Respuesta :
Answer:
The value of the equilibrium constant is [tex]2.7\times 10^2[/tex].
Explanation:
[tex]Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)[/tex]
Initially
[tex]1.1\times 10^{-3}[/tex] [tex]7.8\times 10^{-4}[/tex] 0
At equilibrium :
[tex](1.1\times 10^{-3}-x)[/tex] [tex](7.8\times 10^{-4}-x)[/tex] x
The concentration of [tex]SCN^{-}[/tex] at equilibrium is given ,x=[tex]1.6\times 10^{-3}[/tex]
The expression of an equilibrium constant is given as;
[tex]K_c=\frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]}[/tex]
[tex]K_c=\frac{x}{(1.1\times 10^{-3}-x)\times (7.8\times 10^{-4}-x)}[/tex]
Putting value of x:
[tex]K_c=\frac{1.6\times 10^{-3}}{(1.1\times 10^{-3}-(1.6\times 10^{-3}))\times (7.8\times 10^{-4}-(1.6\times 10^{-3}))}[/tex]
The value of the [tex]K_c[/tex]
[tex]K_c=274.53\approx 2.7\times 10^2[/tex]
The value of the equilibrium constant is [tex]2.7\times 10^2[/tex].
