CJ Artates, Tyler Lange, Xavier Cornelius
Part I: Preparation of a standard absorption curve for FeSCN+2
| Standard |
0.20M Fe(NO3)3 |
0.0020 M KSCN |
0.100M HNO3 |
[FeSCN+2] |
Absorbance |
| A
|
10.0 mL |
0.0 mL |
15.0 mL |
0.0 M |
0.00 |
| B
|
10.0 mL |
1.0 mL |
14.0 mL |
8.0 x 10^-5 |
0.39 |
| C
|
10.0 mL |
1.5 mL |
13.5 mL |
1.2 x 10^-4 |
0.50 |
| D
|
10.0 mL |
2.0 mL |
13.0 mL |
1.6 x 10^-4 |
0.56 |
| E
|
10.0 mL |
2.5 mL |
12.5 mL |
2.0 x 10^-4 |
0.79 |
| F
|
10.0 mL |
3.0 mL |
12.0 mL |
2.4 x 10^-4 |
1.03 |
EQUATION: y = 4017.9x + 0.0093 R2 = 0.974
Part 2: Measuring Equilibrium
| Test Solution |
0.0020 M Fe(NO3)3 |
0.0020 M
KSCN |
0.10 M
HNO3 |
Initial [Fe+3] |
Initial [SCN–] |
Absorbance |
Equilibrium
[FeSCN+2]* |
| I
|
5.0 mL |
0 |
5.0 mL |
0.001 |
0 |
0.00 |
|
| II
|
5.0 mL |
1.0 mL |
4.0 mL |
0.001 |
0.0002 |
0.16 |
3.6 x 10^-5 |
| III
|
5.0 mL |
2.0 mL |
3.0 mL |
0.001 |
0.0004 |
0.34 |
8.2 x 10^-5 |
| IV
|
5.0 mL |
3.0 mL |
2.0 mL |
0.001 |
0.0006 |
0.52 |
1.3 x 10^-4 |
| V
|
5.0 mL |
4.0 mL |
1.0 mL |
0.001 |
0.0008 |
0.91 |
2.2 x 10^-4 |
| VI
|
5.0 mL |
5.0 mL |
0.0 mL |
0.001 |
0.001 |
1.04 |
2.6 x 10^-4 |
* To be determined from the standard graph equation.
ANALYSIS:
Use your graph equation to calculate the equilibrium concentrations of FeSCN+2.
- Prepare and ICE chart for each test solution (II – VI) and calculate the value of Keq for each of your 5 tests solutions.
ICE CHARTS
| Test Solution
Keq = 2.30 x 10^2 |
Fe3+ + SCN– ⇄ FeSCN2+ |
| I
|
0.001 |
0.0002 |
0 |
| C
|
– 3.6 x 10^-5 |
– 3.6 x 10^-5 |
+ 3.6 x 10^-5 |
| E
|
9.64 x 10^-4 |
1.64 x 10^-4 |
3.64 x 10^-5 |
| Test Solution
Keq = 2.81 x 10^2 |
Fe3+ + SCN– ⇄ FeSCN2+ |
| I
|
0.001 |
0.0004 |
0 |
| C
|
-8.2 x 10^-5 |
-8.2 x 10^-5 |
+ 8.2 x 10^-5 |
| E
|
9.18 x 10^-4 |
3.18 x 10^-4 |
8.2 x 10^-5 |
| Test Solution
Keq = 3.18 x 10^2 |
Fe3+ + SCN– ⇄ FeSCN2+ |
| I
|
0.001 |
0.0006 |
0 |
| C
|
-1.3 x 10^-4 |
-1.3 x 10^-4 |
+1.3 x 10^-4 |
| E
|
8.70 x 10^-4 |
4.70 x 10^-4 |
1.3 x 10^-4 |
| Test Solution
Keq = 4.86 x 10^2 |
Fe3+ + SCN– ⇄ FeSCN2+ |
| I
|
0.001 |
0.0008 |
0 |
| C
|
-2.2 x 10^-4 |
-2.2 x 10^-4 |
+ 2.2 x 10^-4 |
| E
|
7.80 x 10^-4 |
5.80 x 10^-4 |
2.2 x 10^-4 |
| Test Solution
Keq = 4.75 x 10^2 |
Fe3+ + SCN– ⇄ FeSCN2+ |
| I
|
0.001 |
0.001 |
0 |
| C
|
-2.6 x 10^-4 |
-2.6 x 10^-4 |
+ 2.6 x 10^-4 |
| E
|
7.40 x 10^-4 |
7.40 x 10^-4 |
2.6 x 10^-4 |
Comment on your Keq values. Do your results convince you that Keq is a constant value regardless of the initial concentrations of the reactants? Why or why not? CONCLUSION AND EVALUATION:
Keq is constant. Our average value recorded from our data wasn’t very far off from the reported value of Keq. If we had done the lab more precisely, our Keq values for each solution would be the same. We had some possible cross contamination with each solution that might’ve affected our average Keq.
- Calculate the average value of Keq from your five trials. The actual value of Keq for this reaction at 25oC is reported as 280. Calculate (should you use all of your values?) the percent difference of your average value from the reported value:
% difference = (experimental value – reported value) x 100%
(358 – 280)/280 x 100% = 28% Error
Reported value
