You have to look at the 2nd equation in the reverse manner.

2NO₂ (g) → 2NO (g) + O₂ (g) , ∆H = +113.1 kJ/mol

Then you'll need to halve the equation.
NO₂ (g) → NO (g) + ½O₂ (g) , ∆H = +56.55 kJ/mol ②

For the 3rd equation, we should halve it as well.
2N₂O (g) → 2N₂ (g) + O₂ (g) , ∆H = -163.2 kJ/mol
N₂O (g) → N₂ (g) + ½O₂ (g) , ∆H = -81.6 kJ/mol ③


For the 1st equation, use it as shown.
N₂ (g) + O₂ (g) → 2NO (g) , ∆H = +180.7 kJ/mol ①


Putting them together,

③N₂O (g) → N₂ (g) + ½O₂ (g), ∆H = -81.6 kJ/mol
②NO₂ (g) → NO (g) + ½O₂ (g), ∆H = +56.55 kJ/mol
①N₂ (g) + O₂ (g) → 2NO (g), ∆H = +180.7 kJ/mol


After cancelling common terms, this gives you an overall equation of :

N₂O (g) + NO₂ (g) →3NO (g)


∆H(reaction)
= ∆H of ③ + ∆H of ② + ∆H of ①
= -81.6 kJ/mol + 56.55 kJ/mol +180.7 kJ/mol
= +155.65 kJ/mol
= +156 kJ/mol (3s.f)

Alternatively, double the first reaction.

2N₂O (g) → 2N₂ (g) + O₂ (g) , ∆H = -163.2 kJ/mol ③

2NO₂ (g) → 2NO (g) + O₂ (g) , ∆H = +113.1 kJ/mol ②

2N₂ (g) + 2O₂ (g) → 4NO (g) , ∆H = +361.4kJ/mol ①



After cancelling common terms, we get an overall equation of :

2N₂O (g) + 2NO₂ (g) →6NO (g)


∆H(reaction)

= ∆H of ③ + ∆H of ② + ∆H of ①
= -163.2 kJ/mol + 113.1 kJ/mol + 361.4kJ/mol
= +311.3 kJ/mol


Since this equation is double of the required one, halve the ∆H(reaction)

+311.3 kJ/mol ÷ 2 = +156.65 kJ/mol ≈ +156 kJ/mol (3s.f)