A Fe-P liquid alloy is in equilibrium with a solid solution and a gas phase at 1560 C. The solid solution behaves ideally and contains two components, CaO and Ca3P2O8 (3CaO.P2O5). The P content in the solid solution is 12.1 wt.%. The gas phase includes 1.610^-10 atm oxygen. The activity of P in the alloy, with respect to the 1 wt.% in Fe standard state ((1 % )) is 26.01.

The given data follows:

3 CaO (s) + P2O5 (g) = Ca3P2O8 (s) = 1083210 + 237.5 (J/mol)

P2 (g) = P(1 wt% in Fe) = 122200 19.2 (J/mol)

a) Calculate for the reaction P2 (g) + 5/2 O2 (g) = P2O5 (g) at 1560 C.

b) Calculate the activity of P in the alloy with respect to the Henrian std. state (hP) at 1560 C.

c) Calculate the activity of P in the alloy with respect to the Raoultian std. state (aP) at 1560 C (if = . 116).

d) Calculate the activity of Fe in the alloy (aFe) at 1560 C (Fe obeys Raoults Law).

(The atomic weight: O = 16.0, P = 31.0, Ca = 40.0, Fe = 55.9 g) (R = 8.314 J/mol.K)

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To calculate the mole fractions of Fe CaO and Ca3P2O8 in the alloy at 1560C we first need to determi... View the full answer