Question

3.1 mole of \[FeC{l_3}\] and \[3.2mole\] of \[N{H_4}SCN\] are added to one litre of water. At equilibrium, \[3.0mol\] of \[FeSC{N^{2 + }}\] are formed. The equilibrium constant of \[{K_c}\] of the reaction: \[F{e^{3 + }} + \;SCN \rightleftharpoons FeSC{N^{2 + }}\]will be:
(A) \[6.66 \times {10^{ - 3}}\]
(B) \[0.30\]
(C) \[3.30\]
(D) \[150\]

Answer 1

Hint:We need to know what is equilibrium and equilibrium constant. Equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time. This happens when the rate of forward reaction is equal to the rate of reverse reaction.
For example, let we consider an equilibrium reaction,
$aA + bB \rightleftharpoons cC + dD$
For this reaction,
${K_c} = \dfrac{{{{\left[ C \right]}^c} \times {{\left[ D \right]}^d}}}{{{{\left[ A \right]}^a} \times {{\left[ B \right]}^b}}}$
Kc= Equilibrium constant
A and B are reactants of the reaction
C and D are products of the reaction
a, b, c and d are the number of moles of A, B, C and D respectively.

Complete step by step answer:
According to the question,
\[3.1mole\] of \[FeC{l_3}\] and \[3.2mole\] of \[N{H_4}SCN\] are added to one litre of water. This means that the initial concentrations of the reactants are \[3.1mole\] of \[FeC{l_3}\] and \[3.2mole\] of \[N{H_4}SCN\] before the reaction took place or before the equilibrium.
At equilibrium, \[3.0mol\] of \[FeSC{N^{2 + }}\] are formed. This means the concentration of the product formed after the reaction took place and reached an equilibrium state is \[3.0mol\] of \[FeSC{N^{2 + }}\].
We do not know the concentrations of the reactants at the equilibrium stage in order to calculate the equilibrium constant.
$F{e^{3 + }} + SCN \rightleftharpoons FeSC{N^{2 + }}$
\[\begin{array}{*{20}{c}}
  {{\text{Initial moles}}\left( {{t_0}} \right)}&{{\text{ }}3.1}&{{\text{ }}3.2}&{{\text{ }}0}
\end{array}\]
$\begin{array}{*{20}{c}}
  {{\text{Equilibrium concentration}}\left( {{t_{eq}}} \right)}&{\dfrac{{0.1}}{1}}&{\dfrac{{0.2}}{1}}&{{\text{ }}\dfrac{3}{1}}
\end{array}$
Hence the equilibrium constants can be calculated as
\[{K_{\mathbf{c}}} = \dfrac{{[FeSC{N^{2 + }}]}}{{[F{e^{3 + }}][SCN]}}\]
On substituting the known values we get,
$ \Rightarrow {K_c} = \dfrac{3}{{0.1 \times 0.2}}$
On simplification we get,
  \[{K_c} = 150\]
Hence the correct option is option (D).

Note:
It must be noted that equilibrium constant value has no units as it is a constant. An equilibrium constant is equal to the ratio of the forward and reverse rate constants. The magnitude of the equilibrium constant indicates the extent to which a reaction will proceed: If K is a large number, it means that the equilibrium concentration of the products is large. In this case, the reaction as written will proceed to the right (resulting in an increase in the concentration of products).