Nitric oxide (no) can be formed from nitrogen, hydrogen and oxygen in two steps. in the first step, nitrogen and hydrogen react to form ammonia: n2(g) + 3h2(g) →2nh3(g)δh=−92.kjin the second step, ammonia and oxygen react to form nitric oxide and water: 4nh3(g) + 5o2(g) → 4no(g) +6h2o(g)δh=−905.kjcalculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen from these reactions. round your answer to the nearest kj.

ΔH  = - 272 kJ

Explanation:

We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:

N2(g) +         3H2(g) →          2NH3(g)                 ΔH=−92.kJ  Multiplying by 2:      

2N2(g) +       6H2(g) →          4NH3(g)                        ΔH=− 184 kK

plus

4NH3(g) +     5O2(g) →          4NO(g) +6H2O(g)        ΔH=−905.kJ

2N2(g) +   6H2(g) + 5O2(g)→  4NO(g)  + 6H2O(g)      ΔH = (-184 +(-905 )) kJ

                                                                                     ΔH =    -1089 kJ

Notice how the intermediate NH3 cancels out.

As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH  for the formation  of one mol NO:

-1089 kJ/4 mol NO  x 1 mol NO =  -272 kJ (rounded to nearest kJ)