Chemistry

# LAB: MEASURING Keq

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 Keq 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 Keqis a constant value regardless of the initial concentrations of the reactants? Why or why not?

Because the Keqequals 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 Keq.

1. Calculate the average value of Keqfrom 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%

Reported value

Average value for Keq= (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 