Answer:
The ΔH°rxn for this reaction is -905.9 kJ
Explanation:
<u>Step 1:</u> Balanced reaction:
4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g)
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<u>Step 2</u>: Given data
ΔH°f [NH3(g)] = –45.9 kJ/mol
ΔH°f [NO(g)] = 90.3 kJ/mol
ΔH°f [H2O(g)] = –241.8 kJ/mol
ΔH°f of O2 is defined to be zero
<u>Step 3</u>: Calculate ΔH°rxn
ΔH°rxn = ΔH°products - ΔH°reactants
ΔH°rxn = ((4*90.3) + 6*-(241.8)) - 4*(-45.9) = -905.9 kJ
The ΔH°rxn for this reaction is -905.9 kJ