Phase Equilibria I

Abstract

In the previous chapter, it was discussed the thermodynamics principles that are the basis of thermochemistry. It was shown how to calculate the energy involved in any chemical or physical transformation. Additionally, it was established the Gibbs free energy as a criterion to know whether or not such transformations would take place under a given set of conditions.

Example of Calculations

Example 1 Gibbs rule of phases

Part (A)

An equimolar mixture of H₂ and O₂ is contained in a pressurized cylinder. A spark causes a reaction resulting in the formation of water droplets. List the degrees of freedom before and after the water forming reaction.

Solution

In this case we have a gaseous mixture (1 phase, two components) transforming chemically into 1 phase and one component.

Gibbs rule of phases: $F = C - P + 2$

H2 + 0.5 O2			H2O
C	P	F	C	P	F
2	1	3	1	1	2

Part (B)

Evaluate P, C and F for each of the following systems/reactions:

• a. A mixture of five gases enclosed in a cylinder

• b. 2HI = H₂ + I₂ in vapor phase, starting with various mixtures of hydrogen and iodine

• c. H₂, O₂ and H₂O enclosed in a vessel at room temperature

• d. 2H₂O = 2H₂ + O₂ at high temperature starting with water

• e. Solid carbon, CO, CO₂ and O₂ in equilibrium at 600  °C

• f. (CH₃ COOH)_{2 liquid} = (CH₃ COOH)_{2 gas}

Solution

System	C	P	F	Comments
a	5	1	6	This is a gaseous solution with 5 components. No chemical interaction between components
b	2	1	3	This systems considers the decomposition of HI into H2 and I2 gases; therefore there are two components mixed in one phase
c	3	2	3	At room temperature liquid water is the most stable phase. This is in equilibrium with a gaseous mixture made of H2 and O2
d	1	1	2	In this case water is expected to be in the gaseous phase, therefore there are 3 components mixed together in a single phase
e	4	2	4	Two phases (1 solid, 1 gas) coexist. 3 components are present in the gas and 1 component in the solid
f	1	2	1	There is a single component existing in two distinct phases

Part (C)

At high temperature, magnesium carbonate dissociates and can exist in equilibrium with the decomposition products:

$$ {\text{MgCO}}_{3} = {\text{MgO}} + {\text{CO}}_{2} $$

• (a) State the number of phases and components in this system

• (b) Would the number of components be changed by adding CO₂ from an external source?

• (c) Would the number of phases be altered?

Solution

• (a) At equilibrium, there are 3 phases, 2 solids (MgCO₃ and MgO) and one gas (CO₂). The number of components is also 3, the carbonate, the oxide and the gas.

• (b) No it would not change, since CO₂ is already in the system.

• (c) No, because CO₂ is in the gas phase and already is in the system.

Part (D)

Ammonium chloride vaporizes upon heating.

• (a) State the number of components in a system consisting of solid ammonium chloride and its vapor

• (b) Would the number of components be changed by admitting ammonia gas into the system?

• (c) Would the number of phases be altered?

• (d) Would equilibrium be still possible?

Solution

• (a) There is one component (NH₄Cl) and two phases (1 solid, 1 gas)

• (b) Yes, there would be 2 components (NH₄Cl, NH₃)

• (c) No, there would still be a gaseous phase and a solid one

• (d) Yes, it would.

Example 2 Clausius—Clapeyron equation

The vapor pressures of sodium are written as:

$$ \log P = 10.304 - \frac{5603}{T} $$

(E2.1)

and

$$ \log P = 9.710 - \frac{5377}{T} $$

(E2.2)

Where P is in Pa and T in K. Which of the two equations is for solid sodium?

Solution

We need to convert log P into ln P $(\ln P = 2.303\log P)$ in order to be able to use Clausius–Clapeyron equation:

$$ \begin{aligned} \ln P & = 2.303\left[ {10.304 - \frac{5603}{T}} \right] \\ \ln P & = 23.730 - \frac{12903.709}{T} \\ \end{aligned} $$

(E2.1a)

and,

$$ \begin{aligned} \ln P & = 2.303\left[ {9.710 - \frac{5377}{T}} \right] \\ \ln P & = 22.362 - \frac{12383.231}{T} \\ \end{aligned} $$

(E2.2a)

Using Clausius–Clapeyron Eq. (2.11) witgh the form:

$$ \frac{{{\text{d}}\ln P}}{{{\text{d}}T}} = \frac{{\Delta H_{\text{i}}^{\circ} }}{{{\text{RT}}^{2} }} $$

(E2.3)

Applying equation (E2.3) to expressions (E2.1a) and (E2.2a) results in:

$$ \begin{aligned} \frac{{{\text{d}}\ln P}}{{{\text{d}}T}} & = \frac{\text{d}}{{{\text{d}}T}}\left( {23.730 - \frac{12903.709}{T}} \right) \\ \frac{{{\text{d}}\ln {\text{P}}}}{{{\text{d}}T}} & = \frac{12903.709}{{T^{2} }} = \frac{{\Delta H_{1} }}{\text{RT}} \\ \end{aligned} $$

(E2.1b)

and

$$ \begin{aligned} \frac{{{\text{d}}\ln P}}{{{\text{d}}T}} & = \frac{\text{d}}{{{\text{d}}T}}\left( {22.362 - \frac{12383.231}{T}} \right) \\ \frac{{{\text{d}}\ln P}}{{{\text{d}}T}} & = \frac{12383.231}{{T^{2} }} = \frac{{\Delta H_{2} }}{{{\text{R}}T}} \\ \end{aligned} $$

(E2.2b)

From expressions (E2.1b) and (E2.2b) it is evident that ΔH ₁ > ΔH ₂, therefore, equation (E2.1) corresponds to solid sodium.

Example 3 Use of Clapeyron equation

Calculate the pressure that must be applied to silverquery to increase its melting temperature by 10 and 20 K, using the following data:

Property	Value
Atomic mass (M)	107.87 g mole−1
Melting point (Tfus)	961.8 °C
Solid density (ρs)	10.50 g cm−3
Liquid density (ρl)	9.32 g cm−3
Heat of melting ($\Delta H_{{{\text{fus}}}}^{{\circ }}$)	11280 J mole−1s

Solution

We need to calculate the molar volume of both solid and liquid silver:

$V_{\text{molar}} = \frac{M}{\rho }$; for each phase we have:

$$ \begin{aligned} V_{{{\text{m}},{\text{solid}}}} & = \frac{{{\text{M}}_{\text{Ag}} }}{{\rho_{{{\text{Ag}},{\text{s}}}} }} = \frac{{107.87\tfrac{\text{g}}{\text{mole}}}}{{10.50\tfrac{\text{g}}{{{\text{cm}}^{3} }}}} = 10.27\frac{{{\text{cm}}^{3} }}{\text{mole}} \\ V_{{{\text{m}},{\text{liquid}}}} & = \frac{{{\text{M}}_{\text{Ag}} }}{{\rho_{{{\text{Ag}},{\text{l}}}} }} = \frac{{107.87\tfrac{\text{g}}{\text{mole}}}}{{ 9. 3 2\tfrac{\text{g}}{{{\text{cm}}^{3} }}}} = 11.57\frac{{{\text{cm}}^{3} }}{\text{mole}}. \\ \end{aligned} $$

From these molar volumes, we can calculate the volume change due to silver melting:

$$ \Delta {V} = {V}_{{{\text{m}}.\,{\text{liquid}}}} - {V}_{{{\text{m}},\,{\text{solid}}}} = 11.57 - 10.27 = 1.3\frac{{{\text{cm}}^{3} }}{\text{mole}}\quad {\text{or}}\quad 1.3 \times 10^{-6} \frac{{{\text{m}}^{3}}}{\text{mole}} $$

We can integrate Clapeyron’s equation between limits 1.01325 × 10⁵ Pa at T _fus and P at T:

$$ \begin{array}{*{20}l} {\frac{{{\text{d}}P}}{{{\text{d}}T}} = \frac{\Delta H}{\Delta V}\frac{1}{T}} \hfill \\ {\int\limits_{{1.01325\, \times \,10^{5} }}^{\text{P}} {{\text{d}}P} = \frac{\Delta H}{\Delta V}\int\limits_{1234.8}^{\text{T}} {\frac{{{\text{d}}T}}{T}} } \hfill \\ {P = \frac{\Delta H}{\Delta V}\ln \left( {\frac{T}{1234.8}} \right) + 1.01325 \times 10^{5} } \hfill \\ {P = \frac{11280}{{1.3 \times 10^{ - 6} }}\ln \left( {\frac{T}{ 1 2 3 4. 8}} \right) + 1.01325 \times 10^{5} } \hfill \\ \end{array} $$

Substituting the corresponding temperature values (in K) in the last equation, we can calculate the pressure increase needed to melt silver:

T (°C)	P (atm)
961.8	1
971.8	691.7
981.8	1376.9
991.8	2056.6

The relationship between the temperature (°C) increase and the corresponding pressure (atm) increment is shown in the figure below (Fig. 2.8).

Fig. 2.8

Pressure increase to augment the meltin1 g point of silver

Example 4 Calculation of phase boundary line

The dissociation of Na₂O at 1000 °C can be represented by:

$$ 2{\text{Na}}_{2} {\text{O}}_{{({\text{solid}})}} = 4{\text{Na}}_{{({\text{gas}})}} + {\text{O}}_{{2,({\text{gas}})}} $$

For this reaction, ΔG°₁₂₇₃ = 450.77 kJ. Graphically determine under which conditions, sodium oxide would not decompose:

Solution

For this problem, ΔG° is given, therefore we can relate the equilibrium constant to the partial pressures of the gases resulting from the oxide decomposition

$$ \begin{array}{*{20}l} {\Delta G_{1273}^{\text{o}} = - {\text{RT}}\ln (K)} \hfill \\ {K = \exp \left( { - \frac{{\Delta G_{1273}^{\text{o}} }}{\text{RT}}} \right) = \exp \left( { - \frac{450770}{8.314 \times 1273}} \right)} \hfill \\ {K = 3.18 \times 10^{ - 19} } \hfill \\ \end{array} . $$

On the other hand, form the dissociation reaction we can express the equilibrium constant in terms of partial pressure of sodium and oxygen:

$$ \begin{aligned} & K = {\text{p}}_{{{\text{O}}_{2} }} \times {\text{p}}_{\text{Na}}^{4} \\ & 3.18 \times 10^{ - 19} = {\text{p}}_{{{\text{O}}_{2} }} \times {\text{p}}_{\text{Na}}^{4} . \\ \end{aligned} $$

The logarithm of the previous equation, allow us to express the partial pressure of sodium as a function of that of the oxygen evolving from the oxide:

$$ \begin{aligned} & \log (3.18 \times 10^{ - 19} ) = \log ({\text{p}}_{{{\text{O}}_{2} }} ) + 4\log ({\text{p}}_{\text{Na}} ) \\ & \log ({\text{p}}_{\text{Na}} ) = \frac{{ - 18.5 - \log ({\text{p}}_{{{\text{O}}_{2} }} )}}{4}. \\ \end{aligned} $$

We can assign values to the partial pressure of oxygen so we can calculate the equilibrium value for that of sodium and thus define the equilibrium line between the oxide and its dissociation products; this is illustrated in the figure below (Fig. 2.9).

Fig. 2.9

Phase stability in the dissociation of Na₂O

Example 5 Determination of the heat of sublimation of water using the heat of vaporization and that of melting along with Clasius–Clapeyron equation

The heats of vaporization and of fusion of water are 45.054 kJ mole⁻¹ and 6.01 kJ mole⁻¹ at 0 °C, respectively. The vapor pressure of water at 0 °C is 0.006 atm. Calculate the sublimation pressure of ice at −15 °C, assuming that the enthalpies of transformation do not change with temperature.

Solution

The heats of transformation are related as indicated in the sketch below. According to this figure, we can calculate the heat of sublimation of water at 0 °C by combining the heats of evaporation and melting at the same temperature. With that heat, then we can use Clausius–Clapeyron equation to calculate the pressure for transforming solid water into vapor at −15 °C (Fig. 2.10).

Fig. 2.10

Relationship between the standard heat of evaporation, sublimation and fusion

To calculate the heat of sublimation at 273 K, we can use Eq. (2.18):

$$ \begin{aligned} \Delta H_{{{\text{sub}},273}}^{\circ} & = \Delta H_{{{\text{evap}},273}}^{\circ} + \Delta H_{{{\text{fus}}, 2 7 3}}^{\circ} \\ \Delta H_{{{\text{sub}},273}}^{\circ} & = 45.054 + 6.010\quad {\text{kJ}}\,{\text{mole}}^{ - 1} \\ \Delta H_{{{\text{sub}},273}}^{\circ} & = 51.064\quad {\text{kJ}}\,{\text{mole}}^{ - 1} . \\ \end{aligned} $$

Assuming that this value is independent of temperature, we can calculate the sublimation pressure at a lower temperature:

$$ \ln \left( {\frac{{P_{2} }}{{P_{1} }}} \right) = - \frac{{\Delta H_{\text{sub}}^{\circ} }}{R}\left( {\frac{1}{{T_{2} }} - \frac{1}{{T_{1} }}} \right). $$

Solving for P ₂, and substituting the corresponding values:

$$ \begin{aligned} P_{2} & = P_{1} \exp \left[ { - \frac{{\Delta H_{\text{sub}}^{\circ} }}{R}\left( {T_{2}^{ - 1} - T_{1}^{ - 1} } \right)} \right] \\ P_{2} & = 0.006\exp \left[ { - \frac{51064}{8.314}(258^{ - 1} - 273^{ - 1} )} \right] \\ P_{2} & = 0.0017\,{\text{atm}} .\\ \end{aligned} $$

As temperature drops 15 °C, the pressure to transform solid water into vapor decreases as well. This is observed in the P-T plot below (Fig. 2.11).

Fig. 2.11

Solid–vapor equilibrium for water

Example 6 Estimation of vapor pressure by combining Trouton’s rule and Clausius–Clapeyron equation

Mercury evaporates at 357 °C (630 K) under 1 atm of pressure. Calculate its approximate vapor pressure at 100 °C, assuming that it follows Trouton’s rule.

Solution

First, we need to find ΔH°_vap from Trouton’s rule:

$$ \begin{aligned} \frac{{\Delta H_{\text{vap}}^{\circ} }}{{T_{\text{evaporation}} }} & = 88\quad {\text{J}}\,{\text{mole}}^{ - 1} {\text{K}}^{ - 1} ; \\ \Delta H_{{{\text{vap}},{\text{Hg}}}}^{\circ} & = 88(T_{{{\text{evaporation}},{\text{Hg}}}} ) \\ \Delta H_{{{\text{vap}},{\text{Hg}}}}^{\circ} & = 88(630) \\ \Delta H_{{{\text{vap}},{\text{Hg}}}}^{\circ} & = 55440\quad {\text{J}}\,{\text{mole}}^{ - 1} . \\ \end{aligned} $$

Substituting this ΔH°_{vap, Hg} value, into Clausius–Clapeyron equation, and solving for the gas pressure at the desired temperature:

$$ \begin{array}{*{20}l} {\ln \left( {\frac{{P_{2} }}{{P_{1} }}} \right) = - \frac{{\Delta H_{{{\text{vap}}}}^{^\circ } }}{R}\left( {\frac{1}{{T_{2} }} - \frac{1}{{T_{1} }}} \right){\text{ }}} \hfill \\ {P_{2} = P_{1} \times {\mkern 1mu} \exp \left[ { - \frac{{\Delta H_{{{\text{vap}}}}^{^\circ } }}{R}\left( {\frac{1}{{T_{2} }} - \frac{1}{{T_{1} }}} \right)} \right]{\text{ }}} \hfill \\ {P_{2} = 1 \times {\mkern 1mu} \exp \left[ { - \frac{{55440}}{{8.314}}\left( {\frac{1}{{373}} - \frac{1}{{630}}} \right)} \right]{\text{ }}} \hfill \\ {P_{2} = \exp [ - 7.293]{\text{ }}} \hfill \\ {P_{2} = 0.0007\quad {\text{atm}};\quad \quad P_{2} = 70.9\quad {\text{Pa}}.} \hfill \\ \end{array} $$

Example 7 Estimation of the heat of transformation from raw temperature and pressure data

Calculate the mean heat of vaporization of palladium from the following data:

T (°C)	P (Pa)
1314	1.00 × 10−4
1351	2.15 × 10−4
1568	7.50 × 10−3

Solution

To find the heat of vaporization, we need to work out the raw data to obtain the logarithm of the pressure and the inverse of the temperature in absolute scale, such manipulation results in:

P (Pa)	ln P	T (°C)	T (K)	103/T (1/K)
1.00 × 10−4	−9.210	1314	1587	0.630
2.15 × 10−4	−8.445	1351	1624	0.616
7.50 × 10−3	−4.893	1568	1841	0.543

Accordingly with Clausius–Calpeyron equation, plotting ln (P) versus 10³/T, would result in a straight line whose slope is equivalent to $- \tfrac{{\Delta {\text{H}}_{\text{vap}}^{\circ} }}{\text{R}}$. By taking 10³/T, the resulting heat of evaporation will be expressed in kJ mole⁻¹. This is shown in (Fig. 2.12).

Fig. 2.12

Vapor pressure of palladium as a function of the inverse of temperature

From the slope, the heat of evaporation of palladium becomes:

$$ \begin{aligned} \Delta H_{\text{vap}}^{\circ} & = - {\text{slope}} \times R \\ \Delta H_{\text{vap}}^{\circ} & = 49.428 \times 8.314 \\ \Delta H_{\text{vap}}^{\circ} & = 411\,{\text{kJ}}\,{\text{mole}}^{ - 1} . \\ \end{aligned} $$

Example 8 Use of Clapeyron and Clausius–Clapeyron equations to solve every day problems

Part (A)

How much a speed skater, who is 70 kg by weight, decreases the melting point of ice? The skating iron is 1 by 400 mm. Assume all his weight is on one skate at a time. Data: ΔS°_fus = 21.97 J mole⁻¹ K⁻¹; ΔV _fus = −1.49 cm3 mol⁻¹.

Solution

For this problem, we need to use Clapeyron equation

$$ \begin{aligned} & \frac{{{\text{d}}P}}{{{\text{d}}T}} \approx \frac{\Delta P}{\Delta T} = \frac{{\Delta S_{\text{melting}}^{\circ} }}{{\Delta V_{\text{melting}} }} \\ & \Delta T = \Delta P\frac{{\Delta V_{\text{melting}} }}{{\Delta S_{\text{melting}}^{\circ} }} \\ \end{aligned} $$

(E7.1)

It is necessary to calculate the pressure exerted by the skater on the ice surface:

$$ \Delta P = \frac{F}{A} = \frac{70 \times 9.8}{{10^{ - 3} \times 0.4}} = 1.715 \times 10^{6} \;{\text{Pa}}. $$

(E7.2)

Substituting this pressure value into equation (E7.1):

$$ \Delta T = 1.715 \times 10^{6} \times \frac{{ - 1.49 \times 10^{ - 6} }}{21.97};\quad \Delta T = - 0.12\,{\text{K}} . $$

Therefore, the melting point of ice drops by 0.12 K

Part (B)

The spouse of an engineer showed scientific interest in asking:

• (i) What is the temperature in our pressure cooker?

• (ii) Why does it cook faster than in an ordinary saucepan?

Given: Pressure regulator weighs 78 g and sits on a 2.618 mm outer diameter opening; ΔH°_vap for water at 1 atm and 100 °C is 40585 J mole⁻¹. What would your answers be for (i) and (ii) above?

Solution

For this problem, we have to calculate the pressure inside the cooker using Clausius–Clapeyron equation. To do so, we need to multiply the weight of the regulator by 9.8 to find the force that it exerts over the area defined by the gage opening:

$$ P_{\text{cooker}} = \frac{F}{A} = \frac{{78 \times 10^{ - 3} \times 9.8\quad [{\text{N}}]}}{{0.25\uppi \times (2.618 \times 10^{ - 3} )^{2} [{\text{m}}^{2} ]}} = 1.42 \times 10^{5} \,{\text{Pa}} . $$

With the cooker pressure, we can fix the following conditions: P ₁ = 1.01325 × 10⁵ Pa (1 atm), P ₂ = 1.42 × 10⁵ Pa, T ₁ = 373 K (100 °C), and T ₂ = cooker temperature at the cooker pressure P ₂. These values along with that of the heat of vaporization can be inserted now in Clausius–Clapeyron equation and solve for T ₂:

$$ \begin{aligned} \ln \left( {\frac{{P_{2} }}{{P_{1} }}} \right) & = - \frac{{\Delta H_{\text{vap}}^{\circ} }}{R}\left( {\frac{1}{{T_{2} }} - \frac{1}{{T_{1} }}} \right) \\ \ln \left( {\frac{1.42}{1.01325}} \right) & = - \frac{40585}{8.314}\left( {\frac{1}{{T_{2} }} - \frac{1}{373}} \right) \\ & \quad - \frac{0.3375}{4881.53} = \frac{1}{{T_{2} }} - \frac{1}{373} \\ T_{2} & = 383\,{\text{K}};\quad T_{2} = 110^{{^\circ }} {\text{C}} .\\ \end{aligned} $$

Thus answering to the engineer’s spouse we have that (i) the cooker reaches 110 °C and (ii) since the pressure increases so does the boiling temperature, then meals are cooked faster.