[dtomasiewicz]

Alright so I have a chemistry exam coming up in a couple days and I'm reviewing old exams as practice. I've come across a question that I believe *should* be easy but I just can't find the right answer to.

Given that the standard molar enthalpy of formation for nitrogen dioxide gas, NO₂(g), and for dinitrogen tetraoxide gas, N₂O₄(g), is 33.85 kJ*mol^-1 and 9.66 kJ*mol^-1, respectively, the nitrogen-to-nitrogen, N-N, bond dissociation enthalpy in dinitrogen tetraoxide, O₂N-NO₂ (the dimer of NO₂), is

a) 9.7 kj*mol^-1
b) 24.2 kj*mol^-1
c) 33.8 kj*mol^-1
d) 58.0 kj*mol^-1
e) 77.4 kj*mol^-1

Thanks in advance! Glad I came across this forum, lots of useful info here.

[Astrokel]

Hey dtomasiewicz, you have to show some attempts first.

[dtomasiewicz]

Hey dtomasiewicz, you have to show some attempts first.

Sorry, my bad.

The only result I can understand is

b) 24.2 kJ*mol^-1

As that would be 33.85 - 9.66. The answer is actually d, which would be 2(33.85) - 9.66, but I don't understand where the 2 comes from. Is it because we are looking at the dimer?

[Astrokel]

              33.85
1/2N₂ + O₂  NO₂

              9.66
N₂ + 2O₂   O₂N-NO₂  ---- (2)

                                                      9.66
(2) can also be expressed as: N₂ + 2O₂      2NO₂ + BE of N-N

If you apply hess' law it would work to be 2(33.85) - 9.66. There is a 2 because of 2 NO₂ molecules as shown above.

When dealing with enthalpy change problem, it is good to write out chemical equations as you are able to see it clearly.