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!

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.

the answer is d. 74.02