My aim is to show how substrate concentration affects the rate of reaction for an enzyme controlled reaction.

Equation:


Catalase

Hydrogen Peroxide (2H O ) Water (2H O ) + Oxygen (O )

Introduction:

Enzyme is a protein produced by a living organism that acts as a catalyst in a specific reaction by reducing the activation energy. A globular protein, enzymes have a specific three dimensional shape which is determined by the sequence of amino acids. The region which makes the enzyme specific and functional is known to be the active site. As result of this only the substrate which is complementary to the enzyme active site will be able to fit and form an enzyme-substrate complex.

ACTIVE SITE

SUBSTRATE

ENZYME

Enzyme + Substrate

SUBSTRATE FITS INTO ACTIVE SITE OF ENZYME

ENZYME

Enzyme-Substrate Complex

Lock and Key Theory:

This theory outlines the fact that an enzyme is specific for a substrate. This is shown by using an example of lock and key. This states that as a key is specific in shape it fits into only one lock in order to function. Hence the substrate will only fit into one active site provided by a specific enzyme. As result the shape of the substrate (key) exactly fits the active site of a specific enzyme (lock).

Induced Fit Theory:

This theory suggests that the enzyme doesn’t have to be specific for a complementary substrate as it is stated in the lock and key theory. This theory states that the enzyme is flexible hence has the ability to change its shape in order to allow it to fit the shape of the substrate. Also as the enzyme alters the shape, the enzyme puts on the substrate molecule as result of this it lowers the activation energy.

Activation Energy:

Activation energy is the energy that must be provided to make a reaction take place, enzymes reduce the activation energy required for a substrate to change into a product. This is because enzymes act as a biological catalyst as result of this there are more frequent collisions between particles due to greater kinetic energy hence faster moving particle. As result of more frequent collision there are more enzyme-substrate complex and product produced. So therefore less activation energy is required for a successful collision.

Activation energy without enzyme.

Free

Energy

Activation Energy with enzyme

Energy level of substrate

Lower

activation energy

Energy level

of the products

Time during reaction

Factors that affect enzyme activity:

The factors that affect the enzyme activity and rate of reaction are:

* Enzyme concentration

* Substrate concentration

* Temperature

* pH

* Inhibitors

Enzyme concentration:

As the enzyme concentration increases there will be more enzymes hence more active sites which can be occupied by the complementary substrate. Hence as result of this there will be a greater probability of an enzyme active site being occupied by the substrate. So therefore more products will be produced in a given time due to the fact that more enzymes lead to a faster rate of reaction. However once the substrate has occupied the active sites then excess enzyme concentration won’t make a difference as they are not going to be occupied as there are enough active sites already to be occupied. Hence the substrate will limit the rate of reaction and as result of this the rate of reaction will remain constant after the point when all substrate has occupied the enzyme active site.

Initial

Rate of

Reaction

Enzyme concentration

Substrate concentration:

As the substrate concentration increases the rate of reaction increases. This is because a greater number of substrate will mean that there are more chances for a substrate to collide successfully with an enzyme in order to form a product. As result of this there is a greater rate of reaction which means that more products formed in a given time. However this increase in rate of reaction is limited by the enzyme. This is because when all active sites of enzymes are occupied and there is excess substrate then it will not be able to fit in the active site, this point is called V-max. This is because an enzyme active site is available for one complementary substrate at a time. Hence after this point when all enzymes are occupied, the rate of reaction is constant.

Initial

Rate of

Reaction

Temperature:

As the temperature increases, the rate of reaction increases. This is because increase in temperature provides kinetic energy hence particles move faster. As result of faster moving particles there will be more frequent collisions between enzymes and substrate which will lead to form more products in a given time. Every 10�C increase in temperature doubles the rate of reaction. The optimum temperature is 40�C when the rate of reaction is at its peak. This is because greater heat energy provided will mean more energy gained by particles. As result of this particles will move faster due to greater kinetic energy. Hence faster moving particles will collide more frequently and more products formed in a given time.

However after 40�C the rate of reaction decreases. This is due to the fact that the hydrogen bonds which hold the enzyme 3-d structure are broken due to great heat energy provided hence the enzyme shape is distorted. As result of this the active site is not specific for complementary substrate hence no product formed. Therefore the rate of reaction will decrease if there is an increase in temperature after 40�C.

Initial

Rate of

Reaction

Temperature

pH:

pH is a measure of the concentration of hydrogen ions in a solution. So therefore if the pH decreases, then the concentration of hydrogen ions will increase. Different enzymes work best at particular pH but most work in a fairly neutral condition around pH7. There are also some enzymes which work in an acidic condition such as protease pepsin, which is found in the acidic conditions of the stomach. Enzymes as such have a different optimum pH as certain enzyme work best in specific conditions to carry out a function.

However a change to pH can alter the shape of the enzyme. As result of this the active site I distorted so no substrate can fit into the active site. So therefore pH change alters the shape of the enzyme resulting in the enzyme being denatured as hydrogen and ionic bonds broken. Hence rate of reaction will decrease as there is a change in pH.

Rate

of

Reaction

Inhibitors:

There are two types of inhibitors:

1. Competitive inhibitor

Competitive inhibitors are inhibitors which form the shape of the substrate that is complementary to the enzyme active site. Hence it binds to the enzyme active site. As result of this the substrate cannot bind to the active site. Therefore this slows the rate of reaction hence it lowers the amount of product produced in a given time. However competitor inhibitor is temporary as it binds to an active site for a period of time and then leaves it hence it is reversible.

Competitive inhibitor

Substrate Active site Competitive inhibitor

occupies active site Substrate

temporarily

ENZYME

ENZYME

2. Non- Competitive inhibitor

Non-competitive inhibitors don’t bind to the enzyme active site as competitive inhibitors do, but the do bind to the allosteric site (anywhere on an enzyme but not the active site) of the enzyme. This alters the shape of the enzyme hence distorting the active site, which means that the complementary substrate cannot bind to the active site. As result of this the product produced in a given time decreases therefore the rate of reaction decreases as result. These types of inhibitors can be permanent for some reaction but also can be temporary hence can be reversed.

Substrate

Substrate Active site distorted hence

Active site substrate cannot bind

ENZYME

Non-competitive inhibitor binds

Non- competitive inhibitor to allosteric site

Rate of

Reaction

From all the factors above, my aim is to concentrate on the affect of the substrate concentration on the rate of reaction for an enzyme controlled reaction.

Preliminary:

I carried out a preliminary experiment in order to know how to carry out this experiment. Also I did a preliminary in order to observe the errors which were made due to the equipment used, so I could change the equipment in order to get more accurate and reliable results. The preliminary helped me identify the areas in which errors were made and gave me a chance to improve the method to lower the number of errors. In the preliminary I carried out an experiment which could be used to measure the volume of gas produced, when different concentrations of hydrogen peroxide were catalysed by the enzyme catalase.

Errors and Improvements:

First of all, there was an error caused by the measuring cylinder which I used to collect the volume of oxygen. This is mainly because the scale division were as accurate as +/-1 cm� but I could read as accurate as +/-0.5 cm�. However this scale of +/-0.5 cm� was still too small to gain an accurate reading. This is because the larger the scale, the lower the percentage error.

For example at the temperature of 20�C after 30 seconds the volume of oxygen I collected was on average 0.85 cm� (see results preliminary results table). Hence the percentage error was:

Percentage error = Smallest division � 100

Measured quantity

Smallest division of the measuring cylinder = +/-0.5cm�, this is the scale to which I could read as accurate to.

Measured quantity is the volume of oxygen = 0.85 cm� on average collected.

Hence percentage error = 0.5 � 100

0.85

= 58.82352941%

= 58.8%

This percentage error of 58.8% is too high as the scale division was as accurate as to +/-1 cm�. Hence a huge percentage error suggests that the e measuring cylinder with a division scale of as accurate as +/-1 cm� is not the right equipment to use in order to collect oxygen, this mainly because the measuring cylinder will give a great percentage error resulting in an unreliable and inaccurate results.

So therefore I will be using a burette instead of a measuring cylinder, mainly because the scale division is larger. This is because, the larger the scale division, the lower the percentage error which will result in providing me with more accurate and reliable results. The burette has a scale division as accurate to +/-0.1 cm� hence the scale can be read as accurate to +/-0.05 cm�.

So if I use the same example which I used for measuring cylinder then the percentage error which will be given by using a burette will be:

Percentage error = Smallest division � 100

Measured quantity

Smallest division of the burette = +/-0.05cm�, this is the scale to which I will be able to read as accurate to.

Measured quantity is the volume of oxygen = 0.85 cm� on average collected.

Hence percentage error = 0.05 � 100

0.85

= 5.882352941%

= 5.88%

This shows that the percentage error of the burette is 10 times less than the percentage error of the measuring cylinder. This is because:

Percentage error of measuring cylinder = 58.82352941%

Percentage error of burette = 5.882352941%

Hence: the difference = 58.82352941% (Percentage error of measuring cylinder)

5.882352941% (Percentage error of burette)

= 10

Therefore the percentage error given by the burette will be 10 times less than the percentage error given by the measuring cylinder when collecting oxygen. As result of this I will be using a burette to collect the volume of oxygen as a burette will provide more accurate results.

Another error in the preliminary experiment was that the temperature. This was because the temperature wasn’t kept constant because if the initial temperature of water was 40�C then by the end of the experiment, the temperature will decrease gradually. This is because I did not use specific equipment which would have maintained the temperature, which would have allowed me to gain more accurate and reliable result. This is because temperature increases the rate of reaction increase due to the kinetic energy gained by the hydrogen peroxide molecules and catalase enzyme. As result of greater kinetic energy, the particles move faster and faster moving particles collide more frequently, resulting in greater product, water and oxygen in this case being produced in a given time. So therefore a gradual decrease in temperature will mean an unfair test as temperature was a variable that should be controlled as the rate of reaction is dependant on the temperature.

So therefore in order to overcome this error, I will be using a thermostatically controlled water bath. This will help me maintain a temperature and keeping it constant throughout the experiment. This very important as the rate of reaction is dependant on temperature hence I will have to make sure that temperature is controlled. So therefore in order to limit the error by maintaining temperature at constant I will be using a thermostatically controlled water bath.

Another error which I determined by doing a preliminary experiment was that the syringe was used to insert the hydrogen peroxide into the boiling tube. This left a small gap which allowed some of the volume of oxygen produced to escape. As result of this I didn’t get the actual volume of oxygen produced.

So therefore, in order to solve this problem I will be using plasticine around the syringe and delivery tube, this I can make sure that the top of the bung is air tight, and gas escaping is as minimum as possible. Hence by doing this the volume of gas escaping through the gap will be much lower, As result of this a more accurate reading of the actual volume of oxygen produced.

Preliminary Results:

Results at 20�C:

Time (seconds)

volume of oxygen collected (cm�)

volume of oxygen collected (cm�)

Average volume of oxygen collected (cm�)

Rate of reaction (Cm3/s)

0

0

0

0

0

30

0.8

0.9

0.85

0.0616

60

1.5

1.7

1.6

0.0266

90

2.4

2.4

2.4

0.0266

120

3.2

3.1

3.15

0.0262

150

4.1

3.9

4

0.0266

180

4.9

4.4

4.65

0.0258

Result at 30�C:

Time (seconds)

volume of oxygen collected (cm�)

volume of oxygen collected (cm�)

Average volume of oxygen collected (cm�)

Rate of reaction (Cm3/s)

0

0

0

0

0

30

1.1

1.0

1.05

0.035

60

2.0

1.9

1.95

0.0325

90

2.9

2.8

2.85

0.0316

120

3.8

3.6

3.7

0.0308

150

4.5

4.2

4.35

0.029

180

5.6

5.3

5.45

0.0303

Results at 40�C:

Time (seconds)

volume of oxygen collected (cm�)

volume of oxygen collected (cm�)

Average volume of oxygen collected (cm�)

Rate of reaction (Cm3/s)

0

0

N/A

0

0

30

9.5

N/A

9.5

0.4

60

12

N/A

12

0.241

90

14.5

N/A

14.5

0.188

120

17

N/A

17

0.158

150

19

N/A

19

0.14

180

21

N/A

21

0.131

For 20�C the line of regression: y = 0.026x + 0.0429

For 30�C the line of regression: y = 0.0294x + 0.1179

For 40�C the line of regression: y = 0.1036x + 3.9643

Prediction:

I predict that as the substrate concentration, the hydrogen peroxide concentration increases, the rate of reaction will also gradually increase. This is mainly because more hydrogen peroxide means that there is a greater chance of a substrate to collide with the enzyme, catalase which has the specific active site, which is complementary to hydrogen peroxide. Hence greater chance of frequent collision between catalase and hydrogen peroxide will mean that there will be more enzyme-substrate complex and more products produced in a given time. So therefore as hydrogen peroxide concentration increases, the affect on rate of reaction will be that it will also increase.

However I also predict that the rate of reaction will be constant at a point, known as V-max. This is because all active sites will be occupied hence excess hydrogen peroxide cannot bind to an active site of catalase. As result of this the enzyme, catalase will be the limiting factor resulting in a constant rate of reaction.

Lower concentration of Hydrogen Peroxide

Hydrogen Peroxide

Active site

CATALASE

The diagram above shows at a lower concentration there are a lower number of hydrogen peroxide molecules. As result of this there are less frequent collisions hence less product produced in a given time.

High concentration of Hydrogen Peroxide

Hydrogen Peroxide

Active site

CATALASE

The diagram above shows that at a high concentration there are a greater number of hydrogen peroxide molecules. As result of this there are more frequent collisions hence more products produced in a given time.

Predicted Graphs:

V-max

Rate of Reaction

(cm�/s)

Substrate concentration (mol/dm�)

The graph above shows that as the H2O2 concentration increases, the rate of reaction also increases. This is because if there is more substrate, H2O2 then it will mean that there will be a greater chance of frequent collisions between the H2O2 and catalase. As result of more frequent collisions, there will be more products, water and oxygen produced in a given time. However the optimum rate of reaction is at V-max, this is when all the catalyse active sites are occupied and as result of this no substrate can bind to the active site. Hence the enzyme concentration will be the limiting factor causing a constant rate of reaction.

Volume of

oxygen

(cm�)

Time (seconds)

The graph above shows that as the time increases, the H2O2 decreases. This is due to the fact that the catalase is breaking down the substrate, H2O2. As result of more H2O2 broken down, there will be more product, water and oxygen produced.

Variables:

Variables are factors which would affect the experiment. Some variables can be controlled such as temperature whereas some variables cannot be controlled such as rate of reaction.

Variable

What are the affects of the variables?

How can it be controlled?

Substrate

Concentration

Substrate concentration affects the volume of oxygen produced.

Increase in Hydrogen peroxide concentration will result in an increase the amount of oxygen produced until all enzyme active sites are occupied, V-max. Hence the enzyme concentration will limit the rate of reaction.

Dilute Hydrogen peroxide to specific extents and make sure that concentration of Hydrogen peroxide is accurate. Hence use a dilution table.

Temperature

(Controlled)

Temperature affects the rate of reaction. This is because increase in temperature provides more kinetic energy to the Hydrogen peroxide molecules and enzyme, catalase. As result of this particles move faster and faster moving particles collide more frequently. This will result in more product, water and oxygen produced in a given time. Hence rate of reaction will increase as temperature increases.

Use a thermostatically controlled water bath to maintain the constant temperature of 35�C because a gradual change in temperature will give a great error.

Carry out the whole experiment in the same conditions.

Enzyme concentration

(Controlled)

The affect of enzyme concentration is that it will increase the rate of reaction. This is because as enzyme concentration increases then there is a greater probability of collision between catalase and hydrogen peroxide. Hence more products produced in a given time. However the rate of reaction is constant when all substrate, hydrogen peroxide molecules have occupied the enzyme, catalase active site. Hence the substrate concentration will limit the rate of reaction.

Use the same concentration of catalase throughout the experiment. This is done by using 10 (1 mm) potato discs in each experiment.

Use the same volume of catalase for each concentration of hydrogen peroxide.

Reaction time

(Controlled)

The time will affect the amount of product produced. This is because the more time given to an experiment, the greater the amount of product produced. This is mainly because it will give enzymes and substrate to collide rarely but still get more product as more time is provided.

Use a stopwatch to time

experiment for a specific

duration.

PH

(Controlled)

The affect of pH is that that a change in pH will distort the active site and the enzyme will be denatured. This is mainly because enzymes have preferable working condition in which they work. So if an enzyme works best at neutral conditions, pH7 and pH is changed to be acidic then the enzymes will denature as hydrogen and ionic bonds break. This will result in less product formed.

Maintain a constant pH, by not using a highly acidic or alkali solution throughout the experiment. I will therefore maintain pH by using a pH7 buffer solution which will be used in order to keep the potato discs in a moist condition.

Hydrogen peroxide is relatively weak and neutral.

Apparatus:

The equipment which I will require in order to carry out the experiment is:

Equipment

Quantity

Why is it required?

Boiling tube

12

Used for the reactants to take place in (acts as reaction chamber). Also it is used for diluting hydrogen peroxide into different concentrations.

Boiling tube rack

2

Safely hold the boiling tubes.

Bung with a gas delivery tube

1

Cover boiling tubes and allow the transfer of oxygen from the boiling tube to the burette where the volume of oxygen can be measured.

Watch glass

1

Used to put pH7 buffer and potato pieces (1mm thick) to keep them moist

Thermostatically controlled water bath

1

It is used in order to maintain constant temperature of 35�C throughout the experiment.

Stopwatch

1

Measure the intervals and time taken for reaction.

Measuring cylinder – 10 cm3

2

Measure required volumes of reactants and distilled water.

Distilled water

50 cm�

Wash measuring cylinders and beakers. Also it can be used to dilute the hydrogen peroxide.

Beaker/container – 1 litre

1

It is required to fill water and the invert the burette to collect oxygen.

Syringe – 10 cm�

2

Use to transfer hydrogen peroxide into the boiling tube. This is because the 20 cm� syringe will have a smaller scale hence a larger percentage error. Therefore I will use a 10 cm� syringe because the percentage error caused will be lower.

Burette – 50 cm3

1

Measure the volume of oxygen produced.

Clamp and stand

1

Hold burette in straight in large beaker so the reading taken is accurate and reliable.

2.0 M Hydrogen peroxide solution

50 cm�

Use to dilute to produce the required concentrations. Also is the substrate that is catalysed.

Plasticine

Use to make top of bungs airtight so oxygen can’t escape.

Rubber bungs

12

Use to cap dilutions of hydrogen peroxide solution until use.

Potato

120-130 1mm thick discs

Contains catalase hence used to catalyse hydrogen peroxide.

Cork borer

1

To gain the same diameter of potato, this is smaller in diameter of the boiling tube diameter.

Knife (scalpel)

1

It is required in order to cut the potato, 1mm thick.

pH7 buffer solution

20 cm�

Used to keep the potato moist

Method:

* Firstly get all the equipment required.

* Wear goggles and follow other safety precautions.

* First of all, using a cork borer (a little smaller in diameter than the boiling tube), cut a cylinder of potato.

* Use the knife (scalpel) to cut the potato cylinder into 1mm thick discs.

* Place 3 cm� of pH7 buffer solution in a watch glass.

* Place the potato discs in the watch glass, which contains ph7 solution. This will keep the potato discs moist.

* Place ten potato discs in a boiling tube with 3 cm� of pH7 solution and cover it with a bung with a gas deliver tube.

* Place the boiling tube containing potato discs and pH7 solution in the thermostatically controlled water bath, at the temperature of 35�C. This will allow all discs to reach the correct temperature and remain at constant temperature throughout the experiment.

* Get a 250 cm� beaker/container and fill it up to around 200 cm� with water.

* Get a 50 cm� burette and fill it up with water as close to 0.00 cm� as possible.

* Set up clamp and stand so the burette is fixed and secure. This will hold the burette still hence easier to take readings.

* Take the initial reading of where the burette is filled up to.

* Place the burette inverted and allow the gas deliver tube from the boiling tube to insert the burette.

* Make sure that the delivery tube is inserted in the burette.

* Place Hydrogen peroxide in a boiling tube and then in the thermostatically controlled water bath so it also reaches the same temperature.

* By using a 10 cm� syringe covered by plasticine place 10 cm� of hydrogen peroxide of 2.0M into the boiling tube (containing pH7 buffer solution and potato discs) through the bung.

* Start the stopwatch.

* Take reading from the burette after every 20 seconds.

* Record the volume of oxygen produced by 2.0M hydrogen peroxide after every 20 seconds for 3 minutes.

* Repeat the experiment again by using the dilution table below at different concentrations.

* Repeat one more time for each of the concentrations.

Method of producing Hydrogen peroxide:

Hydrogen peroxide

General

Systematic name

hydrogen peroxide

Other names

hydrogen dioxide

hydroperoxide

Molecular formula

H2O2

Molar mass

34.01 g/mol

pH

4.5

Appearance

clear liquid

CAS number

[7722-84-1]

Properties

Density and phase

1.4 g/cm3, liquid

Solubility in water

miscible

Melting point

-11 �C (262 K)

Boiling point

141 �C (414 K)

Acidity (pKa)

11.65

Viscosity

1.245 cP at 20 �C

Structure

Molecular shape

skewed

Dipole moment

2.26 D

Hazards

EU classification

Oxidant (O)

Corrosive (C)

Reference: http://en.wikipedia.org/wiki/Hydrogen_Peroxide

The method of producing Hydrogen peroxide to a concentration of 2 mol/dm�:

Chemical Formula = H2O2.

Hence Molar Mass = (1 � 2) + (16 � 2)

= 2 + 32

= 34.0 g/mol

Therefore 2 moles of Hydrogen peroxide has to be dissolved in 1 dm� (1000 cm�) of water in order to get a concentration of 2 mol/dm�. Hence because 2 moles is:

Number of moles = Mass

Molar Mass

Hence: 2 moles = Mass

34.0g/mol

Therefore Mass = 2 moles � 34.0g/mol

= 68.0g

So therefore in order to make a hydrogen peroxide to a concentration of 2 mol/dm�, the mass of hydrogen peroxide required is 68.0g which is then dissolved in 1000 cm� water.

This calculation can also be proved by the following equation:

Number of moles = Concentration � Volume

1000

= 2 mol/dm� � 1000 cm�

1000

= 2 mol/dm� � 1 dm�

= 2 moles

Dilution Table:

I will be provided with hydrogen peroxide, with a molarity (concentration) of 2.0mol/dm� (2M). I will be using 20 cm� as total volume of hydrogen peroxide an water together. The dilution table below will show the quantity of water and hydrogen peroxide used in the experiment. This is because I will have to dilute the provided 2mol/dm� to gain different concentrations of hydrogen peroxide which I will use in the experiment.

Percentage of Hydrogen peroxide concentration (%)

Concentration of Hydrogen Peroxide (M)

Hydrogen peroxide (cm�)

Distilled water (cm�)

10

0.2

2

18

40

0.8

8

12

50

1.0

10

10

60

1.2

12

8

80

1.6

16

4

100

2.0

20

0

Measurements and errors:

Burette:

A burette is required in order to collect the volume of oxygen which is produced by the reaction in which hydrogen peroxide is broken down by catalase to produce water and oxygen. Hence the burette is needed to determine the volume of oxygen produced. The scale on the burette is accurate to +/- 0.1 cm� hence I would be able to read between the scales which will make the reading to 0.05 cm�. So therefore because I will be able to read as accurate as +/- 0.05 cm� it will mean that the percentage error will be:

Percentage error = Smallest division � 100

Measured quantity

Hence percentage error of the burette will be: Percentage error = 0.05 � 100

Volume of oxygen

Measuring cylinder:

A measuring cylinder is needed in order to measure the volume of water. This is because I will be doing the experiment on different concentrations of hydrogen peroxide. Hence I will need to dilute the hydrogen peroxide to different concentrations before using it in an experiment. As result of this a certain volume of water will be required in order to dilute the hydrogen peroxide to specific extent.

Syringe:

I will need a syringe for this experiment because I have to measure 5 cm� of hydrogen peroxide. This is because 5 cm� of hydrogen peroxide and its dilutions will be added to the potato which contains the enzyme catalase that will break down the hydrogen peroxide. Hence to make it a fair and reliable test, equal amount or volume of hydrogen peroxide (of different concentrations) should be added each time.

Safety Precautions:

* Wear goggles for safety.

* Hydrogen peroxide is corrosive hence gloves should be worn while handling with solution.

* Hydrogen peroxide is an oxidizing agent.

* Tables should be clear to allow maximum area for the set up of apparatus.

* Chairs/stools should be tucked in under the table.

* If any accidents report to teacher straight away.

Rate of Reaction:

Concentration of Hydrogen Peroxide (mol/dm�)

Calculations

Initial Rate of Reaction (cm�/s)

0.20

5 cm� �141 sec = 0.035460992

0.0355

0.80

5 cm� � 78 sec = 0.064102564

0.0641

1.00

5 cm� � 60 sec = 0.083333333

0.0833

1.20

5 cm� � 45 sec = 0.111111111

0.111

1.60

5 cm� � 28.5 sec = 0.175438596

0.175

2.00

5 cm� � 24 sec = 0.208333333

0.208

Regression Line

“The least square regression line is drawn through the points representing the observations in such a way that the sum of the squares of the vertical distances between the line and the points plotted is minimised.”

Reference: Edexcel- Statistics 1

x = mean value of x

y = mean value of y

yi = Values of y

xi= Values of x

n = Number of Values or Samples

r = Correlation Co Efficient

S = Sample

? = Sum of

? yi = Sum of values of y

? xi = Sum of values of x

? xiyi = Sum of xy

? yi � = Sum of y�

? xi � = Sum of x�

Formulas

Sxy = ? xiyi – ? xiyi

n

Sxx = ? xi � – (? xi �)

n

Syy = ? yi � – (? yi �)

n

r = Sxy

SxxSyy

OR

Sxy = ? (xi – x) (yi – y)

Sxx = ? (xi – x) �

Syy = ? (yi – y) �

r = Sxy = ? (xi – x) (yi – y)

V (SxxSyy) V ? (xi – x) � ? (yi – y) �)

The least squares regression line is a line of best fit which ensures that the sum of the squares of the vertical distance between the line and the plotted points is minimised

By using advanced pure mathematics it can be worked out that the equation of the regression line of “y” and “x” given by:

y = � + �x

Where � = Sxy and � = y – �x

Sxx

Where x and y are the mean values for “x” and “y”

x = ? xi and y = ? yi

n n

� is the slope of the line.

� is the intercept (the value of y when x = 0)

x is the independent variable.

y is the dependent variable.

The equation for the least square regression line for the rate of reaction is:

x

y

xy

x�

0.20

0.0355

0.0071

0.04

0.80

0.0641

0.05128

0.64

1.00

0.0833

0.0833

1.00

1.20

0.111

0.1332

1.44

1.60

0.175

0.28

2.56

2.00

0.208

0.416

4.00

?x = 6.8

?y = 0.6769

?xy = 0.97088

?x� = 9.68

x = ? xi

n

Hence x = 6.8

6

= 1.133333333

= 1.13

y = ? yi

n

So y = 0.6769

6

= 0.112816666

= 0.113

� = Sxy

Sxx

So therefore � = 0.97088

9.68

= 0.10029752

= 0.100

� = y – �x

Hence � = 0.113 – (0.100 � 1.13)

= 0.113 – 0.113

= 0

Line of regression: y = 0.100x – 0.00

Analysis:

In my hypothesis I predicted that the rate of reaction would increase with increase of Hydrogen peroxide concentration. The graphs that have been plotted were on the y-axis the rate of reaction in cm�/s and on the x-axis concentration of hydrogen peroxide in mol/dm�. The graph is a straight line hence there is a continuous increase in the rate of reaction.

The initial rates of reaction were found by finding the gradient of the graph at 0 seconds. The initial rates of reaction were found at 0.20 mol/dm�, 0.80 mol/dm�, 1.00 mol/dm�, 1.20 mol/dm�, 1.60 mol/dm� and 2.00 mol/dm�. At 0.20 mol/dm the rate of reaction was only 0.0355 cm�/sec however when the concentration of Hydrogen

peroxide was 2.00 mol/dm�, the maximum concentration used for the experiment, the rate of reaction was 0.208 cm�/sec. This establishes my initial hypothesis that the rate of reaction increases with increase in hydrogen peroxide concentration.

So therefore the graph and the results support my prediction as I had predicted that an increase in the substrate concentration, Hydrogen peroxide will mean an increase in the rate of reaction. This is because more hydrogen peroxide means that there is a greater chance of a substrate to collide with the enzyme, catalase which has the specific active site, which is complementary to hydrogen peroxide. Hence greater chance of frequent collision between catalase and hydrogen peroxide will mean that there will be more enzyme-substrate complex and more products, water and oxygen produced in a given time. This can be analysed by the rate of reaction at 0.2 mol/dm� and 2.00 mol/dm�, because the difference between the rates of reaction is 0.1725 cm�/s, which means that an increase in concentration has lead to the increase in the rate of reaction. This is mainly due to the fact that there are a greater number of hydrogen peroxide molecules as concentration increases, which leads to more frequent collisions as there is a greater probability of colliding and binding to the catalase active site. This will mean that more enzyme-substrate complex and more products, oxygen and water produced in a given time. So therefore as the concentration of Hydrogen peroxide increases, the rate of reaction will also increase. So therefore this shows that the results, which I have obtained, show some form of reflection to what I had predicted.

Overall, my results and graph support what I had predicted to a moderate extent I had outlined in my prediction that as there is an increase in hydrogen peroxide concentration, the rate of reaction will also increase. This is due to the fact that an increase in concentration means that there will be an increase in the number of hydrogen peroxide molecules. This will therefore result in more frequent collision, leading to more products, oxygen and water produced in the provided time.

So therefore as the concentration of Hydrogen peroxide increases, it enables more molecules to collide more frequently in the right orientation in the time provided. Hence the increase in concentration would mean an increase in the rate of reaction.

Collision Theory:

Collision Theory explains how chemical reactions take place and how the rates of reactions are affected. For a reaction to occur, particles must collide successfully for which they have to collide and hit the solid particles in the right place at the right speed. If the collision causes a chemical change it is referred to as a fruitful collision. The fruitful collisions have enough energy (activation energy) at the moment of collision to break the existing bonds and form new bonds, consequential in the products of the reaction. Increasing the concentration of the reactants and raising the temperature bring about more collisions and therefore more fruitful collisions, increasing the rate of reaction.

Also a chemical reaction can only occur between particles when they collide (hit each other). Particles may be atoms, ions or molecules. There is a minimum amount of energy which colliding particles need in order to react with each other. If the colliding particles have less than this minimum energy then they just bounce off each other and no reaction occurs. This minimum energy is called the activation energy. The faster the particles are going, the more energy they have. Fast moving particles are more likely to react when they collide. You can make particles move more quickly by heating them up (raising the temperature).

Overall the collision theory simply outlines that rate of reaction is depends on how often and how hard the reacting particles collide with each other. Therefore the result of a collision can only be obtained if the particles collide which would enable them to react, and they have to collide hard enough in order to make a successful collision. Hence more collisions can increase the rate of reaction and there are four factors, which can cause more collision and increase in rate of reaction. These four factors are temperature, concentration, surface area and catalyst, all four of these factors speed up the rate of collision hence increase then rate of reaction.

Evaluation:

I am able to observe from my graphs that I have gained a few anomalous results. This outlines that there was an error of some kind, which resulted in the production of anomalous results. There could have been several factors such as the hydrogen peroxide not diluted to the exact extent, which resulted into the anomalous results. This could have been the case due to the human error made in reading the scale on the measuring cylinder when measuring the volume of distilled water which was to be used in order to dilute the Hydrogen peroxide.

The percentage error caused by the measuring cylinder is:

Concentration of Hydrogen peroxide (mol/dm�)

Volume of distilled water required for dilution (cm�)

Calculation

Percentage error (%)

0.20

18

(0.5 � 18) � 100

2.78

0.80

12

(0.5 � 12) � 100

4.17

1.00

10

(0.5 � 10) � 100

5

1.20

8

(0.5 � 8) � 100

6.25

1.60

4

(0.5 � 4) � 100

12.5

As you can see from the table above, there is an error in measuring the volume of water using a 50 cm� measuring cylinder which may have lead to the anomalous results.

Another reason for the anomalous result could be the measurement of the volume of oxygen collected in the burette. This is mainly because there could have been an error in reading the measurement which results in anomalous results.

The percentage error caused by the burette is:

Concentration of Hydrogen peroxide (mol/dm�)

Average volume of oxygen collected(cm�)

Calculation

Percentage error (%)

0.20

4.60

(0.05 � 4.60) � 100

1.09

0.80

7.00

(0.05 � 7.00) � 100

0.714

1.00

11.70

(0.05 � 11.70) � 100

0.427

1.20

16.80

(0.05 � 16.80) � 100

0.298

1.60

24.10

(0.05 � 24.10) � 100

0.207

2.00

31.00

(0.05 � 31.00) � 100

0.161

As you can see from the table above, there is an error in measuring the volume of oxygen collected using a 50 cm� burette which may have lead to the anomalous results.

Minor Errors

Effect on results

Improvements

Justification

Initial concentration of Hydrogen peroxide.

If the initial concentration of the hydrogen peroxide was lower than 2.00 mol/dm�, then it will mean that there are a fewer number of hydrogen peroxide molecules. As result of this the rate of reaction will not be as fast because lower number of molecules means fewer successful collisions and fewer products produced in a given time.

We can prepare our own hydrogen peroxide to a concentration of 2.00 mol/dm�.

OR

Witness the hydrogen peroxide being prepared for us by the technician.

In order to be aware of the fact that the hydrogen peroxide is prepared to the actual extent of 2.00 mol/dm� which is required in order to carry out the experiment.

Temperature

If the room temperature increases then the rate of reaction will increase. This is because the temperature provides energy to molecules hence the molecules move faster. Faster moving molecules due to kinetic energy collide more frequently causing more products, oxygen and water being produced in a given time. Hence increase in temperature will increase the rate of reaction.

Use a ‘quartz thermometer’, which has a scale division accuracy of 0.5 degrees Celsius.

OR

Use a water bath which will make the temperature remain constant.

This would have allowed me to measure the temperature correctly at which the hydrogen peroxide and the catalse are. Also larger scale hence low percentage error.

Dilution of Hydrogen peroxide

The hydrogen peroxide was not diluted exactly to a certain concentration due to a huge percentage error caused by the 50cm� measuring cylinder. So therefore if the hydrogen peroxide was diluted to a greater extent than required then there will be a low concentration of hydrogen peroxide, which will result in fewer collisions and less products, oxygen and water produced in a given time.

In order to lower the percentage error I will be using a 20 cm� measuring cylinder.

I will be using a 20 cm� measuring cylinder becaue it has a larger scale hence a lower percentage error.

The reaction had started before the boiling tube was covered by the bung.

This resulted in the oxygen escaping before the stop watch was started. This is because the time taken to cover the boiling tube and starting the stopwatch allowed the oxygen to escape due to the fact that the reaction had started as catalase had begun to break down hydrogen peroxide. So therefore less gas was collected. Hence a low rate of reaction.

Also the bung had a hole from which the oxygen could have escaped once the reaction started. Despite the fact that the hole was covered by plasticine there was still a gap for the oxygen to escape. Hence not all the oxygen was collected.

In order to avoid this I will use a gas delivery tube which has bung without the hole.

This will minimise the volume of oxygen escaping during the experiment.

Major Errors

Effect on results

Improvements

Justification

Measuring volume of oxygen produced in a burette.

The volume of gas escaping coming from the delivery tube to the burette which will result in an inaccurate reading.

Use a 50 cm� gas syringe in order to collect the volume of oxygen.

There will be a smaller percentage error due to a large scale and also the oxygen gas will not escape. Hence using a gas syringe will minimise the volume of oxygen escaping during the experiment.

Catalase from different potatoes.

There are different concentrations of catalase in each potato.

Use catalase from the same potato.

So the concentration of catalase is the same used for each test.

Reliability:

In order to make sure that my results are reliable I will be repeating the experiment for all concentrations twice. This would outline the difference or similarity between the two pair of results obtained from each concentration. After I have obtained to results for each of the concentration of hydrogen peroxide then I will proceed in my investigation and add the two results for each concentration and divide it by 2 to obtain the volume of oxygen collected in cm�.

I will also carry out the experiment under the same conditions so there is no variation in temperature or anything that will affect the rate of reaction.

Fair Test:

There are certain variables, which I would need to keep the same in order to carry out the experiment because the change of those particular variables may lead to unfair results. Those variables, which are required to be controlled throughout the experiment, are the volume of catalase and hydrogen peroxide used. I will only use 10 cm� of catalase and 20 cm� of hydrogen peroxide throughout the experiment in order to obtain reliable results and also allow the same volume of reactants to react with the at different concentrations of hydrogen peroxide. This is important because the more catalyse used the faster the rate of reaction hence the concentration of the catalase and the volume of catalase used for each concentration is to remain the same.

Blank Results Table:

A blank result table to show the volume of oxygen collected (cm�) when different concentrations of hydrogen peroxide are catalysed by catalase.