Question

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.

Answer 1

-272.25 kJ

Step-by-step 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

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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