Notes For Lab Experiment 3 On Chemical Kinetics Spring 2 2018 Sample Reaction Ti
ID: 303968 • Letter: N
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Notes For Lab Experiment 3 On Chemical Kinetics Spring 2 2018 Sample Reaction Times Table 1 Determinations (s) Table 2 Determinations (s) Table 3 Determinations (s) 1. 32 2. 68 7obo 5. 40 6. 76 7. 164 Ice Bath 800 Room Temperature 80 Heat Bath 20 3. 126 4. 248 8. 308 Students should do the following calculations for this lab: 1. Calculate the reactant concentrations of S,032, I, and S,O32 in the reaction mixtures for each of the determinations 2. Determine the order of reaction with respect to I and S,O and the rate law for the reaction 3. Calculate the rate constant for each determination 1 8 and take the overall average 4. Calculate the reaction rate for each determinationExplanation / Answer
[I-]
1. Calculate the reactant concentrations of S2O32-, I- and S2O82- for each determination.
[S2O32-] forDeterminations 1-4:
Det 1: Flask A total volume = 7.0 ml
We know that initial concentration of [S2O32-] is [S2O32-]i = 1.2x10-2 M
Initial volume of S2O32- = Initial volume of Na2S2O3,Vi = 2.0 ml
Final volume = Total volume of Flask A, Vf = 7.0 ml
Using the formula, [S2O32-]ixVi =[S2O32-]fxVf
1.2x10-2 M X 2.0 ml = [S2O32-]f x 7.0 ml
[S2O32-]f = (1.2x10-2 M x 2.0 ml)/7.0 ml = 3.4x10-3 M
The volume and concentration did not change for determinations 2-4, so the [S2O32-] for 2-3 would be same as 1.
[S2O32-]forDeterminations 5-8:
We know that initial concentration of [S2O32-] is [S2O32-]i = 1.2x10-2 M
Initial volume of S2O32- = Initial volume of Na2S2O3,Vi = 2.0 ml
Using the formula, [S2O32-]ixVi =[S2O32-]fxVf
Final volume = Total volume of Flask A, Vf = 11.0 ml
1.2x10-2 M X 2.0 ml = [S2O32-]f x 11.0 ml
[S2O32-]f = (1.2x10-2 M x 2.0 ml)/11.0 ml = 2.2x10-3 M
Since the volume and concentration did not change for determinations 6-8, the [S2O32-] would be same as 5.
[S2O32-]forDeterminations 9-12:
We know that initial concentration of [S2O32-] is [S2O32-]i = 1.2x10-2 M
Initial volume of S2O32- = Initial volume of Na2S2O3,Vi = 2.0 ml
Using the formula, [S2O32-]ixVi =[S2O32-]fxVf
Final volume = Total volume of Flask A, Vf = 11.0 ml
1.2x10-2 M X 2.0 ml = [S2O32-]f x 11.0 ml
[S2O32-]f = (1.2x10-2 M x 2.0 ml)/11.0 ml = 2.2x10-3 M
Since the volume and concentration did not change for determinations 9-12, the [S2O32-] would be same as 8.
[I-] forDeterminations 1-4:
Det 1: Flask A total volume = 7.0 ml
We know that initial concentration of [I-] is [I-]i = 0.20M
Initial volume of I- = Initial volume of KI,Vi = 4.0 ml
Final volume = Total volume of Flask A, Vf = 7.0 ml
Using the formula, [I-]ixVi =[I-]fxVf
0.20 M X 4.0 ml = [I-]f x 7.0 ml
[I-]f = (0.20 M X 4.0 ml)/7.0 ml = 0.11M
The volume and concentration of [I-] did not change for determinations 2-4, so the [I-]f for 2-3 would be same as 1.
[I-] forDeterminations 5-8:
Det 5: Flask A total volume = 11.0 ml
We know that initial concentration of [I-] is [I-]i = 0.20M
Initial volume of I- = Initial volume of KI,Vi = 8.0 ml
Final volume = Total volume of Flask A, Vf = 11.0 ml
Using the formula, [I-]ixVi =[I-]fxVf
0.20 M X 8.0 ml = [I-]f x 11.0 ml
[I-]f = (0.20 M X 8.0 ml)/11.0 ml = 0.15M
Det 6: Flask A total volume = 11.0 ml
We know that initial concentration of [I-] is [I-]i = 0.20M
Initial volume of I- = Initial volume of KI,Vi = 4.0 ml
Final volume = Total volume of Flask A, Vf = 11.0 ml
Using the formula, [I-]ixVi =[I-]fxVf
0.20 M X 4.0 ml = [I-]f x 11.0 ml
[I-]f = (0.20 M X 4.0 ml)/11.0 ml = 0.073M
Det 7: Flask A total volume = 11.0 ml
We know that initial concentration of [I-] is [I-]i = 0.20M
Initial volume of I- = Initial volume of KI,Vi = 2.0 ml
Final volume = Total volume of Flask A, Vf = 11.0 ml
Using the formula, [I-]ixVi =[I-]fxVf
0.20 M X 2.0 ml = [I-]f x 11.0 ml
[I-]f = (0.20 M X 2.0 ml)/11.0 ml = 0.036M
Det 8: Flask A total volume = 11.0 ml
We know that initial concentration of [I-] is [I-]i = 0.20M
Initial volume of I- = Initial volume of KI,Vi = 1.0 ml
Final volume = Total volume of Flask A, Vf = 11.0 ml
Using the formula, [I-]ixVi =[I-]fxVf
0.20 M X 1.0 ml = [I-]f x 11.0 ml
[I-]f = (0.20 M X 1.0 ml)/11.0 ml = 0.018M
[I-] forDeterminations 9-12:
Det 9 : Flask A total volume = 11.0 ml
We know that initial concentration of [I-] is [I-]i = 0.20M
Initial volume of I- = Initial volume of KI,Vi = 4.0 ml
Final volume = Total volume of Flask A, Vf = 11.0 ml
Using the formula, [I-]ixVi =[I-]fxVf
0.20 M X 4.0 ml = [I-]f x 11.0 ml
[I-]f = (0.20 M X 4.0 ml)/11.0 ml = 0.072M
The volume and concentration of [I-] did not change for determinations 10-12, so the [I-]f would be same as 9.
[S2O82-] forDeterminations 1-4:
Det 1: Flask B total volume = 12.0 ml
We know that initial concentration of [S2O82-] is [S2O82-]i = 0.20M
Initial volume of [S2O82-]= Initial volume of Na2S2O8,Vi = 8.0 ml
Final volume = Total volume of Flask B, Vf = 12.0 ml
Using the formula,[S2O82-]ixVi =[S2O82-]fxVf
0.20M X 8.0 ml = [S2O82-]f x 12.0 ml
[S2O82-]f = (0.20M x 8.0 ml)/12.0 ml = 0.13M
Det 2: Flask B total volume = 12.0 ml
We know that initial concentration of [S2O82-] is [S2O82-]i = 0.20M
Initial volume of [S2O82-]= Initial volume of Na2S2O8,Vi = 4.0 ml
Final volume = Total volume of Flask B, Vf = 12.0 ml
Using the formula,[S2O82-]ixVi =[S2O82-]fxVf
0.20M X 4.0 ml = [S2O82-]f x 12.0 ml
[S2O82-]f = (0.20M x 4.0 ml)/12.0 ml = 0.067M
Det 3: Flask B total volume = 12.0 ml
We know that initial concentration of [S2O82-] is [S2O82-]i = 0.20M
Initial volume of [S2O82-]= Initial volume of Na2S2O8,Vi = 2.0 ml
Final volume = Total volume of Flask B, Vf = 12.0 ml
Using the formula,[S2O82-]ixVi =[S2O82-]fxVf
0.20M X 2.0 ml = [S2O82-]f x 12.0 ml
[S2O82-]f = (0.20M x 2.0 ml)/12.0 ml = 0.033M
Det 4: Flask B total volume = 12.0 ml
We know that initial concentration of [S2O82-] is [S2O82-]i = 0.20M
Initial volume of [S2O82-]= Initial volume of Na2S2O8,Vi = 1.0 ml
Final volume = Total volume of Flask B, Vf = 12.0 ml
Using the formula,[S2O82-]ixVi =[S2O82-]fxVf
0.20M X 1.0 ml = [S2O82-]f x 12.0 ml
[S2O82-]f = (0.20M x 1.0 ml)/12.0 ml = 0.017M
[S2O82-] forDeterminations 5-8:
Det 5: Flask B total volume = 8.0 ml
We know that initial concentration of [S2O82-] is [S2O82-]i = 0.20M
Initial volume of [S2O82-]= Initial volume of Na2S2O8,Vi = 4.0 ml
Final volume = Total volume of Flask B, Vf = 8.0 ml
Using the formula,[S2O82-]ixVi =[S2O82-]fxVf
0.20M X 4.0 ml = [S2O82-]f x 8.0 ml
[S2O82-]f = (0.20M x 4.0 ml)/8.0 ml = 0.10M
Since the volume and concentration did not change, the final concentrations of det.5-8 would remain the same.
[S2O82-] forDeterminations 9-12:
Det 9: Flask B total volume = 8.0 ml
We know that initial concentration of [S2O82-] is [S2O82-]i = 0.20M
Initial volume of [S2O82-]= Initial volume of Na2S2O8,Vi = 2.0 ml
Final volume = Total volume of Flask B, Vf = 8.0 ml
Using the formula,[S2O82-]ixVi =[S2O82-]fxVf
0.20M X 2.0 ml = [S2O82-]f x 8.0 ml
[S2O82-]f = (0.20M x 2.0 ml)/8.0 ml = 0.05M
Since the volume and concentration did not change, the final concentrations of det.10-12 would remain the same.
Flask A Flask A FlaskA FlaskB Flask B Flask B Conc1 Conc2 Conc3 Det. Starch 12x10-2MNa2S2O3 0.20 M KI 0.20M KNO3 0.20M (NH4)2S2O8 0.20M (NH4)2SO4 [S2O32-](M) [S2O82-][I-]
1 1.0 ml 2.0 ml 4.0 ml 4.0 ml 8.0 ml 0.0 ml 3.4x10-3 0.13M 0.11M 2 1.0 ml 2.0 ml 4.0 ml 4.0 ml 4.0 ml 4.0 ml 3.4x10-3 0.067M 0.11M 3 1.0 ml 2.0 ml 4.0 ml 4.0 ml 2.0 ml 6.0 ml 3.4x10-3 0.033M 0.11M 4 1.0 ml 2.0 ml 4.0 ml 4.0 ml 1.0 ml 7.0 ml 3.4x10-3 0.017M 0.11M 5 1.0 ml 2.0 ml 8.0 ml 0.0 ml 4.0 ml 4.0 ml 2.2x10-3 0.10M 0.15M 6 1.0 ml 2.0 ml 4.0 ml 4.0 ml 4.0 ml 4.0 ml 2.2x10-3 0.10M 0.073M 7 1.0 ml 2.0 ml 2.0 ml 6.0 ml 4.0 ml 4.0 ml 2.2x10-3 0.10M 0.036M 8 1.0 ml 2.0 ml 1.0 ml 7.0 ml 4.0 ml 4.0 ml 2.2x10-3 0.10M 0.018M 9 1.0 ml 2.0 ml 4.0 ml 4.0 ml 2.0 ml 6.0 ml 2.2x10-3 0.05M 0.072M 10 1.0 ml 2.0 ml 4.0 ml 4.0 ml 2.0 ml 6.0 ml 2.2x10-3 0.05M 0.072M 11 1.0 ml 2.0 ml 4.0 ml 4.0 ml 2.0 ml 6.0 ml 2.2x10-3 0.05M 0.072M 12 1.0 ml 2.0 ml 4.0 ml 4.0 ml 2.0 ml 6.0 ml 2.2x10-3 0.05M 0.072MRelated Questions
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