What is the enthalpy change for this reaction?

Question

Ammonia, NH3 ( Hf=  –46.2 kJ), reacts with oxygen to produce water ( Hf= –241.8 kJ) and nitric oxide, NO ( Hf= 91.3 kJ), in the following reaction:

Answer 1

You neglected to put down the "following reaction".  I'll assume it is 

2NH₃(g) + O₂(g) ==> 3H₂O(l) + 2NO(g)  ...  BALANCED EQUATION

 

 

Given the ∆H^º_formation of the reactants and products, you can calculate ∆H_reaction (note ∆H_f O₂ is zero):

• ∑∆H_products - ∑∆H_reactants = ∆H_reraction
• ∆H_reaction = [(3x-241.8) + 2(91.3)] - [(2x-46.2)] = (-725.4 + 182.6) - (-92.4) 
• ∆_Hreaction = -542.8 - (-92.4) = -450.4 kJ