# Why is 2H2O2 -> 2H2O + O2 a first order reaction?

My book says that the reaction $\ce{2H_2O_2 -> 2H_2O + O_2}$ is a first order reaction and that it's rate equation is $\ce{k[H_2O_2][I^{-}]}$.

However i don't see why that would be first order? Isn't the order of the reaction supposed to be the sum of powers of the concentration of the reactants?

• I- is catalyst not a reactant – JM97 Mar 31 '16 at 6:54
• Since the concentration of iodide ion remains constant its concentration can be taken under equilibrium constant( just as we do in pseudo 1st order reactions) so the order will be 1st – JM97 Mar 31 '16 at 7:01
• It is better to say "a first order reaction in $\ce{H2O2}$". That would be right and unambiguous. – Ivan Neretin Mar 31 '16 at 7:40
• @IvanNeretin Actually the textbook statement is "_This reaction is first order with respect to both H_2 O_2 and I^- " . So, what is meant by term with respect to in this context? – JM97 Mar 31 '16 at 16:11
• It simply means that the reaction rate is proportional to the first power of $\ce{[H2O2]}$. – Ivan Neretin Mar 31 '16 at 17:09