Hello everyone again,
This is a problem similar to the one I posted before, my answer doesn't match the answer in the book again. Could you please help me find the mistake?
Find the formation heat of AlCl3 at 500 К, if the standard heat of its formation is -697,4 kJ/mol, and the molar heat capacities are as following:
Cp (АlСl3 (s)) = 55,44 + 117,2*10–3Т J/(mol*К);
Cp (Аl (s)) = 20,67 + 12,39*10–3Т J/(mol*К);
Cp (Cl2 (g)) = 36,69 + 1,05*10-3Т – 2,52*105Т–2 J/(mol*К),
The point of АlСl3 fusion is 465,6 К и and the heat of fusion is 35,48 kJ/mol, the molar heat capacity of liquid АlСl3 130,5 J/(mol*К).
The answer from the book: –653,8 kJ/mol.
My attempt at a solution:
Al + 1.5Cl2 = AlCl3
Δa = 55.44-20.67-1.5*36.69= - 20.265= - 20.26
Δb= 117.5*10-3 - 12.39* 10-3 - 1.5*1.05*10-3 = 0.103235= 0.1032
Δc' = 1.5*2.52*105 = 378000
ΔCp = -20,26 + 0,1032T +37800T-2
ΔH = -697,4*103 - 20,26(500-298) + (1/2)*0,1032 (5002-2982) + 378000 (1/298-1/500) = - 692662,35 J = -693000 J.
130,5*(500-465,6) = 4489,2 J (the energy needed to heat up the liquid AlCl3)
-693000 + 4489,2 + 35,48*103 = - 653030 kJ/mol = 653 kJ/mol