Question

The following sequence of reactions occurs in the commercial production of aqueous nitric acid. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) ΔH = −1166.0 kJ/mol 2 NO(g) + O2(g) → 2 NO2(g) ΔH = −116.2 kJ/mol 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) ΔH = −137.3 kJ/mol Determine the total energy change (in kJ) for the production of one mole of aqueous nitric acid by this process. _____ kJ

Answer 1

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)

Multiply equation (1) * 3, (2) * 6, (3) * 4:

12 NH₃(g) + 15 O₂(g) → 12 NO(g) + 18 H₂O(l)..................... ΔH = −3498.0 kJ/mol ..................(4)

12 NO(g) + 6 O₂(g) → 12 NO₂(g).......................................... ΔH = −697.2 kJ/mol ....................(5)

12 NO₂(g) + 3 H₂O(l) → 6 HNO₃(aq) + 3 NO(g) .....................ΔH = −549.2 kJ/mol ...................(6)

Add up all three equations (4), (5), (6):

12 NH₃(g) + 21 O₂(g) → 14 H₂O(l) + 8 HNO₃(aq) + 4 NO(g) ΔH = −4744.4 kJ

(1 mole HNO₃(aq)) x (−4744.4 kJ/ 8 mol HNO₃(aq)) = −593.05 kJ

The total energy change for the production of one mole of aqueous nitric acid by this process = 593 kJ