Chemistry Investigation

Chemistry Investigation

An investigation to find the optimum temperature that will produce the

maximum amount of oxygen from hydrogen peroxide using the enzyme

catalase

Introduction

Hydrogen peroxide is poisonous to all living things if it builds up in

the cells but when it is introduced to the enzyme catalase the react

between the two makes water and oxygen as byproducts. There are many

factors witch affect the production of oxygen these are listed below.

For this experiment the enzyme will be held in potato.

Prediction

I predict that the enzyme catalase will have a temperature at which it

will work best this is called its optimum but this could be anything,

I have chosen temperatures at a ten degrees difference starting at ten

going up to eighty degrees. If the temperature gets too low the

reaction will slow down until the reaction stops this is because of

collision theory this tells us that the more energy a partial has the

fast it goes and there for there is more chance of a successful

collision. At a higher temperature the reaction will speed up but the

enzymes active site will change shape so the enzyme and the partial do

not fit in just right like this

[IMAGE][IMAGE][IMAGE]The enzyme at optimum: just above:

far above

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and eventually the enzymes will denature. To find out the best

conditions I need to consider more then just temperature. These are

possible factors that determine how much oxygen is given off.

Independent Variables

1) The amount of enzyme used (fixed)

2) The concentration of enzyme in the potato (fixed)

3) The surface area of the potato (fixed)

4) Temperature (experimental)

5) Amount of hydrogen peroxide (fixed)

6) The concentration of the hydrogen peroxide (fixed)

7) Freshness of the vegetables (fixed)

8) The pH (fixed)

The amount of enzyme used

I must use the same amount of enzyme each time I do an experiment so I