Question

Two moles of ${\text{N}}{{\text{H}}_3}$ when put into previously evacuated vessel (one litre), partially dissociated into ${{\text{N}}_2}$ and ${{\text{H}}_2}$. If at equilibrium one mole of ${\text{N}}{{\text{H}}_3}$ is present, the equilibrium is:
A.$3/4{\text{ mo}}{{\text{l}}^2}{\text{litr}}{{\text{e}}^{ - 2}}$
B.$27/64{\text{ mo}}{{\text{l}}^2}{\text{litr}}{{\text{e}}^{ - 2}}$
C.$27/32{\text{ mo}}{{\text{l}}^2}{\text{litr}}{{\text{e}}^{ - 2}}$
D.$27/16{\text{ mo}}{{\text{l}}^2}{\text{litr}}{{\text{e}}^{ - 2}}$

Answer 1

In equilibrium reaction, both reactant and product remain in the reaction mixture. Calculation of moles of entities can be calculated by stoichiometric coefficients. And equilibrium constant can simply be calculated by using concentration of entities at equilibrium conditions.
Formula Used:
For the reaction: ${\text{aA}} + {\text{bB}} \to {\text{cC}} + {\text{dD}}$
${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{\left[ {\text{C}} \right]}^{\text{c}}}{{\left[ {\text{D}} \right]}^{\text{d}}}}}{{{{\left[ {\text{A}} \right]}^{\text{a}}}{{\left[ {\text{B}} \right]}^{\text{b}}}}}$

Complete step by step answer:
In case of reversible reactions, when equilibrium is reached; rate of forward reaction becomes equal to rate of backward reaction. That is the rate or speed at which reactant converting into product becomes equal to the rate at which product converting back into reactant. The chemical reaction of decomposition of Ammonia ${\text{N}}{{\text{H}}_3}$ into Nitrogen ${{\text{N}}_2}$ and Hydrogen${{\text{H}}_2}$ in balanced form can be written as:
${\text{2N}}{{\text{H}}_3} \to {{\text{N}}_2} + {\text{3}}{{\text{H}}_2}$
This can be read as 2 moles of ${\text{N}}{{\text{H}}_3}$ when decomposes, it produces 1 mole of ${{\text{N}}_2}$ and 3 moles of ${{\text{H}}_2}$. Now as mentioned in the question initially 2 moles of ammonia are present in the vessel (1 litre). And when the equilibrium reaches, one mole of ${\text{N}}{{\text{H}}_3}$ is left (or 1 mol of ammonia decomposes).
By unitary method, for every 1 mol of ammonia; $\dfrac{1}{2}$mole of ${{\text{N}}_2}$ and $\dfrac{3}{2}$ mole of ${{\text{H}}_2}$ will produce.
In order to find equilibrium constant, we need to find the concentration of each entity present in the reaction mixture. [Concentration of entity= mole of the entity/ volume of vessel]
Concentration of ${{\text{N}}_2}$ at equilibrium $= \dfrac{{0.5}}{1}$
Concentration of ${{\text{H}}_2} = \dfrac{{1.5}}{1}$
Concentration of ${\text{N}}{{\text{H}}_3} = \dfrac{1}{1}$.
The equilibrium constant is equal to product or multiplication of concentration of product raise to the power of their stoichiometric coefficients to the product of concentration of reactant raise to the power of their stoichiometric coefficients.
This whole can be represented as:
$2{\text{N}}{{\text{H}}_3} \to {{\text{N}}_2}\;\; + \;\;3{{\text{H}}_2}$
Initially number of moles: 2 0 0
Number of moles at equilibrium: $2 - 1 = 1$ $\dfrac{1}{2} = 0.5\;$ $\dfrac{3}{2}$
Concentration at equilibrium: ${\text{1}}$ $\dfrac{{\text{1}}}{2}$ $\dfrac{3}{2}$
${{\text{K}}_{\text{c}}} = \dfrac{{\dfrac{1}{2} \times {{\left( {\dfrac{3}{2}} \right)}^3}}}{1} = \dfrac{{27}}{{16}}$

So, the answer is option D.
Note:
To find the number of moles of other entities with respect to the given entity, reaction must be written in balanced form. As all entities are present in the same reaction mixture, so is the volume of each entity taken as the volume of the vessel in which the reaction is going on.
(For gases, volume of gas is equal to the volume of the vessel in which they are present).