METOENCE 434

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LECTURE 5

• METO/ENCE 434
• AIR POLLUTION
• RUSSELL R. DICKERSON

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• Wark and Warner Chapter 7,8
• Finlayson Pitts Pitts, Appendix 1
• II. Thermodynamics (continued)
• a) Enthalpy, Bond Energy, and Entropy
• From the last lecture, you will remember the
  concept of enthalpy or heat. There is the heat
  of formation of substances from their elemental
  parts, and the heat of specific reactions leading
  to complete oxidation of the reactants is called
  the heat of combustion.
• EXAMPLE
• Remember the combustion of isooctane (C8H18), an
  approximation of gasoline
• C8H18 12.5 O2 (47N2) ? 8CO2 9H2O (47N2)
• The heat of combustion,
• is about 1240 kcal/mole.

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• Remember also the Air-Fuel-Ratio. If the AFR is
  higher than stoichiometry dictates (here 15) than
  the mixture is lean if it is less than 15, then
  the mixture is said to be rich and pollutants
  such as CO or H2CO must be formed. For example
  in the combustion of natural gas, methane
  primarily, if insufficient oxygen is present the
  following reaction can occur
• But there is more energy to be had from this
  reaction
• This shows that burning rich not only produces
  pollutants, but cheats you out of some of the
  available heat (and energy).

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• Students calculate the heat of combustion of
  methane. How does it compare the sun of the
  heats of the two above reactions? How is this
  related to ?H being an equation of state?
• How do we find ?Hf for new compounds?
• BOND ENERGIES
• Example water, H-O-H
• Look at the table and see an O-H bound requires
  about 462 kJ/mole
• ( 110.6 kcal/mole) to break. Two such bonds must
  be broken to turns H2O into
• 2H O.
• ?Hf H2O ?
• H - O - H ? H H O
• H H ? H2
• O ? ½ O2
• NET H2O ? H2 ½ O2
• Check against the table for the reverse reaction
  ?Hf (H2O) -242, pretty close

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• Enthalpy is a powerful tool, but is it the
  criterion of feasibility? There are reactions
  that proceed in spite of the ?H being positive,
  such as air expanding to fill a vacuum. These
  processes are driven by another force and the
  Second Law of Thermodynamics defines that force.
  Entropy is the degree of mixing or disorder in a
  system. Systems tend toward the highest state of
  entropy. The products of the combustion of
  isooctane are favored by entropy because there is
  much more order to isooctane and oxygen than to
  eight CO2s and H2Os.
• Entropy is defined as dS DQ/T
• The total entropy of a system plus surroundings
  always increases, that is ?S 0.0. Entropy has
  units of kcal/moleK of kJ/moleK.

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• By combining the First and Second Laws of
  Thermodynamics, we can derive the criterion of
  feasibility Gibbs Free Energy.
• ?G ?H T?S
• Units are kcal/mole or kJ/mole. If ?G 0.0
  reaction will not proceed spontaneously without
  the input of energy. This does not mean that all
  reactions with ?G 0.0 proceed immediately,
  otherwise the gas tank on your car would explode,
  of course the potential is there.

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• Calculating the ?G of a reaction is the same
  as calculating the ?H of a reaction. Consider
  the following particularly important reaction.
• This reaction can go spontaneously, that is
  without the input of energy. If n1 we have
  formaldehyde combustion. But consider n6 then
  (H2CO)6 is glucose and when glucose is oxidized
  2870 kJ/mole (686 kcal/mole) of free energy are
  released. This process is respiration. But we
  know that this reaction proceeds in the opposite
  direction as well. This process is
  photosynthesis. Does this violate the second law
  of thermodynamics? No, because six photons are
  involved in the synthesis of glucose molecule.
• Thermodynamics tells us that respiration can
  provide energy to an organism, but that
  photosynthesis requires energy from outside the
  system it even tells how much energy. The
  problem is that these processes are much more
  complicated than the reaction written above.
  Both respiration and photosynthesis require about
  100 steps. Fortunately, ?H and ?G are equations
  of state and therefore are path independent.

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• Problem for students Calculate the ?Gº for the
  complete combustion of isooctane. (-5229 kJ/mole)