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Analysing; The Reaction of Hydrogen Peroxide and Iodide ions

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Introduction

Skill 3: Analysing; The Reaction of Hydrogen Peroxide and Iodide ions

Table of Results

Experiment

Volume of KI used (dm3)

Volume of water used (dm3)

Concentration of I- ions

Time taken for colour to appear (S)

A

5.0 x 10-3

0.020

0.02

9:56

B

0.010

0.015

0.04

5:36

C

0.015

0.010

0.06

4:07

D

0.020

5.0 x 10-3

0.08

3:27

E

0.025

0.000

0.1

2:40

We calculate the concentration of I- ions by following these stages (taking the hypothetical value of experiment A)

Number of moles of KI contained         = volume x concentration

                                        = 5.0 x 10-3 x 0.1

= 5.0 x 10-4 moles

                                        = Number of I- ions contained in the solution

Concentration of solution                = number of mols

                                        Volume of solvent

                                        = 5.0 x 10-4 / (25/1000)

                                        = 0.02 mol/dm3

Thus I have calculated the other values, which appear on the table

Analysis

Derived from the rate equation

Log (1/time) = n log (volume of KI) + constant

We can calculate the rate

...read more.

Middle

-2.53

C

-1.82

247

-2.39

D

-1.70

207

-2.31

E

-1.60

160

-2.20

Thus, we could derive the gradient of the graph, by drawing it, as seen upon the nest page.  


We could also use the statistical method of finding the equation of the regression line, which is:

Y = a +bx (where y and x are the axis's)

B =         (xI - x)(yI -y)

(xI - x)2

a = y - Bx

Where  y = average of all y values

x = average of all x values

xI = x value

yI = y value

Therefore, when an equation is formed, we can mathematically derive the gradient, as the gradient of the graph, mathematically is the constant in front of x, providing y has no constants.  

Thus B =         (xI - x)(yI -y)

(xI - x)2

        = 0.796

        = 0.80

therefore, the order of reaction, with respect to the iodide ions is 0.80.

Maximum error of reading an apparatus        = maximum error x 100

                                                       Amount read

Maximum error of 5 cm3 pipette        = 0.1/ 5 x 100

                                        = 2%

Maximum error of 10 cm3 pipette        = 0.1/ 10 x 100

                                        = 1%

Maximum error of burette         = 0.15/25 x 100

                                = 6.0 x 10-3 %

Maximum error of measuring cylinder        = 0.5/25 x 100

                                                = 2%

...read more.

Conclusion

There could have also been substances that were left in the beaker or pipette, thus the actual experiment that was performed did not include these leftovers.  A way of avoiding this is to ensure that they would be "washed" and when transferring solution from a pipette, to ensure all the solution has come out, it would also be suggested to touch the surface of the solution with the pipette.

One way in which should not have affected the reaction, but probably did, is our concentration span, while waiting for the solution to turn colour, we became increasingly distracted by the surroundings, and hence, at times may have missed the actual time of which the solution changed colour, thus could have affected the results, even though in a very minor way.

        Other ways of improving this experiment would include the use of a pipette instead of a measuring cylinder; even a burette would increase accuracy of the reaction, as a measuring cylinder is quite inaccurate.  Another way of improving the reaction is by using a higher concentration of the solutions.

...read more.

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