Question

The unit of ${{\rm{K}}_{\rm{c}}}$ for the decomposition of solid ammonium bisulphide are--.
A. ${\rm{mol}}{{\rm{e}}^{\rm{2}}}{\rm{/}}{{\rm{L}}^{\rm{2}}}$
B. ${{\rm{L}}^{\rm{2}}}/{\rm{mol}}{{\rm{e}}^{\rm{2}}}$
C. ${\rm{mole/L}}$
D. ${\rm{L}}/{\rm{mole}}$

Answer 1

Hint:${{\rm{k}}_{\rm{C}}}$is the equilibrium constant. It is determined for the chemical reaction which is in equilibrium. The concentration of all reactants and products are taken in molarity. Molarity is defined as the mole of solute dissolved in per litter of the solution.

Formula used: ${{\rm{k}}_{\rm{C}}}{\rm{ = }}\,{\left( {{\rm{mol/L}}} \right)^{\Delta {\rm{n}}}}$

Complete answer:
The equilibrium constant is defined as the product of the concentration of products having a stoichiometric coefficient as power divided by the product of the concentration of reactants having a stoichiometric coefficient as power.
Consider a general reaction as follows:
${\rm{aA}}\, + \,{\rm{bB}}\,\, \rightleftharpoons {\rm{cC}}\,{\rm{ + }}\,{\rm{dD}}$
The equilibrium constant expression for the above reaction is as follows:
${{\rm{k}}_{\rm{C}}}{\rm{ = }}\,\dfrac{{{{\left[ {\rm{C}} \right]}^{\rm{c}}}\,\,{{\left[ {\rm{D}} \right]}^{\rm{d}}}}}{{{{\left[ {\rm{A}} \right]}^{\rm{a}}}{{\left[ {\rm{B}} \right]}^{\rm{b}}}}}\,$
Where, $\left[ {} \right]$represents the concentration.
Concentration is taken in molarity. The unit of molarity is mol/L. so, the unit of ${{\rm{K}}_{\rm{c}}}$is,
${{\rm{k}}_{\rm{C}}}{\rm{ = }}\,\dfrac{{{{\left[ {{\rm{mol/L}}} \right]}^{{\rm{c + d}}}}}}{{{{\left[ {{\rm{mol/L}}} \right]}^{{\rm{a + b}}}}}}\,$
${{\rm{k}}_{\rm{C}}}{\rm{ = }}\,{\left( {{\rm{mol/L}}} \right)^{\Delta {\rm{n}}}}$
Where,
\[{\rm{\Delta }}{{\rm{n}}_{\rm{g}}}\] is the difference between the sum of stoichiometric coefficients of gaseous products and gaseous reactants.
The ${\rm{\Delta n}}$is determined as follows;
\[{\rm{\Delta n}}\,{\rm{ = }}\,\sum {{{\rm{n}}_{\rm{g}}}{\rm{(product)}}} - \sum {{{\rm{n}}_{\rm{g}}}{\rm{(reactant)}}} \]

The chemical equation for the decomposition of solid ammonium bisulphite is given as follows:
${\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{HS}}\,{\rm{(s)}} \rightleftharpoons \,{\rm{N}}{{\rm{H}}_{\rm{3}}}{\rm{(g)}}\,{\rm{ + }}\,{{\rm{H}}_{\rm{2}}}{\rm{S(g)}}$
We will determine the $\Delta \,{\rm{n}}\,$for above reaction as follows:
\[{\rm{\Delta n}}\,{\rm{ = }}\,\left( {1 + 1} \right) - 0\]
\[\Rightarrow {\rm{\Delta n}}\,{\rm{ = }}\,2\]
So, the unit of equilibrium is,
${{\rm{k}}_{\rm{C}}}{\rm{ = }}\,{\left( {{\rm{mol/L}}} \right)^2}$
${{\rm{k}}_{\rm{C}}}{\rm{ = }}\,{\rm{mo}}{{\rm{l}}^2}{\rm{/}}{{\rm{L}}^2}$
So, the unit of ${{\rm{K}}_{\rm{c}}}$ for the decomposition of solid ammonium bisulphide is ${\rm{mol}}{{\rm{e}}^{\rm{2}}}{\rm{/}}{{\rm{L}}^{\rm{2}}}$.

Therefore, option (A) ${\rm{mol}}{{\rm{e}}^{\rm{2}}}{\rm{/}}{{\rm{L}}^{\rm{2}}}$, is correct.

Note:
When the concentrations are given in pressure, the equilibrium constant is given as ${{\rm{K}}_{\rm{p}}}$. The relation between ${{\rm{K}}_{\rm{p}}}$ and ${{\rm{K}}_{\rm{c}}}$ is, ${{\rm{K}}_{\rm{p}}}\,{\rm{ = }}\,{{\rm{K}}_{\rm{c}}}{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}}$. The unit of ${{\rm{K}}_{\rm{p}}}$is ${\left( {{\rm{atm}}} \right)^{{\rm{\Delta n}}}}$ and unit of ${{\rm{K}}_{\rm{c}}}$ is \[{\left( {{\rm{mol}}\,\,{{\rm{L}}^{ - 1}}} \right)^{{\rm{\Delta n}}}}\]. The equilibrium reaction in which all species are in the same phase is known as homogeneous equilibrium. The equilibrium reaction in which all species are in different phases is known as heterogeneous equilibrium. $\Delta \,{\rm{n}}$ is calculated only for gaseous species.