N2(g) + 3h2(g) → 2nh3(g) =δh−92.kj in the second step, ammonia and oxygen react to form nitric oxide and water: 4nh3(g) + 5o2(g) → 4no(g) + 6h2o(g) =δh−905.kj calculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen from these reactions. round your answer to the nearest kj.

-272.25 kJ

Explanation:

For the Hess' Law, when a reaction happens in more than 1 step, the steps can be summed and simplified for a global reaction, and the enthalpies must be summed too. If a change is done in the reaction (such as multiplying for a number) the enthalpy must suffer the same change.

N₂(g) + 3H₂(g) → 2NH₃(g)  ΔH = -92 kJ (x2)

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

2N₂(g) + 6H₂(g) → 4NH₃(g)    ΔH = -184 kJ

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

2N₂(g) + 6H₂(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) ΔH = -184 - 905 = -1089 kJ

So,

4 moles of NO -1089 kJ

1 mol of NO     x

By a simple direct three rule:

4x = -1089

x = -272.25 kJ