar volume of a gas is much larger than the molar volume of a liquid, we can write \Delta_\mathrm{vap}V_m = V_m(g) = RT/p (ideal gas). Then

\displaystyle \frac{dp}{dT} = \frac{\Delta_\mathrm{vap}H_m p}{RT^2}

Rearrangement leads to the Clausius-Clapeyron equation \displaystyle \frac{1}{p}\frac{dp}{dt} = \frac{d\ln p}{dT} = \frac{\Delta_\mathrm{vap}H_m}{RT^2}

Integration leads to

\displaystyle \int_{p^*}^{p} d\ln p = \frac{\Delta_\mathrm{vap}H_m}{R} \int_{T^*}^{T} \frac{dT}{T^2}

After simple manipulation

\displaystyle \ln \left(\frac{p}{p^*}\right) = -\frac{\Delta_\mathrm{vap}H_m}{R}\left( \frac{1}{T} - \frac{1}{T^*} \right)

we get the final formula p = p^* e^{-\kappa} where

\displaystyle \kappa = \frac{\Delta_\mathrm{vap}H_m}{R} \left( \frac{1}{T} - \frac{1}{T^*} \right)

Phase transistions (pure substances)

Let’s see the change of H and V during a phase transition \alpha \rightarrow \beta
\displaystyle \left( \frac{\partial\mu_\beta}{\partial p}\right)_T - \left( \frac{\partial\mu_\alpha}{\partial p}\right)_T = V_{m,\beta} - V_{m,\alpha}} = \Delta_{\mathrm{trs}}V_m

\displaystyle \left( \frac{\partial\mu_\beta}{\partial T}\right)_p - \left(\frac{\partial\mu_\alpha}{\partial T}\right)_p = -S_{m,\beta} + S_{m,\alpha}} = \frac{\Delta_{\mathrm{trs}}H_m}{T_{\mathrm{trs}}}

For a first order transition, the changes in molar enthalpy and molar volume are nonzero. Therefore, the first derivatives of chemical potentials are discontinuous. For example, the constant pressure heat capacity, defined by C_p = \partial H/\partial T} is infinite during the transition (infinitesimal change in temperature, but finite change in H). The supplied heat then drives the transition and does not raise the temperature. We encounter this type of phase transition during our experiments.

For a second order phase transition, the first derivatives are continuous and second derivatives are discontinuous

Incongruent melting

The fact that a substance has an incongruent melting point means only, that this substance is not stable as a liquid phase, i.e. that the liquid phase does not exist. One of examples may be the system Na-K. A substance \mathrm{Na_2K} exists, but in the region corresponding to its liquid phase it decomposes into liquid K and liquid Na. Sodium sulfate decahydrate is another example of the incongruent melting point (it decomposes into water and sodium sulfate).

Temperature dependence of phase equilibria

This part makes all the derivations that are partially omitted in the introductory paper for the laboratory experiment (Eqs. 4-15 and 17-20). It won’t be documented more than necessary.

For component A in system with 2 phases \alpha, \beta in equilibrium \mu_A^\alpha = \mu_A^\beta. Chemical potential of this component can be written (we assume ideal behavior) \mu_A^\nu = \mu_A^{*,\nu} + RT \ln x_A^\nu and in equilibrium \mu_A^{*,\beta} + RT \ln x_A^\beta = \mu_A^{*,\alpha} + RT \ln x_A^\alpha.

At the temperature T we write \mu_A^\alpha(T) = \mu_A^\beta(T). At the temperature T + dT it follows that

\mu_A^\alpha(T+dT) = \mu_A^\beta(T+dT)

If we expand this equality in the variable T and neglect everything beyond linear term, the equation reads

\displaystyle \mu_A^\alpha(T) + \left(\frac{\partial \mu_A^\alpha}{\partial T}\right)_p\,dT= \mu_A^\beta(T) + \left(\frac{\partial \mu_A^\beta}{\partial T}\right)_p\,dT

Since the first terms on both sides are equal (definition of an equilibrium), we obtain the expression for the so called “running equilibrium”

\displaystyle \left(\frac{\partial \mu_A^\alpha}{\partial T}\right)_p\,dT= \left(\frac{\partial \mu_A^\beta}{\partial T}\right)_p\,dT

We differentiate the equation for the chemical potential of a component A with respect to T and also rearrange the same equation:

original equation \mu_A^\nu = \mu_A^{*,\nu} + RT \ln x_A^\nu

differentiated equation \displaystyle \left(\frac{\partial \mu_A^\nu}{\partial T}\right)_p = \left(\frac{\partial \mu_A^{*,\nu}}{\partial T}\right)_p + RT\left(\frac{\partial \ln x_A^\nu}{\partial T}\right)_p + R \ln x_A^\nu

rearranged equation \displaystyle R \ln x_a^\nu = \frac{\mu_A^\nu - \mu_A^{*,\nu}}{T}

Putting all together, we obtain the final expression for the change of chemical potential of component A in phase \nu

\displaystyle \left(\frac{\partial \mu_A^\nu}{\partial T}\right)_p = \left(\frac{\partial \mu_A^{*,\nu}}{\partial T}\right)_p + \frac{\mu_A^\nu - \mu_A^{*,\nu}}{T} + RT\left(\frac{\partial \ln x_A^\nu}{\partial T}\right)_p

Inserting this result into the equation for the “running equilibrium”
\displaystyle \left(\frac{\partial \mu_A^{*,\alpha}}{\partial T}\right)_p + \frac{{\color{red}\mu_A^\alpha} - \mu_A^{*,\alpha}}{T} + RT\left(\frac{\partial \ln x_A^\alpha}{\partial T}\right)_p = \left(\frac{\partial \mu_A^{*,\beta}}{\partial T}\right)_p + \frac{{\color{red}\mu_A^\beta} - \mu_A^{*,\beta}}{T} + RT\left(\frac{\partial \ln x_A^\beta}{\partial T}\right)_p

Noting the fact that the chemical potential of a component A in equilibrium has to be the same throughout all phases leads to the following simplification
\displaystyle \left(\frac{\partial \mu_A^{*,\alpha}}{\partial T}\right)_p - \frac{\mu_A^{*,\alpha}}{T} + RT\left(\frac{\partial \ln x_A^\alpha}{\partial T}\right)_p = \left(\frac{\partial \mu_A^{*,\beta}}{\partial T}\right)_p - \frac{\mu_A^{*,\beta}}{T} + RT\left(\frac{\partial \ln x_A^\beta}{\partial T}\right)_p

To finish this discussion we need one more famous expression, namely the Gibbs-Helmholtz equation. From the definition of Gibbs free energy G = H - TS. We’ve also shown previously that dG = V\,dp - S\,dT. We can, therefore, write

\displaystyle S = \frac{H-G}{T} and \displaystyle\left(\frac{\partial G}{\partial T}\right)_p = S

\displaystyle\color{blue}\left(\frac{\partial G}{\partial T}\right)_p = \frac{G-H}{T}

Now we note that

\displaystyle\left(\frac{\partial}{\partial T}\left(\frac{G}{T}\right)\right)_p = \frac{1}{T}\left(\frac{\partial G}{\partial T}\right)_p + G \left(\frac{\partial}{\partial T}\left(\frac{1}{T}\right)\right)_p = \frac{1}{T}\left( {\color{blue}\left(\frac{\partial G}{\partial T}\right)_p} - \frac{G}{T}\right)

and after substitution we arrive at the Gibbs-Helmholtz equation

\displaystyle \displaystyle\left(\frac{\partial}{\partial T}\left(\frac{G}{T}\right)\right)_p = \frac{1}{T}\left( \frac{G-H}{T} - \frac{G}{T}\right) = -\frac{H}{T^2}

We now just need the expressions for molar entropy of a component in the system

\displaystyle S_{m,i} = -\left(\frac{\partial \mu_i}{\partial T}\right)_p

and the relation between the chemical potential and molar enthalpy of this component (which we obtain from the expression for S given at the beginning of this section)

\displaystyle \left(\frac{\partial \mu_i}{\partial T}\right)_p = \frac{\mu_i - H_{m,i}}{T}

Rearrangement leads to an equation, that we will need in the following discussion.

\displaystyle H_{m,i} = \mu_i - T \left(\frac{\partial \mu_i}{\partial T}\right)

We have to insert the previous expression into the last equation discussed before the Gibbs-Helmholtz. After some reformulation, we see that

\displaystyle RT \frac{\partial}{\partial T} \left(\ln \frac{x_A^\alpha}{x_A^\beta}\right) = \frac{\mu_A^{*,\alpha}}{T} - \left(\frac{\partial \mu_A^{*,\alpha}}{\partial T}\right)_p - \frac{\mu_A^{*,\beta}}{T} + \left(\frac{\partial \mu_A^{*,\beta}}{\partial T}\right)

\displaystyle \frac{\partial}{\partial T} \left(\ln \frac{x_A^\beta}{x_A^\alpha}\right) = \frac{1}{RT^2}\left( \mu_A^{*,\beta} - T\left(\frac{\partial \mu_A^{*,\beta}}{\partial T}\right) \right) - \frac{1}{RT^2}\left( \mu_A^{*,\alpha} - T\left(\frac{\partial \mu_A^{*,\alpha}}{\partial T}\right) \right)

It clearly follows

\displaystyle \frac{\partial}{\partial T} \left(\ln \frac{x_A^\beta}{x_A^\alpha}\right) = \frac{1}{RT^2}\Delta_{\alpha\rightarrow\beta}H_A^*

Now we define, that \alpha and \beta are liquid and solid phase, respectively. Then the enthalpy change in the previous equation corresponds to solidification enthalpy, or its negative value is the melting enthalpy of the pure substance A. The solid phase consists only of the component A (recall that in the text we deal only with substances that are fully immiscible in the solid phase), and therefore

\displaystyle x_A^\beta = x_A^s = 1 and \displaystyle x_A^\alpha = x_A^l

\displaystyle \frac{\partial}{\partial T} \left(\ln x_A^l\right) = \frac{1}{RT^2}\Delta_{\mathrm{fus}}H_A^*

Expanding the differential on the left side and rearranging yields the final equation

\displaystyle \frac{1}{x_A^l}\frac{\partial x_A^l}{\partial T} = \frac{1}{RT^2}\Delta_{\mathrm{fus}}H_A^*

\displaystyle \frac{\partial T}{\partial x_A^l} = \frac{RT^2}{x_A^l \Delta_{\mathrm{fus}}H_A^*}

The enthalpy of fusion can be, therefore, easily determined from the melting temperature of a pure substance. When the temperature changes further, we have to take the temperature dependence of the enthalpy into account.

\displaystyle \Delta_{\mathrm{fus}}H_A^*(T) = \Delta_{\mathrm{fus}}H_A^*(T_{\mathrm{fus}}) + \left\int_T^{T_{\mathrm{fus}}}C_p^{*,s}\,dT + \left\int_{T_{\mathrm{fus}}}^T C_p^{*,l}\,dT = \Delta_{\mathrm{fus}}H_A^*(T_{\mathrm{fus}}) + \left\int_{T_{\mathrm{fus}}}^T \Delta C_p^*\,dT

where \Delta C_p^* is the difference of heat capacities of the substance in liquid and solid phase

\Delta C_p^* = C_p^{*,l} - C_p^{*,s}

Please note that the expression in the introductory text is wrong - the liquid and solid heat capacities are swapped. The latest version of the introductory text has this error corrected.

Freezing point depression, boiling point elevation

We will split the discussion of the reasons behind the freezing point depression and boiling point elevation into two parts. These two phenomena belong to the so called colligative properties. We discuss the situation, when a solute B is added to pure solvent A.

Chemical potential vs. temperature in pure solvent and in solutionsIn the first case, the solute stays only in the liquid phase, i.e., there is no direct influence of the solute on standard chemical potentials of the solvent A in the solid and gaseous phase. As can be seen in the picture om the right (source: Wikipedia), this results in the decrease of the chemical potential of the solvent only in the liquid phase. As a result, the melting and boiling temperatures are decreased and increased, respectively. The decrease is larger than the increase due to much smaller slope of the line showing the chemical potential of the solid when compared to the gas-curve.

Mathematically, we can write
\displaystyle \mu_A^{*,g} = \mu_A^{*,l} + RT \ln x_A^l

For a mixture and the pure solvent, the following expressions are valid (when we neglect the small temperature dependence of H and S on the temperature)

\displaystyle \ln (1-x_B^l) = \frac{\mu_A^{*,g} - \mu_A^{*,l}}{RT} = \frac{\Delta_\mathrm{vap}G^*}{RT} = \frac{\Delta_\mathrm{vap}H^*}{RT} - \frac{\Delta_\mathrm{vap}S^*}{R}

\displaystyle \ln 1 = \frac{\Delta_\mathrm{vap}H^*}{RT^*} - \frac{\Delta_\mathrm{vap}S^*}{R}

Where the T^* denotes the vaporization temperature for the pure solvent. Subtracting these two equations yields

\displaystyle \ln (1-x_B^l) = \frac{\Delta_\mathrm{vap}H^*}{R} \left(\frac{1}{T} - \frac{1}{T^*}\right)

For very small mole fractions of the solute (x_B \ll 1), the logarithm can be expanded

ln (1 - x_B^l) = -x_B^l

which leads to

\displaystyle x_B^l = \frac{\Delta_\mathrm{vap}H^*}{R} \left(\frac{1}{T^*} - \frac{1}{T}\right) = \frac{\Delta_\mathrm{vap}H^*}{R} \frac{T-T^*}{T^*T} = \frac{\Delta_\mathrm{vap}H^*}{R} \frac{\Delta T}{(T^*)^2}

The last equation utilizes the fact that usually the values of T \mathrm{and} T^* are close to each other. For the boiling point elevation, it then follows (with K being the ebulioscopic constant)

\displaystyle\Delta T = x_B^l \frac{R(T^*)^2}{\Delta_\mathrm{vap}H^*} = x_B^l K

Using similar arguments for melting

\displaystyle \mu_A^{*,g} = \mu_A^{*,l} + RT \ln x_A^l

we arrive at the expression for the melting point depression (K being the cryoscopic constant)
\displaystyle\Delta T = x_B^l \frac{R(T^*)^2}{\Delta_\mathrm{fus}H^*} = x_B^l K

In the situation when the solute is present also at the other phases, the situation is a little bit more complex (but anyway the expressions are quite easy to derive). The chemical potential of a solvent decreases in all phases, depending on the molar ratios of the solvent and solute in the respective phase.

\displaystyle \mu_A^{*,g} + RT \ln x_A^g= \mu_A^{*,l} + RT \ln x_A^l

\displaystyle \ln\left(\frac{x_A^g}{\ln x_A^l}\right) = \frac{\Delta_\mathrm{vap}H^*}{RT} - \frac{\Delta_\mathrm{vap}S^*}{R}

Differentiating this equation with repspect to temperature yields

\displaystyle \frac{d}{dT}\,\ln\left(\frac{x_A^g}{x_A^l}\right) = - \frac{\Delta_\mathrm{vap}H^*}{RT^2}

We now integrate this result from the pure solvent to the mixture of an arbitrary composition

\displaystyle \left\int_{{\color{red}x_A=1}}^{x_A^g, x_A^l} d\ln \left(\frac{x_A^g}{x_A^l}\right) = - \frac{\Delta_\mathrm{vap}H^*}{R} \left\int_{T*}^T \frac{dT}{T^2}

This leads to (the logarithm of the molar ratio of a pure substance is equal to zero)

\displaystyle \ln\left(\frac{x_A^g}{x_A^l}\right) = \frac{\Delta_\mathrm{vap}H^*}{R} \left(\frac{1}{T} - \frac{1}{T^*} \right)

We can use the results of the previous subsection to write

\displaystyle\Delta T = \ln\left(\frac{x_A^g}{x_A^l}\right) \frac{R(T^*)^2}{\Delta_\mathrm{vap}H^*}

and

\displaystyle\Delta T = \ln\left(\frac{x_A^l}{x_A^s}\right) \frac{R(T^*)^2}{\Delta_\mathrm{fus}H^*}

for boiling and melting temperature change, respectively.

Comments? Remarks? Questions?

I hope this small overview was be useful for you. If you have any questions or comments, or if you find any typos/mistakes, please let me know.