The following reaction: 2HBr(g)

H

_{2}(g) + Br

_{2}(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/T

_{2}-1/T

_{1})

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!