Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: long-live-mercutio on November 15, 2017, 04:37:56 AM
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Hello everyone!
This is the problem:
Find the heat of AgCl formation at 150°С, if Δ f H (AgCl(s)) = -126,8 kJ/mol, taking into acount the following molar heat capacities:
Cp (Ag)= 23,97 + 5,28*10–3Т – 0,25*105Т –2 J/(mol*К);
Cp (Cl2) = 36,69 + 1,05*10–3Т – 2,52*105Т –2 J/(mol*К);
Cp (AgCl (s)) = 62,26 + 4,18*10–3Т – 11,30*105Т –2 J/(mol*К).
The answer in the textbook: –125,35 kJ/mol.
This how I tried to solve it:
Ag + (1/2)Cl2 = AgCl
Δ a = 62.26-23.97-0.5*36.69=36.445=36.44
Δb=4.18*10-3 - 5.28*10-3 - 0.5*1.05*10-3= -1.625*10-3 = -1.63*10-3
Δ c' = -11,30*105+0,25*105+0,5*2,52*105 = -979000
Δ Cp = 36.44-1.63*10-3T-979000T-2
ΔH = -126800 + 36.44(423-298)-0.5*1.63*10-3*(4232-2982)-979000(1/298-1/423) = -123.3 kJ/mol.
Could you please help me to find the mistake?:( maybe something's wrong with my rounding up?
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If ΔCp = A + BT + CT-2, you cannot say that Δ(ΔH) is AΔT + BΔ(T2) + CΔ(T-1). You have to integrate ΔCpdT over the temperature range. What does that give you?
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If ΔCp = A + BT + CT-2, you cannot say that Δ(ΔH) is AΔT + BΔ(T2) + CΔ(T-1). You have to integrate ΔCpdT over the temperature range. What does that give you?
But I have integrated ΔCpdT over the temperature range and got this: ΔHp =A(T2-T1) + (1/2)B(T22-T12) + C(T1-1-T2-1)= -126800 + 36.44(423-298)-0.5*1.63*10-3*(4232-2982)-979000(1/298-1/423) = -123.3 kJ/mol.
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(I must admit I missed the 0.5 in front of BT2, but) What's the integral of T-2dT?
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I would be very happy if I could find enthalpies of formation accurate to 1kJ/mol.
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(I must admit I missed the 0.5 in front of BT2, but) What's the integral of T-2dT?
It's (-2+1)(1/T2- 1/T1), it's just that I kept the minus inside and changed the position of 1/T2 and 1/T1 so that I got 1*(1/T1-1/T2). It should yield the same result as -1(1/T2-1/T2), shouldn't it?
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I would be very happy if I could find enthalpies of formation accurate to 1kJ/mol.
So is it possible that there's a mistake in the answer given by the author of the book?
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Sorry, my mistake, I didn't read your expression carefully enough. The integration is right. But there is a mistake in your calculation of Δa:
Δ a = 62.26-23.97-0.5*36.69=36.445=36.44
Is this right?
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Sorry, my mistake, I didn't read your expression carefully enough. The integration is right. But there is a mistake in your calculation of Δa:
Δ a = 62.26-23.97-0.5*36.69=36.445=36.44
Is this right?
Thank you! :)