Phase Equilibria II

Abstract

In the last chapter, interactions between components forming mixtures were discussed. As more constituents were added to the system, the equations describing the thermodynamic quantities become more complex. However, it was shown in the examples section that graphical representation of such equations is very useful in determining those quantities.

Examples of Calculations

Example 1. Phase predomination diagram

Part (a) Using the data below, construct a phase predominance diagram for the Mg–Cl–O system at 300 °C

	Reactions @ T = 573 K	ΔG [J]
1	Mg + 0.5 O2(g) = MgO	−539667
2	Mg + Cl2(g) = MgCl2	−550524
3	Cl2(g) + MgO = MgCl2 + 0.5O2(g)	−10858
4	Cl2(g) + MgO2 = MgCl2 + O2(g)	−15156
5	MgO2 = MgO + 0.5 O2(g)	−4298
6	Mg(ClO4)2 = 3O2(g) + Cl2(g) + MgO2	−459339
7	Mg(ClO4)2 = MgCl2 + 4O2(g)	−474495

Solution

To obtain the phase predominance diagram for the set of given equations, we need to express the respective equilibrium constant in terms of the partial pressures of oxygen and chlorine gas. The other species should be considered as condensed and pure, thus their activity values 1.

After this first step, we need to express the equilibrium constants in terms of a logarithm function; these steps are shown in the table below: (Table 5.4).

Table 5.4 Equilibrium constants and its logarithmic form for chemical reactions in the Me–Cl₂–O₂ system

The next step in solving this problem is to plot the logarithmic expressions shown in the table. It can be seen that every one of these expressions represents straight lines

Once the different lines are plotted, we should mark the actual phase boundaries. To do so, we have to join the interceptions where three equilibrium lines meet. After this is done, we need to label the respective phase fields:

Part (b)

The Zn–S–O system at 500 °C can be represented by means of the attached predominance area diagrams

These diagrams show the same equilibria, but with different coordinate axis.

Use these diagrams to estimate the equilibrium constant for the following reaction:

½ S₂ + O₂ = SO_2.

What is the value of ΔG° for this reaction at 773 K?

Solution

In the P _SO2–P _O2 diagram you have the equilibrium

$$ \begin{array}{*{20}l} {{\text{Zn}} + {\text{SO}}_{ 2} + {\text{O}}_{ 2} = {\text{ZnSO}}_{ 4} } \hfill \\ {K_{ 1} = \frac{ 1}{{P_{{{\text{SO}}_{ 2} }} \times P_{{{\text{O}}_{ 2} }} }}} \hfill \\ \end{array} $$

(1)

For reaction (1), log P _SO2 = 0, log P _O2 = −16

In the P _S2–P _O2 diagram you have the equilibrium

$$ \begin{array}{*{20}l} {{\text{Zn}} + 0. 5 {\text{S}}_{ 2} + 2 {\text{O}}_{ 2} = {\text{ZnSO}}_{ 4} } \hfill \\ {K_{ 2} = \frac{ 1}{{\sqrt {P_{{{\text{S}}_{ 2} }} } \times P_{{{\text{O}}_{ 2} }}^{ 2} }}} \hfill \\ \end{array} $$

(2)

For reaction (2), log P _S2 = −7, logP _O2 = −16.

Subtracting reaction (1) from (2)

$$ \begin{array}{*{20}l} {{\text{Zn}} + 0. 5 {\text{S}}_{ 2} + 2 {\text{O}}_{ 2} = {\text{ZnSO}}_{ 4} } \hfill & {{\text{logK}}_{ 2} = - 0. 5\log P_{{{\text{S}}_{2} }} - 2\log P_{{{\text{O}}_{2} }} } \hfill \\ {{\text{ZnSO}}_{ 4} = {\text{Hg}} + {\text{SO}}_{ 2} + {\text{O}}_{ 2} } \hfill & {{ \log }K_{ 1} = \log P_{{{\text{SO}}_{ 2} }} + \log P_{{{\text{O}}_{ 2} }} } \hfill \\ \end{array} $$

$$ \begin{array}{*{20}l} { 0. 5 {\text{S}}_{ 2} + {\text{O}}_{ 2} = {\text{SO}}_{ 2} } \hfill & {\log K_{ 3} = \log P_{{{\text{SO}}_{ 2} }} - \log P_{{{\text{O}}_{ 2} }} - 0. 5\log P_{{{\text{S}}_{ 2} }} } \hfill \\ \end{array} $$

(3)

From reaction (3)

$$ \begin{aligned} { \log }K_{ 3} & = \log P_{{{\text{SO}}_{ 2} }} - \left( {\log P_{{{\text{O}}_{ 2} }} + 0. 5\log P_{{{\text{S}}_{ 2} }} } \right) \\ { \log }K_{ 3} & = 0- ( - 1 6- 0. 5\times 7) \\ {\log} {K}_{ 3} & = 19.5 \\ \end{aligned} $$

$$ K_{ 3} = 1 0^{ 1 9. 5} $$

$$ \begin{aligned} {\Delta} G^{ \circ } & = - {\text{RT}}\,\ln K_{3} \\ \Delta G^{\circ} & = - 8.314 \times 773 \times \ln 10^{19.5} \\ {\Delta} G^{ \circ } & = - 289\;{\text{kJ}}/{\text{mole}} \\ \end{aligned} $$

Example 2. Calculation of liquidus and solidus lines in phase diagrams

Part (a) When mixtures of silicon and germanium are heated, melting occurs over a range of temperatures:

N	XSi	Tstart melting [°C]	XSi	Tfinish melting [°C]
1	1	1414	1	1414
2	0.95	1408	0.97	1391
3	0.885	1395	0.9325	1366
4	0.85	1388	0.8925	1340
5	0.81	1377	0.875	1329
6	0.785	1369	0.7993	1281
7	0.765	1364	0.7625	1259
8	0.6975	1346	0.6875	1215
9	0.665	1335	0.595	1168
10	0.6275	1320	0.5375	1135
11	0.59	1307	0.43	1083
12	0.5525	1294	0.345	1045
13	0.495	1270	0.29	1027
14	0.4475	1250	0.2375	1005
15	0.385	1221	0.1575	977
16	0.35	1206	0.1075	962
17	0.32	1188	0.0575	950
18	0.2625	1157	0.015	940
19	0.2325	1142	0	938
20	0.18	1109
21	0.165	1098
22	0.1275	1074
23	0.105	1056
24	0.08	1038
25	0.06	1016
26	0.045	1001
27	0.03	983
28	0.0225	972
29	0.0125	955
30	0	938

• (i) With the data provided, construct the corresponding phase diagram

• (ii) Calculate the solidus and liquidus lines for this system (ΔH _fusion,Si = 50.2 kJ/mole, ΔH _fusion,Ge = 36.9 kJ/mole)

• (iii) Compare the calculated diagram with the experimental data, comment on your findings.

Solution

We need to plot the data provided to find the experimentally determined phase diagram. The data reveals that the Ge–Si system exhibits complete solubility in both the liquid and solid states, therefore ideal solution in each of these phases can be assumed. Further, to calculate the corresponding solidus and liquidus lines, we need to use Eqs. 5.24a and 5.24b, respectively. The resulting diagram is shown in Fig. 5.18.

Fig. 5.18

Comparison between experimental data and calculated one on the solidification of Si–Ge alloys

In the figure it can be noticed the experimental data (full symbols solidus line, empty symbols liquidus line), along with the calculated equilibrium lines. It is clear that the computed solidus line is in better agreement with the provided data than the estimated liquidus line. In both cases there is good agreement between experimental data and the calculated ones. The more pronounced deviation exhibited by the liquid phase can be attributed to the expected higher entropy associated to the liquid state. Such entropy term is not present in the corresponding liquidus line equation.

Part (b). The following data was found on the heating of different Ag–Cu mixtures upon melting:

N	XCu	T [°C]	N	XCu	T [°C]	N	XCu	T [°C]
1	0	962	28	0.425	793	55	0.775	940
2	0.0175	953	29	0.4375	800	56	0.7875	947
3	0.03	947	30	0.4475	803	57	0.795	950
4	0.0575	930	31	0.46	807	58	0.805	957
5	0.07	923	32	0.4675	810	59	0.815	963
6	0.1	907	33	0.475	813	60	0.8225	967
7	0.115	900	34	0.4975	820	61	0.835	977
8	0.1425	887	35	0.51	823	62	0.85	987
9	0.16	877	36	0.5175	827	63	0.865	993
10	0.1875	863	37	0.53	830	64	0.8825	1003
11	0.205	857	38	0.5425	833	65	0.8925	1010
12	0.2275	847	39	0.5525	837	66	0.9025	1017
13	0.235	843	40	0.585	850	67	0.915	1027
14	0.2575	837	41	0.595	853	68	0.9225	1033
15	0.275	830	42	0.6025	857	69	0.9375	1043
16	0.285	827	43	0.6175	860	70	0.9525	1057
17	0.295	823	44	0.63	867	71	0.9675	1063
18	0.3075	817	45	0.6475	873	72	0.97	1067
19	0.3175	813	46	0.66	880	73	0.9775	1072
20	0.33	810	47	0.6775	890	74	0.9825	1077
21	0.35	803	48	0.6925	897	75	0.9875	1080
22	0.36	800	49	0.705	903	76	0.9925	1083
23	0.3725	793	50	0.715	910	77	1	1084
24	0.3825	790	51	0.725	917
25	0.3925	787	52	0.74	923
26	0.4	780	53	0.75	927
27	0.41	790	54	0.7625	933

Independently, Hultgren et al. published the following activity data for liquid Ag–Cu alloys at 1400 K (1127 °C).

XAg	AAg	γAg	XCu	ACu	γCu
1	1.000	1.000	0	0.000	0.000
0.9	0.918	1.020	0.1	0.221	2.210
0.8	0.852	1.065	0.2	0.348	1.740
0.7	0.770	1.100	0.3	0.468	1.560
0.6	0.678	1.130	0.4	0.600	1.500
0.5	0.615	1.230	0.5	0.675	1.350
0.4	0.552	1.380	0.6	0.738	1.230
0.3	0.485	1.617	0.7	0.791	1.130
0.2	0.394	1.970	0.8	0.846	1.058
0.1	0.250	2.500	0.9	0.912	1.013
0	0.000	0.000	1	1.000	1.000

• (i) With the data provided, plot the liquidus lines corresponding to the silver and copper components in the system

• (ii) Calculate the solidus and liquidus lines for this system (ΔH _fusion,Ag = 11.3 kJ/mole, ΔH _fusion,Cu = 13.1 kJ/mole) assuming no terminal solution in the solid state and also by assuming a regular solution in the liquid (Ω= 15 kJ/mole)

• (iii) Compare the calculated liquidus lines with the data provided, comment on your findings.

Solution

For part (i), simply we need to plot the data provided (temperature versus x_Cu). To determine the liquidus line assuming no solid state solutions in part (ii), we need to use Eq. (5.25). Use the silver and copper activities provided by Hultgren et al. to carry out the computation of the liquidus line. As indicated by (5.25) the liquidus temperature is an implicit function, thus to find its value we need to use a numerical approach to promptly solve the problem. In this case the bisection method with a convergence criterion of 0.001 was used.

Comparison between the data provided and that estimated with Eq. (5.25) is shown in Fig. 5.19. Finally, for part (iii), Eqs. (5.26a, 5.26b) were used to determine the corresponding liquidus lines. Similar to Eq. (5.25), the set of Eqs. (5.26) are implicit functions of T, thus they were solved numerically as in part (ii). The resulting lines are also plotted in Fig. 5.19.

Fig. 5.19

Calculation of liquidus lines in the Ag–Cu system, using different models

As seen in the figure, it is clear that assuming regular solution in the liquid state offers a better fitting in relation to experimental data than the assumption of non-solid solutions. Regarding to the computation in part (ii), the data provided was taken at 1127 °C, which is far higher than the actual liquidus temperature for the different compositions therefore is expected that the activities of either copper or silver would be different when equilibrium with a minimum portion of solid within the system. Also from the data provided, it was found that the Ω parameter was of 13.5 kJ/mole. When using this value in expressions (5.26) there is considerable departure from the experimental data. The best fitting was found when the interaction parameter (Ω) has a value of 15 kJ/mole. (Figure 5.20) and (Fig. 5.21).

Fig. 5.20

Ellingham diagram for the CO, XO, YO, and ZO species

Fig. 5.21

Ellingham diagram for some oxides

Example 3

Part (a) With the attached Ellingham diagram

Answer

• (i) The temperature at which Ni, NiO, and oxygen at 10⁻¹² atm are in equilibrium

• (ii) The temperature at which Si, SiO₂, CO, and CO₂ (1 atm) are in equilibrium together

• (iii) The CO₂ pressure in equilibrium with that of CO in (ii)

• (iv) The mean values of ΔH° and ΔS° for the reactions

  • a. 2Mn + O₂ = 2MnO

  • b. 4/3Al + O₂ = 2/3Al₂O₃

• (v) Determine the ΔG° for the reaction: 3/2Si + Al₂O₃ = 2Al + 3/2SiO₂ at 1000 °C

• (vi) Will the reaction in (v) proceed to the right or to the left?

• (vii) At what temperature, if any, will the reactants and products in part (v) be in equilibrium

• (viii) By reference to the Ellingham diagram, comment on the feasibility of

  • a. Reducing MnO with CO

  • b. Reducing Al₂O₃ with CO

  • c. Reducing Al₂O₃ with Ca

Solution

Part (i)

We need to draw a line from point O in the scale at the left side of the diagram to P _O2 = 10⁻¹². At the point in which this line intercepts the Ni–NiO equilibrium, we need to draw a second line perpendicular to the temperature axis from that interception to the temperature axis and then directly read form the diagram the temperature. In this case the equilibrium temperature is approximately 1075 °C (1348 K).

Part (ii)

For this part, the temperature can be found at the intersection between lines 6 and 12. At that intersection, we need to draw a straight line from that point to the temperature axis and read directly. In this case, the temperature is roughly 1680 °C.

Part (iii)

From part (ii) the equilibrium we are looking for is Si + 2CO₂ = SiO₂ + 2CO (line 12–6 in the diagram). The equilibrium constant is K = (P _CO/P _CO2)². The equilibrium CO/CO₂ ratio in equilibrium with the reaction we are looking for can be read from the diagram; in this case that ratio is 10⁻² (draw a line from point C in the left side scale that passes through the Si–CO₂–SiO₂–CO intersection and extend it to the CO/CO₂ ratio scale and read directly). Since the total pressure is 1 atm, P _CO + P _CO2 = 1. Thus solving the following system:

$$ \begin{array}{*{20}l} {\left( {\frac{{P_{\text{CO}} }}{{P_{{{\text{CO}}_{2} }} }}} \right)^{ 2} = 1 0^{ - 2} } \hfill \\ {P_{\text{CO}} + P_{{{\text{CO}}_{ 2} }} = 1} \hfill \\ \end{array} $$

This results in P _CO2 = 0.91 atm.

Part (iv)

Extending the Mn/MnO and Al/Al₂O₃ (lines 10 and 14 respectively) to the left side scale of the diagram directly yields the mean enthalpy change (ΔH°). On the other hand direct calculation of the slope of these lines gives the mean entropy term (ΔS°). This results in ΔH°_Mn/MnO = −775 kJ/mole; ΔH°_Al/Al2O3 = −1140 kJ/mole; ΔS°_Mn/MnO = 0.16 kJ/mole; ΔS°_Al/Al2O3 = 0.22 kJ/mole.

Parts (v, vi & vii)

The equilibrium we are looking for is: 3/2Si + Al₂O₃ = 2Al + 3/2SiO₂. To get this reaction we need to combine reactions (12) and (14). The Gibbs free energy changes for these reactions at 1000 °C are −680 kJ/mole and −850 kJ/mole, respectively. To obtain the desired equilibrium we must multiply both reactions by 3/2 and subtract reaction (14) from reaction (12). Consequently the Gibbs free energies values have to be multiplied by the 3/2 factor. This results in 3/2 × −680 + 3/2 × 850 = 255 kJ/mole.

The positive Gibbs free energy change indicates that the reaction would take place to the left, this means that silicon cannot reduce aluminum oxide; however, aluminum can reduce silicon oxide.

By examining lines 12 and 14 in the diagram, they are almost parallel over the entire temperature interval, and also line 14 lies beneath line 12; this means that at any temperature, aluminum is able to reduce silicon oxide; but the silicon never can reduce aluminum oxide. This means that the products and reactants in the proposed reaction never reach the equilibrium condition.

Part (viii)

Manganese oxide can be reduced with CO at least at 1400 °C. Below that temperature that is not possible, since ΔG° for Mn/MnO is below that of CO/CO₂. Aluminum oxide can only be reduced by CO gas above 2000 °C. Finally, aluminum oxide can be reduced by calcium at any temperature indicated in the diagram since the Gibbs free energy line of formation of CaO is always below that of the aluminum oxide.

Part (b) The following Gibbs free energies of formation are given in kJ/mole of oxygen.

T [°C]	CO	Xo	YO	ZO
0	−251	0	−628	−1046
500	−342	50.2	−523	−941
1000	−439	105	−418	−837
2000	−628		−209	−628

What information can be obtained from this data set regarding to the extraction of metals X, Y, and Z?

Solution

By plotting the data provided (DG° versus T) for each of the components shown in the table, it results in the following diagram:

From this graph, it is clear that over the entire temperature interval, XO cannot exist since its energy of formation is always positive, furthermore it can be reduced by any of the other oxides. YO can only be reduced by CO above 900 °C, below this temperature that is not possible since the Gibbs free energy line of YO lies below that of CO under 900 °C. The same can be concluded for ZO, but for this species, the equilibrium temperature is reached at 2000 °C.