Why is ΔH°rxn -1172 kJ for This Reaction?

[Drakkith]

Homework Statement

Given the data in the table below, ΔH°_rxn for the reaction

4NH₃ (g) + 5O₂ (g) → 4NO (g) + 6H₂O (l)

is ________ kJ.

Substance ΔH∘f(kJ/mol)
H₂O (l) -286
NO (g) 90
NO₂ (g) 34
HNO₃ (aq) -207
NH₃ (g) -46

Homework Equations

ΔH°_rxn = ΔH°_products - ΔH°_reactants

The Attempt at a Solution

ΔH°_rxn = [(4x90)+(6x-286)] - [(4x-46)+(X)]

ΔH°_rxn = (-1356) - (-184+X)

ΔH°_rxn = -1356 + 184 - X

ΔH°_rxn = -1172 + X

At first I selected, D.) The ΔH°f of O2 (g) is needed for the calculation.
However, apparently the answer is -1172, but I have no idea why. How can -1172 be the answer here? And why do they give me the enthalpy of formation for NO₂ and HNO₃ if neither of those compounds are in my reaction?

 

[Drakkith]

Okay, I just realized the ΔH°f of O₂ is defined to be zero...
Now it all makes sense.

 

[Borek]

I just realized the ΔH°f of O2 is defined to be zero...

And you answer - that it is needed - wasn't wrong in general. Defining ΔH°f of O₂ as zero means you know its value ;)

 

[gracy]

And why do they give me the enthalpy of formation for NO2 and HNO3 if neither of those compounds are in my reaction?

why?

 

[Borek]

Either to confuse him, or because it was copy/pasted from another problem and someone didn't bother to leave just what is important for the question.