For parts II and III find the equilibrium concentrations and Kc For part iii exp
ID: 1017267 • Letter: F
Question
For parts II and III find the equilibrium concentrations and Kc
For part iii explain why the Kc changes
Exp 29 (1) [Protected View] - Excel HOME INSERT PAGE LAYOUT FORMULAS DATA REVIEW VIEW B41 Group 1 ar 581 nm y 850.03x R2 = 0.9749 3 Concentration (M) Absorbance 0.0017 0.0013 0.001 0.0005 0.0002 1.455 0.719 0.399 Series! Linear (Series1) 13 0 0.0005 0.001 0.0015 0.002 18 Solution FeSCN Abs Vol Fe (L) Vol SCN (L) [Fe]i SCN]eq [Fe(SCN)Jeq Kc 0.01 0.008 0.024 0.022 0.033 0.036 0.048 0.057 0.001 0.002 0.005 0.005 0.005 0.005 0.01 0.001 0.005 0.005 0.002 1.00E-04 1.00E-04 2.00E-04 5.00E-04 5.00E-04 5.00E-04 5.00E-04 2.00E-04 5.00E-04 1.00E-02 5.00E-04 1.30E-03 1.00E-03 5.00E-04 1.30E-02 5.00E-04 8.82E-05 8.82E-05 1.17643E-05 1511.04 0.013 0.005 0.005 6 28 Part III FeSCN Abs Vol Fe (L) Vol SCN (L) [Fe]i SCN]eq [Fe(SCN)Jeq Kc 0.037 0.632 0.494 0.126 0.052 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 8.00E-04 8.00E-04 8.00E-04 8.00E-04 8.00E-04 8.00E-04 8.00E-04 8.00E-04 8.00E-04 8.00E-04 or parts Il and III find the eauilibrium concentrations and Kc Sheet1 Sheet2 Sheet3 READY + 100% 9:22 PM 7/21/2016 I'm Cortana. Ask me anythingExplanation / Answer
Absorbance = molar absortivity x path length x concentration
slope = molar absorptivity
path length = 1 cm
[FeSCN-]eq = absorbance/molar absorptivity (slope)
[Fe]eq = [Fe]i - [FeSCN-]eq
[SCN-]eq = [SCN-]i - [FeSCN-]eq
Part II
Soln [Fe]eq [SCN-]eq [FeSCN]-eq Kc
1 8.83x10^-5 8.83x10^-5 1.17x10^-5 1513.42
2 1.88x10^-4 4.88x10^-4 9.41x10^-6 102.57
3 4.88x10^-4 4.88x10^-4 2.82x10^-5 118.41
4 4.88x10^-4 2.88x10^-4 2.59x10^-5 184.28
5 4.88x10^-4 8.83x10^-5 3.88x10^-5 900.43
6 4.88x10^-4 1.29x10^-3 4.23x10^-5 67.19
7 9.88x10^-4 4.88x10^-4 5.64x10^-5 117.12
8 0.013 4.88x10^-4 6.70x10^-5 10.56
Part III
Soln [Fe]eq [SCN-]eq [FeSCN]-eq Kc
1 7.56x10^-4 7.56x10^-4 4.35x10^-5 76.11
2 5.70x10^-5 5.70x10^-5 7.43x10^-4 2.29 x 10^5
3 2.19x10^-4 2.19x10^-4 5.81x10^-4 12114.01
4 6.52x10^-4 6.52x10^-4 1.48x10^-4 348.15
5 7.39x10^-4 7.39x10^-4 6.12x10^-5 112.06
For Part III, Kc differs because the absorbance for the complex formed is different for each run.
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