Chemical Equilibrium and Beer’s LawCatherine Lozier04/16/2013Chem 1310 Section C02TA Ryan BucherLab Partners:Chelsey Arnold Sunya MorinTran GHonor Pledge: I did not copy this work from any others student(s), current students in lab or old lab reports ________________________________________________SignatureDATA AND OBSERVATIONSλmax = 468.6 nmID Volume (mL) [Fe(SCN)2+]i(M)AbsFe3+SCN-HNO3Total1 1.0 5.0 4.0 10.0 0.000001 0.0572 2.0 5.0 3.0 10.0 0.000002 0.1213 3.0 5.0 2.0 10.0 0.000003 0.2004 4.0 5.0 1.0 10.0 0.000004 0.3175 5.0 5.0 0.0 10.0 0.000005 0.4475.00E-07 1.00E-06 1.50E-06 2.00E-06 2.50E-06 3.00E-06 3.50E-06 4.00E-06 4.50E-06 5.00E-06 5.50E-0600.050.10.150.20.250.30.350.40.450.5f(x) = 97600x − 0.06R² = 0.98Absorbance v. Concentration of Fe(SCN)2+Concentration of Fe(SCN)2+AbsorbanceID Volume (mL) Abs Fe3+init(M)SCN-init(M)Fe(SCN)2+ (M)Fe3+eq(M)SCN-eq(M)KFe3+SCN-HNO3Tot6 1.0 1.0 5.0 7.0 0.150 3.57E-4 3.57E-4 1.537E-6 3.56E-4 3.56E-4 12.1547 1.0 1.5 4.5 7.0 0.178 3.57E-4 5.36E-4 1.824E-6 3.55E-4 5.34E-4 9.6148 1.0 2.0 4.0 7.0 0.258 3.57E-4 7.14E-4 2.643E-6 3.55E-4 7.12E-4 10.4789 1.0 2.5 3.5 7.0 0.348 3.57E-4 8.93E-4 3.566E-6 3.54E-4 8.89E-4 11.34010 1.0 3.0 3.0 7.0 0.434 3.57E-4 1.07E-3 4.447E-6 3.53E-4 1.07E-3 11.81611 2.0 1.0 4.0 7.0 0.248 7.14E-4 3.57E-4 2.541E-6 7.12E-4 3.55E-4 10.06812 2.0 1.5 3.5 7.0 0.363 7.14E-4 5.36E-4 3.719E-6 7.11E-4 5.32E-4 9.83913 2.0 2.0 3.0 7.0 0.480 7.14E-4 7.14E-4 4.918E-6 7.09E-4 7.09E-4 9.77314 2.0 2.5 2.5 7.0 0.650 7.14E-4 8.93E-4 6.660E-6 7.08E-4 8.86E-4 10.62015 2.0 3.0 2.0 7.0 0.737 7.14E-4 1.07E-3 7.551E-6 7.07E-4 1.06E-3 10.043Average Keq = 10.575 ± 0.899SAMPLE CALCULATIONS1. y = 97600x-0.064497600 = εb97600 = ε*1 cmε = 976 2. SCNinitial = 0.0025M*1 mL/7 mLSCN = 3.57E-43. Fe3+initial = 0.025M * 1 mL/7mLFe3+initial = 3.57E-4 M4. A = εbc0.150 = 976 * 1 cm * Fe(SCN)Fe(SCN) = 1.537E-6 M5. Feeq = Fe3+initial – Fe(SCN)Feeq = 3.57E-4 – 1.537E-6Feeq = 3.56E-4 M6. SCNe = SCNinitial – Fe(SCN)SCNe= 3.57E-4 – 1.537E-6SCNe = 3.56E-4 M7. K = [Fe(SCN)] / [SCN][Fe]K = 1.537E-6 / 3.56E-4 * 3.65E-4K = 12.154DISCUSSIONThe purpose of this experiment is to determine the value of K for the iron thiocyanate reaction, using Beer’s law to relate concentration and wavelength absorbed. In part A, the molar absorptivity of the Fe(SCN)2+ was 976 parts per mole. In part B, the average Keq was calculated to be 10.575 with a standard deviation of ± 0.899. Because this number is greater than 10, it indicates that the products of the reaction were in higher concentration than the reactants. Some sources of error in this experiment could be human error in the dilution of the Fe3+ in part A. Another source of error could be contaminated equipment, as well as dirty cuvettes. These errors could give a falsely high absorbency reading.DISCUSSION QUESTIONS1. It was assumed that the concentration of Fe3+is low enough that the other reactants are in excess and all of the iron reacts fully.2. The concentrations of the stock solutions were known, and because the volumes were known, the initial concentrations could be calculated. By using Beer’s law it was possible to determine the concentration of the Fe(SCN)2+, and by subtracting the concentration of Fe(SCN)2+ from the initial concentrations of Fe3+ and SCN-, it was possible to determine the final concentrations.3. The calculated equilibrium constant K was 10.575 with a standard deviation of 0.899. This is a high value for K, although not an unrealistic number. A K value of greater than 10 indicates that the products of the reaction are in much higher concentrations than the reactants. In this case, it would mean the concentration of Fe(SCN)2+ is much higher than either Fe3+ or SCN-.REFERENCES Lab ManualChemistry for Engineering Students, Second Edition. Lawrence S. Brown, Thomas A.