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Chapter 4 (CIC) and Chapter 5 (CTCS)

Chapter 4 (CIC) and Chapter 5 (CTCS). Read in CTCS Chapter 5.6,7, pgs 172-3 Problems in CTCS: 49, 51, 53, 57, 59, 61, 63, 65, 67, 75. Hess’s Law. H = sum of the individual enthalpies of each step 2 H 2 (g) + O 2 (g)  2 H 2 O (g) H = -483.6 kJ

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Chapter 4 (CIC) and Chapter 5 (CTCS)

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  1. Chapter 4 (CIC) and Chapter 5 (CTCS) • Read in CTCS Chapter 5.6,7, pgs 172-3 • Problems in CTCS: 49, 51, 53, 57, 59, 61, 63, 65, 67, 75

  2. Hess’s Law • H = sum of the individual enthalpies of each step • 2 H2(g) + O2(g) 2 H2O(g) H = -483.6 kJ • 2 H2O(g) 2 H2O(l) H = -88.0 kJ • 2 H2(g) + O2(g) 2 H2O(l) H = -571.6 kJ • It is very difficult to get the enthalpy value for the incomplete combustion of graphite to CO, but we can use Hess’s Law to get this information

  3. C(g) + O2(g) CO2(g) H = -393.5 kJ CO(g) + 1/2O2(g) CO2(g) H = -283.0 kJ Q: What is Hºf of NO3-(aq)? Hint: HNO3 is a strong acid A: -206.6 kJ

  4. Activation Energy C6H12O6 + 6 O2 6 CO2 + 6 H2O • In our bodies, enzymes get this reaction to go • On the bench, KClO3 and H2SO4 are required Ea Reactants Hrxn Enthalpy Products Reaction Progress

  5. Photosynthesis • Ultimate source of all fossil fuels is the sun • 6 CO2 + 6 H2O  C6H12O6 + 6 O2 ΔH = +2800 kJ • Catalyst is chlorophyll • Wood • Coal • Solid • Little Oxygen but fair amount of Sulfur • Petroleum • Liquid • Less Oxygen and Sulfur • Must be refined

  6. Reformulated Gasolines (RFG’s) • Get rid of PbEt4 • Must be 2% by mass oxygen • Use MTBE or EtOH Q: If MTBE is C5H12O and gasoline is C8H18, What percentage of MTBE must be in RFG? A: 14%

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