Question

Nitric oxide (NO) can be formed from nitrogen, hydrogen and oxygen in two steps. In the first step, nitrogen and hydrogen react to form ammonia: N2(g) + 2 H_2(g) rightarrow 2 NH_3 (g) Delta H = -92. kJ In the second step, ammonia and oxygen react to form nitric oxide and water: 4 NH_3(g) + 5 O_2(g) rightarrow 4NO(g) + 6 H_2O(g) Delta H = -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.

Answer 1

ΔHr = -272 kJ

Step-by-step 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!