Examine the conditions under which a liquid Fe-Mn alloy can be in equilibrium with an FeO-MnO liquid solution in an atmosphere containing oxygen at 1800C. For and for Mn(1) + AG (i) = O(g) MnO(1) -344, 800 + 55.90T J Fe(1) + O(g) FeO(1) AG(ii) = 232700 + 45.13T J Part (a) What is the K for reaction: Part (b) Given XFe, what is the value of XFeO Part (c) Plot a phase diagram that shows the po2 (oxygen isobar), i.e. the tie-line connecting the compositions of the equilibrated metallic and oxide alloy FeO(1) + Mn(1) MnO(1) + Fe(1) Part (d) Increasing the partial pressure of oxygen to 1.79 10-11 atm moves the state of the system. In this state, a metallic alloy of XFe = 0.6 (at a) is in equilibrium with an oxide solution of XFeO = 0.0053 (at c). Using the lever rule, find the fraction of the fraction of the system occurring as the metallic alloy in state a, and the fraction occurring as the oxide solution in state c.