Im stuck on a chemistry problem

Question

Dinitrogen trioxide is the product of the reaction shown below. When dissolved in water, it forms nitrous acid, which has many applications in industrial chemistry.    

NO (g) + NO2 (g) → N2O3 (g) ΔH = ?

Calculate the change in enthalpy for the above reaction using Hess’s law and the data given below

2N2 (g) + 3O2 (g) → 2N2O3 (g)    ΔH = 167.4 kJ

N2 (g) + O2 (g) → 2NO (g)    ΔH = 180.4 kJ

N2 (g) + 2O2 (g) → 2NO2 (g)    ΔH = 66.4 kJ

Answer 1

This is a problem involving Hess' Law (read about it so you understand what's going on)

You have a target equation and several given equations. We need to rearrange the given equations to arrive at the target equation. This is done as follows:

NO (g) + NO2 (g) → N2O3 (g) ΔH = ? ... TARGET EQUATION

Given:

eq.1 2N2 (g) + 3O2 (g) → 2N2O3 (g)   ΔH = 167.4 kJ

eq.2 N2 (g) + O2 (g) → 2NO (g)   ΔH = 180.4 kJ

eq.3 N2 (g) + 2O2 (g) → 2NO2 (g)   ΔH = 66.4 kJ

Process:

reverse eq.2 and ÷ by 2 NO(g) --> 1/2N2(g) + 1/2O2(g) ∆H = +90.2 kJ

reverse eq.3 and ÷ by 2 NO2(g) --> 1/2N2(g) + O2(g) -33.2 kJ

copy eq.1 and ÷ by 2 N2(g) + 3/2O2(g) --> N2O3(g) 83.7 kJ

Add these up to get...

NO(g) + NO2(g) + N2(g) + 3/2O2(g) --> 1/2N2(g) + 1/2O2(g) + 1/2N2(g) + O2(g) + N2O3(g)

Combine and cancel appropriate terms to get...

NO(g) NO2(g) --> N2O3(g) TARGET EQUATION

So now we just add up the ∆H values to obtain the ∆H for the target equation:

90.2 kJ + 83.7 kJ - 33.2 kJ = 140.7 kJ