Question

The molar enthalpy of fusion of solid copper is 13.0 kJ mol^-1, and the molar entropy of fusion is 9.58 J K^-1 mol^-1.

(a) Calculate the Gibbs free energy change for the melting of 1.00 mol of copper at 1.39×10³ K.

kJ

(b) Calculate the Gibbs free energy change for the conversion of 4.27 mol of solid copper to liquid copper at 1.39×10³ K.

kJ

(c) Will copper melt spontaneously at 1.39×10³ K? _____Yes No


(d) At what temperature are solid and liquid copper in equilibrium at a pressure of 1 atm?

K

Answer 1

Q1.

a)

dG for 1 mol of copper melting... T = 1.39*10^3 K

note that this specific temperature is the normal melting point of copper, so no need to adjust temperature for dG

so

dG = dH - T*dS

substitute data; change entropy units from J to kJ via 10^-3

dG = (13) - (1.39*10^3)*(9.58*10^-3)

dG = -0.3162 kJ/mol

b)

for n = 4.27 mol of copper:

form previous data

dG = -0.3162 kJ/mol

dG total = (4.27 mol) (-0.3162) kJ/mol = -1.351 kJ/mol

c)

This is most likely the case, since the dG value is NEGATIVE, which implies there is enough free energy to give a forward process, i.e. solid copper to liquid copper (melting)

d)

find equilibrium:

dG = dH - T*dS

substitute for equilibirum, dG = 0

dH - T*dS = 0

13*10^3 J/mol - T * (9.58) J/molK = 0

solve for T

13*10^3 = T*9.58

T = (13*10^3)/(9.58)

T = 1356.99 K