Question

You are provided with the following thermodynamic data for the solidification of silver:

Ag(l) = Ag(s)

Melting Temperature: TM = 1235 K

Heat of Solidification:

   ΔHrxn= -11.28 kJ/mol

a. At T=TM, what is the driving force (ΔG) for solidification?

ΔGrxn=

b. What is the entropy of solidification for this reaction?

ΔSrxn=

c. If liquid silver is cooled 100C below its melting temperature, what is the driving force (ΔG) for solidification?

ΔGrxn=

Answer 1

a) At T = Tm , that is at the meling point the value G = 0 . This is true for all phase changes at their phase change temperature.

b) S is given by,

H = -11.28 KJ/mol = -11280 J/mol

Tm = 1235 K

S = H/ Tm = -11280/ 1235

S = - 9.134 J/molK

c) For temperatures below melting , the heat solidification remains almost constant, so we will be taking the same values of H and S as at Tm.

T = 1235 - 100 = 1135 K

G = H - TS

= -11280 - (-1135)( - 9.134)

G = -912.91 J/mol