To obtain [FeSCN2+]eq in tube 5, you made the assumption that 100% of the ions S

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)

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) 466 345 317 388 379

Explanation / Answer

Clearly, if K > 1, this means it does favour products, that is:

K = [products] /[ reactants]

so

K = [Fe(SCN)2+]/[Fe+3][SCN-]

meaning that this is:

K> 1 so

[Fe(SCN)2+] > [Fe+3][SCN-]

Now...

If K >>1, then we can assume almost all is turned to product, that is [Fe(SCN)2+]

then, the assumption is correct, in reality there is no more than 5% error range, which is acceptable

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