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
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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]
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