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
B	10.0 mL	1.0 mL	14.0 mL	0.00008M	0.312
C	10.0 mL	1.5 mL	13.5 mL	0.00012M	0.503
D	10.0 mL	2.0 mL	13.0 mL	0.00016M	0.725
E	10.0 mL	2.5 mL	12.5 mL	0.0002M	1.025
F	10.0 mL	3.0 mL	12.0 mL	0.00024M	1.045

EQUATION:     y=0.2167x-0.1569                                                                        R²_=0.9711______

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	0.0010M	0	0	0M
II	5.0 mL	1.0 mL	4.0 mL	0.0010M	0.00020M	0.206	4.9×10^-5M
III	5.0 mL	2.0 mL	3.0 mL	0.0010M	0.00040M	0.348	7.9×10^-5M
IV	5.0 mL	3.0 mL	2.0 mL	0.0010M	0.00060M	0.558	1.2×10^-4M
V	5.0 mL	4.0 mL	1.0 mL	0.0010M	0.00080M	0.782	1.7×10^-4M
VI	5.0 mL	5.0 mL	0.0 mL	0.0010M	0.0010M	0.815	1.8×10^-4M

 

* To be determined from the standard graph equation.

ANALYSIS:

1. Use your graph equation to calculate the equilibrium concentrations of FeSCN⁺².
2.  Prepare and ICE chart for each test solution (II – VI) and calculate the value of K_eq for each of your 5 tests solutions.

SAMPLE ICE CHART

Test Solution Keq =343.9	Fe3+               +                SCN–                    ⇄            FeSCN2+
I	0.0010M	0.00020M	0
C	-x	-x	+x
E	0.00095M	0.00015M	4.9×10^-5M

SAMPLE ICE CHART

Test Solution Keq =268.3	Fe3+               +                SCN–                    ⇄            FeSCN2+
I	0.0010M	0.00040M	0
C	-x	-x	+x
E	0.00092M	0.00032M	7.9×10^-5M

 

SAMPLE ICE CHART

Test Solution Keq =284.0	Fe3+               +                SCN–                    ⇄            FeSCN2+
I	0.0010M	0.00060M	0
C	-x	-x	+x
E	0.00088M	0.00048M	1.2×10^-4M

 

SAMPLE ICE CHART

Test Solution Keq =325.1	Fe3+               +                SCN–                    ⇄            FeSCN2+
I	0.0010M	0.00080M	0
C	-x	-x	+x
E	0.00083M	0.00063M	1.7×10^-4M

 

SAMPLE ICE CHART

Test Solution Keq =267.7	Fe3+               +                SCN–                    ⇄            FeSCN2+
I	0.0010M	0.0010M	0
C	-x	-x	+x
E	0.00082M	0.00082M	1.8×10^-4M

 

CONCLUSION AND EVALUATION:

1. Comment on your Keq values.   Do your results convince you that K_eqis a constant value regardless of the initial concentrations of the reactants? Why or why not?

Because the K_eqequals the concentration of products divided by the concentration of the reactants. If we change the initial concentrations of the reactants, it changes the equilibrium concentration of the reactants. So it also changes the K_eq.

2. Calculate the average value of K_eqfrom your five trials. The actual value of K_eq  for this reaction at 25^oC 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%

Reported value

Average value for K_eq= (343.9+268.3+284.0+325.1+267.7)/5=297.8

1: % difference=100%(343.9-280.0)/280.0=22.9%

2: % difference=100%(268.3-280.0)/280.0=4.18%

3: % difference=100%(284.0-280.0)/280.0=1.43%

4: % difference=100%(325.1-280.0)/280.0=16.1%

5: % difference=100%(267.7-280.0)/280.0=4.39%

 

 

The graph for part one