The standard heat of formation of the water is -286 kJ/mol
For most chemistry problems involving #ΔH_f^o#, you need the following equation:
#ΔH_(rxn)^o = ΣΔH_f^o(p) - ΣΔH_f^o(r)#,
where #p# = products and #r# = reactants.
Your chemical equation is
2NH₃(g) + ⁷/₂O₂(g) → 2NO₂(g) + 3H₂O
#ΔH_(rxn)^o# = -698 kJ
#ΣΔH_f^o(p)# = 2 mol NO₂ × #(+34"kJ")/(1"mol NO₂")# + 3 mol H₂O × #(x"kJ")/(1"mol H₂O")# =
68 kJ – 3#x# kJ = (68 - #3x#) kJ
#ΣΔH_f^o(r)# = 2 mol NH₃ × #(-46"kJ")/(1"mol NH₃")# + ⁷/₂ mol O₂ × #(0"kJ")/(1"mol O₂")# = -92 kJ
#ΔH_(rxn)^o = ΣΔH_f^o(p) - ΣΔH_f^o(r)#; so
-698 kJ = (68 - 3#x#) kJ – (-92) kJ
-698 = 68 - 3#x# + 92
3#x# = -698 - 68 - 92 = -858
#x# = #-858/3# = -286
#ΔH_f^o#(H₂O) = #x# kJ/mol = -286 kJ/mol
