Calculations for determining the rate law experiment 24

Exercise 1: Determining the Rate Laws of the Reaction between HCl and Na2S2O3

Data Table 1. Varying Concentrations of 1.0 M HCl.

Concentrations

# of drops

# of drops

# of drops

Stock Solution

Stock Solution

Reaction (after mixing in well)

Reaction (after mixing in well)

Reaction Time

(seconds)

Reaction

Rate

(sec-1)

Well #

HCl

H2O

Na2S2O3

HCl

Na2S2O3

HCl

Na2S2O3

Trial 1

Trial 2

Average

C1, D1

12

0

8

1M

0.3M

0.60

0.12

24

25

24.5

0.04

C2, D2

6

6

8

1M

0.3M

0.3

0.12

58

60

59

0.017

C3, D3

4

8

8

1M

0.3M

0.2

0.12

96

107

101.5

0.0099

INSERT A PICTURE OF YOUR WELL PLATE WITH SOME OF THE WELLS COMPLETE. INCLUDE A PIECE OF PAPER WITH YOUR NAME AND THE DATE ON IT.

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Data Table 2. Varying Concentrations of 0.3 M Na2S2O3.

Concentrations

# of drops

# of drops

# of drops

Stock Solution

Stock Solution

Reaction (after mixing in well)

Reaction (after mixing in well)

Reaction Time

(seconds)

Reaction

Rate

(sec-1)

Well #

HCl

H2O

Na2S2O3

HCl

Na2S2O3

HCl

Na2S2O3

Trial 1

Trial 2

Average

C4, D4

8

0

12

1M

0.3M

0.4

0.18

32

29

30.5

0.033

C5, D5

8

6

6

1M

0.3M

0.4

0.09

40

46

43

0.023

C6, D6

8

8

4

1M

0.3M

0.4

0.06

33

31

32

0.03

Questions

A. Determine the Reaction Order for HCl using calculations described in the Background section. Show your work. Note that your answer will probably not be an even whole number as it is in the examples, so round to the nearest whole number.

Rate of Experiment 1 = (k [HCl]m[Na2S2O3]n ) of Experiment 1

Rate of Experiment 2 (k [HCl]m[Na2S2O3]n ) of Experiment 2

0.04 = k [0.6]m[Na2S2O3]n

0.0017 = k [0.3]m[Na2S2O3]n

Constants would be cancelled on both sides:

2.35 = [2]m

ln [2.35] = ln [2m]

ln [2.35] = m* ln [2]

m = ln [2.35]

ln [2]

m= 1.23 = 1

B. Determine the Reaction Order for Na2S2O3 using calculations described in the Background section. Show your work. Note that your answer will probably not be an even whole number as it is in the examples.

Rate of Experiment 1 = (k [HCl]m[Na2S2O3]n ) of Experiment 1

Rate of Experiment 2 (k [HCl]m[Na2S2O3]n ) of Experiment 2

0.033 = k [HCl]m[0.18]n

0.023 = k [HCl]m[0.09]n

Constants would be cancelled on both sides:

0.143 = [0.5]n

ln [0.143] = ln [0.5n]

ln [0.143] = n* ln [0.5]

n = ln [0.143]

ln [2]

n= 2.8 = 3

C. Write the rate law for the reaction between HCl and Na2S2O3.

Rate = k [A]m[B]n

Rate = k [HCl][Na2S2O3]3

D. Using the rate law that you determined for the reaction in (C), predict the rate of the reaction if the concentration of HCl was changed to 0.8M and the concentration of Na2S2O3 is 0.12M.

First, the value of k would be found using known values:

Rate = k [HCl][Na2S2O3]3

0.04 = k [0.60][0.12]3

k = 0.04/0.001

k = 40

Using the value of k, rate can be calculated as;

Rate = k [HCl][Na2S2O3]3

Rate = 40 [0.8][0.12]3

Rate = 2.76 m/s

E. Using the following rate law, and the experimental values given, calculate k:

Experiment

[F2] (M)

[ClO2] (M)

Initial rate (M/s)

1

0.5

0.5

0.3 M/s

2

0.8

0.8

0.768 M/s

3

0.5

0.8

0.48 M/s

Rate = k [F2] [CLO2]

k = Rate / [F2] [CLO2]

Using data from first set of experiment;

k = 0.3 / [0.5] [0.5]

k = 1.2

22

Rate = k[F][ClO]