= 3801D. DO Experiment 23 Spec #4 Data and Calculations: Determination of the Eq

Question

= 3801D. DO Experiment 23 Spec #4 Data and Calculations: Determination of the Equilibrium Constant for a Chemical Reaction Method II Depth in mm Volume in ml 200 x 10-J M Fed(NO), 5.00 Volume in ml 200 x 10 M Volume in ml, Method I 5.00 5.00 5.00 5.00 1.00 2.00 3.00 4.00 5.00 0290 0.3 OnL 0.42 10-4 M x 104 M om If Method II was used, FeSCNg10 M; [FeSCN '1 in Mixtures 1 to 5 is found by Processing the Data A. Calculation of K assuming the reaction: This cakculation is most easily done by following Sieps I through 5 in the discussion. Results are to be entered in the table on the following page. If you are using Excel, set Fe (aq) + SCN (aq)FeSCN (aq) up the table as we have, and follow the directions that follow Step 1 Find the initial number of moles of Fe and SCN in the mistures in test tubes I through 5. Use Equation 3 and enter the values in the first two columns of the table. Step 2 Enter the experimentally determined value of [FeSCN* at equilibrium for each of the mixtures in the next to last column in the table. Use Equation 3 to find the number of moles of FeSCN2* in each of the mixtures, and enter the values in the fifth column of the table. Note that this is also the number of moles of Fe and SCN that were used up in the reaction. (continued on following page)

Explanation / Answer

Calculation of equilibrium constant

equilibrium concentration [FeSCN2+] = absorbance/3801

3810 M-1.cm-1 is slope of calibration curve

For Mixture 1-5,

initial Fe3+ = 0.002 m x 5 ml = 0.010 mmol

initial [Fe3+] = 0.010 mmol/10 ml = 0.001 M

For Mixture 1,

initial SCN- = 0.002 m x 1 ml = 0.002 mmol

initial [SCN-] = 0.002 mmol/10 ml = 0.0002 M

equilibrium [FeSCN2+] = 0.073/3801 = 1.9 x 10^-5 M

equilibrium [SCN-] = 0.0002 - 1.90 x 10^-5 = 1.81 x 10^-4 M

equilibrium [Fe3+] = 0.001 - 1.90 x 10^-5 = 9.81 x 10^-4 M

Kc = 1.9 x 10^-5/(1.81 x 10^-4)(9.81 x 10^-4) = 107.0

For Mixture 2,

initial SCN- = 0.002 m x 2 ml = 0.004 mmol

initial [SCN-] = 0.004 mmol/10 ml = 0.0004 M

equilibrium [FeSCN2+] = 0.195/3801 = 5.1 x 10^-5 M

equilibrium [SCN-] = 0.0004 - 5.1 x 10^-5 = 3.5 x 10^-4 M

equilibrium [Fe3+] = 0.001 - 5.1 x 10^-5 = 9.5 x 10^-4 M

Kc = 5.1 x 10^-5/(3.5 x 10^-4)(9.5 x 10^-4) = 154

For Mixture 3,

initial SCN- = 0.002 m x 3 ml = 0.006 mmol

initial [SCN-] = 0.006 mmol/10 ml = 0.0006 M

equilibrium [FeSCN2+] = 0.290/3801 = 7.6 x 10^-5 M

equilibrium [SCN-] = 0.0006 - 7.6 x 10^-5 = 5.24 x 10^-4 M

equilibrium [Fe3+] = 0.001 - 7.6 x 10^-5 = 9.23 x 10^-4 M

Kc = 7.6 x 10^-5/(5.24 x 10^-4)(9.23 x 10^-4) = 157

For Mixture 4,

initial SCN- = 0.002 m x 4 ml = 0.008 mmol

initial [SCN-] = 0.008 mmol/10 ml = 0.0008 M

equilibrium [FeSCN2+] = 0.369/3801 = 9.7 x 10^-5 M

equilibrium [SCN-] = 0.0008 - 9.7 x 10^-5 = 7.03 x 10^-4 M

equilibrium [Fe3+] = 0.001 - 9.7 x 10^-5 = 9.03 x 10^-4 M

Kc = 9.7 x 10^-5/(7.03 x 10^-4)(9.03 x 10^-4) = 153

For Mixture 5,

initial SCN- = 0.002 m x 5 ml = 0.010 mmol

initial [SCN-] = 0.010 mmol/10 ml = 0.001 M

equilibrium [FeSCN2+] = 0.443/3801 = 1.2 x 10^-4 M

equilibrium [SCN-] = 0.001 - 1.2 x 10^-4 = 8.8 x 10^-4 M

equilibrium [Fe3+] = 0.001 - 1.2 x 10^-4 = 8.8 x 10^-4 M

Kc = 1.2 x 10^-4/(8.8 x 10^-4)(8.8 x 10^-4) = 155

average Kc = 145.2

standard deviation = sq.rt.[(sum x-mean)^2/5]

                               = 19.2

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