In my Bio 1 lab we studied the way catalase catalyzes the decomposition of H2O2 and at the end it asked me what conditions would affect the rate of the reaction. It was the type of question that obviously wants you to answer in a specific way, in this case, that a high pH or high enough temperatures would denaturalize catalase since we discussed that in class. I turned it in with that but now I have the question of whether LeChatelier's principle applies here. The decomposition is not a reversible function, so it doesn't really reach an equilibrium, but I couldn't help but think that a high enough atmospheric O2 pressure would "push down" on the H2O2 and not allow it to release any more oxygen.
1 Answer
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The simple answer is that the equilibrium of this reaction is shifted too far towards the products to reverse with any ordinary amount of pressure.
For a very rough back-of-the envelope estimate, the decomposition of hydrogen peroxide has $\ce{\Delta G} = -120\ \rm{kJ/mol}$ at standard temperature and pressure. Oxygen is not an ideal gas, but let's just use that as an approximation for the moment. Let's also pretend that the $\ce{H2O2}$ and $\ce{H2O}$, being liquids, are completely unaffected by the pressure change.
Increasing the pressure of an ideal gas by a factor of 10 changes its Gibbs free energy by $nRT \ln(10) = 5.2\ \rm{kJ/mol}$. So you would have to increase the pressure by a factor of more than $10^{23}$ before the reaction became endergonic. That's somewhere in the stellar core or white dwarf pressure range. At those pressures, some unwanted side reactions might happen... specifically, unwanted nuclear side reactions.
Clearly we've left the realm of validity of our assumptions from the prior paragraph. Still, it's safe to conclude increasing the pressure is not a practical method of reversing the decomposition of hydrogen peroxide, even to a small extent.
