
In which of the following value of ${{\text{K}}_{\text{p}}}$ is less than ${{\text{K}}_{\text{c}}}$?
A) ${{\text{H}}_{\text{2}}}\, + \,{{\text{I}}_2}\, \rightleftharpoons \,{\text{2 HI}}$
B) ${{\text{N}}_{\text{2}}}\, + \,3\,{{\text{H}}_2}\, \rightleftharpoons \,{\text{2}}\,{\text{N}}{{\text{H}}_3}$
C) ${{\text{N}}_{\text{2}}}\, + \,{{\text{O}}_2}\, \rightleftharpoons \,{\text{2}}\,{\text{NO}}$
D) ${\text{CO}}\, + \,{{\text{H}}_2}{\text{O}}\, \rightleftharpoons \,{\text{C}}{{\text{O}}_{\text{2}}}\,{\text{ + }}\,{{\text{H}}_2}$
Answer
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Hint: ${{\text{K}}_{\text{p}}}$ and ${{\text{K}}_{\text{c}}}$ are the equilibrium constants. The ${{\text{K}}_{\text{p}}}$ is the product of ${{\text{K}}_{\text{c}}}$, gas constant and temperature raised stoichiometric difference in the power. The stoichiometric difference is determined as the sum of the stoichiometry of all reactants subtracted from the sum of the stoichiometry of all products.
Formula used: ${{\text{K}}_{\text{p}}}\,{\text{ = }}\,{{\text{K}}_{\text{c}}}{\text{R}}{{\text{T}}^{{{\Delta n}}}}$
Complete answer
The relation between equilibrium constant expressed in term of pressure and equilibrium constant expressed in term of concentration is as follows:
${{\text{K}}_{\text{p}}}\,{\text{ = }}\,{{\text{K}}_{\text{c}}}{\text{R}}{{\text{T}}^{{{\Delta n}}}}$
Where,
${{\text{K}}_{\text{p}}}$ is the equilibrium constant when the amount of reactant and products are taken in form of pressure.
${{\text{K}}_{\text{c}}}$ is the equilibrium constant when the amount of reactant and products are taken in form of concentrations.
${\text{R}}$ is the gas constant.
${\text{T}}$ is the temperature.
\[{{\Delta n}}\] is the difference between the sum of stoichiometric coefficients of products and reactants.
Take natural log as,
${\text{ln}}\,\,{{\text{K}}_{\text{p}}}\,{\text{ = ln}}\,\,{{\text{K}}_{\text{c}}}{\text{ + }}\,{{\Delta n}}\,\,{\text{ln}}\,\,{\text{R}}\,{\text{T}}$
The ${{\Delta n}}$ is determined as follows;
\[{{\Delta n}}\,{\text{ = }}\,\sum {{{\text{n}}_{\text{g}}}{\text{(product)}}} - \sum {{{\text{n}}_{\text{g}}}{\text{(reactant)}}} \]
For \[{{\Delta n}}\,{\text{ = }}\,{\text{ + }}\], ${{\text{K}}_{\text{p}}}\, > \,{{\text{K}}_{\text{c}}}$
For $\Delta \,{\text{n}}\,{\text{ = }}\, - \,$, ${{\text{K}}_{\text{p}}}\, < \,{{\text{K}}_{\text{c}}}$
For $\Delta \,{\text{n}}\,{\text{ = }}\,0$ , ${{\text{K}}_{\text{p}}}\, = {{\text{K}}_{\text{c}}}$
We will determine the $\Delta \,{\text{n}}\,$ for each reaction as follows:
${{\text{H}}_{\text{2}}}\, + \,{{\text{I}}_2}\, \rightleftharpoons \,{\text{2 HI}}$
\[{{\Delta n}}\,{\text{ = }}\,2\, - \left( {1 + 1} \right)\]
\[{{\Delta n}}\,{\text{ = }}\,0\]
${{\text{N}}_{\text{2}}}\, + \,3\,{{\text{H}}_2}\, \rightleftharpoons \,{\text{2}}\,{\text{N}}{{\text{H}}_3}$
\[{{\Delta n}}\,{\text{ = }}\,2\, - \left( {3 + 1} \right)\]
\[{{\Delta n}}\,{\text{ = }}\, - 2\]
${{\text{N}}_{\text{2}}}\, + \,{{\text{O}}_2}\, \rightleftharpoons \,{\text{2}}\,{\text{NO}}$
\[{{\Delta n}}\,{\text{ = }}\,2\, - \left( {1 + 1} \right)\]
\[{{\Delta n}}\,{\text{ = }}\,0\]
${\text{CO}}\, + \,{{\text{H}}_2}{\text{O}}\, \rightleftharpoons \,{\text{C}}{{\text{O}}_{\text{2}}}\,{\text{ + }}\,{{\text{H}}_2}$
\[{t{\Delta n}}\,{\text{ = }}\,\left( {1 + 1} \right)\, - \left( {1 + 1} \right)\]
\[{{\Delta n}}\,{\text{ = }}\,0\]
The $\Delta \,{\text{n}}\,{\text{ = }}\, - \,$ for ${{\text{N}}_{\text{2}}}\, + \,3\,{{\text{H}}_2}\, \rightleftharpoons \,{\text{2}}\,{\text{N}}{{\text{H}}_3}$. So, for ${{\text{N}}_{\text{2}}}\, + \,3\,{{\text{H}}_2}\, \rightleftharpoons \,{\text{2}}\,{\text{N}}{{\text{H}}_3}$ reaction, ${{\text{K}}_{\text{p}}}\, < \,{{\text{K}}_{\text{c}}}$.
Therefore, option (B) ${{\text{N}}_{\text{2}}}\, + \,3\,{{\text{H}}_2}\, \rightleftharpoons \,{\text{2}}\,{\text{N}}{{\text{H}}_3}$, is correct.
Note: Ionic compounds have ionic bonds and covalent compounds have covalent bonds. The covalent compounds are made up of non-metals only. Fajan’s rule determines the covalent character in the ionic bond only, not the ionic character in the covalent bond. The covalent character is directly proportional to the solubility in non-polar solvents and inversely proportional to the solubility in the polar solvent.
Formula used: ${{\text{K}}_{\text{p}}}\,{\text{ = }}\,{{\text{K}}_{\text{c}}}{\text{R}}{{\text{T}}^{{{\Delta n}}}}$
Complete answer
The relation between equilibrium constant expressed in term of pressure and equilibrium constant expressed in term of concentration is as follows:
${{\text{K}}_{\text{p}}}\,{\text{ = }}\,{{\text{K}}_{\text{c}}}{\text{R}}{{\text{T}}^{{{\Delta n}}}}$
Where,
${{\text{K}}_{\text{p}}}$ is the equilibrium constant when the amount of reactant and products are taken in form of pressure.
${{\text{K}}_{\text{c}}}$ is the equilibrium constant when the amount of reactant and products are taken in form of concentrations.
${\text{R}}$ is the gas constant.
${\text{T}}$ is the temperature.
\[{{\Delta n}}\] is the difference between the sum of stoichiometric coefficients of products and reactants.
Take natural log as,
${\text{ln}}\,\,{{\text{K}}_{\text{p}}}\,{\text{ = ln}}\,\,{{\text{K}}_{\text{c}}}{\text{ + }}\,{{\Delta n}}\,\,{\text{ln}}\,\,{\text{R}}\,{\text{T}}$
The ${{\Delta n}}$ is determined as follows;
\[{{\Delta n}}\,{\text{ = }}\,\sum {{{\text{n}}_{\text{g}}}{\text{(product)}}} - \sum {{{\text{n}}_{\text{g}}}{\text{(reactant)}}} \]
For \[{{\Delta n}}\,{\text{ = }}\,{\text{ + }}\], ${{\text{K}}_{\text{p}}}\, > \,{{\text{K}}_{\text{c}}}$
For $\Delta \,{\text{n}}\,{\text{ = }}\, - \,$, ${{\text{K}}_{\text{p}}}\, < \,{{\text{K}}_{\text{c}}}$
For $\Delta \,{\text{n}}\,{\text{ = }}\,0$ , ${{\text{K}}_{\text{p}}}\, = {{\text{K}}_{\text{c}}}$
We will determine the $\Delta \,{\text{n}}\,$ for each reaction as follows:
${{\text{H}}_{\text{2}}}\, + \,{{\text{I}}_2}\, \rightleftharpoons \,{\text{2 HI}}$
\[{{\Delta n}}\,{\text{ = }}\,2\, - \left( {1 + 1} \right)\]
\[{{\Delta n}}\,{\text{ = }}\,0\]
${{\text{N}}_{\text{2}}}\, + \,3\,{{\text{H}}_2}\, \rightleftharpoons \,{\text{2}}\,{\text{N}}{{\text{H}}_3}$
\[{{\Delta n}}\,{\text{ = }}\,2\, - \left( {3 + 1} \right)\]
\[{{\Delta n}}\,{\text{ = }}\, - 2\]
${{\text{N}}_{\text{2}}}\, + \,{{\text{O}}_2}\, \rightleftharpoons \,{\text{2}}\,{\text{NO}}$
\[{{\Delta n}}\,{\text{ = }}\,2\, - \left( {1 + 1} \right)\]
\[{{\Delta n}}\,{\text{ = }}\,0\]
${\text{CO}}\, + \,{{\text{H}}_2}{\text{O}}\, \rightleftharpoons \,{\text{C}}{{\text{O}}_{\text{2}}}\,{\text{ + }}\,{{\text{H}}_2}$
\[{t{\Delta n}}\,{\text{ = }}\,\left( {1 + 1} \right)\, - \left( {1 + 1} \right)\]
\[{{\Delta n}}\,{\text{ = }}\,0\]
The $\Delta \,{\text{n}}\,{\text{ = }}\, - \,$ for ${{\text{N}}_{\text{2}}}\, + \,3\,{{\text{H}}_2}\, \rightleftharpoons \,{\text{2}}\,{\text{N}}{{\text{H}}_3}$. So, for ${{\text{N}}_{\text{2}}}\, + \,3\,{{\text{H}}_2}\, \rightleftharpoons \,{\text{2}}\,{\text{N}}{{\text{H}}_3}$ reaction, ${{\text{K}}_{\text{p}}}\, < \,{{\text{K}}_{\text{c}}}$.
Therefore, option (B) ${{\text{N}}_{\text{2}}}\, + \,3\,{{\text{H}}_2}\, \rightleftharpoons \,{\text{2}}\,{\text{N}}{{\text{H}}_3}$, is correct.
Note: Ionic compounds have ionic bonds and covalent compounds have covalent bonds. The covalent compounds are made up of non-metals only. Fajan’s rule determines the covalent character in the ionic bond only, not the ionic character in the covalent bond. The covalent character is directly proportional to the solubility in non-polar solvents and inversely proportional to the solubility in the polar solvent.
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