Answer:
The total energy change for the production of one mole of aqueous nitric acid is −494 kJ
Explanation:
<u>Step 1</u>: 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
<u>Step 2:</u> 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
<u>Step 3:</u> 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
<u>
Step 4:</u> 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
