In: Chemistry
Use an ICE table to calculate the equilibrium amounts of all 3 chemicals if the initial amounts are 0.20M of each.
Kc=1.59x102 Fe3+(aq)+SCN-(aq)-----> (Fe(SCN)2+) |
Kc = 1.59x102 = 159
Fe3+(aq) + SCN-(aq) -----> Fe(SCN)2+
Initial 0.2 M 0.2 M 0.2 M
At equilibrium 0.2 -x 0.2 -x 0.2 + x
Kc = [Fe(SCN)2+] /[ Fe3+(aq)+] [SCN-(aq)]
159 = (0.2 + x) / (0.2 - x) (0.2 - x)
159 (x2- 0.4 x + 0.04 ) = 0.2 + x
159 x2- 63.6 x + 6.36 = 0.2 + x
159 x2- 64.6 x + 6.16 = 0
On solving,
x = 0.153
Therefore, equilibrium concentrations are
[ Fe3+(aq)+] = 0.2 - x = 0.2 - 0.153 = 0.047 M
[SCN-(aq)] = 0.2 - x = 0.2 - 0.153 = 0.047 M
[Fe(SCN)2+] = 0.2 + x = 0.2 + 0.153 = 0.353 M