pHYSICAL cHEMISTRY - Iv

Experiment 3

Aim of the Experiment

To investigate the order of reaction between KBrO3( Potassium bromate) & KI (Potassium iodide) [a = b ]

Principle

The “clock reaction” is a reaction famous for its dramatic colorless-to-blue color change, and is often used in chemistry courses to explore the rate at which reactions take place. The color change occurs when I2 reacts with starch to form a dark blue iodine/starch complex. The ability to record the time at which the blue complex appears allows the rate of reaction to be determined accurately with a stopwatch.

In this experiment, the rate law for a reaction is determined using the method of initial rates. The effect of concentration on the rate of this reaction is determined by measuring the initial reaction rate at several reactant concentrations.

The Clock Reaction

The primary reaction to be studied is the oxidation of I- by Bro3- (bromate) in aqueous solution:

6HCl + 6KI + KBrO3 → 3H2O + 3I2 + 6KCl + KBr (slow, rate determining) Equation 1

This reaction will be run in the presence of a known amount of S2O3 2- (thiosulfate), which reacts very rapidly with I2. As long as S2O3 2- is present, I2 is consumed by S2O3 2- as fast as it is formed. This competing reaction prevents the I2 produced from our reaction of interest from reacting with starch, so no color change is observed until the thiosulfate is completely used up. The "clock" reaction is the reaction of a very small amount of S2O3 2- (thiosulfate) with the I2 produced in the primary reaction:

I2(aq) + 2S2O3 2-(aq) ----> 2I- (aq) + S4O6 2-(aq) (fast) Equation 2

The “clock” reaction will signal when the primary reaction forms a specific amount of I2. The amount of I2 formed before the color change can be calculated from the known amount of S2O3 2- added using the molar ratio in Equation 2.

Overall reaction:

KBr03 + HCl ---> HBr03 + KCl

KI + HCl -----> HI + KCl

HBr03 + HI -----> I2 + HBr + H20

Materials required

0.1 N KBrO3, 0.1 N KI , 0.1 N HCl,0.01 N Na2S2O3 (Sodium thiosulphate),Starch solution (indicator)

Procedure

1.Fill 0.01 N Sodium thiosulphate in Burette.Remove air bubble from Burette tio pressing pinch-cork.Again fill 0.01 N Sodium thiosulphate in Burette adjusting at '0' mark.

2.Make a water bath with water.

3.For Part-1

Take 25 mL 0.1 N Potassium bromate in stopper Bottle-1 and take 25 mL o.1 N KI + 100 mL 0.1 N HCl + 100 mL water in Bottle-2

4. Place both bottles in water bath for 10 minutes to attain same temperature of both solution.

5.Mix the content of Bottle-2 in Bottle -1

6.Wait for 5 minutes .Gradually the colour of solution appeared dark.

7. Suck 10 mL of this solution with pipette and transfer it to conical flask.Few pieces of ice are added to stop the reaction. Add few drops of starch is added to the solution.The colour of solution will become blue then titrate against 0.01 N Sodium thiosulphate.Here the end point is blue to colurless.

8.Reading is noted in observation table.Find out reading by titration running for 10,15,20 and 25 minutes by using stop watch.

9.Calculate a-x , 1/ (a-x) and K1 .and determine average K value.also calculate a, b and x1 as shown in calculation part.

10.For Part-2Part-2Part-2Part-2

11.Take 25 mL diluted Potassium bromate (25 mL Potassium bromate diluted to50 mL with D.W.) in stopper Bottle-1 and take diluted KI ( 25 mL 0.1 N KI diluted to 50 mL with D.W.) + 100 mL 0.1 N HCl + 100 mL Distilled water in Bottle-2 and follow the same procedure as Part-1. Readings are noted in observation table.

12.Find out reading by titration running for 10,15,20 and 25 minutes by using stop watch.

9.Calculate a-x , 1/ (a-x) and K2 .and determine average K value.also calculate a, b and x2 as shown in calculation part.

10. Plot the graph-1 x against t and graph-2 x against t. Put the value of x1 from part-1 will give t1 and x2 from part-2 will give t2. Take ratio of t2 and t1.

This will give reaction order.

11. Plot the graph of 1 / (a-x) against t for Part-1 and 1 / (a-x) against t for Part-2. Take a slope which will give you K1 and K2.

Observation table

Calculations

Part -1Calculations of a,b and X1 :

1.N1V1 = N2V2

0.1 * 25 = N2 * 250

N2 = 0.01 N

now,

N1V1 = N2V2

0.01 N* 25 =0.01 N * V2

V2= 25 mL = a mL

2.

N1V1 = N2V2

0.1 * 25 = N2 * 250

N2 = 0.01 N

now,

N1V1 = N2V2

0.01 N* 25 =0.01 N * V2

V2= 25 mL = b mL

3.X1 =b / Convinient no.

= mL

Part -2 Calculations of a,b and X1 :

1.N1V1 = N2V2

0.05* 25 = N2 * 250

N2 = 0.005 N

now,

N1V1 = N2V2

0.005 N* 25 =0.01 N * V2

V2= 12.5 mL = a mL

2.

N1V1 = N2V2

0.05 * 25 = N2 * 250

N2 = 0.005 N

now,

N1V1 = N2V2

0.005 N* 25 =0.01 N * V2

V2= 12.5 mL = b mL

3.X2 =b / Convinient no.

= mL

Result

1. From graph , t2 /t1 = _________ indicates reaction is ______________ order.

2. The values of K1 = ___________ & K2 = _____________ from calculation.

3. The values of K1 = __________ & K2 = _____________ from plot (graph).

Questions

1.What is order of reaction?

2.Why can't molecularity of any reaction be equal to zero?

Reference Material

Practical Physical Chemistry Google EBOOK Link.