5 0441 Evaluating the Equilibrium Constant for the Reaction of Iron(II) Ion with
ID: 557084 • Letter: 5
Question
5 0441 Evaluating the Equilibrium Constant for the Reaction of Iron(II) Ion with Thiocyanate Iorn Prepured by Conrad H. Bergo, Michael P. Deherty, William M. Loffredo and Robert F. Schramm, East Stroudsburg University PURPOSE OFTHE EXPERIMENT Use spectrophotometric techniques to evaluate the equilibrium constant for the reaction of iron(III) ion with thiocyanate ion. Determine the equilibrium constants for five equilibrium mixtures prepared using solutions of these two ions with varying concentrations. Compare the five constants to deter- mine a mean equilibrium constant for this reaction. Treat reaction mixtures for safe disposal BACKGROUND INFORMATION Chemists often need to be able to predict the mass of products that could form from the employ a limiting reagent calculation. These calculations are based on the reactants in a particular chemical reaction. To do so, we ofter that the reaction proceeds until the limiting reagent is completely converted into products. These calculations often produce inaccurate results. This is because, rather than proceeding to completion, many reactions are reversible. In reversible reactions, while reactants are forming products, some of the products are reverting into reactants. We symbolize reversibility in chem- ical equations using double arrows ( In reversible reactions, the rate of the forward reaction depends on the concentrations of reactants; the rate of the reverse reaction depends on the concentrations of the products. At first, the high reactant concentrations cause a rapid forward reaction. As the forward reaction proceeds, however, reactant concentrations decrease as reactants change into products. There- fore, the forward reaction rate gradually decreases. At the same time, the increasing product concentrations cause the reverse reaction rate to increase. Eventually, the rates of the forward and reverse reactions become LearningExplanation / Answer
Volume of flasks used = 10 mL
For Experiment no. 2 (E2)
Initial [Fe3+] = 0.001 M
Initial [SCN-] = 0.002 M
Final equilibrium [FeNCS2+] = 0.5 x 10-3 M
Fe3+ + SCN- --> FeNCS2+
(0.001– x) + (0.002 – x) --> x
x = 0.5 x 10-3 M
At final equilibrium,
[Fe3+] = 0.001 – x = 5 x 10-4 M
[SCN-] = 0.002 – x = 1.5 x 10-3 M
Keq = [FeNCS2+]/([Fe3+] * [SCN-])
= 0.5 x 10-3 / (= 5 x 10-4 * 1.5 x 10-3)
= 6.67 x 102
Completing the table similarly we get,
Exp No.
Initial conc. (M)
Final equilibrium conc. (M)
Keq
[Fe3+]
[SCN-]
[FeNCS2+]
[Fe3+]
[SCN-]
E2
0.001
0.002
5.0E-04
5.0E-04
1.5E-03
666.67
E3
0.001
0.004
3.5E-05
9.7E-04
4.0E-03
9.15
E4
0.001
0.006
7.0E-05
9.3E-04
5.9E-03
12.69
E5
0.001
0.08
2.1E-04
7.9E-04
8.0E-02
3.33
E6
0.001
0.001
2.6E-04
7.4E-04
7.4E-04
474.80
Mean Keq = (666.67 + 9.15 + 12.69 + 3.33 + 474.8)/5 = 233.3
Exp No.
Initial conc. (M)
Final equilibrium conc. (M)
Keq
[Fe3+]
[SCN-]
[FeNCS2+]
[Fe3+]
[SCN-]
E2
0.001
0.002
5.0E-04
5.0E-04
1.5E-03
666.67
E3
0.001
0.004
3.5E-05
9.7E-04
4.0E-03
9.15
E4
0.001
0.006
7.0E-05
9.3E-04
5.9E-03
12.69
E5
0.001
0.08
2.1E-04
7.9E-04
8.0E-02
3.33
E6
0.001
0.001
2.6E-04
7.4E-04
7.4E-04
474.80
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