Heat-driven chemical reaction and thermodynamics

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Discussion Overview

The discussion centers around the efficiency of heat-driven chemical reactions, specifically the process of cracking water into hydrogen and oxygen using heat and a catalyst. Participants explore the thermodynamic implications and energy losses associated with this process.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions how the efficiency of cracking water can be less than 100%, suggesting that any energy not converted into chemical energy must revert to heat.
  • Another participant argues that efficiency depends on the definition used, noting that cooling the hydrogen-oxygen mixture to room temperature results in energy losses due to heat capacity.
  • A third participant references the zinc-zinc oxide cycle and presents calculations regarding the energy content of the products, questioning the apparent energy deficit when comparing the energy released during combustion to the energy calculated from the products.
  • A later reply mentions that the exothermic reaction of zinc with water releases heat to the surroundings, implying that this is a significant source of energy loss.

Areas of Agreement / Disagreement

Participants express differing views on the definition of efficiency and the sources of energy loss in the process. There is no consensus on the overall efficiency or the implications of the energy calculations presented.

Contextual Notes

Participants have not fully resolved the assumptions regarding energy definitions, the role of heat capacity, and the specifics of the chemical reactions involved.

vemvare
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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?
 
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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.
 
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?
 
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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