What is the calculation of the predicted rate for experiment #4? Time Data for K
ID: 1037831 • Letter: W
Question
What is the calculation of the predicted rate for experiment #4?
Time Data for Kinetic Reactions (10 points)
Experiment
Temperature
Trial #1 time
Trial #2 time
Average time
#1
25oC
6:06:26 min
5:51:70 min
5:78:98 min
#2
25oC
3:12:12 min
3:06:09 min
3:09:10 min
#3
25oC
2:53:68 min
2:51:48 min
2:52:58 min
#4
25oC
4:35:06 min
5:04:12 min
4:69:59 min
Low Temp.
17.5oC
11:59:76 min
12:58:00 min
12:08:88 min
High Temp.
37.5oC
2:48:08 min
2:46:32 min
2:47:20 min
Catalyst
25oC
0:38:62 min
Experimental Rates and Determination of the Rate Law (25 points)
Experiment
[I-]initial
[S2O82-]initial
Avg. ?t
Rate (M/s)
#1
0.031 M
0.015 M
5:78:98 min
6.6 x 10-6 M/s
#2
0.031 M
0.031 M
3:09:10 min
1.3 x 10-5 M/s
#3
0.062 M
0.015 M
2:52:58 min
1.4 x 10-5 M/s
#4
0.023 M
0.023 M
4:69:59 min
8.1 x 10-6 M/s
Low Temp.
0.031 M
0.015 M
12:08:88 min
3.4 x 10-6 M/s
High Temp.
0.031 M
0.015 M
2:47:20 min
1.5 x 10-5 M/s
Catalyst
0.031 M
0.015 M
0:38:62 min
6.6 x 10-5 M/s
Determine the specific form of the rate equation. (10 points)
Rate = k [KI]x [(NH4)2S2O8]y. You must solve for x and y.
Value for x =
1
Value for y =
1
Specific form of the rate law:
k [KI] [(NH4)2S2O8]
Calculation of the rate constants (k) (10 points)
Experiment
Value for k (use correct units!)
#1 (room temp)
3.3 x 10-4 s-1
#2 (room temp)
6.5 x 10-4 s-1
#3 (room temp)
7.0 x 10-4 s-1
Average (1-3)
5.6 x 10-4 s-1
#1 (low temp)
1.7 x 10-4 s-1
#1 (high temp)
7.5 x 10-4 s-1
#1 (catalyst)
3.3 x 10-3 s-1
Calculation of the predicted rate for experiment #4. Show or explain your work! (10 points)
(You may scan and attach a handwritten version.)
Predicted value for the experiment #4 rate:
Actual value for the experiment #4 rate:
4.1 x 10-4 s-1
Please help
Thank you
Experiment
Temperature
Trial #1 time
Trial #2 time
Average time
#1
25oC
6:06:26 min
5:51:70 min
5:78:98 min
#2
25oC
3:12:12 min
3:06:09 min
3:09:10 min
#3
25oC
2:53:68 min
2:51:48 min
2:52:58 min
#4
25oC
4:35:06 min
5:04:12 min
4:69:59 min
Low Temp.
17.5oC
11:59:76 min
12:58:00 min
12:08:88 min
High Temp.
37.5oC
2:48:08 min
2:46:32 min
2:47:20 min
Catalyst
25oC
0:38:62 min
Explanation / Answer
#1
0.031 M
0.015 M
5:78:98 min
6.6 x 10-6 M/s
#2
0.031 M
0.031 M
3:09:10 min
1.3 x 10-5 M/s
#3
0.062 M
0.015 M
2:52:58 min
1.4 x 10-5 M/s
If we see the data of experiment 1, 2 and three...there is the initial concentration of both the reactants are .031 and .015M and compare these values with second experiemnt where first reactant concentration is same but the second one is double but time is half so the reaction rate = 1.3 x 10-5 M/s......now compare the values of experiemtnt 1 and 3..........here first reactant concentration is double but the second one remains same and the time is almost same as the second experiment also the rate is almost equal to the second experiement.
With this description we see that the rate is depending on concentrations of both the reactant and you also have shown the value of x = 1 and y = 1 so why the rate constants unit has been used s-1. because this unit is of first order reaction and the given reaction and the data it should be more than 1 or should be 2. may be second order or third order. all calculations and units should be checked from strating again.
#1
0.031 M
0.015 M
5:78:98 min
6.6 x 10-6 M/s
#2
0.031 M
0.031 M
3:09:10 min
1.3 x 10-5 M/s
#3
0.062 M
0.015 M
2:52:58 min
1.4 x 10-5 M/s
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