Question

The following sequence of reactions occurs in the commercial production of aqueous nitric acid: 4NH3(g)+5O2(g)⟶4NO(g)+6H2O(l)ΔH=−907kJ 2NO(g)+O2(g)⟶2NO2(g)ΔH=−113kJ 3NO2+H2O(l)⟶2HNO3(aq)+NO(g)ΔH=−139kJ Determine the total energy change for the production of one mole of aqueous nitric acid by this process.

Answer 1

Answer:

The total energy change for the production of one mole of aqueous nitric acid is −494 kJ

Explanation:

Step 1: Data given

4NH3(g)+5O2(g)⟶4NO(g)+6H2O(l)ΔH=−907kJ 2NO(g)+O2(g)⟶2NO2(g)ΔH=−113kJ 3NO2+H2O(l)⟶2HNO3(aq)+NO(g)ΔH=−139kJ

Step 2: Multiply equations

Multiply the first equation by 3:

12 NH3(g) + 15 O2(g) → 12 NO(g) + 18 H2O(l) ΔH = −2721 kJ

Multiply the second equation by 6:

12 NO(g) + 6 O2(g) → 12 NO2(g) ΔH = −678 kJ

Multiply the third equation by 4:

12 NO2(g) + 4 H2O(l) → 8 HNO3(aq) + 4 NO(g) ΔH = −556 kJ

Step 3: Get the equations together

12 NH3(g) + 15 O2(g) + 12 NO(g) + 6 O2(g) + 12 NO2(g) + 4 H2O(l) →

12 NO(g) + 18 H2O(l) + 12 NO2(g) + 8 HNO3(aq) + 4 NO(g)

ΔH = −2721 kJ − 678 kJ − 556 kJ

We can simplify as followed:

12 NH3(g) + 21 O2(g) → 14 H2O(l) + 8 HNO3(aq) + 4 NO(g) ΔH = −3955 kJ

Step 4: Determine the total energy change for the production of one mole of aqueous nitric acid by this process:

−3955 kJ/8 moles HNO3= −494 kJ

The total energy change for the production of one mole of aqueous nitric acid is −494 kJ