Question

Silver and gold atoms mix together randomly on the same
lattice to form a homogeneous alloy. If you have 30 g of gold mixed with 50 g of silver, what are the mole fractions of Au and Ag? What is the molar entropy of mixing? What is the total entropy of mixing? Assuming an ideal solution, what is ΔGmix at T = 750 °C?

Answer 1

Given: The silver (Ag) and gold (Au) mix together randomly on the same lattice to form a homogeneous alloy. Also we have been asked to assume that the resulting alloy is an ideal solution

Molar mass of Au=196.97g/mol

Molar mass of Au=107.87g/mol

Given 30g of Au and 50g of Ag form the solution

Therefore the number of moles of Au, n_Au= 30/196.97 = 0.1523 mol

Number of moles of Ag, n_Ag= 50/107.87 = 0.4635 mol

Therefore mole fraction of Au, X_Au = n_Au/(n_Au+n_Ag)

=0.1523/(0.1523+0.4635) = 0.2473

X_Au=0.2473

Mole fraction of Ag, X_Ag = 1-X_Au

=1-0.2473 = 0.7527

X_Ag = 0.7527

Therefore the mole fractions of Au and Ag are 0.2473 and 0.7527 respectively

Assume 1mol of alloy

Molar entropy of mixing, $\Delta$s_mixing = -R$\sum$X_i ln(X_i)

Therefore $\Delta$s_mixing = -R ( X_Au ln(X_Au) + X_Ag ln(X_Ag) )

=-8.314 $\times$ ( 0.2473 $\times$ ln(0.2473) + 0.7527 $\times$ ln(0.7527))

=4.650 J/mol.K

Therefore molar entropy of mixing = 4.650 J/mol.K

Total number of moles in the alloy, n_T = n_Au + n_Ag

= 0.1523 + 0.4635

=0.6158 moles

Therefore total entropy of mixing, $\Delta$S_mixing = n_T $\times$ molar entropy of mixing $\Delta$s_mixing

= 0.6158 $\times$ 4.650

=2.8635 J/K

Therefore total entropy of mixing = 2.8635 J/K

Gibbs free energy of mixing, $\Delta$G_mixing = $\Delta$H_mixing - T$\Delta$S_mixing

But for ideal solutions $\Delta$H_mixing = 0

Therefore $\Delta$G_mixing = - T$\Delta$S_mixing where T=750^oC = 1023K

$\Delta$G_mixing = - 1023 $\times$ 2.8635

= - 2929.3605 J

Therefore $\Delta$G_mixing = -2.929 kJ. The negative sign is characteristic of an ideal solution which indicates that the mixing of ideal solutions is always spontaneous.