HOW DOES THE CONCENTRATION OF HYDROGEN PEROXIDE AFFECT THE TIME TAKEN, IN SECONDS, TO PRODUCE 50 CM3 OF OXYGEN WHEN ADDED TO 5G OF CATALASE?

Benjamin Atkins
Background information.

Our body produces Hydrogen Peroxide (H2O2) as a by-product of respiration. It is produced in every single one of our living cells, due to its harmful nature Hydrogen Peroxide must be removed as soon as it is produced in the cell. Therefore, cells produce the enzyme catalase to remove the harmful substance. 1

I hypothesise that the higher the concentration of the Hydrogen Peroxide the less time it would take to produce 50 cm3 when I add the Hydrogen Peroxide to 5g of Catalase. This is because the higher the concentration the more molecules of Hydrogen Peroxide are present and therefore more collisions between the catalase enzyme and the Hydrogen Peroxide molecules will occur causing a faster reaction rate.

Apparatus List

- Conical Flask (100cm3)
- Stopwatch – To the nearest second
- Clamp
- Boss
- Electronic Balance – To the nearest milligram
- Measuring Cylinder (25cm3)– To the nearest 0.5cm3
- Gas Syringe (100cm3) – To the nearest 0.5 cm3
- 20 cm3 Hydrogen Peroxide (20 vol)
- Distilled Water
- Pureed Potato (5g)
- Sieve

I chose a gas syringe, instead of placing a measuring cylinder in water, because there was a smaller margin for error and so made the results more precise as there was no chance of oxygen being lost. Also from a preliminary experiment I found that 5g of potato puree was the optimum amount because it causes a good trend to be found.

Hydrogen Peroxide is found in bleaches and so is harmful to the skin and eyes, therefore, throughout the experiment, Lab Coats and Safety Goggles were worn.
Catalase can be found in pureed potatoes and so for the experiment we used
When disposing of the Hydrogen Peroxide we ensured that it all went down the sink and didn’t harm any plants or animals.

Variables

The Independent Variable for this investigation was the Concentration of Hydrogen Peroxide. We were given a concentration of 20 vol and so had to use the distilled water and the measuring cylinder to make different concentrations. The chosen concentrations were: measured in vol and were 10, 12.5, 15, 17.5, and 20.

The Dependant Variable was the time it took to produce 50 cm3 of Oxygen. The stopwatch was started as soon as the Hydrogen Peroxide was added and stopped as soon as the gas syringe was at 50 cm3.

There were a series of controlled variables to ensure that the validity of the experiment was kept, these included:
The volume of Hydrogen Peroxide (20cm3) will be measured in the measuring cylinder before added to the potato puree. If the volume was different then there would be more Hydrogen Peroxide molecules in some experiments and the experiments with a higher volume would produce 50cm3 of Oxygen in less time.

The mass of potato puree (5g) – will measure on the electronic balance before I add the Hydrogen Peroxide. If there was a higher mass of potato puree there would be more of the enzyme catalase to react with the Hydrogen Peroxide and the rate of reaction would increase. However, if there was a smaller mass of potato puree there would be less of the enzyme catalase to react with the Hydrogen Peroxide and so the rate of reaction would be lower.

The surface area of the conical flask – the size of the flask (100cm3) With a higher surface area there would be more opportunity for the Hydrogen Peroxide molecules to collide with the catalase enzyme and so the reaction would be quicker. However, if there was a smaller surface area there would be less space for the molecules to collide and so the rate of reaction would decrease.

The temperature of the experiment (room temperature) With a higher temperature the molecules would have more kinetic energy and so would be moving faster and therefore would collide more often increasing the rate of reaction. However, if the