• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

THE CATALYTIC DECOMPOSITION OF HYDROGEN PEROXIDE

Extracts from this document...

Introduction

Chemistry Internal Assessment: THE CATALYTIC DECOMPOSITION OF HYDROGEN PEROXIDE Introduction: For this internal assessment, I will be looking into the catalytic decomposition of hydrogen peroxide. I will be conducting an experiment to show how the mass of catalyst used affects the rate at which hydrogen peroxide decomposes. My research question will be, "How does the mass of Manganese (V) Oxide used affect the rate of decomposition of hydrogen peroxide. A catalyst is a substance which alters the rate of a chemical reaction but is chemically unchanged at the end of the reaction. This means that the amount of catalyst at the start and the end of the reaction will remain constant. Catalysts alter the rate of a chemical reaction and this is the reason why I have chosen to conduct an experiment to find out to what extent a catalyst could affect the rate of a chemical reaction. Aim: To investigate how the mass of catalyst (manganese (V) oxide) used affects the speed of decomposition of hydrogen peroxide, H2O2. Hypothesis: My hypothesis for this experiment is that the more catalyst I used, the faster the decomposition of the hydrogen peroxide. ...read more.

Middle

oxide: 0.05/3 X 100% = 1.67% 2/46.7 X 100% = 4.3% 1.67% + 4.3% =5.97% = 6% Calculating the uncertainties for the test using 4 g of manganese (V) oxide: 0.05/4 X 100% = 1.25% 2/43.30 X 100% = 4.6% 1.25% + 4.6% = 5.85% = 5.9% Calculating the uncertainties for the test using 5 g of manganese (V) oxide: 0.05/5 X 100% = 1% 2/42.6 X 100% = 4.7% 1% + 4.7% = 5.7% Table of Uncertainties: Experiment with Variable (g) % Uncertainties for mass of manganese (V) oxide % Uncertainties for Time taken for the reaction to stop Total Uncertainties (Found by adding the two uncertainties) 1 5% 3.3% 8.3% 2 2.5% 3.9% 6.4% 3 1.67% 4.3% 6% 4 1.25% 4.6% 5.9% 5 1% 4.7% 5.7% Graph of Mass of Manganese (V) Oxide used against Average Time Taken for Reaction to Stop(S): Comments on the Graph: As predicted in my hypothesis, the more manganese (V) oxide I used, the faster the rate of decomposition of the hydrogen peroxide. There would also be a point where anymore catalyst added would not make the reaction go any faster and this happens between 4-5 grams of catalyst added and onwards. ...read more.

Conclusion

oxide used. Another flaw is that I used very little hydrogen peroxide (10cm3). This is a very minute amount and this could also mean that the amount of catalyst needed to totally overcome activation energy is much less and thus I cannot conclude exactly what value of catalyst would cause a peak in the rate of decomposition. A solution for this is to use various values of hydrogen peroxide and find the average mass of catalyst added relative to volume of hydrogen peroxide used and I could then come up with a general equation to find the peak decomposition rates for specific volumes. The last flaw in my experiment, which I briefly mentioned in the conclusion, is the fact that I did not use enough variables for the mass of manganese (V) oxide. I only used values of 1 gram - 5 grams and thus I could not ascertain results for variables where the mass is more than that of 5 grams. Had I expanded my range from 5 grams to 10 grams, I could have a wider range of data and even might be able to achieve the perfectly straight line in my graph which I hypothesised. Thus, I can say that this experiment has many flaws and could be improved in many areas especially in data recording. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Chemistry essays

  1. To determine the concentration of hydrogen peroxide

    x 26.6(obtained from titration) = 0.5301mol. * Now we know that 0.5301 mol is 2/5th of the number of mols of H2O2. Therefore 0.5301/2= 0.26505mol 0.26505 x 5= 1.325mol Therefore, number of mols of H2O2 = 1.325 mol. * But we have to divide this by the actual amount of H2O2 used.

  2. Investigate the rate of reaction of luminol in various factors. The objective was to ...

    2, Using rate-concentration graph Draw tangent for each of the concentration in the concentration-time graph, the gradient of each tangent represents the initial rate of each concentration. Plot rate- concentration graph, i.e. rate in the x-axis and concentration in the y-axis.

  1. Aim: Using an iodine clock reaction to find the order of hydrogen peroxide and ...

    An example of this is while changing the volume of H2O2; the other reactants' volumes should be held constant, and H2O2's volume should increase by 2 for each experiment. 10. Carry out the experiment with volumes of 2, 4, 6, 8 and 10 for both H2O2 and CH2COOH while keeping

  2. DCP+CE Analysis of a Hydrogen Peroxide Solution

    � (0.0750ml + 0.0750ml) = 21.50ml �0.15ml Table 5. Processed Data - Average of 1st Titration and 2nd Titration Titration Calculation Average 1st Amount of moles in each trials for Na2C2O4 Average amount of Na2C2O4 is 0.0024mols � 1.44x10-8mols 2nd As amount in H2O2 was constant, this does not matter

  1. To determine the standard enthalpy of formation of Magnesium Oxide using Hess Law.

    Once 120 seconds were up, the calorimeter was emptied and cleaned so that another trial of the same experiment could be performed. Part Y ? With Magnesium Oxide Powder 1. The calorimeter was cleaned thoroughly ensuring no chemicals were left behind which would hinder in the reaction in part Y.

  2. Reaction Rate

    Furthermore, particles of higher concentration H2SO4 collided and reacted with the magnesium particles to produce magnesium sulfate and hydrogen gas at a quicker rate. The heat developed in the test tube also varied significantly with more concentrated solutions radiating warmer temperatures due to higher levels of friction and energy in the reaction.

  1. Investigating Factors that Affect the Rate of Reaction of the Decomposition of Hydrogen ...

    Distilled water - 15 mL A matt/cover that is fire resistant 700 mL of room temperature water from a sink A one-hole rubber stopper with stem Two test tube holders Two 10 mL graduated cylinders A bunsen burner Two solid rubber stopper Plastic tubing containing two Luer-lock connectors A one-hole

  2. The chemistry of atmospheric and water pollution.

    and then into the river, there would be increased levels of calcium, magnesium, sulfate and chloride ions. If the river was situated near a limestone deposit or cave, the water would leach the limestone and thus there would be a high concentration of calcium and carbonate ions.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work