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H2O2 vs. KIkinetics Practical

This experiment investigates the kinetics of the reaction between hydrogen peroxide and hydroiodic acid. Two sets of experiments are performed where the concentrations of hydrogen peroxide and potassium iodide are varied. The rate of reaction is measured by titrating samples taken at various time intervals against a sodium thiosulfate solution. Graphs of concentration versus time and the integrated rate law are plotted and used to determine the order of the reaction and rate constants. The results show the reaction is second order with rate constants of k1 and k2 for the first and second sets respectively.

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Mehul Khimani
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1K views3 pages

H2O2 vs. KIkinetics Practical

This experiment investigates the kinetics of the reaction between hydrogen peroxide and hydroiodic acid. Two sets of experiments are performed where the concentrations of hydrogen peroxide and potassium iodide are varied. The rate of reaction is measured by titrating samples taken at various time intervals against a sodium thiosulfate solution. Graphs of concentration versus time and the integrated rate law are plotted and used to determine the order of the reaction and rate constants. The results show the reaction is second order with rate constants of k1 and k2 for the first and second sets respectively.

Uploaded by

Mehul Khimani
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Aim: To investigate the kinetics of the reaction between Hydrogen peroxide and hydroiodic acid.

Requirements: 0.1 N H2O2, 0.1 N KI, 0.1 N H2SO4 , 0.002 N Na2S2O3 and starch as indicator.

Procedure: Perform two experiments as follows

Set 1. Take 10 ml 0.1 N H2O2 in one reagent bottle and in another reagent bottle 10 ml of 0.1 N KI
+ 40 ml 0.1 N H2SO4 + 50 ml 0.1 N distilled water.

Set 2. Take 5 ml 0.1 N H2O2 in one reagent bottle and in another reagent bottle 5 ml of 0.1 N KI +
40 ml 0.1 N H2SO4 + 50 ml 0.1 N distilled water.

Take 10 ml 0.1 N H2O2 in one reagent bottle and 10 ml 0.1 N KI + 40 ml 0.1 N H2SO4 + 40 ml water
in another reagent bottle. Keep them in water bath for about 10 minutes to attain constant
temperature. After that mix both the solutions, note this time and shake it well. After 10 minutes
withdraw 10 ml of the reaction mixture in a conical flask containing crushed ice. Titrate this
reaction mixture against 0.01 N Na2S2O3 solution using starch as indicator. The titration is carried
out at 10,20,30,40 and 50 minutes from the time of mixing.

The second set is also done in the same way.

Reaction:

Rate of reaction = K [I − ]P [H2 O2 ]q [H + ]r


In reaction concentration of H+ ion is constant

∴ Rate of reaction = K[I − ]P [H2 O2 ]q

Record your results in tables as follows:


1st Set 2nd Set
a=50 ml a=25 ml
b=50 ml b=25 ml

Time Titration a-x 1/(a-x) K1 Time Titration a-x 1/(a-x) K1


‘t’ reading ‘t’ reading
min (x ml) min (x ml)
10 10
20 20
30 30
40 40
50 50
60 60

Calculation of ‘a’ and ‘b’ in first set

1st set: Total volume=100 ml

Let the initial concentration of H2O2 is ‘a’ and that of KI is ‘b’

10 ml 0.1 N H2O2 in diluted to 100 ml


Normality of H2O2 in the reacting mixture is
N 1 V 1 = N1 V 2
0.1 N x 10 ml = NH2 O2 x 100
NH2 O2 = 0.01 N

Now 10 ml reacting mixture is taken for titration against 0.002 N Na2S2O3


NH2 O2 X VH2 O2 = NNa2 S2 O3 x VNa2 S2 O3
VNa2 S2 O3 = 0.01 x 10 / 0.002 = 50 ml = a

This is the concentration of H2O2 in terms of 0.002 N Na2S2O3. Similarly, concentration of KI in


100 ml solution is [KI] = b = 50 ml in terms of 0.002 N Na2S2O3

Calculation of ‘a’ and ‘b’ in 2nd set

5 ml 0.1 N H2O2 in diluted to 100 ml

Thus, Normality of H2O2 in the reacting mixture is 0.1 N x 5 ml = NH2 O2 x 100


NH2 O2 = 0.005 N
Now 10 ml reacting mixture is taken for titration against 0.002 N Na2S2O3
NH2 O2 X VH2 O2 = NNa2 S2 O3 x VNa2 S2 O3
0.005 x 10 = 0.002 x VNa2 S2 O3
VNa2 S2 O3 = 25 ml = a

This is the concentration of H2O2 in terms of 0.002 N Na2S2O3. Similarly, the concentration of KI
in 100 ml solution is [KI] = b= 25 ml

Characteristics of second order reactions


1. Draw a graph of x vs t for both the sets. It will be smooth curve.

2. In order to determine the order of reaction (n) Divide the initial concentration of both sets
by a convenient number (say 10)

x1 = a/10 = 50/10 = 5

and corresponding time is t1


Similarly, in 2nd set

x2 = a/10= 25/10 = 2.5

and corresponding time is t2

n = t2/t1

It should be two in this case.

3. Plot 1/(a-x) vs t. It will be straight line in both sets.

The slopes will give values of k1 and k2

Results:
1. The graph of x vs t is found to be smooth curve for both the sets.
2. The value of n=t2/t1 is found to be 2 for both the sets. Hence reaction is second order.
3. The mean value of k1 and k2 obtained from calculations are k1= and k2=
4. The graph of 1/a-x vs t is found to be straight line for both the sets indicating second
order reaction.
5. The values of k1 and k2 from 1/a-x vs t plot are: k1= and k2=

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