The Effect of Changing the Concentration of the Enzyme Catalyst on the Rate of Reaction on Hydrogen Peroxide

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The Effect of Changing the Concentration of the Enzyme Catalyst on the Rate of Reaction on Hydrogen Peroxide


Planning
--------

Hydrogen Peroxide Water + Oxygen

2H2O2 2H2O + O2


Introduction - Background Information
-------------------------------------

To help me understand what a chemical reaction is or involves I did
some research.

A chemical reaction is a process by which atoms or groups of atoms are
redistributed, resulting in a change in the molecular composition of
substances.

Enzymes are a biological catalyst, which controls a cellular reaction,
they are proteins that act as a catalyst. A catalyst is a substance
that speeds up a reaction but does not get used up. It works by
reducing the Activation Energy, which is the minimum energy needed for
a reaction to happen. A catalyst can make a reaction occur even if it
would not happen other wise. Enzymes only affect the speed at which a
product is formed, not how much is produced. Enzymes are specific,
this means that once an enzyme has acted on one substance it will not
act on a different one.

There are two reactions which involve enzymes they are anabolic and
catabolic. An anabolic reaction is a build up of smaller molecules
into larger molecules. A catabolic reaction is the speeding up of the
reaction time but is the breaking down of a substance. I will be
focusing on the catabolic reaction. To explain the reaction there is a
method called the lock and key theory:

The Collision Theory - this is when reacting molecules collide with
each other with enough energy to react. Surface area, temperature,
concentration and the use of an enzyme affect the collision theory.

All of these factors increase the rate of reaction if you increase
them. If there is a larger surface area then the molecules are exposed
and are more available to react as there are more of them. This means
there will be more collisions so a faster reaction.

If you increase the temperature this will make the molecules move more

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quickly, so they will collide more often with greater energy, so there
will be more successful collisions, therefore increasing the rate of
reaction.

If you increase the concentration the molecules will be closer
together therefore colliding with each other more often. This will
mean an increase in the rate of reaction.

A catalyst speeds up the rate of reaction, but does not get used up.

Factors that affect how well an enzyme works

o The temperature.

o The pH of an enzyme.

o The concentration/amount/volume of the enzyme.

o The concentration/amount/volume of the substrate.

The pH and temperature of which an enzyme works is called its optimum.

If the temperature increases it will increase the rate of which the
enzyme works therefore increasing the rate of reaction. If the
temperature decreases it will slow down the rate of which the enzyme
works therefore decreasing the rate of reaction. This is because if
you increase the temperature the molecules will move quickly with
greater energy causing more collisions and therefore increasing the
rate of reaction.

Most changes in the pH of an enzyme are reversible. An enzyme that was
working at a low pH will resume its normal activity once it starts at
its optimum activity again. However if there is a very high pH then
the enzyme may become irreversible.

If the amount/ volume of an enzyme is high then it will increase the
rate of reaction. This is because if there is a greater surface area
of an enzyme then there are more molecules exposed and available for
collisions. This will increase the amount of collisions therefore
increasing the rate of reaction.

If there is a high concentration/amount/volume of the substrate the
rate of reaction will increase. This is because if there is a greater
concentration in the same area then there will be a greater amount of
molecules resulting in more collisions. This will increase the rate of
reaction.

Variables

Input variable - Concentration of enzyme, the amount of food; liver,

potato, pea.

Output variable - Volume of gas produced in cylinder. The rate of how

much oxygen is produced.


Controlled variables

o Temperature - to control this variable I will keep a water bath at

room temperature which is 23°C.

o pH - to control the pH of the solution I will keep the concentration

the same throughout the experiment. A buffer of pH7 was

added to the H2O2 solution to keep the pH the same.

o Volume of H2O2 - to control this variable I will use a measuring

cylinder to measure out 20cm3 of H2O2

solution.

o Concentration of H2O2 - to keep this variable constant I will keep

the concentration the same throughout

the experiment. The concentration is 1.8

mol/dm3.

Prediction

If the surface area of the food doubles, then I will expect the rate
of reaction to double. So if I used a surface area of 2cm2 I would
expect to produce 10cm3 rather than 5cm3 if I was using a surface area
of 1cm2.

This is because if you double the amount of molecules in a certain
area then there will be twice as many collisions therefore the rate of
reaction will increase. This is the shape I predict my final graph to
look like.

Preliminary work

To determine what apparatus and food I should use I carried out
various experiments.

I tried two different methods of measuring the rate of oxygen
production. One method was to use a boiling tube, delivery tube and a
gas burette, and measure the amount of gas produced in the gas
burette.

Another method I did was to count the bubbles that were produced from
the reaction between the food and the Hydrogen Peroxide.

I also decided to use the gas burette with the more accurate readings
on it. This is to make my results accurate.

I used 20cm3 of Hydrogen Peroxide, as this seems a reasonable amount
to use.

I then tried three different foods:

1. Liver - this produced more than 19cm3 of oxygen in less than 10
seconds.

2. Pea -

Time (s)

Amount of oxygen produced

30

0.1

60

0.1

90

0.3

120

0.4

150

0.4

180

0.6

3. Potato -

Time (s)

Amount of oxygen produced

30

5.0

60

7.3

90

9.1

120

11.0

150

33.8

180

14.5


Main Experiment Method

I have decided to use the method, which includes the apparatus:
boiling tube, delivery tube and a gas burette. This is because they
are the most sensible and more accurate set of apparatus compared to
the method of counting the bubbles that were produced. It was very
inaccurate because the bubbles were produced to fast in some cases.

I will use the potato as the pea was too slow and the liver was too
fast too take accurate readings. I will use 1cm lengths of potato to
increase the surface area and increase the number each time. I will
use 20cm3 of Hydrogen Peroxide. I will take 7 readings at intervals of
30 seconds, this gives enough time to read the result and for the
reaction to take place. I will take two repeats of each number of 1cm
potatoes; this will make sure I will have some reliable readings. I
will be measuring the amount of oxygen produced. 1 will take all the
cylinders of potato from the same potato, as all potatoes are
different in concentration and enzymes etc. so I kept mine the same.

Method

1. Prepare a water bath for the test tube that will contain the
Hydrogen Peroxide and one for where the gas will be collected via
the delivery tube.

2. Measure out 20cm3 of Hydrogen Peroxide and pour it into the test
tube.

3. Fill up the gas burette with water and then place in the second
water bath without letting gas into the burette.

4. Measure 1cm of potato and then place into the test tube, and then
place the bung over the test tube.

5. Wait until the first bubble appears from the delivery tube into
the gas burette then start the stopwatch.

6. Time for three minutes recording the volume of gas produced every
30 seconds.

7. Rinse out the test tube and repeat steps 2 - 6 increasing the
number of potatoes in step 4 until you have used at least 5
different surface areas.

I will wear safety goggles throughout the experiment, as the Hydrogen
Peroxide is corrosive. The potato is a biological hazard as it is an
organism so I will wash my hands after the experiment. I should also
be careful, as I will be using glass apparatus, not to break any or to
cut myself.

Results Tables

Results for the surface area of 4.10cm2.

Temperature - 24°C.

Volume of oxygen produced (cm3)

Time (s)

First results

Repeat 1

Repeat 2 (23°C)

0

0.00

0.00

0.00

30

0.20

0.20

0.10

60

0.40

0.45

0.35

90

0.60

0.60

0.55

120

0.80

0.85

0.75

150

1.00

1.00

1.00

180

1.45

1.20

1.35

Results for the surface area of 8.11cm2.

Temperature - 24°C

Volume of oxygen produced (cm3)

Time (s)

First results

Repeat 1

Repeat 2

0

0.00

0.00

0.00

30

0.70

0.50

0.55

60

1.45

1.00

1.35

90

2.30

1.55

1.75

120

3.10

2.10

2.35

150

3.80

2.50

2.95

180

4.65

3.00

3.55

Results for the surface area of 12.21cm2

Temperature - 24°C

Volume of oxygen produced (cm3)

Time (s)

First results

Repeat 1

Repeat 2 (23°C)

0

0.00

0.00

0.00

30

0.80

1.20

1.10

60

1.60

2.35

1.95

90

2.55

3.60

2.70

120

3.70

4.90

3.80

150

4.60

6.25

4.85

180

5.50

7.30

5.90

Results for the surface area of 16.31cm2

Temperature - 24°C

Volume of oxygen produced (cm3)

Time (s)

First results

Repeat 1

Repeat 2

0

0.00

0.00

0.00

30

1.10

1.10

1.15

60

2.20

2.20

2.20

90

3.30

3.25

3.20

120

4.65

4.20

4.35

150

5.70

5.40

5.55

180

6.45

6.15

6.20

Results for the surface area of 20.41cm2

Temperature - 24°C

Volume of oxygen produced (cm3)

Time (s)

First results

Repeat 1

Repeat 2

0

0.00

0.00

0.00

30

2.50

2.30

2.60

60

3.70

3.75

3.70

90

6.65

5.80

6.15

120

7.40

7.50

7.45

150

8.90

9.05

9.20

180

10.60

10.70

10.60

Results for the surface area of 24.51cm2

Temperature - 24°C

Volume of oxygen produced (cm3)

Time (s)

First results

Repeat 1

Repeat 2

Repeat 3

0

0.00

0.00

0.00

0.00

30

2.70

2.50

1.40

2.60

60

5.00

4.95

3.85

5.20

90

7.55

7.40

4.20

7.35

120

9.95

10.10

5.70

10.00

150

12.00

11.95

6.80

12.65

180

14.10

13.50

7.90

13.75

For this particular experiment I wasn't happy with my second set so I
decided to take another set. I will ignore the incorrect set when I
come to work out the average rates.

Average rates

For surface area 4.10cm2 the average rate =

0.36 + 0.39 + 0.42

3

= 0.39 cm/s

For surface area 8.11cm2 the average rate =

1.00 + 1.20 + 1.50

3

= 1.23 cm/s

For surface area 12.21cm2 the average rate =

1.80 +1.90 + 2.40

3

= 2.03 cm/s

For the surface area 16.31cm2 the average rate =

2.05 + 2.15 + 2.30

3

= 2.17 cm/s

For the surface area 20.41cm2 the average rate =

3.70 + 4.00 + 4.20

3

=3.97 cm/s

For the surface area 24.51cm2 the average rate =

4.80 + 5.00 + 5.20

4

= 5.00 cm/s

Analysis and Conclusion

From my graph of the average rate against surface area, I can see that
it is directly proportional. On my predicted graph I drew a line that
finished with a curve. On my actual graph all my graphs have a
best-fit line. For the graph of surface area against the average rate,
the best-fit line does not pass through the origin. On my predicted
graph I predicted that the line would pass through the origin. As I
can see this does not happen. This is because if I made my line pass
through the origin then it would not be best fit. So I have drawn it
as close as I could.

From my first set of graphs I can see that as the surface area of the
potato increases the volume of oxygen also increases. I can see this
on my final graph more clearly as it shows the rate of reaction
against surface area. When the surface area is 5cm2 the average rate
is 0.45cm/s. When the surface area is 10cm2 the average rate is
1.6cm/s. This shows the rate of reaction of the potatoes does increase
with the surface area.

This is because of the collision theory. If there is a larger surface
area the molecules are more exposed and so there are more available to
react.

Evaluation

I feel that my experiment went well. I don't think that there were any
major problems with my method.

I have one anomalous result, which I have circled on my graph. This
could have been caused by

I feel my results are quite accurate. However the error bars are
fairly big. It is reasonably close to the best-fit line, although it
does not follow the same line. The best-fit line is steeper than the
predicted line. To work out how accurate my results are I can work out
the % error and % accuracy:-

% Error = Difference in rates x 100

Average rate

% Accuracy = 100- the % error


Errors within my experiment

The potato pieces were sticking together which results in a lower
surface area. I could not be exact in cutting the potato pieces
exactly the same size every time. Also I had to do the experiment over
a period of two days. The first day I did the experiment the
temperature was hotter than the second day. So on the first day the
reaction would have been quicker due to the collision theory.


Improvements that could be made

To reduce the risk of the potato pieces sticking together I could have
used a larger boiling tube. This would have meant that the potato
pieces were further away from each other. To make sure they were equal
sizes I could have used a cutter that was set to 1cm. To keep the
temperature the same I could allow a longer time to do my experiment
in, so I would finish it on the same day that I started it.


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