Only need help with Bottom part

ΔHr = -272 kJ

Explanation:

The steps of formation of NO are:

(1) N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = -92 kJ

(2) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) ΔH = -905 kJ

The sum of (1)/2 + (2)/4 is:

¹/₂N₂(g) + ³/₂H₂(g) + ⁵/₄O₂(g) → NO(g) + ⁶/₄H₂O(g)

By Hess's law, it is possible to sum in the same way the change in enthalpies to obtain the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen. Thus:

ΔHr = -92 kJ/2 + -905 kJ/4 = -272 kJ

I hope it helps!