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


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

Please send any comments, corrections, or suggestions to svanbram@science.widener.edu.

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Last Updated Wednesday, May 08, 2002 4:23:33 PM