Questions About the Formation of Hydrochloric Acid

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    Acid Formation
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The discussion centers on the formation of hydrochloric acid through the photochemical reaction of hydrogen and chlorine, focusing on the exothermic energy released and the activation energy required. It highlights that the energies involved are experimentally derived and influenced by factors like pressure and temperature. To determine activation energy, one should calculate the bond energies of H2 and Cl2 and consider the bond energy of HCl for exothermic energy. The electrolytic process for separating H2 and Cl2 from HCl involves understanding free energy and voltage values. Resources like the CRC handbook and reduction tables are recommended for further data.
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Got a question about the energy dissipated during the formation of hydrochloric acid, upon the reaction of hydrogen and chlorine (photochemical reaction). Whats the exothermic energy involved and the intensity of light needed to activate this reaction, in other words the activation energy. And what would be the enrgy required to separate H2 and Cl2 from a solution of HCl by electrolytic process?
thnx let me know soon.
 
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are you sure this is a physics question? I believe that this is more general chemistry.
the endothermic and exothermic energies are all derived experimentally. this means they are not always the same. There are too many factors that affect it like pressure and temperature (eg. when it's 110*C out you need to add less heat than when it's -30 degrees out). You could try the CRC handbook to get these data. :)
 
You simply need to find the bond energies of Cl_2, H_2. Add these two up, I believe that it'll be the activation energy. What light energy corresponds to this energy; you should definitely know this.

To find the exothermic energy, you'll need to incorporate the bond energy of HCl. Draw the process out. You've got diatomic gases, you dissociate them, then these species reassociate.

The electrolytic aspect pertains to free energy. You can find the voltage values of each of the reactions I mentioned and work from there.
 
thanx a lot I am sure that will help
 
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