Respuesta :
The question is incomplete, here is the complete question:
Consider the following reaction: [tex]Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)[/tex] A solution is made containing an initial [tex][Fe^{3+}][/tex] of 1.2×10⁻³ M and an initial [SCN⁻] of 8.0×10⁻⁴ M . At equilibrium, [FeSCN²⁺]= 1.8×10⁻⁴ M.
Calculate the value of the equilibrium constant (Kc).
Answer: The value of [tex]K_c[/tex] for above equation is 284.63
Explanation:
We are given:
Initial concentration of [tex][Fe^{3+}]=1.2\times 10^{-3}M[/tex]
Initial concentration of [tex][SCN^{-}]=8.0\times 10^{-4}M[/tex]
Equilibrium concentration of [tex][FeSCN^{2+}]=1.8\times 10^{-4}M[/tex]
The given chemical equation follows:
[tex]Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)[/tex]
Initial: [tex]1.2\times 10^{-3}[/tex] [tex]8.0\times 10^{-4}[/tex]
At eqllm: [tex](1.2\times 10^{-3}-x)[/tex] [tex](8.0\times 10^{-4}-x)[/tex] x
Equilibrium concentration of [tex][Fe^{3+}]=(1.2\times 10^{-3})-(1.8\times 10^{-4)=1.02\times 10^{-3}M[/tex]
Equilibrium concentration of [tex][SCN^{-}]=(8.0\times 10^{-3})-(1.8\times 10^{-4)=6.2\times 10^{-4}M[/tex]
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]}[/tex]
Putting values in above equation, we get:
[tex]K_c=\frac{1.8\times 10^{-4}}{(1.02\times 10^{-3})\times (6.2\times 10^{-4})}\\\\K_c=284.63[/tex]
Hence, the value of [tex]K_c[/tex] for above equation is 284.63
