Effect of Hydrogen Peroxide Concentration on Decomposition

✅ Paper Type: Free Essay	✅ Subject: Chemistry
✅ Wordcount: 9516 words	✅ Published: 23rd Sep 2019

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My research question:

How does the change in concentration of hydrogen peroxide affect the initial rate of decomposition of hydrogen peroxide, catalysed by catalase from yeast?

1. Introduction and Background

Enzymes are proteins, therefore are a long chain of amino acids that took a specific three- dimensional shape. In the three- dimensional shape of the enzyme we can find an area, called the “activate site”, that is designed to match with a specific molecule which is known as the “enzyme’s substrate”. This procedure will allow enzyme to catalyse chemical reactions. In addiction enzymes will function as a catalyst several times, which means that it will speed up reactions without getting used up. Enzymes will provide at the same time a reduction of the activation energy, which is the energy needed for the reaction to occur. The chemicals produced from this reaction are called “products”. Furthermore, enzymes, as proteins, will have a denaturation point which is above about 45°C. A common analogy to describe the enzyme-substrate activity is the “lock and key model”. In this analogy the lock is the enzyme and the key is the substrate, because the complex three dimensional shape of the lock is specific, just one key will fit in it. Enzymes work in the same way: they are specific to each other.

Image 1: Lock and key model

In the human body there is an indirect production in the thyroid gland, gut and lungs of hydrogen peroxide. For this reason, enzymes are required to catalase the decomposition of it. Catalase is a common enzyme found in the majority of the organisms. For example, catalase in found in the liver, and it helps to breakdown the hydrogen peroxide. Hydrogen peroxide (H2O2) is a chemical used in our immune system to fight against diseases and pathogens but at the same time it’s a chemical highly toxic and can’t remain in the human body. For this reason, it gets decomposed by the catabolic process to not lead any damage to our body.

2H₂O₂  O₂ + 2H₂O

The experiment chosen was for me a great opportunity to improve my knowledge regards how catabolism works in our human body. For this reason, I was captivated by the idea of how a change of how a change in the amount of catalyse could affect this chemical process. Furthermore, I found this experiment extremely interesting because I considered to study Biochemistry at university in one-year time.

2. Pilot experiment

For my plot experiment I set up my equipment as shown in the picture above. My necessity was to stablish which source of catalase was the more appropriate, therefore I decided to try the experiment with random amounts of catalase (potatoes, yeast or leaver) to roughly comprehend how the ultimate result of my final experiment should look like.

I started with potatoes, but since the beginning of the plot experiment I realized how hard is to cut potatoes into similar pieces in order to have a similar surface area and weight to each other. Another problem I found with potatoes (1.5o g) was the amount of time needed to see an effective reaction in 20V pure hydrogen peroxide concentration.

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For this reason, I decided to try another source of catalase: liver. As with potatoes I didn’t know a specific amount of liver to use for the experiment, therefore I randomly chose 1.00 g of it. Because the previous experiment was slow in reacting I selected 20V pure Hydrogen peroxide concentration to put the liver into. Even though I hurried to close the flask and measure the reaction in the smallest amount of time I could, by I the time I was ready to start collecting the measurements the reaction was already over. In 20 seconds, the amount of O₂ in the measuring cylinder was already up to 15 mm. My last source of catalase available was yeast, and after the pervious unsuccessful tries I was a bit concerned to not be able to find the right source of catalase to set up my experiment. For this trial I used 0.30 g of year and again 20V pure hydrogen peroxide concentration. In 60 second the amount of O₂ in the measuring cylinder was up to 1 cm and in 120 second it was up to 4cm.

After the plot experiment I decided to use yeast as my source for catalase in the reaction because I found it a good balance between liver and potatoes for the time needed to react with the hydrogen peroxide. Additionally, I decided to change every time the volume of hydrogen peroxide contained in the final concentration (always 20V) by mixing it with a specific concentration of water each time.

The amount of oxygen produced was measured within 6 minutes every 60 seconds, for each experiment.

3. Health and Safety issues

The substrate chosen for my experiment (hydrogen peroxide) is a toxic chemical, therefore I had to follow several precautions to avoid any possible danger. For this reason, I had to wear safety glasses and gloves to prevent skin burns or blurred vision. I also needed to set my experiment in a ventilate space to prevent a possible inhalation of the chemical and consequently a possible shortness of breath.

4. Experiment and Method

– I decided to use yeast as my source for catalase for the reasons previously explained (please look at page 5/ section 2. Plot experiment). Additionally, I decided to change the volume of hydrogen peroxide contained in the final concentration (always 20V) by mixing it with a specific concentration of water ( V) each time. – The different volumes of hydrogen peroxide contained in the final concentration are: 0V of hydrogen peroxide + 20V of water; 4V of hydrogen peroxide + 16V of water; 8V of hydrogen peroxide + 12V of water; 12V of hydrogen peroxide + 8V of water; 16V of hydrogen peroxide + 4V of water; 20V of hydrogen peroxide + 0V of water. – I got those concentrations by measuring the volume of the two substances with a measuring cylinder and by mixing them up in a conical flask. – With the help of a measuring cylinder I had the possibility to measure the amount of O₂ produced by the chemical reaction in mm – The amount of oxygen produced was measured within 6 minutes every 60 seconds.

My hypothesis for my experiment is: The quantity of O₂ produced by decomposition oh hydrogen peroxide by catalase will be influenced by a change in concentration of hydrogen peroxide in the solvent of the reaction.

Selective variables:

Independent variable

Volume of H₂0₂ (V) in the solvent of the reaction

Dependent variable

The quantity of O₂ formed by the decomposition of hydrogen peroxide in the reaction

Control variables

- Concentration of catalase: same mass of yeast used for each experiment

- The temperature of the room in which I based my experiment should always be stable (22°C ±1°C)

4. Equipment

• A balance to measure the mass of yeast • Conical Flasks 100cc • Hydrogen peroxide • Water • Measuring cylinders 10cc • boss and clamp • thermometer for room temperature and for the solvent of the reaction

4. Method of the experiment

– I organized my working place and my equipment (please look at point 4. EQUIPMENT) – I diluted in a conical flask a specific amount of hydrogen peroxide (V) with different volumes of water (V) for each investigation. The overall volume of the solvent was always 20V (please look at point 5. Quantitative and Qualitative data) – I measured 0.30 g of yeast with the help of a balance – Then I placed, with the help of the boss and clamp, the measuring cylinder in the bath. This will help me to measure the volume of oxygen produced during the reaction. -Therefore, I connect the bung of the conical flask with the delivery tube of the measuring cylinder -After I open again the bung and drop in the previously weight 0.30 g of yeast -Quickly I close the bung of the conical flask -The reaction starts and therefore I begin to measure the oxygen produced every 60 seconds within 6 minutes with the help of a timer – I repeated this method for the following volume of H₂0₂ (V): 0V; 4V; 8V; 12V; 16V; 20V.

5.Quantitative and Qualitative data

The results of the experiment:

Volume of H₂0₂ (V) (± 0.5V)

Volume of H₂O

(V) (± 0.5V)

Final H₂0₂

Concentration (V) (± 0.5V)

Volume of 0₂ (cm3) (±0.5cm³) by 0.30g of yeast (catalase) after time (s) (±1s)

60s

120s

180s

240s

300s

360s

0.00

1.00

1.50

1.25

0.00

0.50

1.00

1.50

1.25

2.00

3.00

2.50

4.00

5.00

4.50

6.00

6.50

6.25

0.00

1.00

0.50

1.25

2.00

3.50

6.00

8.00

9.00

8.50

0.00

0.50

2.50

2.00

2.25

5.00

4.50

4.75

7.00

6.50

6.75

9.00

11.0

12.0

11.5

0.00

1.00

3.00

4.00

3.50

5.00

6.00

5.50

7.00

8.00

7.50

9.00

10.0

9.50

11.0

12.0

11.5

Calculating the mean:

Mean=

Volume of H₂0₂ (V) (± 0.5V)	Volume of H₂O (V) (± 0.5V)	Final H₂0₂ Concentration (V) (± 0.5V)	Volume of 0₂ (cm3) (±0.5cm³) by 0.30g of yeast (catalase) after time (s) (±1s)
			0s	60s	120s	180s	240s	300s	360s
0	20	0	0.00	0.00	0.00	0.00	0.00	0.00	0.00
4	16	4	0.00	0.00	0.00	0.00	0.00	0.00	1.25
8	12	8	0.00	0.00	0.50	1.25	2.50	4.50	6.25
12	8	12	0.00	0.00	0.92	2.00	3.50	6.00	8.50
16	4	16	0.00	0.50	2.25	4.75	6.75	9.00	11.5
20	0	20	0.00	1.00	3.50	5.50	7.50	9.50	11.5

My observations:

The reaction has started in the exact time yeast touched the hydrogen peroxide. Therefore, when I putted the yeast into the flask with just water the reaction was not occurring. When I putted the yeast in the solvent with the smallest Volume of H₂0₂ (V) (± 0.5V) the amount of oxygen produced was tiny, even after 360 seconds. That is because the hydrogen peroxide was too less to produce a big quantity of oxygen in a small slot of time. For this reason, the more of H₂0₂ (V) (± 0.5V) was contained in the solute the more likely a reaction was able to produce a progressive amount of oxygen.

6. Gradients

I represented the data collected from the experiment with a graph and therefore included gradients for each change in Volume of H₂0₂ (V). this will show a connection between the change in Volume of H₂0₂ (V) and the quantity od O₂ produced per second

7. Rate of reaction

Calculation of the rate of reaction:

Volume of H₂0₂ (V) (± 0.5V)

Calculation of the gradient of the tangent

Error calculation

Calculation of the uncertainty

Initial rate of reaction (cm of O₂/sec)

Gradient= $\frac{0.00}{120}$

= 0.00

$\frac{\pm 0.5}{\pm 1}$

$\frac{0.00 \pm 0.5}{120 \pm 1}$

= 0.00 $\pm$

0.00

0.00 $\pm$

0.00

Gradient= $\frac{0.00}{120}$

= 0.00

$\frac{\pm 0.5}{\pm 1}$

$\frac{0.00 \pm 0.5}{120 \pm 1}$

= 0.00 $\pm$

0.005

0.00 $\pm$

0.005

Gradient= $\frac{0.50}{120}$

= 0.004

$\frac{\pm 0.5}{\pm 1}$

$\frac{0.50 \pm 0.5}{120 \pm 1}$

= 0.004 $\pm$

0.03

0.004 $\pm$

0.03

Gradient= $\frac{0.92}{120}$

= 0.008

$\frac{\pm 0.5}{\pm 1}$

$\frac{0.92 \pm 0.5}{120 \pm 1}$

= 0.008 $\pm$

0.03

0.008 $\pm$

0.03

Gradient= $\frac{2.25}{120}$

= 0.019

$\frac{\pm 0.5}{\pm 1}$

$\frac{2.25 \pm 0.5}{120 \pm 1}$

= 0.019 $\pm$

0.05

0.019 $\pm$

0.05

Gradient= $\frac{3.50}{120}$

= 0.030

$\frac{\pm 0.5}{\pm 1}$

$\frac{3.50 \pm 0.5}{120 \pm 1}$

= 0.030 $\pm$

0.05

0.030 $\pm$

0.05

In order to visualise the data, I created a graph that represent the initial rates of reaction for experiments with different volume of H₂O₂ (V).

8. Conclusion and Evaluation

Conclusion:

This experiment gave me the necessary data to come to a conclusion to my answer question “How does the change in concentration of hydrogen peroxide affect the initial rate of decomposition of hydrogen peroxide, catalysed by catalase from yeast?”: the higher the Volume of H₂0₂ (V) is in the solute the quicker will be the rate of decomposition of hydrogen peroxide, consequently we would have a progressive amount of O₂ produced in a small amount of time. However, it was interesting to ascertain that the catalase of yeast gets affected by the concentration of H₂O₂ until it reaches a volume of around 16(V). When the concentration is higher than 16(V) the change in Volume of 0₂ (cm3) in barely visible after the first 120s. This can be shown by the graph containing the gradients (please look at point 6. GRADIENTS). This anomaly may be explained either as an error in collecting data or as the closure of the enzyme’s active sites to being saturated with H₂O₂. In this situation the active sites of the enzymes are occupied, therefore active sites are defined saturated with H₂O₂. this in fact will decrease the rate of reaction of the experiment. All of that information can support my hypothesis that the rate of reaction increases as the volume of H₂O₂ contained in the solute of the reaction increases.

Evaluation:

I found the experiment that measure how the change in concentration of hydrogen peroxide affects the initial rate of decomposition of hydrogen peroxide very fascinating but also challenging for the
accuracy and precision needed. Keeping the room in which, the experiment was based, and the water, used in the solvent of the experiment, with a stable temperature was difficult. Furthermore, I found hard to understand after how many seconds, after the start of the reaction, I had to collect my data regards the amount of O₂ produced. In fact, collecting my data every 60s allowed me to collect them with more accuracy, because I had more time to write down my results. I also understood that taking a photo of the measuring cylinder every 60s was providing me a quicker and more efficient way to collect way results with a small error, because I had no rush to write them down. However, I had small issues with my source of catalase, in fact yeast it’s hard to measure precisely, especially in such a small amount (0.30 g) and also, I didn’t take in consideration that a small quantity of it was just attaching to the internal glass of the conical flask and not touching the solvent. I could have avoided it by inserting the yeast with a laboratory spoon in the conical flask, but un unfortunately my time between the yeast would have touch the solvent and the start of the reaction was limited. Also, I could have decided to measure the reaction for more that 360s, this would have provided a more accurate result. However, I measured the amount of water and hydrogen peroxide in the solvent quite well, it spent a long time in order to do so but it was not particularly difficult. Furthermore, the setting up of the equipment worked without any problematic after the first trial with the pilot experiment. I definitely think there are few factors in the experiment that I should have taken in account, even though I believe their effect on the final result is barely noticeable and therefore I think my result are still reliable and accurate.

9. Sources

( 1) https://saylordotorg.github.io/text_the-basics-of-general-organic-and-biological-chemistry/s21-06-enzyme-action.html

(2) https://www.focuseducational.com/product/decomposition-hydrogen-peroxide-by-catalase/122