[big]

The following reaction: 2HBr(g)  H₂(g) + Br₂(g)
The values of the equilibrium constant temperatures are:

K_c1=1.3*10^-12 @ 500K
K_c2=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(K_c1/K_c2)= :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!

[UG]

C looks right if you ask me 

[Vidya]

yes
everything is fine here

[Zeppos10]

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.