Answer:
The value of the equilibrium constant [tex]K_c[/tex] is [tex] 5.57\times 10^{4}[/tex].
Explanation:
The initial concentration of ferric ions [tex][Fe^{3+}]=0.200 M[/tex]
The initial iodide ion concentration = [tex][I^-]=0.300 M[/tex]
The equilibrium concentration of [tex]I_3^{-}=[I_3^{-}]=x=0.0866 M[/tex]
[tex]2 Fe^{3+}(aq) + 3 I^-(aq)\rightleftharpoons 2 Fe^{2+}(aq) + I_3^{-}(aq)[/tex]
Initially
0.200 M 0.300 M
At equilibrium:
(0.200M-2x) (0.300-3x) 2x x
The expression of equilibrium constant is given by :
[tex]K_c=\frac{[Fe^{2+}]^2[I_3^{-}]}{[Fe^{3+}]^2[I^-]^3}[/tex]
[tex]K_c=\frac{(2x)^2\times x}{(0.200-2x)^2\times(0.300-3x)^3}[/tex]
Putting value of x :
x = 0.0866 M
[tex]K_c=\frac{(2\times 0.0866 )^2\times 0.0866}{(0.200-2\times 0.0866)^2\times(0.300-3\times 0.0866)^3}[/tex]
[tex]K_c=5.57\times 10^{4}[/tex]
The value of the equilibrium constant [tex]K_c[/tex] is [tex] 5.57\times 10^{4}[/tex].
