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Consider the following reaction: Fe3+(aq)+SCN−(aq)⇌FeSCN2+(aq) A solution is made containing an initial [Fe3+] of 1.2×10−3 M...

Consider the following reaction: Fe3+(aq)+SCN−(aq)⇌FeSCN2+(aq) A solution is made containing an initial [Fe3+] of 1.2×10−3 M and an initial [SCN−] of 7.8×10−4 M . At equilibrium, [FeSCN2+]= 1.7×10−4 M .Calculate the value of the equilibrium constant (Kc).

Solutions

Expert Solution

Fe3+(aq)   + SCN−(aq) <--------------------> FeSCN2+(aq)

1.2x10^-3     7.8 x 10^-4                                     0             ----------------> initial

1.2x10^-3-x     7.8 x 10^-4 -x                           x             ---------------->equilibrium

but x = 1.7 x 10^-4 is given

equilibrium concentrations :

[Fe+2] = 1.2x10^-3-x   = 1.03 x 10^-3 M

[SCN-] = 7.8 x 10^-4 -x = 6.1 x 10^-4 M

[FeSCN2+]= 1.7×10^−4 M

equilibrium constant (Kc)   = [FeSCN2+] / [Fe+2] [SCN-]

                                           = 1.7×10^−4 / (1.03 x 10^-3 ) (6.1 x 10^-4)

                                          = 270.6

equilibrium constant (Kc)   = 270.6

queen_honey_blossom answered 2 years ago
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