The following sequence of reactions occurs in the commercial production of aqueous nitric acid. (the overall industrial process produces excess no.) 4 nh3(g) + 5 o2(g) → 4 no(g) + 6 h2o(l) δh = −1166.0 kj/mol (1) 2 no(g) + o2(g) → 2 no2(g) δh = −116.2 kj/mol (2) 3 no2(g) + h2o(l) → 2 hno3(aq) + no(g) δh = −137.3 kj/mol (3) determine the total energy change (in kj) for the production of one mole of aqueous nitric acid by this process.

The total energy change for the production of one mole of aqueous nitric acid is 709,8kJ

Explanation:

The three steps in the industrial production of nitric acid are:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l) ΔH = −1166.0 kJ/mol (1)

2NO(g) + O₂(g) → 2NO₂(g) ΔH = −116.2 kJ/mol (2)

3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g) ΔH = −137.3 kJ/mol (3)

For the total process:

4NH₃(g) + 6O₂(g) + NO₂ → 3NO(g) + 2HNO₃(aq) + 5H₂O(l)

The ΔH is (Hess's law) : -1166,0kJ/mol - 116,2kJ/mol - 137,7kJ/mol = -1419,5 kJ/mol

As the total process produce two moles of nitric acid, the total energy change for the production of one mole of aqueous nitric acid is:

-1419,5 kJ/mol× = 709,8 kJ

I hope it helps!

increased stability in the completion of fluorine's outer shell.