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Chemical Equilibrium: Iron (III) with SCN Lab Report Introduction: The purpose of this lab is to observe the chemical equilibria between the reaction of Fe3 + ions and SCN- ions by looking at absorbance values through a spectrophotometer. A spectrophotometer emits a beam of light through a sample and then measures the intensity of light through the solution. This information can be used to calculate the absorbance and can be then used with Lambert-Beer’s law to calculate the concentration of a solution tested. Those concentration values can help predict what product was formed from a certain reaction, and can help determine equilibrium concentrations of products and reactants + Keq  values. Experimental: No deviations were made in experimental procedures. Data Analysis: [Fe3+  ]i Solution [SCN]i A A/Ai 1 0.1M 0.001M .295 1.000 2 0.01M 0.001M .283 0.959 3 0.005M 0.001M .232 0.786 4 0.0025M 0.001M .157 0.532 5 0.00125M 0.001M .077 0.261 The above table lists the amounts of each ion present in each beaker along with the absorbances and relative absorbances. Potential Product of ( FeSCN)2 2+  If (FeSCN)22+ was the product formed from the reaction, then the equation of the reaction would be as follows: Fe3+ + SCN- ⇌ (FeSCN)22+ ICE charts can be used to determine the equilibrium concentrations of the ions involved in the reaction. An example calculation is below for Solution 1: I Fe3+ (M) SCN- (M) 0.1 0.001 (FeSCN)22+ (M) 0 C -X -X +X E 0.1 - X 0 X This calculation allowed for X to be easily determined as being equal to 1.00x10-3, equivalent to the initial concentration of thiocyanate (we assume it is all consumed, as iron is in excess). Due to this, the equilibrium concentration of Fe3+  was determined to be 0.099 M, while that of [FeSCN]2+ is 0.001M. The following equation can be used with the relative absorption values to determine the concentrations of (FeSCN)22+ in the other beakers: A 1/ A 2 = C 1/ C 2 Example calculation for Solution 2: (1.00)/(.959) = (.001/C2) → C2 = .000959; this was repeated for beakers 3-5... The equilibrium concentrations of Fe3+  and SCN- can be determined by subtracting the values of 2+ [FeSCN] concentrations calculated from Lambert-Beer’s law from their respective initial concentrations. The equilibrium constant, Keq, can be determined using the following equation: Keq = [(FeSCN)2 2+]eq/([Fe3+]eq*[SCN-]eq) Solution A/A1 [(FeSCN)2 2+]eq [Fe3+  ]eq [SCN- ]eq Keq 1 1.00 .001M .099M 0.00 … 2 0.959 0.000959M .00904M .0000410 2590 3 0.786 0.000786M .00421M .000214 872 4 0.532 0.000532M .00197M .000468 577 5 0.261 0.000261M .000989M .000739 357 Stdev of Keq = 1020, 260 (without Solution 2) Mean of Keq = 1099, 602 (without Solution 2) We assumed that all of the SCN- was consumed in Solution 1, and we must test this assumption now: SCN- = [(FeSCN)22+]/Keq(avg)*[Fe3+] = (.001)/(.099)(602) = 1.68*10-5 % Unreacted: (1.68*10-5  )/(.001) x 100 = 1.68% Potential Product of Fe(SCN)3 Below is the chemical equation for equilibrium production of Fe(SCN)3: Fe3+ + 3SCN- ⇋ Fe(SCN)3 ICE charts were again useful for assessing this reaction. An example calculation for the reaction in Solution #1 is seen below: Fe3+ (M) SCN- (M) Fe(SCN)3 (M) I .10 .001 0 C -X -3X +X E .10-X 0 X Given that .001-3X is 0 (an assumption made because Fe3+ was in excess), X = .00033. Thus, the equilibrium concentration of Fe(SCN)3+  is .00033M. The following equation can be used with the relative absorption values to determine the concentrations of Fe(SCN)3  in the other beakers: A 1/ A 2 = C 1/ C 2 Example for Solution 2: (1.000)/(.959) = (.00033)/(C2) → C2 = .000316; this was repeated for beakers 3-5... The equilibrium concentrations of Fe3+  and SCN- can be determined by subtracting the values of Fe(SCN)3 concentrations calculated from Lambert-Beer’s law from their respective initial concentrations. The equilibrium constant, Keq, can be determined using the following equation: Keq = [Fe(SCN)3]/([Fe3+  ]eq*[SCN- ]3 eq) For possible product Fe(SCN)3: Solution A/A1 [Fe(SCN)3]eq [Fe3+  ]eq [SCN- ]eq Keq 1 1.00 .000333M .00967 0.00 … 2 0.959 0.000316M .00968 0.000052 2.32E10 3 0.786 0.000262M .00474 .000921 7.08E7 4 0.532 0.000177M .00232 .000469 7.40E8 5 0.261 0.00008616M .00116 .000742 1.82E10 Stdev of Keq = 1.19E10 Mean of Keq = 1.06E10 Discussion: The mean of equilibrium constants of the possible product FeSCN)22+ is 1020 and the standard deviation is 260 (if we exclude the outlier in solution 2). The mean of the constants for Fe(SCN)3 was 1.06E10 and the standard deviation was 1.19E10. Thus, F  eSCN)22+ has to be the equilibrium product because it has more consistent equilibrium constants. Our setting the value of SCN- as zero in the first solution was accurate because-- after excluding beaker 2-- we obtained a percent unused of 1.68%, a value very close to 0.0%. Therefore, there was very little residual, however, there was some residual thiocyanate left. The additional Fe3+  would not affect the absorbance values, because SCN- was a limiting reagent and the absorbance is dependant on the amount of red product formed by the reaction. Thus, the addition of further surplus iron would not affect the absorbances. Conclusion: From this lab, we applied the principles of Lambert-Beer’s Law and photospectrometry to determine that the actual product of the reaction between iron(III) and thiocyanate to be Fe(SCN)22+ and not Fe(SCN)3. This was because the reactions between iron(III) and thiocyanate yielded a smaller standard deviation in equilibrium constants for potential product validates our conclusion because Fe(SCN)22+ than that of the potential product Fe(SCN)3. This  the same reaction should have similar equilibrium constants. There was some error in this lab, as we assumed all of the thiocyanate reacts in solution #1, however, there was 1.68% residual.