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Topic: Enthalpy of a Reaction  (Read 1867 times)

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Offline big

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Enthalpy of a Reaction
« on: December 01, 2011, 10:54:32 PM »
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!

Offline UG

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Re: Enthalpy of a Reaction
« Reply #1 on: December 02, 2011, 06:31:24 AM »
C looks right if you ask me  ???

Offline Vidya

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Re: Enthalpy of a Reaction
« Reply #2 on: December 03, 2011, 04:46:14 AM »
yes
everything is fine here

Offline Zeppos10

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Re: Enthalpy of a Reaction
« Reply #3 on: December 06, 2011, 06:49:45 AM »
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.


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