# Solution to Ostwald Process Thermodynamics

1. 4 NH3 (g) + 5 O2 (g) -> 4 NO (g) + 6 H2O (g)
2.  Process Amount Compound Equation [delta] H (kJ) unmake 4 moles NH3 4 * 45.94 183.76 unmake 5 moles O2 5 * 0 0 make 4 moles NO 4 * 90.4 361.6 make 6 moles H2O 6 * -241.814 -1450.88 TOTAL -905.52

3. 2 NO (g) + O2 (g) -> 2 NO2 (g)
4.  Process Amount Compound Equation [delta] H (kJ) unmake 2 moles NO 2 * (-90.4) -180.8 unmake 1 mole O2 1 * 0 0 make 2 moles NO2 2 * 33.8 67.6 TOTAL -113.2

5. 3 NO2 (g) + H2O (l) -> 2 HNO3 (aq) + NO (g)
6.  Process Amount Compound Equation [delta] H (kJ) unmake 3 mole NO2 3 * (-33.8) -101.4 unmake 1 mole H2O 1 * 285.83 285.83 make 2 mole HNO3 2*(-207.36) -414.72 make 1 mole NO 1*90.4 90.4 TOTAL -139.89

7. This gives for each step:
8.  Reaction [delta] H (kJ) 4 NH3 (g) + 5 O_2 (g) -> 4 NO (g) + 6 H2O (g) -905.52 2 NO (g) + O2 (g) -> 2 NO2 (g) -113.2 kJ 3 NO2 (g) + H2O (l) -> 2 HNO3 (aq) + NO (g) -139.89

9. Next step is to balance the overall equation: This balances as:
10.  Reaction [delta] H (kJ) 12 NH3 (g) + 15 O2 (g) -> 12 NO (g) + 18 H2O (g) -2716 12 NO (g) + 6O2 (g) -> 12 NO2 (g) -679.2 12 NO2 (g) + 4 H2O (l) -> 8 HNO3 (aq) + 4 NO (g) -559.56 4 H2O (g) -> 4 H2O (l) 4 * (-44.01) -176.06

11. TOTAL: 12 NH3 + 21 O2 -> 8 HNO3 (aq) + 4 NO (g) + 14 H2O -4131 kJ
12.  Process Amount Compound Equation [delta] H (kJ) unmake 12 mole NH3 12 * 45.94 551.28 unmake 21 mole O2 21 * 0 0 make 8 mole HNO3 8 * (-207.36) -1658.9 make 4 mole NO 4*90.4 362 make 14 mole H2O (g) 14 * (-241.814) -3385.39 TOTAL -4131

Scott Van Bramer
Department of Chemistry
Widener University
Chester, PA 19013