Calculate the ΔHrxn for the following 2NOCl(g)→N2(g)+O2(g)+Cl2(g)

Given:
12N2(g)+12O2(g)→NO(g)
ΔH = 89.32

NO(g)+12Cl2→NOCl(g)
ΔH = -30.39

2 NO (g) → N2 (g) + O2 (g)

2 NOCl (g) → 2 NO (g) + Cl2 (g)

2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)

ΔH = [90.3 kJ x 2 x -1] +  [-38.6 kJ x -1 x 2] = -103.4 kJ

The ΔH for the reaction is -103.4 kJ

i just learnt this recently but i need to look through my notes again

pie chart because it is in percents ( % )