Question

Following reaction, \[\text{NH}_{\text{4}}^{\text{+}}\text{(aq)}\,\text{+}\,\text{C}{{\text{N}}^{\text{-}}}\text{(aq)}\,\,\rightleftharpoons \,\text{HCN(aq)}\,\text{+}\,\text{N}{{\text{H}}_{\text{3}}}\text{(aq)}\] Proceed in:
(A) Forward direction
(B) Backward direction
(C)In both side equally
(D) Can’t be predicted

Answer 1

Hint: $\text{N}{{\text{H}}_{\text{3}}}$ in aqueous solution acts as a bronsted base (which has the tendency to gain hydrogen ion).
The direction of an equilibrium reaction is determined by Le-chatelier’s principle.
According to Le-chatelier’s principle if a system at equilibrium is subjected to a change in concentration, pressure and temperature then the equilibrium will shift in such a way as to nullify the effect of change.

Complete step by step answer:
-When $\text{N}{{\text{H}}_{\text{3}}}$ is dissolved in an aqueous solution it acts as a base and accepts the hydrogen ion. So $\text{N}{{\text{H}}_{\text{3}}}$ (a polar molecule) is highly soluble in water (a polar solvent) by making hydrogen bond and form ammonium ion or hydroxyl ion which is also soluble in water. This reaction is represented by following reaction-
$\text{N}{{\text{H}}_{\text{3}}}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{O}\to \,\,\text{NH}_{\text{4}}^{\text{+}}\,\text{+}\,\text{O}{{\text{H}}^{\text{-}}}$
- Cyanide ion (\[\text{C}{{\text{N}}^{\text{-}}}\]) is also soluble in water to form \[\,\text{HCN}\]. In the basic medium \[\,\text{HCN}\] more predominantly found in the \[\text{C}{{\text{N}}^{\text{-}}}\]ion.
-Since, high concentration of \[\text{C}{{\text{N}}^{\text{-}}}\] and $\text{NH}_{\text{4}}^{\text{+}}$ ion present in product side, so the reaction will proceed from right side to left side in order to attain equilibrium.

So, option (B) will be the correct answer.

Note: The change in either pressure or concentration at equilibrium may shift the equilibrium, or more likely change the state of equilibrium, but it cannot change the equilibrium constant because equilibrium constant only depends on temperature.
Rate of forward and backward reaction will change on changing the concentration of the reactant particle. At equilibrium, the rate of forward direction will be equal to the rate of backward direction.