# How does a catalyst make Hydrogen Peroxide's decomposition quicker? What is actually happening?

May 10, 2015

A catalyst makes the decompostition reaction of hydrogen peroxide faster because it provides an alternative pathway with a lower activation energy for the reaction to take.

Activation energy is just a term used to express the minimum energy required in order for a reaction to take place.

If no catalyst is present, hydrogen peroxide will decompose at a very, very slow rate - I think its concentration will drop by 10% per year.

When a catalyst is added, an alternative pathway through which the reaction can form water and oxygen gas is introduced. The speed of a catalyzed reaction will increase because this alternative pathway has a lower activation energy.

Here's what an alternative pathway means. For example, let's say you add potassium iodide, $K I$, to a hydrogen peroxide solution.

Potassium iodide will dissociate completely to give potassium ions, ${K}^{+}$, and iodide ions, ${I}^{-}$. The decomposition reaction will now take place in two steps

$\textcolor{b l u e}{\left(1\right)} : \text{ } {H}_{2} {O}_{2 \left(a q\right)} + {I}_{\left(a q\right)}^{-} \to O {I}_{\left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)}$

$\textcolor{b l u e}{\left(2\right)} : \text{ } {H}_{2} {O}_{2 \left(A q\right)} + O {I}_{\left(a q\right)}^{-} \to {H}_{2} {O}_{\left(l\right)} + {O}_{2 \left(g\right)} + {I}_{\left(a q\right)}^{-}$

An iodide ion will react with a hydrogen peroxide to produce water and a hypoiodite ion, $O {I}^{-}$. Then, a hypoiodite ion will react with another hydrogen peroxide molecule to produce water, oxygen gas, and a iodide ion.

That's why a catalyst is never consumed in a reaction - it is reformed at the end of the multi-step reaction.

Adding equations $\textcolor{b l u e}{\left(1\right)}$ and $\textcolor{b l u e}{\left(2\right)}$ together will get the overall reaction

$2 {H}_{2} {O}_{2 \left(a q\right)} + \cancel{{I}_{\left(a q\right)}^{-}} + \cancel{O {I}_{\left(a q\right)}^{-}} \to \cancel{O {I}_{\left(a q\right)}^{-}} + 2 {H}_{2} {O}_{\left(l\right)} + {O}_{2 \left(g\right)} + \cancel{{I}_{\left(a q\right)}^{-}}$

In this case, the activation energy for the potassium iodide-catalyzed reaction is $\text{56 kJ/mol}$. The activation energy of the uncatalyzed reaction is $\text{75 kJ/mol}$.

Lower activation energy $\to$ faster reaction.