Question

Substance	S(subscript)298, J/K	cp, J/(K mol)
BaO	72.1	55.0
TiO2	50.6	75.0
BaTiO3	107.9	120.0

Calculate the entropy change for the reaction as a function of temperature for the reaction, BaO + TiO2 = BaTiO3

Answer 1

NOTE: assume specific heats, CP are constant in T range

Srxn = Sproducts - Sreactants

dS = n*Cp*ln(T2/T1) for sensible heat

T1 = 298K (since we have heat of formation at T= 298K)

for heat of formation:

Srxn° = Sproducts - Sreactants

Srxn° =  SBaTiO3 - (STiO2 + SBaO)

Srxn° = 107.9 - (50.6+72.1) = -14.8 J/mol

for the temperature:

Srxn =  Sproducts - Sreactants = Cp-BaTiO3*ln(T2/T1) - Cp-TiO2*ln(T2/T1) - Cp-BaO*ln(T2/T1)

Srxn = 120*ln(T2/298) - 50.6*ln(T2/T1) - 72.1*ln(T2/T1)

then

Stotal = Sformation + Srxn =  -14.8 +   120*ln(T2/298) - 50.6*ln(T2/T1) - 72.1*ln(T2/T1)