Question

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.

Answer 1

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

Step-by-step 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×
`(1molReaction)/(2molHNO_(3))` = 709,8 kJ

I hope it helps!