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

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The discussion revolves around calculating the standard enthalpy change (ΔH°rxn) for the reaction of ammonia and oxygen to produce nitric oxide and water. The correct formula for ΔH°rxn is established as the difference between the enthalpy of formation of products and reactants. The value of ΔH°f for O2 is defined as zero, which clarifies the calculation leading to the answer of -1172 kJ. Confusion arises regarding the inclusion of enthalpy values for NO2 and HNO3, which are unnecessary for this specific reaction. Ultimately, the focus is on understanding how the enthalpy of formation values contribute to the overall reaction energy calculation.
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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?
 
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Okay, I just realized the ΔH°f of O2 is defined to be zero...
Now it all makes sense.
 
Drakkith said:
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 ;)
 
Drakkith said:
And why do they give me the enthalpy of formation for NO2 and HNO3 if neither of those compounds are in my reaction?
why?
 
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.
 
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