Question

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 8.0×10−4 M . At equilibrium, [FeSCN2+]= 1.8×10−4 M .

Answer 1

The question is incomplete, here is the complete question:

Consider the following reaction: $Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$ A solution is made containing an initial $[Fe^{3+}]$ 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 $K_c$ for above equation is 284.63

Explanation:

We are given:

Initial concentration of $[Fe^{3+}]=1.2\times 10^{-3}M$

Initial concentration of $[SCN^{-}]=8.0\times 10^{-4}M$

Equilibrium concentration of $[FeSCN^{2+}]=1.8\times 10^{-4}M$

The given chemical equation follows:

                               $Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Initial:                       $1.2\times 10^{-3}$      $8.0\times 10^{-4}$

At eqllm:          $(1.2\times 10^{-3}-x)$   $(8.0\times 10^{-4}-x)$      x

Equilibrium concentration of $[Fe^{3+}]=(1.2\times 10^{-3})-(1.8\times 10^{-4)=1.02\times 10^{-3}M$

Equilibrium concentration of $[SCN^{-}]=(8.0\times 10^{-3})-(1.8\times 10^{-4)=6.2\times 10^{-4}M$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]}$

Putting values in above equation, we get:

$K_c=\frac{1.8\times 10^{-4}}{(1.02\times 10^{-3})\times (6.2\times 10^{-4})}\\\\K_c=284.63$

Hence, the value of $K_c$ for above equation is 284.63