In which of the following value of ${\rm{log}}\,\dfrac{{{{\rm{K}}_{\rm{p}}}}}{{{{\rm{K}}_{\rm{c}}}}}\, + \,\log {\rm{RT}}\,{\rm{ = }}\,{\rm{0}}$ is true relationship for which of the following reaction(s)?
A. ${\rm{PC}}{{\rm{l}}_5}\, \rightleftharpoons \,{\rm{PC}}{{\rm{l}}_3}\, + \,{\rm{C}}{{\rm{l}}_2}$
B. ${\rm{2S}}{{\rm{O}}_{\rm{2}}}\, + \,{{\rm{O}}_2}\, \rightleftharpoons \,{\rm{2}}\,{\rm{S}}{{\rm{O}}_3}$
C. ${{\rm{N}}_{\rm{2}}}\, + \,3\,{{\rm{H}}_2}\, \rightleftharpoons \,{\rm{2}}\,{\rm{N}}{{\rm{H}}_3}$
D. Both (B) and (C)
Answer
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Hint: ${{\rm{K}}_{\rm{p}}}$ and ${{\rm{K}}_{\rm{c}}}$ are the equilibrium constants. The ${{\rm{K}}_{\rm{p}}}$ is the product of ${{\rm{K}}_{\rm{c}}}$, gas constant and temperature raised stoichiometric difference in the power. We will substitute the value of ${{\rm{K}}_{\rm{p}}}$ in the given equation and determine the condition where the equation becomes zero.
Formula used: ${{\rm{K}}_{\rm{p}}}\,{\rm{ = }}\,{{\rm{K}}_{\rm{c}}}{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}}$
Complete answer:
The relation between equilibrium constant expressed in term of pressure and equilibrium constant expressed in term of concentration is as follows:
${{\rm{K}}_{\rm{p}}}\,{\rm{ = }}\,{{\rm{K}}_{\rm{c}}}{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}}$
where,
${{\rm{K}}_{\rm{p}}}$ is the equilibrium constant when the amount of reactant and products are taken in form of pressure.
${{\rm{K}}_{\rm{c}}}$ is the equilibrium constant when the amount of reactant and products are taken in form of concentrations.
${\rm{R}}$ is the gas constant.
${\rm{T}}$ is the temperature.
\[{\rm{\Delta n}}\] is the difference between the sum of stoichiometric coefficients of products and reactants.
We will substitute value of ${{\rm{K}}_{\rm{c}}}$in given formula as follows:
$\Rightarrow {\rm{log}}\,\dfrac{{{{\rm{K}}_{\rm{p}}}}}{{{{\rm{K}}_{\rm{c}}}}}\, + \,\log {\rm{RT}}\,{\rm{ = }}\,{\rm{0}}$
$\Rightarrow {\rm{log}}\,\dfrac{{{{\rm{K}}_{\rm{c}}}{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}}}}{{{{\rm{K}}_{\rm{c}}}}}\, + \,\log {\rm{RT}}\,{\rm{ = }}\,{\rm{0}}$
$\Rightarrow {\rm{log}}\,{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}} + \,\log {\rm{RT}}\,{\rm{ = }}\,{\rm{0}}$
Because ${\rm{log}}\,{\rm{a}} + \,\log {\rm{b}}\,{\rm{ = }}\,{\rm{log}}\,{\rm{a}} \times \,\log {\rm{b}}$
So, ${\rm{log}}\,{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}} + \,\log {\rm{RT}}\,{\rm{ = }}\,{\rm{log}}\,{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}} \times \,\log {\rm{RT}}\,$
$\Rightarrow {\rm{log}}\,{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}} \times \,\log {\rm{RT}}\, = 0$
$\Rightarrow {\rm{log}}\,{\rm{R}}{{\rm{T}}^{{\rm{\Delta n + 1}}}} = 0$
The above equation has two parts, ${\rm{log}}\,{\rm{RT}}$and ${\rm{\Delta n + 1}}$. Because R is constant so, ${\rm{log}}\,{\rm{RT}}$ cannot be zero, so, ${\rm{\Delta n + 1}}$ will be equal to zero.
So, we will determine the reaction for which ${\rm{\Delta n + 1}}$.
The ${\rm{\Delta n}}$ is determined as follows:
\[{\rm{\Delta n}}\,{\rm{ = }}\,\sum {{{\rm{n}}_{\rm{g}}}{\rm{(product)}}} - \sum {{{\rm{n}}_{\rm{g}}}{\rm{(reactant)}}} \]
We will determine the ${\rm{\Delta n + 1}}$ for each reaction as follows:
A. ${\rm{PC}}{{\rm{l}}_5}\, \rightleftharpoons \,{\rm{PC}}{{\rm{l}}_3}\, + \,{\rm{C}}{{\rm{l}}_2}$
\[\Rightarrow {\rm{\Delta n}}\, + 1{\rm{ = }}\,\left( {1 + 1} \right) - 1 + 1\]
\[\Rightarrow {\rm{\Delta n}}\, + 1{\rm{ = }}\,2\]
B. ${\rm{2S}}{{\rm{O}}_{\rm{2}}}\, + \,{{\rm{O}}_2}\, \rightleftharpoons \,{\rm{2}}\,{\rm{S}}{{\rm{O}}_3}$
\[\Rightarrow {\rm{\Delta n}}\, + 1\,{\rm{ = }}\,2\, - \left( {1 + 2} \right) + 1\]
\[\Rightarrow {\rm{\Delta n}}\, + 1 = \,0\]
C. ${{\rm{N}}_{\rm{2}}}\, + \,3\,{{\rm{H}}_2}\, \rightleftharpoons \,{\rm{2}}\,{\rm{N}}{{\rm{H}}_3}$
\[\Rightarrow {\rm{\Delta n}}\, + 1\,{\rm{ = }}\,2\, - \left( {3 + 1} \right) + 1\]
\[\Rightarrow {\rm{\Delta n}}\, + 1{\rm{ = }}\, - 1\]
The \[{\rm{\Delta n}}\, + 1 = \,0\] for ${\rm{2S}}{{\rm{O}}_{\rm{2}}}\, + \,{{\rm{O}}_2}\, \rightleftharpoons \,{\rm{2}}\,{\rm{S}}{{\rm{O}}_3}$. So, ${\rm{log}}\,\dfrac{{{{\rm{K}}_{\rm{p}}}}}{{{{\rm{K}}_{\rm{c}}}}}\, + \,\log {\rm{RT}}\,{\rm{ = }}\,{\rm{0}}$ is true relationship for ${\rm{2S}}{{\rm{O}}_{\rm{2}}}\, + \,{{\rm{O}}_2}\, \rightleftharpoons \,{\rm{2}}\,{\rm{S}}{{\rm{O}}_3}$ reaction.
Therefore, option (B) ${\rm{2S}}{{\rm{O}}_{\rm{2}}}\, + \,{{\rm{O}}_2}\, \rightleftharpoons \,{\rm{2}}\,{\rm{S}}{{\rm{O}}_3}$, is correct.
Note:
The stoichiometric difference is determined as the sum of the stoichiometry of all reactants subtracted from the sum of the stoichiometry of all products. According to the relation of ${{\rm{K}}_{\rm{p}}}$ and ${{\rm{K}}_{\rm{c}}}$, as the value of stoichiometric difference goes from negative to positive, the value of ${{\rm{K}}_{\rm{c}}}$ increases and the value of ${{\rm{K}}_{\rm{p}}}$ decreases. 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.
Formula used: ${{\rm{K}}_{\rm{p}}}\,{\rm{ = }}\,{{\rm{K}}_{\rm{c}}}{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}}$
Complete answer:
The relation between equilibrium constant expressed in term of pressure and equilibrium constant expressed in term of concentration is as follows:
${{\rm{K}}_{\rm{p}}}\,{\rm{ = }}\,{{\rm{K}}_{\rm{c}}}{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}}$
where,
${{\rm{K}}_{\rm{p}}}$ is the equilibrium constant when the amount of reactant and products are taken in form of pressure.
${{\rm{K}}_{\rm{c}}}$ is the equilibrium constant when the amount of reactant and products are taken in form of concentrations.
${\rm{R}}$ is the gas constant.
${\rm{T}}$ is the temperature.
\[{\rm{\Delta n}}\] is the difference between the sum of stoichiometric coefficients of products and reactants.
We will substitute value of ${{\rm{K}}_{\rm{c}}}$in given formula as follows:
$\Rightarrow {\rm{log}}\,\dfrac{{{{\rm{K}}_{\rm{p}}}}}{{{{\rm{K}}_{\rm{c}}}}}\, + \,\log {\rm{RT}}\,{\rm{ = }}\,{\rm{0}}$
$\Rightarrow {\rm{log}}\,\dfrac{{{{\rm{K}}_{\rm{c}}}{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}}}}{{{{\rm{K}}_{\rm{c}}}}}\, + \,\log {\rm{RT}}\,{\rm{ = }}\,{\rm{0}}$
$\Rightarrow {\rm{log}}\,{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}} + \,\log {\rm{RT}}\,{\rm{ = }}\,{\rm{0}}$
Because ${\rm{log}}\,{\rm{a}} + \,\log {\rm{b}}\,{\rm{ = }}\,{\rm{log}}\,{\rm{a}} \times \,\log {\rm{b}}$
So, ${\rm{log}}\,{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}} + \,\log {\rm{RT}}\,{\rm{ = }}\,{\rm{log}}\,{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}} \times \,\log {\rm{RT}}\,$
$\Rightarrow {\rm{log}}\,{\rm{R}}{{\rm{T}}^{{\rm{\Delta n}}}} \times \,\log {\rm{RT}}\, = 0$
$\Rightarrow {\rm{log}}\,{\rm{R}}{{\rm{T}}^{{\rm{\Delta n + 1}}}} = 0$
The above equation has two parts, ${\rm{log}}\,{\rm{RT}}$and ${\rm{\Delta n + 1}}$. Because R is constant so, ${\rm{log}}\,{\rm{RT}}$ cannot be zero, so, ${\rm{\Delta n + 1}}$ will be equal to zero.
So, we will determine the reaction for which ${\rm{\Delta n + 1}}$.
The ${\rm{\Delta n}}$ is determined as follows:
\[{\rm{\Delta n}}\,{\rm{ = }}\,\sum {{{\rm{n}}_{\rm{g}}}{\rm{(product)}}} - \sum {{{\rm{n}}_{\rm{g}}}{\rm{(reactant)}}} \]
We will determine the ${\rm{\Delta n + 1}}$ for each reaction as follows:
A. ${\rm{PC}}{{\rm{l}}_5}\, \rightleftharpoons \,{\rm{PC}}{{\rm{l}}_3}\, + \,{\rm{C}}{{\rm{l}}_2}$
\[\Rightarrow {\rm{\Delta n}}\, + 1{\rm{ = }}\,\left( {1 + 1} \right) - 1 + 1\]
\[\Rightarrow {\rm{\Delta n}}\, + 1{\rm{ = }}\,2\]
B. ${\rm{2S}}{{\rm{O}}_{\rm{2}}}\, + \,{{\rm{O}}_2}\, \rightleftharpoons \,{\rm{2}}\,{\rm{S}}{{\rm{O}}_3}$
\[\Rightarrow {\rm{\Delta n}}\, + 1\,{\rm{ = }}\,2\, - \left( {1 + 2} \right) + 1\]
\[\Rightarrow {\rm{\Delta n}}\, + 1 = \,0\]
C. ${{\rm{N}}_{\rm{2}}}\, + \,3\,{{\rm{H}}_2}\, \rightleftharpoons \,{\rm{2}}\,{\rm{N}}{{\rm{H}}_3}$
\[\Rightarrow {\rm{\Delta n}}\, + 1\,{\rm{ = }}\,2\, - \left( {3 + 1} \right) + 1\]
\[\Rightarrow {\rm{\Delta n}}\, + 1{\rm{ = }}\, - 1\]
The \[{\rm{\Delta n}}\, + 1 = \,0\] for ${\rm{2S}}{{\rm{O}}_{\rm{2}}}\, + \,{{\rm{O}}_2}\, \rightleftharpoons \,{\rm{2}}\,{\rm{S}}{{\rm{O}}_3}$. So, ${\rm{log}}\,\dfrac{{{{\rm{K}}_{\rm{p}}}}}{{{{\rm{K}}_{\rm{c}}}}}\, + \,\log {\rm{RT}}\,{\rm{ = }}\,{\rm{0}}$ is true relationship for ${\rm{2S}}{{\rm{O}}_{\rm{2}}}\, + \,{{\rm{O}}_2}\, \rightleftharpoons \,{\rm{2}}\,{\rm{S}}{{\rm{O}}_3}$ reaction.
Therefore, option (B) ${\rm{2S}}{{\rm{O}}_{\rm{2}}}\, + \,{{\rm{O}}_2}\, \rightleftharpoons \,{\rm{2}}\,{\rm{S}}{{\rm{O}}_3}$, is correct.
Note:
The stoichiometric difference is determined as the sum of the stoichiometry of all reactants subtracted from the sum of the stoichiometry of all products. According to the relation of ${{\rm{K}}_{\rm{p}}}$ and ${{\rm{K}}_{\rm{c}}}$, as the value of stoichiometric difference goes from negative to positive, the value of ${{\rm{K}}_{\rm{c}}}$ increases and the value of ${{\rm{K}}_{\rm{p}}}$ decreases. 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.
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