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Lab Overview

Investigation #20:
"The Kinetics of the Decomposition
of Hydrogen Peroxide"


In this experiment, you will determine that rate of the reaction of the decomposition of hydrogen peroxide. The balanced equation for this decomposition is:

2H2O2 Þ 2H2O + O2

Normally, this is a slow reaction, but we will be using a catalyst to speed up the reaction. A catalyst assists in the rate of a reaction without being used up in the reaction. In other words, the catalyst begins and ends in the same form. Somewhere along the way however, it forms intermediate steps with the reactants to help increase the decomposition of the hydrogen peroxide. The rate can now be written as an equation of the form:

rate = k[H2O2]n[I-]m

In the beginning of this experiment we will first find the rate of this reaction. Later we will compare this rate with rates from two other reactions each with different concentrations of reactants to find n and m.

To find the rate, you will measure the volume of O2 produced in the decomposition per time (minutes). Once you have done this, you should have a chart that is something like this:
 

Time
Volume Reading (mL)
Vol. Coll.
Vol. Corr. for H2O
2:26:35
3.2
0
0
2:26:55
5.4
2.2
  
2:27:15
7.6
4.4 (this number is found by subtracting the first volume, 3.2, from the volume just read, 7.6)
 

 

 Note that in the volume collected column, the volume collected is calculated as the current volume reading - the first volume reading.

To find the volume corrected for H2O, you need to first find the volume that is taken up by the water vapor. To find this volume you will use:

Vw = V[Pw/Pb]

where V is the volume collected, Pw is the partial pressure of water found in Investigation 13, and Pb is the barometric pressure.

Once you have Vw, you can find VO2. VO2 will be equal to the Vtotal (collected) - Vw.

VO2 = Vtotal (collected) - Vw

 Now that you have the VO2 you can graph your data. It theoretically should look something like the following:

Draw a tangent line through the (0,0) the origin approximating the slope of the first few points which should be near linear. Since this line goes through the origin, the slope can easily be found.

slope = rate = m = (y2 - y1) / (x2 - x1) = (y2 - 0) / (x2 - 0) = y2/x2

Now you will have the rate of the reaction.

Repeat this procedure for the other two reactions. Note that in each reaction the concentrations of each reaction is varied. To calculate these concentrations use the relationship:

M1V1 = M2V2

Remember that the total volume in the first one is 20 mL of KI + 20 mL H2O + 10 mL of H2O2 = 50 mL total. So for example, the concentration of KI in the first reaction is:

(0.15 mol/L)(20 mL) = M2(50 mL)

Now you must find n and m in the relationship:

rate = k[H2O2]n[I-]m

To do this you will compare the rates of two different reactions while holding either [H2O2] or [I-] constant. So if we compare the two and divide the one by the other we will get:

R = rate = k[H2O2]n[I-]m and R' = rate = k[H2O2']n[I-]m holding [I-] the same.

Divide R and R' to and get:

R/R' = [H2O2]n/[H2O2']n

Rearranging this equation to solve for n we get:

n = {log (R/R')} / {log {[H2O2]n/[H2O2']n}}

We can repeat this procedure holding the concentrations of hydrogen peroxide the same to find m in exactly the same way.