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Heat-driven chemical reaction and thermodynamics

  1. Feb 22, 2012 #1
    If we use heat and a catalyst to crack water into oxygen and hydrogen, how can the efficiency of the process ever be less than 100%? Every joule that doesn't crack open the bonds and end up in the chemical energy of the products must end up as heat, and then we're back at square one.

    I'm clearly doing something wrong here, but what?
     
  2. jcsd
  3. Feb 23, 2012 #2

    DrDu

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    This depends on your definition of efficiency. The way you look at it, you are right. But, if for example you want to use the hydrogen oxygen mixture to use it e.g. as a fuel, you have to cool it down again to room temperature. Even if you do so avoiding back reaction, you have to carry off a lot of heat stored in the heat capacity of the gases which then is lost.
     
  4. Feb 23, 2012 #3
    http://en.wikipedia.org/wiki/Zinc_zinc-oxide_cycle

    (I know, it's wikipedia but still)

    The products of the process are oxygen at 2173K and hydrogen at 700K, how could these possibly contain 60% of the used energy?

    2173 * 0.8888 (O2 in 1kg of water) * 918 (heat capacity of oxygen) = 1771.4 KJ
    700 * 0.11111 (H2) * 14304 (heat capacity of hydrogen) = 1112.4 KJ
    sum: 2883.8 KJ.

    But H2/O2 when burned releases ~13400 KJ of energy per kilogram of mix, so what happens to the ~17200KJ that is missing from the equation?
     
  5. Feb 23, 2012 #4

    DrDu

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    The reaction of zink with water is exothermic and the heat is given off to the surrounding. I suppose that is where most energy is lost.
     
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