he excess partial molar free energy of iron at 1600°C (1873 K) in Fe-Ni solutions atdifferent compositions isgiven below. xNi lÇ9 : 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 cal/mole : 0 -15 -40 -90 -260 -620 -1330 -2030 -2730 deg/mole: 4.56 2.87 2.20 1.84 1.42 1.04 0.72 0.44 0.21 (^Fe> J//mot : ° ~6Z ~16? ~Z7? ~1088 ~2594 ~5566 ~8494 ~11422) Calculate the excess partial molar free energy of nickel in an equi atomic solution.
Question
he excess partial molar free energy of iron at 1600°C (1873 K) in Fe-Ni solutions atdifferent compositions isgiven below. xNi lÇ9 : 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 cal/mole : 0 -15 -40 -90 -260 -620 -1330 -2030 -2730 deg/mole: 4.56 2.87 2.20 1.84 1.42 1.04 0.72 0.44 0.21 (^Fe> J//mot : ° ~6Z ~16? ~Z7? ~1088 ~2594 ~5566 ~8494 ~11422) Calculate the excess partial molar free energy of nickel in an equi atomic solution.
Solution
To calculate the excess partial molar free energy of nickel in an equiatomic solution, we need to use the Gibbs-Duhem equation. This equation states that in a binary solution, the sum of the changes in the partial molar free energies of the components is equal to zero. Mathematically, it is expressed as:
x_Fe * dGE_Fe + x_Ni * dGE_Ni = 0
Given that the solution is equiatomic, the mole fractions of iron (Fe) and nickel (Ni) are both 0.5 (x_Fe = x_Ni = 0.5).
We are given the excess partial molar free energy of iron (GE_Fe) at different compositions. At an equiatomic solution (x_Ni = 0.5), GE_Fe is -260 cal/mole.
Substituting these values into the Gibbs-Duhem equation gives:
0.5 * (-260) + 0.5 * dGE_Ni = 0
Solving for dGE_Ni, we find that the excess partial molar free energy of nickel in an equiatomic solution is 260 cal/mole.
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