The iron in an Fe-Ni alloy containing 85.4 atom% Fe oxidises more readily than nickel when equilibrated with a gas mixture consisting of 57.5% W and 42.5% H20 by volume at 1 atm (101325 N/m2) pressure and 840°C (1113 K): ^alloy + <H2°) = <Fe0> + (H2> AG° = -3150 + 1.85 Teal (-13180 + 7.74 T J). Calculate the excess partial molar free energy of iron in the alloy, assuming that the FeO formed is insoluble in iron.
Question
The iron in an Fe-Ni alloy containing 85.4 atom% Fe oxidises more readily than nickel when equilibrated with a gas mixture consisting of 57.5% W and 42.5% H20 by volume at 1 atm (101325 N/m2) pressure and 840°C (1113 K): ^alloy + <H2°) = <Fe0> + (H2> AG° = -3150 + 1.85 Teal (-13180 + 7.74 T J). Calculate the excess partial molar free energy of iron in the alloy, assuming that the FeO formed is insoluble in iron.
Solution
To calculate the excess partial molar free energy of iron in the alloy, we need to use the Gibbs free energy equation:
ΔG = ΔG° + RT ln(Q)
where: ΔG is the Gibbs free energy, ΔG° is the standard Gibbs free energy, R is the gas constant, T is the temperature in Kelvin, Q is the reaction quotient.
Given: ΔG° = -3150 + 1.85T cal = -13180 + 7.74T J T = 1113 K R = 8.314 J/(mol·K)
First, convert ΔG° to J/mol: ΔG° = -13180 + 7.74 * 1113 = -13180 + 8610.62 = -4569.38 J/mol
Next, calculate Q. Since the reaction is at equilibrium, Q = K, the equilibrium constant. The equilibrium constant K can be calculated from the partial pressures of the reactants and products. However, the problem does not provide enough information to calculate K or Q.
Without the values of the partial pressures or concentrations of the reactants and products, we cannot calculate the excess partial molar free energy of iron in the alloy.
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