Question

In an aqueous solution, iron(III) ions react with iodide ions to give iron(II) ions and triiodide ions, I3-. Suppose the initial concentration of Fe3+ ions is 0.200 M, the initial I- ion concentration is 0.300 M, and the equilibrium concentration of I3- ions is 0.0866 M. What is the value of Kc? 2 Fe3+(aq) + 3 I-(aq) ⇄ 2 Fe2+(aq) + I3-(aq) Kc =

Answer 1

Answer:

The value of the equilibrium constant $K_c$ is $5.57\times 10^{4}$.

Explanation:

The initial concentration of ferric ions $[Fe^{3+}]=0.200 M$

The initial iodide ion concentration = $[I^-]=0.300 M$

The equilibrium concentration of $I_3^{-}=[I_3^{-}]=x=0.0866 M$

   $2 Fe^{3+}(aq) + 3 I^-(aq)\rightleftharpoons 2 Fe^{2+}(aq) + I_3^{-}(aq)$

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 :

$K_c=\frac{[Fe^{2+}]^2[I_3^{-}]}{[Fe^{3+}]^2[I^-]^3}$

$K_c=\frac{(2x)^2\times x}{(0.200-2x)^2\times(0.300-3x)^3}$

Putting value of x :

x = 0.0866 M

$K_c=\frac{(2\times 0.0866 )^2\times 0.0866}{(0.200-2\times 0.0866)^2\times(0.300-3\times 0.0866)^3}$

$K_c=5.57\times 10^{4}$

The value of the equilibrium constant $K_c$ is $5.57\times 10^{4}$.