Rates of Chemical Reactions

Experimental Procedure

1. We prepared for the lab by obtaining all the necessary materials in the equipment list.

2. With the funnels, we poured 50mL of the following reactants: Acetone, HCL, and Iodine; in their respective burette.

3. In the Erlenmeyer flask, we carefully poured 10mL of Acetone from the burette, 10mL of HCL, and 20mL of distilled water. Once we had those three in the Erlenmeyer flask we poured 10mL of iodine and started the timing as soon as the iodine touched the three solutions. We swirled the reactants and stopped as soon as the yellowish liquid became clear.

4. As soon as the reaction stopped, we stopped the stopwatch and then we carefully poured it in a test tube and measured the temperature. We performed this procedure one more time.

5. We performed four different mixtures and in each mixture was a total of 50 mL (three solutions of 10mL and one solution of 20mL) :

1. The first mixture was 20mL of H2O and 10mL of the other three.
2. Second mixture was 20mL of Acetone and 10mL of the other three.
3. Third mixture was 20mL of HCL and 10mL of the other three.
4. Fourth mixture was 20mL of I2 and 10mL of the other three.

6. For each mixtures we performed two runs.

Reaction Rate Data

Determination of Reaction Orders with Respect to Acetone, H+, and I2

Determination of the Rate Constant k

Equations

1. Rate = k(acetone)m(I2)n(H+)p
2. Rate = - ∆(I2) 

                          ∆t

CALCULATIONS

• Rate=k(acetone)m(I2)n(H+)p

• Rate II=k(1.6M)m(0.0010M)n(0.20M)p
• Rate I=k(0.80M)m(0.0010M)n(0.20M)p

• Rate II = k(1.6M)m(0.0010M)n(0.20M)p

Rate I     k(0.80M)m(0.0010M)n(0.20M)p

            = 2Mm

• Rate II = 1.03 x 10-5

Rate I      4.68 x 10-6

                              =2.2m

• m= Log2.2m/ Log2m

   = 1.14

• Rate III=k(0.80M)m(0.0010M)n(0.40M)p

• Rate III = k(0.80M)m(0.0010M)n(0.40M)p

Rate I     k(0.80M)m(0.0010M)n(0.20M)p

            = 2Mp

• Rate III = 1.11 x 10-5

Rate I      4.68 x 10-6

                              =2.4p

• p= Log2.4p/ Log2p

   = 1.24

• Rate IV=k(0.80M)m(0.0020M)n(0.20M)p

• Rate IV = k(0.80M)m(0.0020M)n(0.20M)p

Rate I      k(0.80M)m(0.0010M)n(0.20M)p

            = 2Mn

• Rate IV = 5.33 x 10-6

Rate I      4.68 x 10-6

                              =1.13n

• n= Log1.13n/ Log2n

   = 0.18

• Rate I: k= 4.68 x 10-6 /(0.80M)1.14(0.0010M)0.18(0.20M)1.24

             = 4.68 x 10-6 /.0285627343

             = 1.64 x 10-4

• Rate II: k= 1.03 x 10-5 /(1.6M)1.14(0.0010M)0.18(0.20M)1.24

               = 1.03 x 10-5 /.0629468461

               = 1.64 x 10-4

• Rate III: k= 1.11 x 10-5 /(0.80M)1.14(0.0010M)0.18(0.40M)1.24

               = 1.11 x 10-5 /.0674647592

               = 1.65 x 10-4

• Rate IV: k= 5.33 x 10-6 /(0.80M)1.14(0.0020M)0.18(0.20M)1.24

               = 5.33 x 10-6 /.0325607537

               = 1.64 x 10-4

CONCLUSION

In the iodination of acetone, we found the order of the reaction with respect to acetone and HCl and find a value for the rate constant, k.  Since the concentrations of acetone and HCl were much higher than that of I2, the concentrations of acetone and HCl changed very little.  Thus the rate was determined by the time needed for iodine to be used up. Since iodine has color we can easily follow changes in iodine concentration visually.  The equation, rate  =  k(A)m(H+)n(I2)p,  can be simplified to rate  =  k[I2]/t  since the values for acetone and HCl essentially remain constant during the course of any run. In mixtures I & IV the constant rate were similar, even though the iodine doubled in mixture IV, it is still the limited reagent. Mixtures II &III were similar do to the doubling of acetone in mixture II and HCL in mixture III.