Calculate ∆H for the reaction (n)

Question

Calculate ∆H for the reaction

2 NOCl(g) → N₂(g) + O₂(g) + Cl₂(g)

given the following data:

Equation	∆H
½ N2(g) + ½ O2(g) → NO(g)	90.3 kJ
NO(g) + ½ Cl2(g) → NOCl(g)	-38.6 kJ

∆H= kJ

Answer 1

Flipping the 2nd equation:

NOCl --> NO + 1/2 Cl₂ delta H = 38.6 kJ (flip the sign on delta H)

Multiplying this equation by 2:

2 NOCl --> 2 NO + Cl2 delta H = 2*(38.6) = 77.2 kJ <-- Let's call this equation (3)

Flipping the first equation:

NO --> 1/2 N₂ + 1/2 O₂ delta H = -90.3 KJ

Multiplying this equation by 2:

2 NO --> N₂ + O₂ delta H = -90.3*(2) = -180.6 kJ <-- Let's call this equation (4)

Adding our new equations (3) and (4) gives you the overall equation. To find the delta H you add the delta H's we calculated for (3) and (4) = 77.2 + -180.6 = -103.4 kJ