DIY 3D Solar Panels

If the alloying component P is chemically active and forms solutions with M in both the reactant and product phases, the equilibria involved are:

To distinguish the two solutions, the metal alloy is termed the melt and the mixed oxide is called the slag. At the temperature of the reaction, the standard free energy changes of the two reactions are AG^ and AG£. Assuming for simplicity that both solutions are ideal, the activities can be replaced by concentrations and the mass action laws for reactions (9.47a) and (9.47b) are:

The solution method starts by dividing the above equations to eliminate the oxygen pressure, yielding:

Allowing for chemical reactivity of both species has not altered the number of components (C = 3) or the number of phases (P = 2), so there are still two degrees of freedom. The first is the temperature and the second is the composition of either one of the two liquid phases. If one is known, the other follows from Eq (9.49), and so is not an independently variable quantity.

To complete the mathematical solution for the compositions of the two phases, the relative amounts of M, P, and O needs to be specified. These mole relationships are most readily followed by combining Eqs (9.46a) and (9.46b) into the single reaction:

Suppose the fixed M/P mole ratio is unity and O/P = 2. This is equivalent to charging one mole each of M and PO2 to the system and not permitting exchange of oxygen with the surroundings. The reaction progress variable 4 is defined as the number of moles of MO2 (slag) and P(melt) at equilibrium. The moles of M remaining in the melt and the moles of PO2 in the slag are both equal to 1 - 4. By this choice of the initial charge, the total number of moles in each phase remains constant at unity as the reaction moves towards equilibrium. Thus, the number of moles in each phase is equal to the mole fraction, and Eq (9.49) becomes:

from which 4 can be determined for specified values of KM and KP. With 4 known, the mole fractions in each phase are fixed. The intrinsic oxygen pressure of the system follows from either of Eqs (9.48).

Contrary to the single metal reactant case, the two-metal reaction system does not exhibit stability diagrams such as those shown in Fig. 9.2, or the modified version for nonreactive component P. An imposed oxygen pressure different from the intrinsic oxygen pressure simply dictates the relative amounts of slag and melt according to Eqs (9.48).

Example Recall that the standard free energy changes for the reactions described by Eqs (9.47a) and (9.47b) refer to complete conversion of the pure metals by O2 at 1 atm pressure to pure oxide products. These free energy changes uniquely determine the oxygen pressure if both product and reactants of each metal are pure and unmixed. The temperature is 1000 K, at which AGM = -200 kJ/mole and AGp = -250 kJ/mole. The initial charge is 1 mole each of M and PO2 (or any combination of the metals and their oxides with element mole ratios M/P = 1 and O/P = 2). From Eq (9.48), the equilibrium constants are Km = 2.8x1010 and Kp = 1.15x1013. Using these values in Eq (9.51) and solving for the reaction progress variable at the equilibrium state gives 4 = 0.047. The mole fractions in the melt and slag at equilibrium are:

Using these compositions in either of Eqs (9.48) gives p0 — 1.7 x 10 atm. This value is about a factoi of 20 lower than the oxygen pressure in equilibrium with the pure M/pure Mo2 couple. The reason is the presence of element P, which has a lower Free energy of formation of its oxide than does element M. Element P binds oxygen more strongly than element M, thereby lowering the equilibrium, or intrinsic oxygen pressure of the system. The stronger oxygen-gettering ability of element P forces 95% of M to remain as a metal while 95% of P forms the oxide and moves into the slag phase.

There is no unique oxygen pressure at which the system is all metal or all oxide, as there is in the M/M02 couple. If the imposed oxygen pressure is greater than 1.7x10-12 atm, more of the metal oxidizes and the slag becomes somewhat richer in Mo2 than in the case above where the restriction was an oxygen-to-total metal mole ratio of unity.

Problems 9.7 - 9.9 explore several variations of the above melt-slag example. Problem 9.4 treats the industrial process for production of uranium metal by a reaction similar to Eqs (9.40) except that fluorides rather than oxides are involved.

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