Question

$3.1$ mole of $Fe{Cl}_3$ and $3.2$ mole of ${NH}_{4}SCN$ are added to one litre of water. At equilibrium, $3.0$ mol of $FeS{CN}^{2+}$ are formed. The equilibrium constant $K_c$ of the reaction:
${Fe}^{3+}+S{CN}^{-}\rightleftharpoons FeS{CN}^{2+}$
will be:

• A. $6.66\times {10}^{-3}$
• B. $0.30$
• C. $3.30$
• D. $150$

Answer 1

Answer: (D)

$150$

$3.1$ mole of $Fe{Cl}_3$ and $3.2$ mole of ${NH}_{4}SCN$ are added to one litre of water. At equilibrium, $3.0$ mol of $FeS{CN}^{2+}$ are formed by the reaction of $3.0$ moles of $Fe{Cl}_3$ with $3.0$ moles of ${NH}_{4}SCN$.

$3.1-3.0=0.1$ moles of $Fe{Cl}_3$ and $3.2-3.0=0.2$ moles of ${NH}_{4}SCN$ will remain. Total volume is 1 L.
${Fe}^{3+}+S{CN}^{-}\rightleftharpoons FeS{CN}^{2+}$
Initial moles $t_0$ $3.1$ $3.2$ $0$
Equilibrium concentration $t_{eq.}$ $\frac{0.1}{1}$ $\frac{0.2}{1}$ $\frac{3}{1}$
The equilibrium constant:

$K_c=\frac{\left[FeS{CN}^{2+}\right]}{\left[{Fe}^{3+}\right]\times \left[S{CN}^{-}\right]}$

$K_c=\frac{3}{0.1\times 0.2}=150$