(no subject)
Anam Mendha
Period 6 1/18/06
Decomposition of Hydrogen Peroxide
I. Data Table
Part Reactants Temperature ْC Initial Rate (kPa/s)
I 4 mL 3.0 % H2O2 + 1 mL .5 M KI 20.80 ± .01 .064096
II 4 mL 3.0 % H2O2 + 1 mL .25 M KI 20.91 ± .01 .01365
III 4 mL 1.5 % H2O2 + 1 mL .5 M KI 20.91 ± .01 .02272
IV 4 mL 3.0 % H2O2 + 1 mL .5 M KI 20.91 ± .01 .025575
II. Establishing the Rate Law Expression
Since the initial concentration of H2O2 in Trials I and II is the same, the rate for those two trials will be used to set up an equation which will allow for the calculation of the order of the reaction with respect to KI.
Rate I = k[H2O2]x[KI]y
Rate II k[H2O2]x[KI]y
Since the values of k and H2O2 are the same in both the numerator and the denominator, they undo each other, simplifying the expression to:
Rate I = [KI]y
Rate II [KI]y
This can be substituted with the values:
.0641 = 0.5y
.0137 .25y 4.696=2y
Log2(4.696) = 2.231, the value of y, which is the order of the reaction with respect to KI.
Since the initial concentration of KI is the same in Part I and Part III, the rate for those two parts will be used to set up the rate equation, which will allow for the calculation of the order of the reaction with respect to H2O2.
Rate I = k[H2O2]x[KI]y
Rate III k[H2O2]x[KI]y
Since the values of k and KI are the same in both the numerator and the denominator, they undo each other, simplifying the expression to:
Rate I = [H2O2]x
Rate III [H2O2]x
This can be substituted with the values:
.0641 = 0.88x
.0227 0.44x 2.821=2x x= Log2(2.821) = 1.496
To determine the value of k, the specific rate constant, the results of Part III will be used to set up the following equation:
Rate III = k[H2O2]1.496[KI]4.696
0.0227 = k[.44]1.496[.50]4.696
K = .364
Therefore, the expression is:
Rate = 0.364[H2O2]1.496[KI] 4.696
III. Presenting the results of Part III.
Time (seconds ± 1) Pressure (kPa ± 0.001)
0 104.503
10 106.321
20 106.321
30 106.094
40 106.321
50 106.321
60 106.548
70 106.775
80 107.002
90 107.230
100 107.457
110 107.684
120 107.911
130 108.138
140 108.365
150 108.820
160 109.274
170 109.501
180 109.956
Taking the 60 second interval from 80 seconds to 140 seconds, the CBL calculated the initial rate of the reaction to be r = .02272t, where t=0 is the starting point of the reaction, and the rate is in kPa per second.
Period 6 1/18/06
Decomposition of Hydrogen Peroxide
I. Data Table
Part Reactants Temperature ْC Initial Rate (kPa/s)
I 4 mL 3.0 % H2O2 + 1 mL .5 M KI 20.80 ± .01 .064096
II 4 mL 3.0 % H2O2 + 1 mL .25 M KI 20.91 ± .01 .01365
III 4 mL 1.5 % H2O2 + 1 mL .5 M KI 20.91 ± .01 .02272
IV 4 mL 3.0 % H2O2 + 1 mL .5 M KI 20.91 ± .01 .025575
II. Establishing the Rate Law Expression
Since the initial concentration of H2O2 in Trials I and II is the same, the rate for those two trials will be used to set up an equation which will allow for the calculation of the order of the reaction with respect to KI.
Rate I = k[H2O2]x[KI]y
Rate II k[H2O2]x[KI]y
Since the values of k and H2O2 are the same in both the numerator and the denominator, they undo each other, simplifying the expression to:
Rate I = [KI]y
Rate II [KI]y
This can be substituted with the values:
.0641 = 0.5y
.0137 .25y 4.696=2y
Log2(4.696) = 2.231, the value of y, which is the order of the reaction with respect to KI.
Since the initial concentration of KI is the same in Part I and Part III, the rate for those two parts will be used to set up the rate equation, which will allow for the calculation of the order of the reaction with respect to H2O2.
Rate I = k[H2O2]x[KI]y
Rate III k[H2O2]x[KI]y
Since the values of k and KI are the same in both the numerator and the denominator, they undo each other, simplifying the expression to:
Rate I = [H2O2]x
Rate III [H2O2]x
This can be substituted with the values:
.0641 = 0.88x
.0227 0.44x 2.821=2x x= Log2(2.821) = 1.496
To determine the value of k, the specific rate constant, the results of Part III will be used to set up the following equation:
Rate III = k[H2O2]1.496[KI]4.696
0.0227 = k[.44]1.496[.50]4.696
K = .364
Therefore, the expression is:
Rate = 0.364[H2O2]1.496[KI] 4.696
III. Presenting the results of Part III.
Time (seconds ± 1) Pressure (kPa ± 0.001)
0 104.503
10 106.321
20 106.321
30 106.094
40 106.321
50 106.321
60 106.548
70 106.775
80 107.002
90 107.230
100 107.457
110 107.684
120 107.911
130 108.138
140 108.365
150 108.820
160 109.274
170 109.501
180 109.956
Taking the 60 second interval from 80 seconds to 140 seconds, the CBL calculated the initial rate of the reaction to be r = .02272t, where t=0 is the starting point of the reaction, and the rate is in kPa per second.