# Le Chatelier's Principle and Rate constants

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?

• 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! – Learner Aug 6 '15 at 5:12