Calculate the enthalpy change for the reaction.?

I got #-"902 kJ"# for the reaction as-written. How would you rewrite this into #"kJ/mol NH"_3#?

The standard molar enthalpy of formation is for the formation of #"1 mol"# of product from its elements as they exist in nature at #25^@ "C"# and #"1 atm"#. For example...

#1/2"N"_2(g) + 3/2"H"_2(g) -> "NH"_3(g)#, #" "DeltaH_f^@("NH"_3(g)) = -"45.9 kJ/mol"#

But that also means formation of elements in their standard states must yield zero enthalpy change:

#1/2"O"_2(g) -> 1/2"O"_2(g)#, #" "DeltaH_f^@("O"_2(g)) = 0#

That means we can simply derive from Hess's law and that fact to get:

#DeltaH_(rxn)^@ = sum_P n_PDeltaH_(f,P)^@ - sum_R n_RDeltaH_(f,R)^@#

where #n# indicates the mols of product #P# or reactant #R#. Here we actually have just summed many formation reactions together that we know have #DeltaH_(rxn)^@ = DeltaH_f^@#.

Here we have:

#color(blue)(DeltaH_(rxn)^@) = ["4 mols NO"(g) cdot DeltaH_(f,NO(g))^@ + "6 mols H"_2"O"(g) cdot DeltaH_(f,H_2O(g))^@] - ["4 mols NH"_3(g) cdot DeltaH_(f,NH_3(g))^@ + "0 kJ/mol"]#

where we have set a zero contribution #"O"_2(g)# right off the bat.

#= ["4 mols NO"(g) cdot "91.3 kJ/mol" + "6 mols H"_2"O"(g) cdot -"241.8 kJ/mol"] - ["4 mols NH"_3(g) cdot -"45.9 kJ/mol" + "0 kJ/mol"]#

#= [-"1085.6 kJ"] - [-"183.6 kJ"]#

#= color(blue)(-"902 kJ")# for the reaction as-written.