This is the question from the textbook.

Ozone O3 can be prepared in the laboratory by passing an electrical discharge through oxygen gas. Assume that an evacuated steel vessel with a volume of 10L is filled with 32 ATM of O2 at 25 degrees Celsius and an electrical discharge is passed through the vessel, causing some of the oxygen to be converted into ozone. As a result, the pressure inside the vessel drops to 30.64 ATM at 25 degrees Celsius .

The solutions manual subtracts 3x from the initial pressure of O2 to determine the final pressure of O2 after the electrical discharge(32-3x=final pressure of O2). X is the amount of pressure lost (32-30.63=1.36=x). My question is, how are we justified in using the stiochiometric ratios to determine the final pressure. In other words, how is (32-3x) justified when finding the final pressure of O2.

If my question is not clear. Please let me know and I'll try my best to rephrase it.


The actual reaction being discussed is $$\ce{3O2 -> 2O3}$$

We also know from the gas law that $$\ce{PV=nRT}$$

In the current experiment the only things changing are the pressure and the number of moles; the volume and temperature remaining constant. In this case the gas law simplifies to $$\ce{\Delta P = \Delta n}$$

Therefor, we know the change in moles for the overall reaction (-3+2=-1) is $\ce{\Delta n}$ and by the above stoichiometry the change in the moles of oxygen will be three times this amount or $\ce{3\Delta n}$ or $\ce{3 \Delta P}$ or $\ce{3x}$.

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  • $\begingroup$ Thanks for the reply but I am a bit confused. From what I understand from your answer is since P=N and N=-1, then P=-1. However, according to my solutions manual P=1.36 as stated in the question above. $\endgroup$ – Amuna Jul 12 '14 at 2:41
  • $\begingroup$ In the stoichiometric equation delta P=delta n=-1; in your question delta P=-1.36 $\endgroup$ – ron Jul 12 '14 at 2:48
  • $\begingroup$ So what you're saying is that since delta P= delta N, and since P=1.36, for every mole, 1.36 atm will be reduced and by the above stoichiometric equation, since three moles of O2 are being decomposed, then 3(1.36) is the total pressure that will be reduced? $\endgroup$ – Amuna Jul 12 '14 at 3:21
  • 1
    $\begingroup$ No, 3(1.36) is the number of moles of O2 consumed. $\endgroup$ – ron Jul 12 '14 at 3:25
  • $\begingroup$ No, thats not correct, 1.36atm should decrease for every 1.36 mol since delta P= delta N. $\endgroup$ – Amuna Jul 12 '14 at 3:27

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