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You use the standard of the reaction and the enthalpies of formation of everything else.
For most chemistry problems involving ##ΔH_f^o##, you need the following equation:
##ΔH_(reaction)^o = ΣΔH_f^o(p) – ΣΔH_f^o(r)##,
where p = products and r = reactants.
The ##ΔH_(reaction)^o## for the oxidation of ammonia
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
is -905.2 kJ. Calculate ##ΔH_f^o## for ammonia. The standard enthalpies of formation are: NO(g) = +90.3 kJ/mol and H₂O(g) = -241.8 kJ/mol.
4NH₃(g)+ 5O₂(g) → 4NO(g) + 6H₂O(g)
##ΔH_(reaction)^o = ΣΔH_f^o(p) – ΣΔH_f^o(r)##
##ΣΔH_f^o(p) = 4 mol NO×(+90.3 kJ)/(1 mol NO) + 6 mol H₂O×(-241.8 kJ)/(1 mol H₂O)## = 361.2 kJ – 1450.8 kJ = -1089.6 kJ
##ΣΔH_f^o(r) = 4 mol NH₃ × (x kJ)/(1 mol NH₃) + 5 mol O₂ × (0 kJ)/(1 mol O₂)## = 4x kJ
##ΔH_(reaction)^o = ΣΔH_f^o(p) – ΣΔH_f^o(r)##; so
-905.2 kJ = -1089.6 kJ – 4x kJ
4x = -184.4
x = -46.1
##ΔH_f^o##(NH₃) = x kJ/mol = -46.1 kJ/mol
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