Question

Ammonia, NH3 (Delta.Hf = –45.9 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:

4 upper N upper H subscript 3 (g) plus 5 upper O subscript 2 (g) right arrow 6 upper H subscript 2 upper O (g) plus 4 upper N upper O (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..
–902 kJ
–104.6 kJ
104.6 kJ
900.8 kJ

Answer 1

ΔHrxn = [6(−241.8) + 4(91.3)] − [4(−45.9) + 5(0)]
= (−1450.8 + 365.2) − (−183.6)
= −1085.6 + 183.6
= −902.0 kJ

Answer: −902 kJ.