Hess' Law chemistry

Question

Given the following equations

H₂(g) + 1/2 O₂(g)  H₂O(l)      ΔH = -285.8 kJ

N₂O₅(g) + H₂O(l)  2 HNO₃(l)      ΔH = -76.6 kJ

1/2 N₂(g) + 3/2 O₂(g) + 1/2 H₂(g)  HNO₃(l)      ΔH = -174.1 kJ

Calculate the ΔH for the following reaction.

2 N₂(g) + 5 O₂(g)  2 N₂O₅(g)

Answer 1

Target equation: 2N₂(g) + 5O₂(g) ==> 2N₂O₅(g)

Given:

Eq.1 H₂(g) + 1/2 O₂(g)  H₂O(l)      ΔH = -285.8 kJ

Eq.2 N₂O₅(g) + H₂O(l)  2 HNO₃(l)      ΔH = -76.6 kJ

Eq.3 1/2 N₂(g) + 3/2 O₂(g) + 1/2 H₂(g)  HNO₃(l)      ΔH = -174.1 kJ

Eq.3 x 4: 2N2 + 6O2 + 2H2 ==> 4HNO3 ∆H = -174.1 x 4 = -696.4 kJ

Eq.2 x 2 and reversed: 4HNO3 ==> 2N2O5 + 2H2O ∆H = +76.6 x 2 = +153.2 kJ

Eq.1 x 2 and reversed: 2H2O ==> 2H2 + O2 ∆H = +285.8 x 2 = +571.6 kJ

Add up the 3 equations:

2N2 + 6O2 + 2H2 + 4HNO3 + 2H2O ==> 4HNO3 + 2N2O5 + 2H2O + 2H2 + O2

Cancel and combine like terms:

2N₂ + 5O₂ ==> 2N₂O₅ = TARGET EQUATION

Now add up the ∆H values: -696.4 kJ + 153.2 kJ + 571.6 kJ = +28.4 kJ