Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: big on December 01, 2011, 10:54:32 PM
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The following reaction: 2HBr(g) ::equil:: H2(g) + Br2(g)
The values of the equilibrium constant temperatures are:
Kc1=1.3*10-12 @ 500K
Kc2=9.0*10-18 @ 300K
What is the :delta: H for this reaction in kJ/mol? (R=8.31*10-3 kJ/mol)
A) -67.8 kJ/mol B) 16.5 kJ/mol C) 74.0 kJ/mol D) 97.2 kJ/mol
I thought this was a pretty straightforward application of an equation, but I got it wrong, so if anyone could tell me if I did anything wrong, I would really, really appreciate it!
ln(Kc1/Kc2)= :delta: H/R (1/T2-1/T1)
ln[(1.3*10-12)/(9.0*10-18)]*(8.31*10-3)/(1/300-1/500)= :delta: H=74 kJ, which is C. But the answer key says the answer is D?
Again, thanks soo much!
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C looks right if you ask me ???
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yes
everything is fine here
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The problem with this problem seems to be that bromine has a boiling point of 331K, and :delta:H(vap)=31 kJ/mol.
According to my data :delta:H for the given reaction is 103 kJ/mol, which is close to the value of 97.2 kJ =answer D.