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

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, #KI#, 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

#color(blue)((1)):" " H_2O_(2(aq)) + I_((aq))^(-) -> OI_((aq))^(-) + H_2O_((l))#

#color(blue)((2)): " " H_2O_(2(Aq)) + OI_((aq))^(-) -> H_2O_((l)) + O_(2(g)) + I_((aq))^(-)#

An iodide ion will react with a hydrogen peroxide to produce water and a hypoiodite ion, #OI^(-)#. 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 #color(blue)((1))# and #color(blue)((2))# together will get the overall reaction

#2H_2O_(2(aq)) + cancel(I_((aq))^(-)) + cancel(OI_((aq))^(-)) -> cancel(OI_((aq))^(-)) + 2H_2O_((l))+ O_(2(g)) + cancel(I_((aq))^(-))#

In this case, the activation energy for the potassium iodide-catalyzed reaction is #"56 kJ/mol"#. The activation energy of the uncatalyzed reaction is #"75 kJ/mol"#.

Lower activation energy #-># faster reaction.