An Investigation of the Order of the reaction of Iodine with Propanone

6. Relevant Equations/Chemical Reactions Involved :

(1) Reaction between propanone and iodine :

   (2) Reaction between iodine and thiosulphate(VI) ion (titration) :

                  I2(aq) + 2S2O32-(aq) → S4O62-(aq) + 2I-(aq)

7.  Chemicals :

    0.02 M iodine solution (Iodine dissolved in potassium iodide solution)

    1 M propanone solution

    1 M sulphuric acid

    0.5 M sodium hydrogencarbonate solution

    0.01 M sodium thiosulphate(VI) solution

    Starch solution

8.  Apparatus and equipment :

9. Procedure :

1.  25.0 cm3 of sulphuric acid and V1 cm3 of propanone solution and V2 cm3 of distilled water were measured by measuring cylinder into a conical flask.

2. 50.0 cm3 OF 0.02 M iodine solution was added by measuring cylinder to the propanone mixture as quickly as possible. Stop watch was started at the same time. And the content was mixed.
3. During the 5 minutes interval , 10.0 cm3 of the 0.5 M sodium hydrogencarbonate solution was measured and then added into another conical flask .
4. After 5 minutes, 10.0 cm3 of the reaction mixture was pipetted into the conical flask prepared in step 4. When it was done, the time should be noted.
5. The solution in step 4 was mixed and then titrated with 0.01 M sodium thiosulphate(VI) solution. Starch solution was used as indicator.
6. After 10, 15, 20, 25 and 30 minutes, 10 cm3 portions of the reaction mixtures in step 2 were withdrawn and the above procedure was then carried out each time.
7. The results were recorded.

10. Observations :

The solution changed from black to pale yellow at the end point.

11. Data, Calculation and Results :

Results of group A :  

12. Conclusion :

It was found that the order of the reaction with respect to iodine was zero.    

13. Discussion :

1.  [ The required graph please refer to Appendix 1.]

2.  a)  The function of the sodium hydrogencarbonate in step 3 is quenching.

   b)  The iodine concentration in the reaction fell throughout the experiment as iodine was consumed by propanone.

   c)  The slope of the straight line in the graph is the rate of equation.

   d)  The iodine concentration changed at a uniform rate throughout the experiment as the slope of the graph is constant.

   e)  The rate of iodine concentration is independent on the iodine concentration because the reaction is zero order with respect to iodine.

       f)  The order of reaction with respect to iodine is zero, i.e. n = 0

       g)  Iodine does not take part in the rate determining step of this reaction.

       h)  The intercept of the graph with the y-axis is the initial concentration of iodine.

       i)  The concentration of the sodium thiosulphate(VI) solution is 0.01M.

   3.  

        [ The required graph please refer to Appendix 2. ]

1. The gradient of the graph in Discussion 1 is directly proportional to the rate of reaction and volume of propanone solution is also directly proportional to its concentration.
   Since a straight line is plotted in the graph above, rate of reaction is directly proportional to the concentration of propanone. Thus, the order of the reaction with respect to propanone is 1.  

     b)  Rate equation :

Rate=k[CH3COCH3]a[I2]b[H+]c

Since b = 0 and [H+] = constant

So, rate = k[CH3COCH3]1