Ammonia, NH sub 3 , Δ⁢H_f=−46.2kJ reacts with oxygen to produce water Δ⁢H_f=−241.8kJ and nitric oxide, NO , Δ⁢H_f=91.3kJ , in the reaction: 4 NH sub 3 g plus 5 cap o sub 2 g rightarrow 6 cap h sub 2 cap o g plus 4 NO g What is the enthalpy change for this reaction? Use Δ⁢H_rxn=∑(Δ⁢H_f_1p⁢r⁢o⁢d⁢u⁢c⁢t⁢s)−∑(Δ⁢H_f_1r⁢e⁢a⁢c⁢tan⁡t⁢s)

Use the formation-enthalpy sum for products minus the sum for reactants. The products total 6(−241.8 kJ) + 4(91.3 kJ) = −1085.6 kJ. The reactants total 4(−46.2 kJ) + 5(0 kJ) = −184.8 kJ, since oxygen in its elemental form has ΔH_f = 0. Thus, ΔH_rxn = −1085.6 − (−184.8) = −900.8 kJ, so the enthalpy change is negative 900.8 kJ.

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