A level Biology Coursework - Effect of substrate concentration on rate of activity of the enzyme catalase
Aim: To investigate the effect of substrate concentration on the rate of activity of the enzyme catalase
NOTE: This is an A* level piece of coursework which achieved full marks.
Catalase is an enzyme which is found in most living organisms. It catalyses the decomposition of Hydrogen Peroxide into water and oxygen.
2H2O2 + Catalase >>> 2H2O + O2
Catalase dramatically reduces the activation energy needed for the reaction. Without catalase the decomposition would take much longer, and would not be fast enough to sustain human life. Hydrogen peroxide is also a dangerous, very potent by-product of metabolism, so it is essential that it is broken down quickly, otherwise it would cause damage to cells.
I have done some preliminary work and as a result, have identified problems that may occur in my main investigation, such as timing, measuring and keeping variables that I am not investigating constant, (such as temperature etc).
In the main procedure I will control the temperature in a water bath in order to create a constant external temperature to dissipate the heat energy. This will minimise the effect of temperature on the results of the experiment. I have decided to do this because in my preliminary procedures, I used a thermometer to measure the temperature of the hydrogen peroxide (when left on the side) at different intervals, and on different days, and found that the temperature of the hydrogen peroxide fluctuated slightly.
By doing this, it will ensure that the test is as fair as I can make it. Although the reaction is exothermic and will give out heat during the reaction anyway, since the heat has been dissipated by being kept in the water bath, the amount of heat given off in the experiment will be relative to the concentration of hydrogen peroxide, but obviously some reactions will take longer than others, so more heat will be produced. However the temperature initially will be kept the same in each case.
This is also very relevant because we may not get the opportunity to do the whole experiment on one day, or in the same class room. This means the room temperature in each class room or on different days will not be the same for each procedure, because of obvious factors such as type of day (very cold or mild etc) and the level of heating within the class rooms.
Temperature directly affects the shape of the active site. At a temperature below the optimum, the molecules have less kinetic energy so the rate of collisions between enzyme and substrate molecules are low and therefore less enzyme-substrate complexes are formed. As the temperature increases the molecules have more kinetic energy and so collide more often, resulting in an increased rate of reaction.
Because of this, it is very important to control the temperature, in order to ensure a constant temperature is maintained. Above the optimum temperature, the thermal energy breaks the hydrogen bonds holding the secondary and tertiary structure together, so the active site changes shape and eventually the reaction can no longer be catalysed.
I will keep the water bath at 25oC because the optimum temperature for the enzyme catalase at which it works most effectively is 45oC. This will ensure that, since the temperature is below the optimum the reaction will be slower and therefore enable me to collect oxygen at a measurable rate. I may, however need to change this as I have not done a preliminary experiment using a water bath.
In my preliminary work I also found that when doing the experiment with 1g of yeast, and 5cm3 of 20 volumeof hydrogen peroxide, the rate of reaction was too fast to collect oxygen at a measurable rate, and therefore made it impossible to gain meaningful results. I consequently reduced the mass of yeast to 0.2grather than the 1.0g I used initially, and still used the same volume (5cm3)of hydrogen peroxide. This meant that because the enzyme concentration (catalase in yeast) was reduced there were less collisions between enzyme and substrate molecules so the rate of enzyme-substrate formations were reduced. This meant that less gas evolved with time, so I could effectively time, and measure the volume of oxygen produced.
Another factor which I had to consider was the surface area of the yeast granules, because each yeast granule has a different surface area the amount of enzyme will differ in each granule, but more importantly the greater the surface area of the yeast the more reactions take place because of the fact that there will be more collisions between the enzyme and substrate molecules.
In my first preliminary experiment I just weighed 1.0g of yeast as it was supplied in its granule form. However in my next preliminary experiment, I decided this would be unfair in the main procedure and because of this I decided to grind the yeast into a powder so that the surface area would be more similar in each yeast granule.
Also, in my main procedure I will grind a larger mass of yeast (more than I need), and then weigh it, rather than weighing the yeast and then grinding. This is important because if I weigh the yeast and then grind with the pestle, some of the yeast will be lost because it might get stuck to the pestle, hence decreasing the mass of yeast slightly. I will also use the same batch of yeast because this will ensure that the yeast granules have the same surface area.
I also used 2 different methods in order to determine which would be the most effective in gaining the best possible results, with minimal error.
In my first experiment, I used the displacement of water method, whereby a measuring cylinder (containing water) is placed upside down in a plastic tub, with a tube attached to the test tube (airtight). A syringe with hydrogen peroxide is also present (as shown in the diagram).
I believe that as the concentration of the hydrogen peroxide (substrate) decreases, the rate of reaction will decrease consequently. This is because there will be fewer collisions between the substrate and enzyme molecules (catalase in yeast) – because there are fewer molecules of hydrogen peroxide, so there will be a decrease in enzyme-substrate complexes formed. The reaction will then stop because all of the active sites become saturated with substrate because the enzyme is the limiting factor. This will result in a decreased volume of oxygen being produced as one of the by-products of this reaction.
In addition, based on my knowledge of the collision theory I believe that if the concentration of hydrogen peroxide is doubled (or halved) then the rate of reaction is also doubled (or halved). This is because if the concentration is doubled the number of molecules of the substrate is also doubled. This means there will be twice as many successful collisions. Therefore it is true to say that in theory: rate µ concentration.
I will investigate if this is true for this reaction.
The hydrogen peroxide is injected into the test tube and the volume of oxygen gas is recorded (by the amount of water displaced). This would determine the rate of reaction. However I decided against this method for several reasons. Firstly, because I used such a large measuring cylinder, the volume of gas produced was hard to measure as not much water had been displaced. Although I could have used a smaller measuring cylinder, I decided that the best possible way I could do the experiment was by measuring the volume of gas directly using a gas syringe, rather than by the displacement of water. Also, because the hydrogen peroxide had to be inserted into the syringe before the reaction could begin, the amount of time it would be out of the water bath (which I intend to use in my main experiment) was longer than necessary. I decided that I could reduce this time by using a different method.
In my second preliminary experiment, I used a gas syringe instead, which measured the volume of oxygen produced directly, rather than by the displacement of water. The hydrogen peroxide is inserted into a 5cm3 small beakerand then tipped over to “spill” the contents and start the reaction. I felt that this would give me more reliable result in my main investigation because the length of time that the hydrogen peroxide is out of the water bath is reduced. Furthermore, the volume of gas is measured directly. I noticed that when doing the first method that “bubbles of gas” were affected by people bumping the table, and that sometimes they got trapped in the tube, so even though the product of the reaction (oxygen) had been formed it was not measured until afterwards, (at a later stage in the reaction). Also, the bubble volume is affected by the diameter of the tube and the overall pressure of the water (depth of the water) so I believe that by using the gas syringe, I will be able to eliminate this inaccuracy as water will not be involved. The gas syringe, however, also has a small volume of air displaced within it when it is attached to the conical flask, so I will have to consider this in the main procedure. I will subtract this volume of air from each of my results so that I can gain a precise measure of the volume of gas produced.
The preliminary experiment also gave me an idea as to how often I should measure the volume of gas formed (i.e. every 5, 10, 15 seconds etc). In my first preliminary experiment the reaction went to fast to collect oxygen at a measurable rate, but in the second preliminary experiment I measured the volume of gas every 10 seconds, but found that the reaction was over before I had enough measurements and that the results I gained would not be sufficient to obtain enough results and make a valid conclusion. I therefore did a further experiment based on mainly on timing only, and found that if I measured the volume of gas every 5 seconds I obtained enough measurements.
However, I do have to take into account that I will be using different concentrations of hydrogen peroxide in my main experiment, so 5 seconds may not be sufficient to measure the volume of oxygen produced in the slower reactions and I may need to change this.
The concentrations of hydrogen peroxide I will use will be 100%, 90%, 80%, 70%, 60% and 50%. I will use these concentrations because I believe that if I were to go any lower than 50% that the rate of reaction will be relatively slow, and will not produce enough results because the substrate concentration (hydrogen peroxide) will be too low in concentration. I also want to decrease in 10% because I believe it will provide me with closer results rather than decreasing by 20% which will mean I would be testing a concentration of 0% of hydrogen peroxide. Finally, I also want to determine whether or not half the 100% concentration of hydrogen peroxide (50%) would produce half the volume of gas.
Bored of revsion? need to read something GOOD?
Want to say thanks for this article? Help me out by checking out this book I helped co-edit. Thanks in advance!
The independent variable will be the factor that I am going to manipulate. This will be the concentration of the hydrogen peroxide. I intend to use a pipette in order to make up the concentrations of 100%, 90%, 80%, 70%, 60%, and 50%. I will do this by making each mixture up to 100cm3, so for example, the 90% concentrated solution will consist of 90cm3 of hydrogen peroxide, and 10cm3 water. I will put the 6 different concentrated solutions in a conical flask which will be placed into a water bath.
Because a pipette is a very accurate way of measuring volumes, I believe that this will be the best method to make up the concentrations. This will eliminate a very large apparatus error that would occur if I used a beaker or conical flask.
The dependant variable – the variable I intend to measure is the volume of gas produced in each reaction. This will vary as a direct result to the different concentration of hydrogen peroxide
Controlled variable – the controlled variable consist of all the other factors which must be kept constant.
This will be the mass of yeast for each experiment (0.2g). I will make sure that I measure 0.2g of yeast as accurately as I can using the balance. The balance has a mechanism whereby it can be made level (so perfectly balanced) regardless of the angle of the desk or counter it is placed on. I have explained this is my method below. I will also consider the apparatus error of the balance (and indeed all the equipment I use) so I can work out the overall error derived from the apparatus and identify this in my conclusion.
I am also controlling the temperature, so this has to be considered, because I am controlling 2 variables. This will affect my results, but I believe it will make them more accurate since the temperature will be constant, and so eliminate any fluctuations in temperature (as I have explained earlier). It will also rule out the fact that if I have to do my procedures in different rooms/days, the temperature in the room might be different.
- Conical flask
- 20 vols hydrogen peroxide
- Gas syringe
- Stop clock
- Clamp stand
- 50cm3 pipette
- 20cm3 pipette
- 25cm3 pipette
- Water bath
- Pestle and mortar
- 5cm3 Beaker
Firstly I am going to measure out the concentrations of hydrogen peroxide that I have decided on. This will be 100%, 90%, 80%, 70%, 60% and 50%.I will do this by adding different volumes of water to make up 100cm3.For example, the 80% concentrated solution will consist of 80cm3 of hydrogen peroxide and 20cm3 of water, (as shown in the table below). Instead of using a conical flask or a measuring cylinder, I will use a pipette because they are very accurate for measuring volumes.
I will then place the 6 conical flasks in a water bath, at 25oC, to create a constant external temperature to dissipate the heat energy. I will do this first to ensure that the mixtures have enough time to reach a constant temperature, rather than only putting them in for a short time because this will mean there may be fluctuations in temperature of the different molecules which I was initially trying to eliminate.
Next, I will grind the yeast into a powder using a pestle and mortar. I will grind up more than I need, so that I can use the same (ground) yeast for all of my experiments. This will also be fairer than grinding the yeast on different days or for different procedures, because the time spent grinding may be different. Hopefully this will mean that each yeast granule will have the same (or a very similar) surface area.
I will then set up my apparatus.
Finally, on setting up the apparatus, I will place the balance on the table, making sure the bubble in the spirit level is in the middle level, thereby eliminating any deviation from the balance position. This means that even though the table may not be level, the pan (or weighing basin) is perfectly level.
I will set up my apparatus this way every time I need to, e.g. when I need another lesson to complete my investigation.
Once the apparatus is set up, I will place a conical flask on the balance, and set the balance to 0, so that I can weigh the yeast only, which I will place in the conical flask using a spatulauntil the right weight (0.2g) is in the conical flask. I will weigh the yeast directly into the conical flask, not a Petri dish, because then I do not have to worry about transferring the yeast from the Petri dish to the conical flask.
Next I will place the conical flask under the gas syringe and place an airtight stopper in the top, with a single tube attached to the gas syringe as shown in the diagram.
I will then take the conical flask with the 100% hydrogen peroxide out of the water bath and measure exactly 5cm3 of the mixture using a syringe, and place it into the 5cm3 small beaker. Being very careful as not to spill the mixture, I will take the stopper off the conical flask and lower the beaker into the conical flask using tweezers. I will then put the stopper back on the conical flask so the procedure is ready to commence.
Using a stop clock I will time from the moment the small beaker is tipped over to when the reaction stops, measuring the volume of gas evolved every 15 seconds. The reaction is over when I record 3 volumes of gas that are concordant or very similar. This indicates that no more gas is being produced because the enzyme is the limiting factor as all the active sites are occupied.
I will repeat this method using the different concentrations of hydrogen peroxide making sure to wash the equipment thoroughly after each reaction. I will carry out each reaction 3times for each to gain an average. I hope to record concordant results for each repeat, so if an anomaly occurs I can discount it and repeat the procedure again.
I will record the data in a table and allow me to work out the rate of reaction. I will then represent the results in a graph. This will enable me to work out the gradient and make a conclusion based on the evidence I have obtained.
Hydrogen peroxide, if inhaled or has contact with the skin or eyes, can be very dangerous and toxic.
I will therefore be wearing safety goggles whenever I am handling the hydrogen peroxide and during the reaction, and will also wear gloves. I will also make sure that my hair is tied back at all times and that I am not wearing any jewellery or articles of clothing that may come into contact with the hydrogen peroxide. I will also clean up any spillages immediately.
(Predict what the graph will show)
I believe that the graph will firstly start of steep in all the reactions, but steepest in the 100% concentration of hydrogen peroxide and gradually decreasing as the concentration of hydrogen peroxide decrease. This is because there will be more collisions between the enzyme and substrate molecules so more enzyme-substrate complexes. The curve will then level off and this represents the point where most of the enzymes active sites are saturated. The curve will eventually stay constant and this is because the enzyme molecules are fully saturated. This is called the maximum velocity of reaction or Vmax. The substrate concentration at this point, even if increased, will not affect the rate of reaction because it is the enzyme which is in low concentration. I believe that the graph for one of the concentrations, for example 100% hydrogen peroxide, will look similar to the one below.
Draw a graph showing what your PREDICTION will be, and write a statement (such as the one below) showing why the graph shows what it does.
I believe that each curve for each concentration will look like the one above, but for each decreased concentration – 90%, 80%, 70%, 60% and 50%, the value of the Vmax will decrease too, as will the initial rate of reaction, This is because of the fact that there will be less substrate molecules in each successive concentration, so fewer collisions between particles that can react with each other. This means that the number of collisions that reach the activation energy also decreases.
This can be explained by the Maxwell-Boltzmann distribution curve:
THEN Draw the graph using your results or the one's in the above table.
"Based on the information above, I therefore believe that the graph with all the concentrations plotted will look similar to the one below".
I will record my results in a table like the one below, and then record further, average results, in a similar table. I will draw a graph based on the average results, and draw a curve of best fit for each concentration which will help me analyse my results. I will then work out the gradient of each curve and plot a further graph of percentage of H2O2 against rate of reaction on the y-axis. I would expect this graph to be linear as this would show that as the concentration increases, time taken for a set volume of gas would decrease - in other words, the rate is proportional to the concentration. I expect this graph to look similar to the one I have drawn previously. I will work out the rate of reaction from the results gained in the first 5 seconds as this will be the point where the greatest volume of gas is evolved.
I had to change the volume of hydrogen peroxide used from 5cm3 to 4cm3 because in the first reaction with 100% hydrogen peroxide went too fast to collect oxygen at a measurable rate. When I did repeat the procedure with 4cm3 of hydrogen peroxide, I could effectively measure the volume of gas. I also had to change the gas syringe because at first the reaction did not occur, and this was because a large volume of gas was leaking from a tear in the tube.
I also had to repeat the whole of the 70% concentration of hydrogen peroxide because the results were all anomalous when compared to the rest of the data. I will talk about why this might have been in my evaluation.
* Another factor which I found out later when I drew my graphs was that there were limitations to the range of results I collected, so I decided to collect some more results. I have explained this later on.
Below is a table of the results I collected, including all the results which I had to repeat. The raw results can be seen in the appendix.
Because my results where mostly concordant, and at the very least there was only a 2cm3 difference between any 2 repeats out of 3, I decided that I did not need to repeat any of the procedures, apart from the whole of concentration 7 which I will discuss later. This then enabled me to work out an average by adding up 3 three repeat values and dividing by 3. For example the 100% concentration average would be (48+49+48) ÷ 3).
Below is a table showing the average results.
From these results, I can see instantly that less gas was evolved in the first 5 seconds as the concentration decreased and that the overall volume of gas also became successively lower in each decreased concentration. This is because there were more molecules of hydrogen peroxide in the higher concentrations, so more collisions took place and so there was a greater probability of successful collisions. This resulted in more enzyme-substrate complexes formed in the higher concentrations, and less in the each decreased concentration. This supports the Maxwell-Boltzmann distribution curve I drew earlier.
I have drawn a graph based on these average results, with a curve of best fit for each concentration which will allow me to identify any anomalies.
(on your graph - draw a curve of best fit )
From the graph I can see that as the concentration of hydrogen peroxide decreased, the volume of oxygen produced decreased as a direct result. This is because as the concentration decreased, the number of molecules of hydrogen peroxide also decreased. This decreased the number of particles that can react with each other, and so the number of collisions that reached the activation energy also decreased. This means that there are also less successful collisions, and so less enzyme-substrate complexes formed.
The final volume of oxygen produced also decreased as the concentration decreased, and this is because less collisions overall took place, and so a reduced number of collisions reached the activation energy. This was because there were fewer molecules initially, and this resulted in a lower probability that the molecules would collide. This means that there were less successful collisions overall.
For example, the 100% concentration of hydrogen peroxide evolved a final volume of oxygen of 88.3cm3, 90% concentration evolved 73.3cm3 of oxygen, 80% - 63.7cm3, 70% - 49.3cm3, 60% - 44.7cm3 and the 50% concentration evolved 37cm3 of oxygen.
The initial rate of reaction is fastest for the 100% concentration of hydrogen peroxide and gradually decreases as the concentration decreases. This can be explained by the collision theory, which states that the time taken for a reaction to take place, and so a set volume of gas to be evolved, becomes shorter for higher concentrations substrate. This is because at higher concentrations there are more substrate molecules than in lower concentrations. Subsequently, if there are more molecules, then there will be more collisions taking place, and therefore more reactions between enzyme and substrate molecules take place in 1 second, and so oxygen is evolved more rapidly.
So at the 100% concentration of hydrogen peroxide, the oxygen was given of more rapidly because there were more substrate and enzyme molecules reacting in a second.
From the curves of best fit, I can also see that there were no anomalous results, only some results which were slightly above or below the curve, though they were not excessively distorted. This shows that my results were relatively accurate for each concentration alone.
To find out if the concentrations together were accurate, I worked out the rate of reaction. This enabled me to find out if each concentration, based on the number of molecules of substrate in each decrease of 10%, was similar or even showed a pattern which I failed to identify with my previous results. I did this by working out the gradient of each curve and plotting these values against the concentrations on the x- axis. The method which I used to do this can be seen on the next page.
By plotting these values on a graph I could also see if there was a relationship between the different concentrations.
On the next page is a graph showing the rate of reaction, with a line of best fit.
(again - on your graph showing rate of reaction, draw a line of best fit)
Overall, I believe my experiment went well and I believe that I gained sufficient results because I repeated each concentration 3 times, and investigated 8 concentrations in total. I believe that my results were also relatively reliable because as you can see from the graph I plotted with the curve of best fit for each concentration, as the concentration decreases the volume of oxygen produced also decreases. For example, the 100% concentration of hydrogen peroxide evolved a final average volume of gas of 77cm3 of oxygen while the 90% concentration evolved a final average volume of 73.3cm3. Also most of the points were on or close to the curve of best fit for each concentration. However there are some factors that I must take into consideration.
Firstly, there were limitations of the apparatus that I used. Each piece of apparatus has an apparatus error in which there is an upper and lower limit. For example the balance had an apparatus error of ±0.01 which means that since I used 0.2g of yeast, this value could either be 0.21g or 0.19g. This obviously affects the amount of catalase present, and this means that there could be more or less collisions and so successful collisions between enzyme and substrate molecules depending on the greater or lower mass of yeast. For example, if there were more molecules of yeast the rate of reaction would increase because there would be more collisions between enzyme and substrate molecules which would result in a greater probability of successful collisions, and therefore more enzyme-substrate complexes being produced. This means that in my results, the volume of gas produced in the first 5 seconds may have been higher than it should have been if I had used exactly 0.2g of yeast. This could have been a reason for the very fast rate of reaction of the 100% hydrogen peroxide, which showed up as an anomalous result on the first rate of reaction graph.
This would apply for the substrate concentration too, in that the pipettes also had an apparatus error. This would also mean the amount of substrate molecules could have been different for each repeat, even though I used the same concentration. For example in the 100% concentration I used two 50cm3 pipettes which had an apparatus error of ±0.01. So in 100cm3 the actual volume could have been either 99.98cm3 of hydrogen peroxide or 100.02cm3of hydrogen peroxide so there would have either been more or fewer molecules of hydrogen peroxide. If there were fewer molecules of hydrogen peroxide, there would have been fewer collisions between molecules of enzyme and substrate, resulting in fewer enzyme-substrate complexes being made.
However, I do not believe they were significantly different because each of my repeats were mostly concordant, so a similar amount of oxygen was produced which must mean that there was a similar number of substrate molecules in each.
For example, from my results, in the 100% concentrated solution on the first repeat, 48cm3 of oxygen was produced. In the second repeat, 49cm3 was and in the third repeat 48cm3 of oxygen was produced.
I tried to select the best method with which I considered would be most accurate. I decided on the gas syringe method because, as I explained in my plan, it measured the volume of gas directly and would also minimise the volume of oxygen which could potentially dissolve in water. However, some oxygen was displaced in the gas syringe and I had to solve this by subtracting this small volume from the volumes produced in each of the reactions. Also, I noticed if the barrel was wet, the syringe often got stuck for a short time before it recorded the volumes of gas. To prevent this I had to dry out the barrel and syringe before commencing the procedure. It was very hard to insert the small 5cm3 beaker into the conical flask, and when it came to tipping it over, some of the substrate was still trapped inside the beaker. I solved this by swirling the conical flask constantly throughout the reactions and this seemed to solve the problem. Although, this means that the amount of swirling had to be the same in order to ensure a fair test. I tried to keep this constant by making sure I swirled the conical flask evenly. The accuracy of the results showed that this factor did not distort the results too much, and so a similar amount of substrate molecules were present in each reaction. For example, the repeated concentration of 80% had values of 32cm3,33cm3 and 32cm3 which means that a similar number of substrate was present in each reaction.
Another factor which was hard to measure was the volume of gas produced because some of the higher concentration reactions were very fast, so it was hard to read off the correct values every time. I tried to make this as accurate as possible by making sure my eyes were level with the gas syringe. Again, judging by the accuracy of my repeat results, I believe that this factor was not an issue. Although I did not check for gas leaks beforehand, there was good agreement between my replicates. In the 60% concentration the repeats at 5 seconds were 20, 21 and 20cm3, which is concordant
If my replicates had not been so close I would have had to change the tube.
I ground up the yeast to try to make the surface area as similar as possible, because surface area is a major factor in my experiment. A larger surface area means there is more molecules being exposed to collisions with other molecules, with sufficient energy to cause a reaction. This means that having the same surface area of yeast in each reaction is very important in ensuring a fair test because the number of molecules exposed to collisions must be the same.
Temperature is a major factor which affects rate of reaction. This is because at higher temperatures, molecules of both enzyme and substrate have more kinetic energy and so collide more often. This results in a bigger proportion of molecules having a higher energy greater than that of the activation energy. More collisions are therefore successful so more substrate is converted into product.
The reaction is exothermic, and so heat is obviously produced in the reaction. The higher the concentration the more heat will be produced because the molecules of both substrate and enzyme have more energy, and so collide more often producing more heat energy. This heat energy is transferred to the environment.
Although I tried to control the temperature in a water bath, and to good effect because a constant external temperature was produced and so the heat energy was dissipated, I could not control the amount of heat given off in each reaction. This would have affected my results for several reasons. Firstly, more oxygen dissolves in water at a lower temperature than at high temperatures and this means that for the reactions involving low concentrations more oxygen would have dissolved than in the higher concentrations because of the decreased amount of heat energy given off. Because the volume of oxygen dissolved in the reaction is not constant for all the reactions, because less oxygen is dissolved in water at higher temperatures, this would have affected my results. This may have been why the difference in the final volume of oxygen produced was not equal, but instead decreased in steps of 3.7cm3, 9.6cm3, 14.4cm3, 4.6cm3 and 7.7cm3.
The different concentrations of hydrogen peroxide that I made up could not have been exactly accurate because this would have meant that the volume of gas evolved would have increased in equal steps which it did not. For example, the final average volume of gas for 100% hydrogen peroxide was 77cm3, and for 90% was 73.3cm3, 80% - 63.7cm3, 70% - 49.3cm3, 60% - 44.7cm3 and 50% it was 37cm3. As I have mentioned earlier, this decreases in steps of 3.7cm3, 9.6cm3, 14.4cm3, 4.6cm3 and 7.7cm3, which is far from equal.
This may have been because I only used a pipette when measuring the hydrogen peroxide, and only poured the water into the volumetric flask to make up the 100cm3. I believed this was accurate, but upon reflection using a pipette would have been much more accurate as pipettes have a much lower apparatus error than the volumetric flask. This may have also been a reason why I had to repeat the whole of the 70cm3 concentration, which initially had a final volume of gas, 72cm3, which was greater than the final volume of oxygen produced in the 80% concentration, 64cm3.
Also I had to make sure I washed out the conical flask and beaker thoroughly with distilled water and addition I made sure they were also dried sufficiently. If I hadn’t I could have risked diluting the solutions more and this would have affected the number of molecules of hydrogen peroxide present which in turn would have affected the number of collisions between enzyme and substrate molecules. For example, if there was still only 1cm3 of water still remaining in the conical flask and beaker combined, then in say the 80% concentration of hydrogen peroxide, the concentration would have been closer to 79%. This can be shown by the simple calculation of 80 ÷ 101 x 100 = 79.2%.
Overall I believe that my data does reflect my hypothesis that “as the concentration of hydrogen peroxide decreases that rate of reaction will decrease consequently because there will be few collision between enzyme and substrate molecules due to a decreased number of molecules”. This is shown by my rate of reaction graph which shows that for the 100% concentration of hydrogen peroxide, the rate of reaction was 8cm3 second-1, and the 90% concentration was only 7.4cm3 second-1.
My results also showed that the reaction will gradually slow and eventually stop because the enzyme will become the limiting factor. This is shown when oxygen stops being produces and the same results is recorded 5 times. So for example, I knew that the 100% concentration of hydrogen peroxide reaction was over because I recorded 88cm3 at least 5 times.
However, I also believed that if I halved the concentration then the rate of reaction (volume of oxygen produced) would also be halved, and so the rate would be proportional to the concentration. This would show that the reaction is a first order reaction. Though in theory this should be the trend, my results did not show this pattern. So, although my results did show a positive correlation, it doesn’t necessarily mean I have an accurate correlation, because my results do not follow specific trends
For example the final value at 50% was 37cm3 whilst the volume of oxygen produced at 100cm3 was 77cm3 which is not double 37. Again the final volume of oxygen produced at 30% was 27.3cm3, whilst the final value produced in the 60% concentration was 44.7cm3 which also not double.
As can be seen from the rate of reaction graph, the concentrations of 50%, 60%, 70%, 80% and 90% are relatively even, and would suggest that the line of best fit should be drawn where I have drawn the line of best fit (1). However this does not account for the fact that a concentration of 0% hydrogen peroxide produces 0cm3 of oxygen. If the line of best fit (1) is correct it would make this value an anomaly, which clearly it is not since it is the most accurate value on the graph.
The line of best fit (2) therefore makes much more sense, because it is crosses through (0,0), and also shows that the concentrations of 50%, 60%, 70%, 80% and 90% are still quite even. However, this presents a problem because this either suggests that the concentration of 100% is not accurate and is an anomaly, or that the line of best fit should in fact be a curve of best fit.
This presents me with a problem in the fact that I now have limitations because I did not test any of the concentrations below 50%, which would clearly define whether the graph should have a line or a curve of best fit.
Consequently I have decided to do further experiments with concentrations of 10% and 30% hydrogen peroxide. I will use exactly the same method as I did previously, and since I still have some yeast left, I can still use the same batch of yeast. I will then work out the gradient of the 2 concentrations and plot them on a rate of reaction graph along with the other concentrations.
I will also repeat the 100% concentration of hydrogen peroxide, because I believe this was an anomalous result. Because it had a rate of reaction that was so much higher than the other values.
Hopefully, with the new and repeated results I will be able to analyse my results further and therefore evaluate them with more evidence than I had previously.
Below is a table of results showing my repeated concentration of 100% and the 2 new concentrations of 10% and 30% hydrogen peroxide.
I will work out the gradient of these new results and plot them on a new rate of reaction graph. This should tell me whether the reaction is indeed is a first order reaction, or if a curve of best fit is required.
Draw a new graph
Now I have carried out the repeats and plotted the points on the rate of reaction graph I can see that the graph is in fact clearly linear. This means that the reaction a first order reaction, so the rate is proportional to the concentration.
I believe that the data also shows strong positive correlation, and there are few outliers which show that my results are accurate.
I have drawn a line of best fit to clearly illustrate this trend. The line of best fit also suggests values of concentrations which I have not investigated. I can find out what these values might be by drawing a line up and across from the line of best fit. So, for example the 40% concentration should have a curve gradient near to the value of 3 (shown by red line on the graph)
Overall, the there is a pattern showing a constant trend in that as the concentration decreases the rate of reaction decreases too, and that the overall volume of gas evolved also decreases. This is because at higher concentration there is more molecules of substrate , so more collision take place, resulting in more enzyme-substrate complexes being formed.
This is shown in the table with all the results I have gained.
Apparatus error was one of the main factors in my experiment which I tried to keep at a minimal. I did this by only using pipettes which have a very small apparatus error when compared to beakers. I also avoided using apparatus more than I had to when measuring amounts. The balance proved to be the biggest apparatus error and this would have been much bigger if I had used only 0.1g rather than 0.2g of yeast.
Below is a summary of all the percentage errors.
Scales ± 0.01
50cm3 pipette ± 0.01
20cm3 pipette ± 0.03
10cm3 pipette ± 0.02
Balance 0.01 ÷ 0.2 x 100 = 5%
Making up concentrations
100% using 2 x 50cm3 pipettes: 0.01 ÷ 50 x 100 = 0.02% x 2
Total - 0.04%
90% using 1 x 50cm3 pipette, 0.01 ÷ 50 x 100 = 0.02%
2 x 20cm3 pipette: 0.03 ÷ 20 x 100 = 0.15% x 2 = 0.3%
Total – 0.32%
- 80% using 1 x 50cm3 pipette,
1 x 20cm3 pipette, 1 x 10cm3 pipette 0.01 ÷ 50 x 100 = 0.02%
0.03 ÷ 20 x 100 = 0.15%
0.02 ÷ 10 x 100 = 0.10%
Total – 0.27%
- 70% using 1 x 50cm3 pipette, 0.01 ÷ 50 x 100 = 0.02%
1 x 20cm3 pipette 0.03 ÷ 20 x 100 = 0.15%
Total – 0.17%
- 60% using 1 x 50cm3 pipette, 0.01 ÷ 50 x 100 = 0.02%
1 x 10cm3 pipette 0.02 ÷ 10 x 100 = 0.02 %
- 50% using 1 x 50cm3 pipette 0.01 ÷ 50 x 100 = 0.02%
Total – 0.02%
Total apparatus error for apparatus used for concentrations = 0.66%
Total error for apparatus – 5.66%
Considering the whole experiment, 5.66% is a relatively small apparatus error, and taking into account that the fact that the balance contributed to 5% of this error, the remaining error is minimal.
Remembering that a apparatus error on the balance is ±0.01, this would mean that the yeast could weigh 2.01g or 1.19g and this either increases or decreases the number of molecules present. I will talk about this in my evaluation.
Bored of coursework?
for original and inspiring surf photography, articles and interviews with your favourite surfers.
More by this Author
- 20A Grade GCSE Physics Experiment— Investigation Into the Effect of Parachute Surface Area in Relation to Its Fall Time
Follow along with my physics experiment and see the conclusions of my investigation.
A grade GCSE chemistry coursework, Rates of reaction, Decomposition of sodium thiosulphate, introduction, method, safety, results, discussion. Sodium thiosulphate decomposition for GCSE, grade a easy.
An investigation into the resistance of a wire. Free GCSE physics coursework essay.