Question

For the reaction:
$2F{{e}^{3+}}\left( aq \right)+{{\left( H{{g}_{2}} \right)}^{2+}}\left( aq \right)\rightleftarrows 2F{{e}^{2+}}\left( aq \right)+2H{{g}^{2+}}\left( aq \right)$
${{K}_{c}}=9.14\times {{10}^{-6}}$ at $25{}^\circ C$.
If the initial concentrations of ion are:$F{{e}^{3+}}=0.5M,{{\left( H{{g}_{2}} \right)}^{2+}}=0.5M,F{{e}^{2+}}=0.03M~$ $H{{g}^{2+}}=0.03M~$ what will be the concentration of ions at equilibrium?
(A) $F{{e}^{3+}}=0.4973M,H{{g}_{2}}^{2+}=0.4987M,\,F{{e}^{2+}}=0.0327M\,~\text{and }\\!\\!~\\!\\!\text{ }H{{g}^{2+}}=0.0327M~$
(B) $F{{e}^{3+}}=0.0327M,H{{g}_{2}}^{2+}=0.0327M,F{{e}^{2+}}=0.497M\,and\,H{{g}^{2+}}=0.4973M$
(C) $F{{e}^{3+}}=0.3973M,H{{g}_{2}}^{2+}=0.3987M,\,F{{e}^{2+}}=0.1327M\,and\,H{{g}^{2+}}=0.1327M$
(D) None of these

Answer 1

To find the concentration of ions at equilibrium, we know that the equilibrium constant (${{K}_{c}}$), is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.

Complete step by step solution:
We know that the equilibrium constant expresses the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit.
So, first of all, we write the reaction.
This below equation represents the concentration before reaction.

& 2F{{e}^{3+}}\left( aq \right)+{{\left( H{{g}_{2}} \right)}^{2+}}\left( aq \right)\rightleftarrows 2F{{e}^{2+}}\left( aq \right)+2H{{g}^{2+}}\left( aq \right) \\\ & \,\,\,\,\,0.5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.03\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.03 \\\ & (0.5-a)\,\,\,\,\,\,\,\,\,\,(0.5-\frac{a}{2})\,\,\,\,\,\,\,\,\,\,\,\,\,(0.03+a)\,\,\,\,\,\,\,\,(0.03+a) \\\ \end{aligned}$$ This equation represents the concentration after reactions. We know that, dissociation constant decrease on reactants sides and increases and products side. We know to write the concentration of ions in the formula of ${{K}_{c}}$. $${{K}_{c}}=9.14\times {{10}^{-6}}=\frac{{{\left[ F{{e}^{2+}} \right]}^{2}}{{\left[ H{{g}^{+2}} \right]}^{2}}}{{{\left[ F{{e}^{3+}} \right]}^{2}}\left[ Hg_{2}^{+2} \right]}$$ Now, we will write the concentration of ions after dissociation in the formula of ${{K}_{c}}$. $$9.14\times {{10}^{-6}}=\frac{{{\left( 0.03+a \right)}^{2}}{{\left( 0.03+a \right)}^{2}}}{{{\left( 0.5-a \right)}^{2}}\left( 0.5-\frac{a}{2} \right)}$$ After calculation we will find value of a=0.0027 Now, we will the value of a in the equations that are formed after concentration. These equations are: $$[F{{e}^{3+}}]=0.5-a=0.5-0.0027=0.4973M$$ $$\left[ Hg_{2}^{2+} \right]=0.5-\frac{a}{2}=0.5-\frac{0.0027}{2}=0.4987M$$ $$[F{{e}^{2+}}]=0.03+a=0.03+0.0027=0.0327M$$ $$[H{{g}^{2+}}]=0.03+a=0.03+0.0027=0.0327M$$ **So, from the above discussion and calculation we can say that option A is correct.** **Note:** Equilibrium constant of a reaction depends on pressure. We know that if we increase the pressure by the addition of inert gas, then neither the composition at equilibrium nor the equilibrium constant is appreciably affected (because the partial pressures remain constant, assuming an ideal-gas behaviour of all gases involved). But we know that the composition at equilibrium will depend on pressure when the pressure is changed by compression or expansion of the gaseous reacting system, and the reaction results in the change of the number of moles of gas in the system.