Exercise 1: Determining the Rate Laws of the Reaction between HCl and Na 2 S 2 O
ID: 878097 • Letter: E
Question
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
27.44
27.40
27.42
C2, D2
6
6
8
1M
0.3M
31.66
31.68
31.67
C3, D3
4
8
8
1M
0.3M
46.90
46.98
46.94
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
19.96
19.64
19.8
C5, D5
8
6
6
1M
0.3M
48.69
48.63
48.66
C6, D6
8
8
4
1M
0.3M
134.62
134.65
134.64
need help with both charts that are empty
Questions
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. From table 1
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
27.44
27.40
27.42
C2, D2
6
6
8
1M
0.3M
31.66
31.68
31.67
C3, D3
4
8
8
1M
0.3M
46.90
46.98
46.94
Explanation / Answer
Solution :-
We are asked to calculate the order with respect to the HCl
We know the drops of HCl used and the molairty of the HCl
So lets first calculate the concentration of the HCl after mixing ( initial concentration of HCl =1.0 M)
In the first trial 12 drops of HCl are mixed with 8 drop of Na2S2O3
So the final concentration of the HCl trial 1 = 12 drop * 1M /20 drop =0.60 M
In the second trial 6 drops of HCl are mixed with 8 drop of Na2S2O3 and 6 drop water
For trial 2 concentration of HCl = 6 drop * 1 M / 20 drop = 0.30 M
In the trial 3 4 drops of HCl are mixed with 8 drop of Na2S2O3 and 8 drops of water
For trial 3 concentration of HCl = 4 drop * 1 M / 20 drop = 0.20 M
Now lets calculate the rate for each trial
Rate = 1/time
Tiral 1 rate =1/27.42 = 0.0365 s-1
Trial 2 rate = 1/ 31.67 = 0.0316 s-1
Trial 3 rate = 1/46.94 = 0.0213 s-1
Now lets calculate the order with respect to HCl
Select the experiment 1 and 2 and find order with respect to HCl
rate 2/ rate 1 = ([HCl]2/[HCl]1)m
lets put the values in the formula
0.0316 s-1 /0.0365 s-1 = (0.30 M /0.60M)m
0.866 = (0.5)m
m= log 0.866 / log 0.5
m = 0.2
so we can round it to zero
therefore order with respect to the HCl is zero order.
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