Computer 10 Chemical Equilibrium: Finding a Constant, Kc of this lab is to exper

Question

Computer 10 Chemical Equilibrium: Finding a Constant, Kc of this lab is to experimentally determine the equilibrium constant, Ke for the following chemical reaction: FescN? (aq) Fes (aq) SCN (aq) thiocyanoiron (II) iron(II) thiocyanate When Fe and SCN- are combined, equilibrium is established between these two ions and on. In order calculate Kc the necessary to know the concentrations to for reaction, it is ofallions le will equilibrium systems containing different concentrations of these three ions. The equilibrium These values will be substituted into the equilibrium constant expression to see ifKeis indeed constant. You will use a Colorimeter or a to determine [FescN +leq. The FescN2+ ion produces solutions with a red color. Because the red solutions absorb blue light very well, so Colorimeter be instructed to use the 470 nm (blue) LED. Spectrometer users will determine an appropriate wavelength based on the absorbance spectrum of the solution. The light striking the detector is reported as absorbance or percent transmittance. By comparing the absorbance of each equilibrium system, A to the absorbance of standard solution, Astd, you can determine (FeSCN leg. The standard solution has a known FesCN concentration. To prepare the standard solution, a very large concentration of Fe will be added to a small initial concentration of SCN Chereafter The [Fe jin standard solution is 100 times larger than [Fe in the equilibrium mixtures. According to LeChatelier's this high forces the reaction far to the right, using up nearly 100% ofthe assumed t FeSCN is Assuming [FesCN2+]and absorbance are related directly (Beer's law), the concentration of FesCN for any of the equilibrium systems can be found by: [FesCN Knowing the [FeSCN leg allows you to determine the concentrations of the other two ions at equilibrium. For each mole of FeSCN tions produced, one less mole of Fe ions will be found in the solution (see the ratio of coefficients in the equation on the previous page). The [Fe can be determined by: Because one mole of SCN is used up for each mole of FescN2+ ions produced, ISCN leq can be determined by [SCN e Knowing the values of [Fex leg, [scN Jeq, and [FescN2 leq, you can now calculate the value of Kc, the equilibrium constant. Chemistry with Vernier 10 1

Explanation / Answer

Initial concentrations,

Test tube 1-5, [Fe3+]i = 0.002 M x 5 ml/10 ml = 0.001 M

Test tube 1 : [SCN-]i = 0.002 M x 2 ml/10 ml = 0.0004 M

Test tube 2 : [SCN-]i = 0.002 M x 3 ml/10 ml = 0.0006 M

Test tube 3 : [SCN-]i = 0.002 M x 4 ml/10 ml = 0.0008 M

Test tube 5 : [SCN-]i = 0.002 M x 5 ml/10 ml = 0.001 M

[FeSCN]2+std = 0.002 M x 2 ml/20 ml = 0.0002 M

Equilibrium concentration,

Test tube 1 : [FeSCN]2+eq = 0.176 x 0.0002/0.329 = 1.07 x 10^-4 M

Test tube 2 : [FeSCN]2+eq = 0.192 x 0.0002/0.329 = 1.17 x 10^-4 M

Test tube 3 : [FeSCN]2+eq = 0.229 x 0.0002/0.329 = 1.39 x 10^-4 M

Test tube 4 : [FeSCN]2+eq = 0.260 x 0.0002/0.329 = 1.58 x 10^-4 M

[Fe3+]eq

Test tube 1 : [Fe3+]eq = 0.001 - 1.07 x 10^-4 M = 8.93 x 10^-4 M

Test tube 2 : [Fe3+]eq = 0.001 - 1.17 x 10^-4 M = 8.83 x 10^-4 M

Test tube 3 : [Fe3+]eq = 0.001 - 1.39 x 10^-4 M = 8.61 x 10^-4 M

Test tube 4 : [Fe3+]eq = 0.001 - 1.58 x 10^-4 M = 8.42 x 10^-4 M

[SCN-]eq

Test tube 1 : [SCN-]eq = 0.0004 - 1.07 x 10^-4 M = 2.93 x 10^-4 M

Test tube 2 : [SCN-]eq = 0.0006 - 1.17 x 10^-4 M = 4.83 x 10^-4 M

Test tube 3 : [SCN-]eq = 0.0008 - 1.39 x 10^-4 M = 6.61 x 10^-4 M

Test tube 4 : [SCN-]eq = 0.001 - 1.58 x 10^-4 M = 8.42 x 10^-4 M

Kc = [FeSCN]2+eq/[Fe3+]eq.[SCN-]eq

Test tube 1 : Kc = 1.07 x 10^-4/(2.93 x 10^-4)(8.93 x 10^-4) = 408.9

Test tube 2 : Kc = 1.17 x 10^-4/(4.83 x 10^-4)(8.83 x 10^-4) = 274.3

Test tube 3 : Kc = 1.39 x 10^-4/(6.61 x 10^-4)(8.61 x 10^-4) = 196.1

Test tube 4 : Kc = 1.58 x 10^-4/(8.42 x 10^-4)^2 = 277.6

Average Kc = 289.2

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