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Consider the reaction: $$\ce{H2O2(g) <=> 2OH^.(g)}\label{a}\tag{1}$$ to be at at equilibrium ($\ce{OH^.}$ is a radical). As the pressure is increased, the reaction moves towards forming $\ce{H2O2}$ as this direction reduces the amount of substance in the system.

However, it is well-established that the forward rate constant of $\ref{a}$ increases with increase in pressure, meaning forming more $\ce{OH}$. This seems to be in conflict to what the Le Chatelier's principle suggests. Could someone help me get my head around this?

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It is true that the forward reaction rate increases as you increase pressure. However, the reverse reaction rate will increase as well, and it will increase more than the forward reaction rate.

The key here is that the system is in equilibrium; the forward and backward reaction rates are equal. Altering the pressure affects both reactions.

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  • $\begingroup$ This makes sense. So, with the increase in pressure, the reverse reaction rate increases and is dominant compared to the forward reaction rate (although the forward rate constant increases with increase in pressure) This definitely confirms with the Le Chatelier's principle then, resulting in the formation of more $\rm H_2O_2$. Thanks! $\endgroup$ – Learner Aug 6 '15 at 5:12
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With increasing pressure volume of system increases and thus active mass increaeas then to keep the constant value of Keq whole setup is disturbed in direction where no of moles of reacting substance decreases. Thus increment in rate of one reaction is preferred over other.

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