how do I use these equations to find the enthalpy change for this reaction

Question

4 NO(g)→2 NO2(g)+N2(g) △H=?kJ

1/2N2(g)+1/2O2(g)+NO(g) △H=+90.4 kJ

2 NO(g)+O2(g)→2 NO2(g) △H=-113.1kJ

Answer 1

This is a problem using Hess's Law. Read up on it for more details.

4NO(g) ==> 2NO2(g) + N2(g) ... TARGET EQUATION

Given:

(1). 1/2 N₂(g) + 1/2 O₂(g) ==> NO(g) ... ∆H = 90.4 kJ

(2). 2NO(g) + O₂(g) ==> 2NO₂(g) ... ∆G = -113.1 kJ

Procedure:

reverse (1) & x 2: 2NO(g) ==> N₂(g) + O₂(g) ... ∆H = -180.4 kJ

copy equation (2): 2NO(g) + O₂(g) ==> 2NO₂(g) ... ∆H = -113.1 kJ

Add together to get ...

2NO(g) + 2NO(g) + O₂(g) ==> N₂(g) + O₂(g) + 2NO₂(g)

Combine/cancel like terms to get the final equation ...

4NO(g) ==> N₂(g) + 2NO₂(g) ... TARGET EQUATION

∆H = -180.4 kJ + -113.1 kJ

∆H = -293.5 kJ