Lecture 11/5/99, Chapter 6, Section 3-8

Announcements

Calculations of /\Hrxn

  1. Production of CCl4
    1. CH4 (g) + 4 Cl2 (g) -> CCl4 (l) + 4 HCl (g)
    2. [delta] Hrxn from [delta] Hf
      Compound [delta] Hof (kJ)
      CH4 (g) -74.8
      Cl2 (g) 0
      CCl4 (l) -135.4
      HCl (g) -92.3
    3. [delta] H for 500 g CCl4
  2. Hydrazine rocked fuel is formed from the reaction: (Ebbing page 252)
    1. N2 (g) + 2H2 (g) -> N2H4 (l)
    2. Calculate [delta] Hrxn based upon the following:
      Reaction [delta] H (kJ)
      N2H4 (l) + O2 (g) -> N2 (g) + 2 H2O (l) -622.2
      H2 (g) + 1/2O2 -> H2O (l) -285.8
    3. How much energy is required to produce 1.00 kg of hydrazine?
  3. Reaction for an anerobic engine, ref Pritchard and Clothier J. Chem. Soc. Chem. Commun. 1986, 1529.
    1. (CH3)3COOC(CH3)3 (g) + O2 (g) -> CO2 (g) + H2O (g)
      Compound [delta] Hof (kJ)
      (CH3)3COOC(CH3)3 (g) -347.3
      O2 (g) 0
      CO2 (g) -393.5
      H2O (g) -241.8
    2. (CH3)3COOC(CH3)3 (g) -> C2H6 (g) + 2 (CH3)2CO (g)
      Compound [delta] Hof (kJ)
      (CH3)3COOC(CH3)3 (g) -347.3
      C2H6 (g) -84.1
      (CH3)2CO (g) -217.1
  4. Radical cleavage of 1 hexene at beta and gamma position
  5. Combustion of gasoline
    1. 2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O
    2. Calculate [delta] Hrxn
    3. How much energy is released by burning 1 gallon of gasoline? (Density - 0.7025 g/cm3)
  6. Example 5.4, Formation of Glucose from CO2 and H2O
    1. 6 CO2 (g) + 6 H2O (l) -> C6H12O6 (s) + 6 O2 (g)
    2. How much energy is required to produce 150 g of glucose ([delta]Hf = -1274 kJ mole-1, Aitkins)?
  7. The Ostwand process for nitric acid is:
    1. Reaction Steps
      1. 4 NH3 (g) + 5 O2 (g) -> 4 NO (g) + 6 H2O (g)
      2. 2 NO (g) + O2 (g) -> 2 NO2 (g)
      3. 3 NO2 (g) + H2O (l) -> 2 HNO3 (aq) + NO (g)
    2. Find [delta] H for each reaction.
    3. Balance total reaction and find [delta] H.
    4. How much energy is required to produce 1000 kg of nitric acid?
    5. Solution
      1. 4 NH3 (g) + 5 O2 (g) -> 4 NO (g) + 6 H2O (g)
        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
      2. 2 NO (g) + O2 (g) -> 2 NO2 (g)
        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
      3. 3 NO2 (g) + H2O (l) -> 2 HNO3 (aq) + NO (g)
        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
      4. This gives for each step:
        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
      5. Next step is to balance the overall equation: This balances as:
        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
      6. TOTAL: 12 NH3 + 21 O2 -> 8 HNO3 (aq) + 4 NO (g) + 14 H2O -4131 kJ
        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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