Concentration of Hydrogen Peroxide

Volume of Oxygen evolved(ml)Average Concentration of Oxygen Evolved (ml)1.

0065.065.063.064.30.

8053.051.051.052.

00.6038.038.039.038.30.

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4028.028.027.028.00.2015.

014.014.014.30.003.

004.003.003.30Calculating the Rate the Reaction.Rate (ml/s) =Volume of oxygen released (ml)Time(s)TABLE SHOWING THE RATE OF REACTION AT DIFFERENT CONCENTRATION OF HYDROGEN PEROXIDEConcentration Of Hydrogen Peroxide(M)Average Concentration of Oxygen EvolvedTime(s)Rate(ml)1.0064.337.

21.730.8052.040.01.

230.6038.342.40.

900.4028.049.30.570.2014.

351.00.280.003.3059.50.55From the results, I have noticed that the volume of the products (Oxygen) decreases with decrease in the substrate (Hydrogen Peroxide) and so does the rate of the reaction. This is why on the graph the rate of the reaction is in proportion to the substrate concentration, ie.

The rate of the reaction decreases as substrate concentration decreases. To explain this, I will take in comparison the highest hydrogen concentration (1.00M) and the least (0.00M).The amount of Oxygen evolved and the rate of reaction in 1.00M is higher than it is in the 0.00M concentration of Hydrogen, this is because in enzyme catalysed reactions, having a high substrate concentration increases the amount of products to be formed due to the fact that there is great chance of enzymes to collide with substrates molecules forming enzyme-substrate complexes.

This results in a fast rate of reaction.Because in this experiment I used a constant enzyme concentration, decrease in substrate concentration results in less or limited number of enzyme- substrate complexes as the enzyme’s active sites are not working to full capacity, this is because there is less chance of enzymes colliding with the substrate molecules. Which is why there is decrease in the volume of the products and the rate of reaction in the 0.00M concentration of Hydrogen Peroxide? Substrate concentration is limiting the rate of this reaction. However, if the substrate concentration was increased to lets say; 2M, I think the graph of “Average Volume Against Hydrogen Concentration” would tail off when it reaches the maximum point called the” V-max “.The reaction would then become constant because the active sites of the enzymes would gradually become fully utilised and adding more substrate would have no effect on the rate of the reaction.I did not expect the 0.00M to have any effect on the Catalase because there were no substrate molecules to be broken down by the enzyme and therefore no products to be formed.

But my results show evidence of a reaction as some oxygen was collected. This could have been due to Insufficient washing of the measuring cylinder.EVALUATIONDuring this experiment, I observed some anomalous results (circled) which show evidence of error during the test, these errors could be as a result of the following:-1.

Apparatus UsedI tried to be as accurate as possible and so I chose to use the syringe, stop clock and measuring cylinder. It is the measuring cylinder that I think was inaccurate of the three because I found read the volumes.Percentage Error of the measuring CylinderThe measuring cylinder measures to about 1cm accuracy in 100cm.So it has % Error = 1 X 100=1%1002. Difficulties in reading the highest volume reached by the froth. The meniscus of the froth was not clear.

3. Inadequate washing of the measuring cylinder-some Hydrogen Peroxide may have remained in the measuring cylinderThe above errors can be improved by:-1. Replacing the measuring cylinder with a more accurate apparatus e.g. Burette.2. Using the method of displacement to collect the Oxygen gas.

It has a good and more accurate way of taking the volume of the oxygen because there is no froth.3. Use of different measuring cylinder for each dilution, this will lessen the errors because there would be no need to rewash the same measuring cylinder.I don’t think the test is reliable because there is too many minor errors which have a great effect on the results. I suggest the use of a manometer would be more reliable and would generally improve the test.