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	0.0M	0.0
B	10.0 mL	1.0 mL	14.0 mL	8.0×10^-5	0.304
C	10.0 mL	1.5 mL	13.5 mL	1.2×10^-4	0.500
D	10.0 mL	2.0 mL	13.0 mL	1.6×10^-4	0.670
E	10.0 mL	2.5 mL	12.5 mL	2.0×10^-4	1.070
F	10.0 mL	3.0 mL	12.0 mL	2.4×10^-4	1.080

EQUATION:         y=4862.1x-0.0443                                    R²=0.9659

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	-0.006	0
II	5.0 mL	1.0 mL	4.0 mL	0.0010M	2.0×10^-4	0.159	4.2×10^-5
III	5.0 mL	2.0 mL	3.0 mL	0.0010M	4.0×10^-4	0.378	8.7×10^-5
IV	5.0 mL	3.0 mL	2.0 mL	0.0010M	6.0×10^-4	0.534	1.4×10^-4
V	5.0 mL	4.0 mL	1.0 mL	0.0010M	8.0×10^-4	0.716	1.6×10^-4
VI	5.0 mL	5.0 mL	0.0 mL	0.0010M	1.0×10^-3	0.820	1.8×10^-4

 

* 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.

ICE CHARTS

Test Solution II Keq = 277.5	Fe3+               +                SCN–                   ⇄            FeSCN2+
I	0.0010M	0.00020M	0
C	-x	-x	+x
E	0.000958	0.000158	4.2×10^-5
Test Solution III Keq = 304.4	Fe3+               +                SCN–                   ⇄            FeSCN2+
I	0.0010M	0.00040M	0
C	-x	-x	+x
E	0.000913	0.000313	8.7×10^-5

 

Test Solution IV Keq = 353.9	Fe3+               +                SCN–                   ⇄            FeSCN2+
I	0.0010M	0.00060M	0
C	-x	-x	+x
E	0.00086	0.00046	1.4×10^-4

 

Test Solution V Keq = 297.6	Fe3+               +                SCN–                   ⇄            FeSCN2+
I	0.0010M	0.00080M	0
C	-x	-x	+x
E	0.00084	0.00064	1.6×10^-4

 

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

 

 

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?

Our Keq values were all fairly close to the reported Keq of 280. The only two Keqs that were out of the ordinary were 304.4 and 353.9. We believe this is because of an error during the actual lab work. In the end when we calculated the percent difference including these values the % was still less than 10, as we got 7.22%. Without using those two values, the percent difference we calculated was less than 1, as we got 0.33%. Our results do convince us that Keq is a constant value regardless of the initial concentrations of the reactants as our percent differences were quite low. Our Keq values individually were also fairly constant being in the range of 267.7 to 353.9. Since both of these point in the direction of being close to the reported value, the Keq must be a constant value regardless of the initial concentrations.

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 using all 5 trials, including Keqs above 300 = (300.22 – 280)  x 100% = 7.22%

280

% difference disregarding the Keqs above 300 = (280.93 – 280)  x 100% = 0.33%

280