To obtain [FeSCN2+]eq in tube 5, you made the assumption that 100% of the ions S
ID: 479206 • Letter: T
Question
To obtain [FeSCN2+]eq in tube 5, you made the assumption that 100% of the ions SCN– had reacted. Now that you know the value for Kc (average in table 5), build an ICE diagram to calculate the true percentage of SCN– reacted in tube 5. Comment on the validity of the assumption. (Make sure you include the ICE diagram in your anwer) Fe3+ (aq) + SCN- (aq) Fe(SCN)2+(aq)
**** PLEASE show the ICE diagram and calculate the true percentage of SCN– in tube 5. THANKS
Table 5-Equilibrium constant, at room temperature Tube Absorbance (au) 5 0.895 0.522 2 0.616 0.748 4 1.030 Initial cocentrations [Fe' ISCN lo (M) 0.180 0.000200 0.00100 0.000400 0.00100 0.000600 0.000800 0.00100 0.00100 0.00100 Concentrations at equilibrium ISCN lea e leq (M 0.000200 0.000117 0.000883 0.000283 0.000138 0.000862 0.000462 0.000167 0.000833 0.000633 0.000230 0.000770 0.000770 Average (MT1) 466 345 317 388 379Explanation / Answer
Let x= drop in concentration of SCN-
ICE table
Fe+3 SCN- FeSCN+2
Initial 0.180 0.0002 0
Change -x -x x
Equilibrium 0.180-x 0.0002-x x
Kc= x/(0.180-x)*(0.0002-x)= 379
When solved using excel, x= 0.0001971
% SCN in the tube= 100*(0.0002-0.0001971)/0.0002=1.45%
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