N2(g) + o2(g) → 2 no(g) δh = +180.7 kj 2 no(g) + o2(g) → 2 no2(g) δh = −113.1 kj 2 n2o(g) → 2 n2(g) + o2(g) δh = −163.2 kj use hess's law to calculate δh for the reaction. n2(g) + 2 o2(g) → 2 no2(g).

N2(g) + 2 O2(g) → 2 NO2(g)

(1) N2(g) + O2(g) → 2 NO(g) Δh = +180.7 kJ
(2) 2 NO(g) + O2(g) → 2 NO2(g) Δh = −113.1 kJ
(3) 2 N2O(g) → 2 N2(g) + O2(g) Δh = −163.2 kJ

You want to rearrange the given reactions to match the reaction with the unknown enthalpy. Looking at what we're given, reaction (2) already has 2NO2 on the product side just like our unknown, so let's start there - it can stay as is.

However, it has 2NO on the reactant side - that's not in our unknown reaction, so we need to get rid of it somehow. There's 2NO in reaction (1), and it's on the product side, so it will cancel the 2NO on the reactants side of reaction (2). This means reaction (1) can stay as it is as well.

If you add these two reactions together, you'll get:
N2(g) + O2(g) + 2 NO(g) + O2(g)→ 2 NO(g) + 2 NO2(g)
N2(g) + 2O2(g) + 2 NO(g) → 2 NO(g) + 2 NO2(g)
N2(g) + 2O2(g) → 2 NO2(g)

The 2 NO cancels out and the individual O2's add together to get 2O2. This gives us the unknown reaction so we actually don't need to use reaction (3) at all to answer this question. In fact, it makes things more difficult because it contains N2O, which cannot be cancelled in any way with the reactions we're given.

To find the enthalpy of the unknown reaction, add the enthalpies of the reactions that work. Since we didn't have to modify either of the the first 2 reactions, you can add their original enthalpies as given:

+180.7 kJ + (−113.1 kJ) = +67.6 kJ

High-energy and unstable.