To investigate one of the variables to show its effect on the rate of the reaction.

GCSE Science

Dhiraj Aery

GCSE Practical Assessment:

Biology Investigation

Aim:

To investigate one of these variables to show its effect on the rate of the following reaction:

Variables to choose between:

The concentration of the substrate molecule (H2O2)
The mass of the potato (indirect way of measuring the concentration of the enzyme)

The Reaction:

Hydrogen Peroxide Water + Oxygen

2H2O2 2H20 + O2

For this experiment I have decided to vary the concentration of the substrate molecule (H2O2)

Background information:

Hydrogen Peroxide is a toxic chemical, produced by many living organisms as a product of the biochemical reactions occurring in the cells. As hydrogen peroxide is poisonous it must be removed quickly. The enzyme Catalase speeds up the reaction that breaks down hydrogen peroxide into water and oxygen.

Catalase can be found in potato cells. When hydrogen peroxide is added to the potato as a liquid, the catalase in the potato breaks down the hydrogen peroxide into water and oxygen – two substances that are essential for our bodies. Therefore catalase is essential in our bodies to break down H2O2.

Prediction:

I predict that as the substrate concentration increases, the rate of reaction will go up at a directly proportional rate until the solution becomes saturated with the substrate hydrogen peroxide. When this saturation point is reached, then adding extra substrate will make no difference or vice versa, as the substrate concentration decreases the rate of the oxygen production will decrease as well.

Another hypothesis of mine is that if I use 2m of Hydrogen peroxide and compared the results to the results of 1m of hydrogen peroxide, then the 2m hydrogen peroxide will have double the rate of reaction that the 1m Hydrogen peroxide has.

This is based on that as the rate steadily increases when more substrate is added because more of the active sites of the enzyme are being used which results in more reactions so the required amount of oxygen is made more quickly. Once the amount of substrate molecules added exceeds the number of active sites available then the rate of reaction will no longer go up. This is because the maximum number of reactions are being done at once so any extra substrate molecules have to wait until some of the active sites become available.

Apparatus:

Equipment List:

Grated potato (remove skin before grating)

Hydrogen Peroxide (different molarities - 0.5m, 1m, 1.5m, 2m, 2.5m)

2 Measuring cylinders (one to measure how much acid to use and the other to collect the carbon dioxide being given off)

Delivery tube and bung

Conical flask

Set top balance (to weigh the grated potato)

Stopwatch

A water basin half full of water

Clamp stand (to hold the measuring cylinder in place)

White tile (to cut potato on)

Trial Method:

This is a preview of the whole essay

Delivery tube and bung

Conical flask

Set top balance (to weigh the grated potato)

Stopwatch

A water basin half full of water

Clamp stand (to hold the measuring cylinder in place)

White tile (to cut potato on)

Trial Method:

For my trial method I decided to use 0.5 molar of Hydrogen Peroxide and used 20ml of it. I also used 3 grams of grated potato. I decided to use the weakest molarity (0.5m) so I could see how slow this reaction would go as it is the weakest molarity, because I want to have the experiment completed under ten minutes so I can do repeats of the experiment to see how much oxygen is being collected. I timed the reaction every 30 seconds for 10 minutes. I set up the apparatus as above.

I also decided to use the strongest molarity as well (2.5m) along with the same amount of Hydrogen Peroxide, so I could see if the reaction would be too quick for me to take down results. Also to see if oxygen displaces the total amount of water in the measuring cylinder even though the reaction is still going on. If this did happen then experiment would be a failure as it is only half completed.

Trial Results: Amount of oxygen collected

Method:

First of all grate a potato up and weigh the grated substance, as closest to 3 grams as possible.

Then measure out 25 ml of 0.5 molar Hydrogen Peroxide. I changed the value of the hydrochloric acid from the trial method because the reaction was happening too slowly.
Fill a water basin half way up with water; also fill the measuring cylinder (with which you will be collecting oxygen) with water, all the way to the top.
Pour the hydrochloric acid into the conical flask and attach the delivery tube and bung to it.
Turn the measuring cylinder upside down into the water basin; making sure that no air is in the measuring cylinder.
Get the delivery tube from the conical flask and place it in the measuring cylinder in the water basin.
Place the grated potato inside the conical flask, this will start the reaction. Start the stopwatch. Place the bung on top of the conical flask so the oxygen isn’t released into the surroundings.
The measuring cylinder will start to bubble and start losing water, (this is the oxygen displacing the air) record the readings and put them into a results table.

I will repeat the reactions 3 times for each concentration so I can get a fairly good average.

The variable I will be changing is the Hydrogen Peroxide concentration. The range of the variable is 0.5molar to 2.5 molar concentration of Hydrogen Peroxide. I will be using the same apparatus for all experiments so I can be sure that my results are fairly accurate.

Variables I need to keep constant:

Enzyme Concentration – Provided there is an excess substrate, an increase in enzyme concentration will lead to a corresponding increase in rate of reaction. Where the substrate is in short supply (i.e. it is limiting) an increase in enzyme concentration has no effect.

Temperature – As temperature increases, molecules move faster as they gain more kinetic energy. In an enzyme catalysed reaction, such as the decomposition of hydrogen peroxide, this increases the rate at which the enzyme and substrate molecules meet and form enzyme-substrate complexes therefore the rate at which the products are formed. All enzymes have an optimum temperature at which they function at their fastest rate. As the temperature continues to rise, however, the hydrogen and ionic bonds, which hold the enzyme molecules in shape, are broken. If the molecular structure is disrupted, the enzyme ceases to function as the active site no longer accommodates the substrate, the enzyme is denatured. So I will be working in room temperature so the enzyme will have the same temperature for all experiments.

pH – Any change in pH affects the ionic and hydrogen bonding in an enzyme and so alters it shape. Each enzyme has an optimum pH at which its active site best fits the substrate. Variation either side of pH results in denaturising of the enzyme and a slower rate of reaction.

Safe Test:

Keeping a safe test is also very important for the safety of me and other people around me. To ensure that my experiment is safe, I will carry out the following precautions:

Goggles should be warn to protect my eyes from the hydrogen peroxide or in case another accident occurred.
Be careful when cutting and grating the potato. Try not to be distracted whilst doing this or you could end up cutting yourself.
I should act in a sensible and safe manner, carefully working with the equipment and being aware of any dangers.

Results:

These are the results I have collected from my experiments:

0.5m Concentration-Amount of Oxygen collected:

1.0m Concentration-Amount of Oxygen collected:

1.5m Concentration-Amount of Oxygen collected:

2.0m Concentration-Amount of Oxygen collected:

2.5m Concentration-Amount of Oxygen collected:

I will now draw a graph to show the average amount of oxygen collected for each concentration against time. See next page for graph.

Conclusion:

After this experiment, looking at the results you can see there is a definite evidence that as you increase the concentration of the hydrogen peroxide, the faster the rate of the reaction. This is because when the hydrogen peroxide is more concentrated the enzyme active sites are more likely to come into contact with the hydrogen peroxide particles. This is true up to a point until the amount of substrate molecules exceeds the number of active sites available; this is why the amount of oxygen decreases towards the end. This is also what I had predicted.

Therefore there will be more collisions per second, so there will be a faster rate of reaction. Also every time a collision takes place, a reaction happens and so a gas is produced, therefore the more collisions the more gas produced. So the faster the rate of reaction and the more concentrated the solution the higher the volume of gas produced will be. As the concentration doubles the rate should double and the amount of product should double, as there will be twice as many molecules present, so collisions will twice as much more regular.

As my graph proves which shows that the 2m hydrogen peroxide rate of reaction is roughly twice as fast as the 1m hydrogen peroxide rate of reaction. Which shows my prediction to be correct.

My conclusion completely matches up with my prediction as I predicted that there would be a higher volume of gas produced the higher the concentration, in my graph showing all of the concentrations, it can be seen that less gas is produced when the concentration is lower. Such as the highest concentration produced 193 cm³ of gas whereas the smallest concentration of acid produced 52.5 cm³ of gas, this is a clear difference that as the concentration decreases so does the volume of gas produced, this is true throughout the experiment.

Evaluation:

I believe the method I used was adequate enough to gain readings accurate enough for this work, although to gain more accurate results I should have taken readings on another mass of potato and should have taken my measurements for five minutes instead of three.
As I wasn’t trusted to use a gas syringe instead of an inverted measuring cylinder my readings weren’t completely accurate as the measuring cylinder was only graded in steps of 5cm³, therefore I had to estimate to the nearest whole number as to what the reading was. The other equipment that I used was otherwise accurate enough as the mass balance was accurate to two decimal places and the measuring cylinder I used to measure out the hydrogen peroxide, as I have said was graded in steps of 5cm³ therefore as I was measuring out 20cm³ that was accurate enough.
A slight experimental error was that there was a small time between the hydrogen peroxide being introduced into the conical flask and the bung creating a seal to the delivery tube, although I don’t believe that enough oxygen could escape in that time to make a real difference to my results. A possible larger error was that parts of the skin of the potato were used in the experiment; this is not good as the skin of the potato doesn’t contain any of the enzyme catalase, and therefore could have affected my results significantly, finally another error could have been that as I had grated the potato not all the subsequent pieces of potato would have been the same size, therefore would have a different surface area and therefore would allow differing amounts of the enzyme to come into contact with the hydrogen peroxide.
To improve the accuracy of the investigation I could use more precise apparatus, also I could take readings of the experiment for 15 minutes or longer instead of just 10 minutes,

To extend on the investigation I could investigate if the surface area of the potato affects the rate of reaction, this would show me if the variable I didn’t control had a drastic effect on my results.