Vanadium melts at1720°C (1993 K). The Raoultian activity coefficient of vanadium at infinite dilution in liquid iron at1620 C (1893 K) is 0.068. Calculate the free energy change accompanying the transfer of the standard state from pure solid vanadium to the infinitely dilute, weight percent solu­ tion of vanadium in pure iron at 1620°C. Given: (a) Heat of fusion of vanadium = 4,500 cal/mole (18,828 J/mol). (b) Atomic weights of vanadium and iron are 50.95 and 55.85 respectively.

To calculate the free energy change, we need to consider two steps: the melting of vanadium and the dissolution of vanadium in iron.

1. Calculate the free energy change for the melting of vanadium:

The free energy change for the melting of vanadium can be calculated using the equation ΔG = ΔH - TΔS, where ΔH is the heat of fusion, T is the temperature in Kelvin, and ΔS is the entropy change.

The entropy change for the melting of a substance can be calculated using the equation ΔS = ΔH/T.

Substituting the given values:

ΔS = 18828 J/mol / 1993 K = 9.45 J/mol-K

Then, calculate ΔG:

ΔG = ΔH - TΔS = 18828 J/mol - 1993 K * 9.45 J/mol-K = 18828 J/mol - 18828 J/mol = 0 J/mol

2. Calculate the free energy change for the dissolution of vanadium in iron:

The free energy change for the dissolution of a solute in a solvent can be calculated using the equation ΔG = RT ln γ, where R is the gas constant, T is the temperature in Kelvin, and γ is the activity coefficient.

Substituting the given values:

ΔG = 8.314 J/mol-K * 1893 K * ln 0.068 = -38500 J/mol

3. Add the free energy changes for the two steps to get the total free energy change:

ΔG_total = ΔG_melting + ΔG_dissolution = 0 J/mol - 38500 J/mol = -38500 J/mol

So, the free energy change accompanying the transfer of the standard state from pure solid vanadium to the infinitely dilute, weight percent solution of vanadium in pure iron at 1620°C is -38500 J/mol.