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Topic: Find the formation heat of AlCl3  (Read 3556 times)

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Offline long-live-mercutio

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Find the formation heat of AlCl3
« on: November 16, 2017, 10:39:02 AM »
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

Offline mjc123

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Re: Find the formation heat of AlCl3
« Reply #1 on: November 16, 2017, 01:01:41 PM »
You shouldn't have (500-298) etc. Integrate ΔCp for the solid up to the melting point, then add the heat of fusion, then ΔCp for the liquid from m.p. to 500K.

Offline long-live-mercutio

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Re: Find the formation heat of AlCl3
« Reply #2 on: November 16, 2017, 11:58:33 PM »
You shouldn't have (500-298) etc. Integrate ΔCp for the solid up to the melting point, then add the heat of fusion, then ΔCp for the liquid from m.p. to 500K.

For solid :
-697400 - 20,26(465,6-298) +0,5*0.1032(465,6^2-298^2) +378000(1/298-1/465,6)= - 693735,24 J

For liquid:
130,5*(500-465,6) - 20,26(500-465,6)+0,5*0,1032(500^2-465,6^2)+378000(1/465,6-1/500)=5509,12


All these results plus 35480 = -652746,12 J/mol, still doesn't match :( or you meant something different?  I'm not sure about the equation for the liquid AlCl3

Offline mjc123

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Re: Find the formation heat of AlCl3
« Reply #3 on: November 17, 2017, 04:57:48 AM »
There isn't an equation for liquid AlCl3. Its Cp is 130.5 J/mol/K - constant. At least that's what you are given.
However, for Al and Cl2, the equations are (presumably) valid up to 500K. So what you need is
Δ(ΔH) = int(298 to 465.6){Cp(AlCl3(s))dT} + ΔHfus(AlCl3) + Cp(AlCl3(l))(500-465.6) - int(298 to 500){[Cp(Al(s)) + 1.5Cp(Cl2)]dT}

Offline long-live-mercutio

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Re: Find the formation heat of AlCl3
« Reply #4 on: November 17, 2017, 06:22:19 AM »
There isn't an equation for liquid AlCl3. Its Cp is 130.5 J/mol/K - constant. At least that's what you are given.
However, for Al and Cl2, the equations are (presumably) valid up to 500K. So what you need is
Δ(ΔH) = int(298 to 465.6){Cp(AlCl3(s))dT} + ΔHfus(AlCl3) + Cp(AlCl3(l))(500-465.6) - int(298 to 500){[Cp(Al(s)) + 1.5Cp(Cl2)]dT}

int(298 to 465.6){Cp(AlCl3(s))dT} = -697400 + 55,45(465,6-298) + 0,5*117,2*10-3(465,62 - 2982) = -680608,7 J/mol

ΔHfus(AlCl3)  = 35480 J/mol

Cp(AlCl3(l))(500-465.6) = 130.5 (500-465.6) = 4489.2 J/mol



int(298 to 500){[Cp(Al(s)) + 1.5Cp(Cl2)]dT}:

 Δa = 20.67 + 1.5*36.69 = 75.705
 Δb = 12.39*10-3+1.5*1.05*10-3 = 0.013965
 Δc' = 1.5*(-2.52*105) = -378000

 ΔCp = 75.705 + 0.013965T - 378000T-2
int(298 to 500){[Cp(Al(s)) + 1.5Cp(Cl2)]dT} = 75.705(500-298) + 0.5*0.013965(5002-2982) - 378000 (1/298 - 1/500) = 15905.5 J/mol


-680608,7 J/mol + 35480 J/mol + 4489.2 J/mol - 15905.5 J/mol = -656545 J/mol, still not the answer the book wants, have I done something wrong?

Offline mjc123

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Re: Find the formation heat of AlCl3
« Reply #5 on: November 17, 2017, 07:31:30 AM »
I wonder if the book answer may be wrong. I did this yesterday, but slightly wrongly, and got the book answer. I integrated ΔCp up to 465.6K, then added ΔHfus(AlCl3), then Cp(AlCl3(l))(500-465.6), but forgot about Cp(Al(s)) + 1.5Cp(Cl2) between 465.6 and 500K. Maybe the book did too. As you have found, these calculations are complicated and it's easy to miss something out!

Offline long-live-mercutio

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Re: Find the formation heat of AlCl3
« Reply #6 on: November 17, 2017, 08:50:00 AM »
I wonder if the book answer may be wrong. I did this yesterday, but slightly wrongly, and got the book answer. I integrated ΔCp up to 465.6K, then added ΔHfus(AlCl3), then Cp(AlCl3(l))(500-465.6), but forgot about Cp(Al(s)) + 1.5Cp(Cl2) between 465.6 and 500K. Maybe the book did too. As you have found, these calculations are complicated and it's easy to miss something out!

Yes, they are complicated :D thank you very much!

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