How does an ice bath change the rate of an reaction?

Question

If I had hydrogen peroxide decompose with the help of manganese dioxide and measured the time taken for the reaction to complete and the volume of oxygen collected. I will get a graph that looks like this:

Now what will happen to the shape of the graph if I placed the reaction vessel in ice?

I thought the following. The graph will rise but at a less rate and less steep than in the graph above. Then it will get to the point where the reaction ceases and doesn't go to completion because there are no more collisions.

Answer 1

Placing the reaction vessel in ice will lower the temperature of the reaction and thus slow down the reaction. Temperature is a measure of the average kinetic energy of the molecules, so lowering the temperature lowers their kinetic energy. In order for reactants to form products, they must collide with a certain amount of energy. Thus, at a lower temperature it will take longer for the same amount of product to be formed.

You also need to keep in mind that at lower temperatures the equilibrium constant will change. Since this is an exothermic reaction, at lower temperatures there will be more product formed (Le Chatelier's principle). There will be more moles of oxygen present in the iced reaction. However, you're measuring the volume of oxygen, and a colder gas will have a lower volume (for the same number of moles of gas). So you've got two opposing trends: the ice will cause more oxygen to be produced (which would normally correlate with a larger volume) but the colder temperature will also cause the gas to shrink. Without more information, such as the temperature of the two reactions, I don't think you can determine whether you will wind up with a larger or smaller volume, so it would be hard to determine the shape of the graph for the colder reaction.