Decomposition of Hydrogen Peroxide LabThis is a featured page

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Sample information. . .
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Introduction:
As hydrogen peroxide decomposes very slowly, an enzyme or another type of catalyst is used to speed up the reaction. In this experiment, potassium iodide (KI) was used as the catalyst. The decomposition of hydrogen peroxide produces water and oxygen gas, as the following formula illustrates:
2H2O2(aq)--> 2H2O(l)+ O2(g)
Materials:
  • Zinc and hydrochloric acid
    In the lab, zinc granules react fairly slowly with dilute hydrochloric acid, but much faster if the acid is concentrated.
    Decomposition of Hydrogen Peroxide Lab - Science is a Verb!Decomposition of Hydrogen Peroxide Lab - Science is a Verb!
    The catalytic decomposition of hydrogen peroxide
    Solid manganese(IV) oxide is often used as a catalyst in this reaction. Oxygen is given off much faster if the hydrogen peroxide is concentrated than if it is dilute.
    Decomposition of Hydrogen Peroxide Lab - Science is a Verb!Decomposition of Hydrogen Peroxide Lab - Science is a Verb!
    The reaction between sodium thiosulphate solution and hydrochloric acid
    This is a reaction which is often used to explore the relationship between concentration and rate of reaction in introductory courses (like GCSE). When a dilute acid is added to sodium thiosulphate solution, a pale yellow precipitate of sulphur is formed.
    Decomposition of Hydrogen Peroxide Lab - Science is a Verb!Decomposition of Hydrogen Peroxide Lab - Science is a Verb!
    As the sodium thiosulphate solution is diluted more and more, the precipitate takes longer and longer to form


Diagram:

Decomposition of Hydrogen Peroxide Lab - Science Learning Resources

Figure 1: Diagram of the lab set up


Sample calcultion:
Two formulas are needed to calculate the molarity of a 3% mass/volume solution of H2O2. This calculation is computed by the following:
number of moles (mol) = mass(g) / molar mass (gmol^-1)
= 3g / 34.02 gmol^-1
= 0.08818 mol
molarity (M) = number of moles (mol)/ volume (L)
= 0.08818 mol / 0.1 L
= 0.8818 M
The molarity of the solution is 0.8818M.



Data Collection:




Pressure (kPa)
Time (s)3% H2O2, 0.5M KI, 20ºC3% H2O2, 0.25M KI, 20ºC1.5% H2O2, 0.5M KI, 20ºC3% H2O2, 0.5M KI, 30ºC
0110.045110.670110.386110.386
10110.329110.897110.613110.613
20111.407111.180110.783110.783
30112.202111.464111.010111.067
40113.961111.634111.237111.634
50114.359111.748111.578113.224
60114.756111.918111.918114.359
70114.983112.088112.486115.323
80115.096112.486112.713116.175
90115.210112.713113.280117.196
100115.494112.997113.848118.445
110116.175113.224114.359119.864
120116.458113.564115.210121.566
130116.685113.848116.231123.496
140116.856114.132117.650125.425
150117.083114.472119.693127.355
160117.310115.777121.623129.341
170117.650116.345123.779131.271
180118.104116.799125.709133.257
Table: Data collected during the experiment


Decomposition of Hydrogen Peroxide Lab - Science Learning Resources
Figure 2: A graph illustrating how pressure varied when hydrogen peroxide decomposes


Decomposition of Hydrogen Peroxide Lab - Science Learning Resources
Figure 3: A graph depicting the slopes of the reactions

PartReactantsTemperature (ºC)Initial rate (kPa/s)
I4ml 3.0% H2O2 + 1ml 0.5M KI24.90.409
II4ml 3.0% H2O2 + 1ml 0.25M KI24.90.312
III4ml 1.5% H2O2 + 1ml 0.5M KI24.90.766
IV4ml 3.0% H2O2 + 1ml 0.5M KI30.01.300
Table 2: A table of the rates of reactions
Conversion:
At this point in the experiment, pressure has been measured in kPa. However, the rate of the reaction must be expressed in molarity/second and therefore, the data collected must be converted to molarity. In order to do this, the formula for molarity and the ideal gas law are combined to produce a formula to calculate the molarity of a gas.

Ideal gas law:Pressure(atm) x volume(L) = number of moles(mol) x 0.08206 x temperature(K)
Formula for molarity:molarity(M) = number of moles(mol) / volume(L)


PV = nRT
Volume = (#moles x 0.08206 x temperature)/ pressure
molarity = #moles/ [(#moles x 0.08206 x temperature)/pressure]
= (pressure x #moles) / (#moles x 0.08206 x temperature)
= pressure / (0.08206 x temperature)

Therefore:Molarity(M) = Pressure(atm) / (0.08206 x Temperature(K) )

Take the first measurement of pressure in kPa as an example. This measured 0.409 kPa at 24.9 degrees Celsius. In order to convert the kPa measurement into molarity, the above formula is employed, as well another another formula, which is shown below.
101.325kPa = 1.000atm
0.409kPa = 0.00404 atm

Molarity = 0.00404atm / (0.08206 x (24.9 + 273)K)
= 0.000165 M


Data Analysis:
It is clear, in table 2, that the decomposition of hydrogen peroxide is most rapid out of all the solutions tested. While solution 4 had a rate of 1.30kPa/s, the rest of the solutions did not exceed 0.766kPa/s. This was most probably because the higher temperature accelerated the rate of the reaction. Take solution 1 as the comparative solution. When the amount of hydrogen peroxide was lowered and potassium iodide kept the same in solution 3, the rate of reaction increased. This is because the same amount of potassium iodide only has to work on accelerating the decomposition of half as much hydrogen peroxide and therefore less time is needed. When the amount of potassium iodide was decreased, on the other hand, the reaction slowed down because the smaller amount of the catalyst had to work on twice as much hydrogen peroxide.
The rate law (a.k.a. rate expression) consists of the following formula:
Rate of reaction = R = k [A]^m x [B]^n
This law can only be fully determined by experimental means, as the order of reaction can only be found out by actually carrying out the experiment.

Orders of the reaction:
As stated above, the rate of reaction can only be determined by experimental means. The following chart helps make this possible.

Part
Initial rate (mol/L-s)
[H2O2] after mixing
[KI] after mixing
i
0.00325
0.706
0.100
ii
0.00247
0.706
0.050
iii
0.00619
0.353
0.100
iv
0.00908
0.706
0.100



In order to calculate concentrations after mixing, the following formulas have been used:
#moles = mass (g) / molar mass (g/mol)
concentration = #moles / volume (L)

Take for example, the concentration of H2O2 in part I.
initial concentration = 0.882M = #moles / 0.004L
#moles of H2O2 = 0.00353mol
final concentration = 0.00353mol / 0.005L
= 0.706M


To determine the orders of the reaction, the concentrations of each solution is compared to the initial rate. Take parts 1 and 2 for example. In part 1, the concentration of KI is twice that of part 1. As a result, the initial rate changes by about 1.5, meaning that the order of this concentration is 1/2. Parts 2 and 4 are not compared because the temperatures at which these reactions occur are different. When parts 1 and 3 are compared, it is clear that the concentration of H2O2 is doubled in part 1. However, the initial rate in part 1 is only half that of part 3. Therefore, the order of H202 must be -1. The rate constant can be determined by plugging in the data above.
0.00325 = k(0.706)^-1 (0.1)^1/2
k = 0.00726

Therefore, the rate of reaction of this experiment at room temperature is the following:
Rate = 0.00726 [H202]^-1 [KI]^1/2
Conclusion:
The hypothesis mentioned is reinforced using figure 3 and table 2 as references for the rates of reaction. The reaction rates changed with varying amounts of hydrogen peroxide, potassium iodide and temperature as was expected. As the amount of hydrogen peroxide decreased, the reaction rate increased and as the amount of potassium iodide decreased, the reaction rate decreased. Using 30 degrees Celsius water also increased the reaction rate as was predicted.
When heat is added to a reaction, the molecules involved in the reaction are given more energy and therefore move about the solution quicker. This then increases the chances of two different molecules colliding with each other and reacting, therefore increasing the reaction rate.

Evaluation:
Several changes could have been made during this experiment in order to attain more dependable results. First of all, I made the mistake of swirling the test-tube for a longer time period than was suggested. This swirling would have given the molecules of the reactants more kinetic energy and the reaction would take place quicker as a result. This affects the precision of the data as the test-tube should have been left stationary in the beaker.

The temperature of the water in the beaker was not as close to 20 degrees Celsius as it could have been. As figure 3 shows, temperature can cause a huge effect on the reaction rate and the imprecision of the water temperature would have altered the data collected. Had more time been taken to adjust the water temperature, the different results may have shown a clearer pattern.

It is due to mankind's inability to measure outexactly4ml or 2ml of each solution. Results may have been slightly altered by the limitations of the human eye, although this is not a change one can make in the near future. More accuracy in the measurements of the solutions, however, would have produced more accurate results.



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