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

4N H 3 (g) + 5 O2 (g) ---> 6 H 2O (g) + 4O (g).

What is the enthalpy change for this reaction?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..
–900.8 kJ
–104.6 kJ
104.6 kJ
900.8 kJ

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Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. in the first step, nitrogen and hydrogen react to form ammonia: (g) (g) (g) in the second step, ammonia and oxygen react to form nitric acid and water: (g) (g) (g) (g) calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions. round your answer to the nearest .

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