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Decomposition of Hydrogen Peroxide

Categories Composition

Essay, Pages 4 (972 words)



Essay, Pages 4 (972 words)

Abstract: This lab was designed to observe the decomposition of hydrogen peroxide into it oxygen and water. The equation for this reaction is H202 H2O + +1/2O2 thus by measuring volume and pressure of O2 generated the amount of O2 generated can be calculated which in turn can be utilized to determine the concentration of water already in the H2O2 solution. The results determined that 3.02% of the solution is composed of H2O2.


This purpose of this experiment was to determine the concentration of liquid water within a given amount of hydrogen peroxide.

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Because this reaction happens so slowly the addition of the catalyst KI was used. The important things of note within this lab are how the gas was collected, the balanced equation, the partial gas law and the ideal gas law. The gas was collected via a full graduated cylinder inverted within a pool of water. This water was then utilized to displace a volume of oxygen generated by a reaction within a sealed Erlenmeyer flask.

The flask emptied into a tube which in turn emptied into the full inverted graduated cylinder.

Essentially as gas bubbles form from the reaction they force the water to move aside and the amount of force they generate as well as the volume they displace can then be measured using the cylinders’s calibrated measurements. The Ionic compound Potassium Iodine acts as a catalyst because it disassociated within the solution to K+ and I- ions. The iodine then acts to attract the oxygen molecules away from the hydrogen peroxide transforming it into water.

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However as more oxygen atoms are freed from water they attract each other strongly creating diatomic oxygen and freeing the Iodine ion to continue catalyzing further reactions. Creating a two step reaction:

H202 (l) + I-(aq) H2O (l) + +OI-(aq)
H202 (l) + OI-(aq) H2O (l) + O2 (g) + I-

Notice that the Iodine is conserved in the reaction and can continue to bond with more oxygen molecules. The partial gas law is a very simple explanation that the total pressure of a given system is a function of the sums of its parts. For example in a gaseous mixture of water and oxygen if oxygen has a pressure of 300 torr and water has a pressure of 300 torr then the total pressure of the system is 600 torr. Pressures are additive to the total pressure which can be measured similarly by knowing the pressures involved a partial amount can be derived. The combined gas law is a law which allows us to calculate the pressure*volume = amount*Gas Constant*temperature. Or PV=nRT. By utilizing this equation with 3 variables known any of the other variables can be derived.



The data measurements at first do not give obvious linkages to the amount of oxygen produced. In order to use the stoichiometric relationships seen in this balanced equation: H202 H2O + +1/2O2 The amount of oxygen first must be derived. To do this the partial pressure of the oxygen is needed. This is calculated by determining the partial pressure of water at the given temperature and subtracting this from the total ambient pressure. Because this was not experimentally derived it was obtained from the lab manual. (Atkinson 2012)

Because the total pressure is a sum of the partial pressures of the system and the total pressure was recorded barometrically as 768.9mmHg. Torr is equivalent to mmHg, Thus the partial pressure of water at 20.8 degrees Celsius is closest to 21 degrees Celssius. Although mathematically a more exact relationship might be derived because there’s not notation of the measurements and no way for one to determine the number of significant figures using math to derive a more exact value is folly.

Thus 768.9 Torr – 18.6 Torr = 750.3 Torr
Thus 750.3 Torr is the Ppartial of the O2 gas.

From this relationship the ideal gas law can then be applied with PV=nRT utilizing the values in table 1.1 P= 750.3 Torr; V=73ml; n=X; r= .082057
L*atm/mol*K; T=20.8 degrees Celsius. First the volumes must be converted to liters and Torr must be converted to atmospheres. 750.3torr/760torr =.987 atm. Similarly 73ml/1000ml = .073liters .987 atm*.073liters/.082057 (liters atm/mols kelvin)* 293.95k) = n Simplify and cancel units:

.987*.073/.082057 (1/mols)* 293.95) = n
n=.002995 mols
n=.00300 mols

Since there are 2 mols of H2O2 per 1 mol of O2 this means that there are .00600 mols of H2O2 has a molar mass of 34.02g. Thus .00600mol * 34.02g/mol = .20412g or .204g of H2O2 Of the total mass used initially of 5.33g approximately .204g were composed o hydrogen peroxide. Dividing .204 by 5.33 multiplied by 100 will give us the percentage concentration of hydrogen peroxide. Thus the concentration of hydrogen peroxide was roughly 3.02% which seems quite accurate compared to the normal concentration of commercial hydrogen peroxide which was used.


The experimental data was actually surprisingly accurate. With 3.02% concentration versus the labels claimed 3% is very close differing only 2% overall. This is a fantastic result and speaks of the thoroughness in the preparation of the lab as well as the accuracy involved in the mathematical calculations derived by previous scienctists.

One source of error in the calculations could have been the opening of the hydrogen peroxide and exposing it to the Air. The ambient decay of hydrogen peroxide to oxygen and water might give a more accurate idea of the remaining concentration from the stock solution. Beyond that a small air bubble was trapped at the top of the graduated cylinder. However this amount of air was smaller than a tenth of the nearest measurement making it hard to identify the appropriate loss of volume with the inversion. This easily could have contributed to the 2% error of the experiment.


Atkinson (2012) Chemistry 228 Lab Manual. Portland State University.

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Decomposition of Hydrogen Peroxide. (2017, Feb 09). Retrieved from https://studymoose.com/decomposition-of-hydrogen-peroxide-essay

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