# Why does the rate of reaction increase less than proportionally to the increase in mass of catalyst?

I conducted an experiment where I wanted to speed up the decomposition of hydrogen peroxide with the use of a catalyst (potassium iodide).

I used 5 different test tubes with 5ml of hydrogen peroxide in each, then added different amounts of the potassium iodide: 0.2g, 0.4g, 0.6g, 0.8g, 1.0g. Then I measured the time it took for the reaction to reach a temperature of 30°C. The results I got was that the more of the catalyst I added, the faster the rate of the reaction became. However, the increase in the reaction rate decreased when more of the catalyst I added, and I don't understand why.

I would think that since temperature and amount of hydrogen peroxide are constant variables and the only variable affecting the rate is the amount of potassium iodide, if I doubled the amount the speed would also be doubled. But why is that not the case? Or does it have something to do with errors in the method?

• Surface area of the catalyst is more important than it's mass. – bon Feb 1 '17 at 17:34
• Per @bon 's point, was the reaction stirred or was the KI allowed to settle and possibly not completely mix/dissolve? – airhuff Feb 1 '17 at 17:46
• Yes, the test tubes were shaken right after the KI was added – Rebecka Feb 1 '17 at 18:21
• @airhuff Doesn't matter as regards the scaling - even with perfect mixing, mass increases proportionately to volume, but geometric surface area increases to only the 2/3 power with volume and thus also with mass. – hBy2Py Feb 1 '17 at 18:31
• Don't think this is just surface area. // What concentration was hydrogen peroxide? – MaxW Feb 1 '17 at 19:46

There is a lot going on here. Part of the problem is poor experimental design. I'll just throw this out without evaluating all the energies. (Too lazy...)

(1) The potassium iodide has to be dissolved into the solution since you added it as a solid. This means there is an energy change and this could be endothermic or exothermic. Adding more salt increases the effect obviously.

(2) It takes some time to dissolve the potassium iodide. So before all the potassium iodide is dissolve some is already reacting with the hydrogen peroxide.

Greater surface area (smaller particles of potassium iodide) would allow the crystals to dissolve faster.

(3) The reactions are:
$\ce{H2O2(aq) + I-(aq) = H2O(l) + OI-(aq)}$
$\ce{H2O2(aq) + OI-(aq)= H2O(l) + O2(g) + I-(aq)}$

Overall the reaction is:
$\ce{2H2O2(aq) = 2H2O(l) + O2(g) + \text{heat}}$

But the heat is the combination of the two reactions.

If $\ce{[H2O2](aq) >> [I-](aq)}$ then both reactions will happen.

If $\ce{[H2O2](aq) << [I-](aq)}$ then only the first reaction will happen. The iodide isn't acting as a catalyst but as a reactant.

3.3 % Hydrogen peroxide is

$0.033 \cdot 15 \approx 0.5$ g $\ce{H2O2}$

Moles $\ce{H2O2}$ = $\dfrac{0.5}{34} = 0.015 \text{moles} = 15 \text{millimoles (mM)}$

0.2 g KI = $\dfrac{0.2}{166}$ = 1.2 mM

1.0 g KI = 6.0 mM

So you're adding so much KI that some might get left as $\ce{OI-}$.

(4) so to do the experiment better:

• cut the KI concentration
• dissolve the KI in 5 ml of water & equilibrate temp to that of 3.3% H2O2
• mix solutions with stirring