Nitrogen dioxide, NO2(g), is an emission resulting from the burning of gasoline in the air in an automobile engine. nitrogen dioxide contributes to the formation of smog and acid rain. it can be converted to dinitrogen tetraoxide as shown below:

Question

2NO2(g) --> N2O4(g)

a) use Hess's law and the following equations to determine the enthalpy change for this reaction.
1) N2(g) +2O2(g) --> 2NO2(g) ΔH^o = 66.4 k/J
2) N2(g) + 2O2(g) --> N2O4(g) ΔH^o = 11.1 k/j

b) write the thermochemical equation for the overall reaction.

Answer 1

a. -1 x (1) 2NO2(g) → N2(g) + 2O2(g) ΔH = -66.4 kJ
1 x (2) N2(g) + 2O2(g) → N4O4(g) ΔH = 11.1 kJ
Total 2NO2(g) + N2(g) + 2O2(g) → N2(g) + 2O2(g) + N4O4(g) ΔH = -55.3 kJ


b. 2NO2(g) → N2O4(g) ΔH = -55.3 kJ
or
2NO2(g) → N2O4(g) + 55.3 kJ

Answer 2

I looked at your earlier post and I was confused. However, here is what you do.
(1).......... N2(g) +2O2(g) --> 2NO2(g) ΔH^o = 66.4 kJ
(2)...........N2(g) + 2O2(g) --> N2O4(g) ΔH^o = 11.1 kJ
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add...........2N2 + 4O2 ==> 2NO2 + N2O4 dH = 77.5 kJ
Now you have a 2NO2 on the right and you don't want that.
reverse(1)......2NO2 ==> N2 + 2O2 dH = -66.4 kJ
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..............N2 + 2O2 ==> N2O4 dH = 77.5 - 66.4 = ? kJ.
Check my thinking.

Answer 3

oops. I didn't get the equation you want. You want 2NO2 ==> N2O4 and I calculated N2 + O2 ==> N2O4 Sorry about that.
BY THE WAY, I LOOED UP AND FOUND +57.3 Kj (ENDOTHERMIC). i'LL PLAY AROUND WITH A LATER. IT'S BED TIME FOR ME.

Answer 4

I looked at some other sites and found the rxn actually is exothermic so a negative sign is correct for dH. Go with Tracy's work.