2. A CHM 126 student performed the\"Determination of Equilibrium Constant\" proc
ID: 694438 • Letter: 2
Question
2. A CHM 126 student performed the"Determination of Equilibrium Constant" procedure as outlined in the CHM 126 lab manual. The objective was to measure the equilibrium constant, K for the following reaction: Fe. (aq) + SCN-(aq) Fe(SCN) 2. (aq) The student performed the experiment by making a mixture of all three components in a beaker. This was done by mixing 10.00 mL of 0.00200 M Fe (a) with 8.00 mL of 0.00200 M of SCN) (aq) and 2.00 mL. of deionized water (H.0). Upon mixing these solutions, total volume of 20.00 mL, the final solution was deep red in color. The absorbance of the this solution was measured and the concentration of Fe(SCN was found to be 3.00 X 105 M with the use of Beer's Law constant, k. Determine the equilibrium constant, K, for the reaction above from these values by completing the "ICE" table below. Be sure to report the final value of K, at the bottom. (8 points) Fe (aq) SCN (aq)Fe(SCN)* Initial (1) Change (C) Equilibrium (E) Show your calculations below: Equilibrium Constant (K)Explanation / Answer
2. For the given equilibrium reaction,
Fe3+(aq) + SCN- (aq) <====> FeSCN2+(aq)
initial [Fe3+] = 0.002 M x 10 ml/20 ml = 0.001 M
initial [SCN-] = 0.002 M x 8 ml/20 ml = 0.0008 M
at equilibrium,
[FeSCN]2+ found from aborbance = 3 x 10^-5 M
So,
ICE chart
Fe3+(aq) + SCN- (aq) <====> FeSCN2+(aq)
I 0.001 0.0008 -
C -3 x 10^-5 -3 x 10^-5 +3 x 10^-5
E 0.00097 0.00077 3 x 10^-5
So,
equilibrium constant Kc = [Fe(SCN)]2+eq/[Fe3+]eq.[SCN-]eq
= (3 x 10^-5)/(0.00097)(0.00077)
= 40.167
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