Heat-driven chemical reaction and thermodynamics

[vemvare]

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?

 

[DrDu]

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.

 

[vemvare]

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?

 

[DrDu]

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.

 

[LudusRex]

I understand your confusion about the efficiency of heat-driven chemical reactions and thermodynamics. The concept of energy conservation is a fundamental principle in thermodynamics, which states that energy cannot be created or destroyed, only transformed from one form to another. In the case of cracking water into oxygen and hydrogen using heat and a catalyst, it is true that any energy that is not used to break the chemical bonds will eventually dissipate as heat.

However, the key factor to consider here is the overall energy balance of the reaction. While some energy may be lost as heat, the total energy input (heat and catalyst) is still less than the total energy output (chemical energy of the products). This is because the energy released from breaking the chemical bonds is greater than the energy required to initiate the reaction. This difference in energy is known as the reaction's Gibbs free energy, which is a measure of the reaction's spontaneity.

In other words, the efficiency of the process may be less than 100% due to the energy loss as heat, but the overall energy balance is still favorable. This is a fundamental principle of thermodynamics and is the basis for many industrial processes, such as the production of hydrogen gas through water electrolysis.

Therefore, it is not necessarily that you are doing something wrong, but rather that the efficiency of a heat-driven chemical reaction must be evaluated in terms of the overall energy balance of the system. I hope this helps clarify your understanding of this concept.