Chemistry lab reports
CHM1046 Lab - REPORT # 3
Last name, First name: Student ID:
1) Date of the experiment:
2) Experiment #:
3) Experiment Title:
4) Purpose of the Experiment –
5) Background/Brief Introduction–
6) Experimental Procedure: (Passive Mode)
7) Data:
Rates of Chemical Reactions
Solution 1. Initial [S2O82-] = 0.05 M; initial [I-] = 0.05 M. Sample calculation of [I-]:
(25mL)(0.2M)/ l00mL = 0.05M (100 mL is the total final volume)
|
Aliquot |
T, sec |
Moles S2O82- consumed |
[S2O82-]* |
log [S2O82-]* |
|
1 (already in sol.) 2 (first one added) 3 4 5 6 7
|
|
2 x 10-4 4 x 10-4 6 x 10-4 8 x 10-4 10 x 10-4 12 x 10-4 14 x 10-4
|
|
|
*For "straight line test" of order (do at instructor's option)
Solution 2. Initial [S2O82-] = 0.10 M; initial [I-] = 0.05 M.
|
Aliquot |
T, sec |
Moles S2O82- consumed |
|
1 2 3 4 5 6 7
|
|
2 x 10-4 4 x 10-4 6 x 10-4 8 x 10-4 10 x 10-4 12 x 10-4 14 x 10-4
|
Solution 3. Initial [S2O82-] = 0.05 M; initial [I-] = 0.10 M.
|
Aliquot |
T, sec |
Moles S2O82- consumed |
|
1 2 3 4 5 6 7
|
|
2 x 10-4 4 x 10-4 6 x 10-4 8 x 10-4 10 x 10-4 12 x 10-4 14 x 10-4
|
Solution 4. Initial [S2O82-] = 0.05 M; initial [I-] = 0.05 M.
|
Aliquot |
T, sec |
Moles S2O82- consumed |
|
1 2 2 4 5 6 7
|
|
2 x 10-4 4 x 10-4 6 x 10-4 8 x 10-4 10 x 10-4 12 x 10-4 14 x 10-4 |
Solve for x and y by the Method of Initial Rates (Exponent Test)
|
Solution |
Initial [S2O82-] |
Initial [I-] |
Rate* |
|
1 2 3 4 |
|
|
|
*The "Rate" above= the slope of the line in your graph divided by the final total volume (0.1 L)
8) Experimental Results:
1. Calculate "'y'' (consider solutions 1 and 3, round off to the nearest whole number or zero).
2. Calculate "x".
3. Write a "rate expression" which is consistent with your data.
4. Compute some values of k from your data.
5. Write a two-step mechanism that is consistent with your data. Label 'the slow step and the fast step.
9) Results and Discussion:
10) Conclusions: