Question

Assume that the decomposition of $\text{ HN}{{\text{O}}_{\text{3 }}}$ can be represented by the following equation:
$\text{ 4HN}{{\text{O}}_{\text{3 }}}(g)\text{ }\rightleftharpoons \text{4N}{{\text{O}}_{\text{2}}}(g)\text{ + 2}{{\text{H}}_{\text{2}}}\text{O}(g)\text{ + }{{\text{O}}_{\text{2}}}\text{ }(g)\text{ }$
The reaction approaches equilibrium at $\text{ 400 K }$ temperature and 30 atm pressure. At equilibrium, the partial pressure of $\text{ HN}{{\text{O}}_{\text{3 }}}$ is 2 atm.
The value of $\text{ }{{\text{K}}_{\text{C }}}(\text{mol/}{{\text{L}}^{\text{3}}}\text{) }$ at $\text{ 400 K }$ is:
(Use $\text{ R = 0}\text{.08 atm L/mol K }$ )

Answer 1

The equilibrium constant can be written in terms of activities , partial pressure , molar concentration or mole fraction. The equilibrium constant in terms of molar concentration and partial pressure is related as follows:
$\text{ }{{\text{K}}_{\text{P }}}=\text{ }{{\text{K}}_{\text{C }}}{{\left( \text{RT} \right)}^{\Delta n}}\text{ }$
Where , $\text{ }{{\text{K}}_{\text{P }}}$ is the partial pressure equilibrium constant ,$\text{ }{{\text{K}}_{\text{C }}}$ is the equilibrium constant in terms of molar concentration , R is gas constant , T is temperature and $\text{ }\Delta \text{n }$ is the difference in the number of moles of product and reactant.

Complete Solution :
Nitric acid $\text{ HN}{{\text{O}}_{\text{3 }}}$ dissociated to give $\text{ N}{{\text{O}}_{\text{2}}}\text{ }$ , $\text{ }{{\text{H}}_{\text{2}}}\text{O }$ and oxygen gas $\text{ }{{\text{O}}_{\text{2 }}}$ four moles of $\text{ HN}{{\text{O}}_{\text{3 }}}$ gives four moles of $\text{ N}{{\text{O}}_{\text{2}}}\text{ }$, two moles of $\text{ }{{\text{H}}_{\text{2}}}\text{O }$and one mole of water. If we start with one mole of $\text{ HN}{{\text{O}}_{\text{3 }}}$ and if $\text{ }\\!\\!\alpha\\!\\!\text{ }$ is the degree of dissociation .Then the dissociation of $\text{ HN}{{\text{O}}_{\text{3 }}}$ is written as follows:
$\,\text{ }\begin{matrix} {} & 4\text{ HN}{{\text{O}}_{\text{3 }}} & \rightleftharpoons & \text{ 4N}{{\text{O}}_{\text{2 }}} & \+ & 2{{\text{H}}_{\text{2}}}\text{O} & \+ & {{\text{O}}_{\text{2}}} \\\ \text{Before} & 1 & {} & 0 & {} & 0 & {} & 0 \\\ \text{After} & \left( 1-\alpha \right) & {} & 4\alpha & {} & 2\alpha & {} & \alpha \\\ \end{matrix}\text{ }$
The equilibrium constant $\text{ }{{\text{K}}_{\text{p }}}$ for the dissociation of $\text{ HN}{{\text{O}}_{\text{3 }}}$ is written as,
$\text{ }{{\text{K}}_{\text{P }}}=\text{ }\dfrac{\text{P}_{\text{N}{{\text{O}}_{\text{2}}}}^{4}\times {{\text{P}}_{{{\text{H}}_{\text{2}}}\text{O}}}\times {{\text{P}}_{{{\text{O}}_{\text{2}}}}}}{\text{P}_{\text{HN}{{\text{O}}_{3}}}^{4}}\text{ }$ (1)
The total pressure of the reaction is calculated as follows,
$\text{ }{{\text{P}}_{\text{Total}}}\text{ = }{{\text{P}}_{\text{HN}{{\text{O}}_{\text{3}}}}}+{{\text{P}}_{\text{N}{{\text{O}}_{\text{2}}}}}+{{\text{P}}_{{{\text{O}}_{\text{2}}}}}+{{\text{P}}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{ }$
Let's write down the pressure of nitrogen dioxide and water in terms of oxygen. We have,
$\begin{aligned} & \text{ }{{\text{P}}_{\text{N}{{\text{O}}_{\text{2}}}}}=4{{\text{P}}_{{{\text{O}}_{\text{2}}}}} \\\ & \text{ }{{\text{P}}_{{{\text{H}}_{\text{2}}}\text{O}}}=2{{\text{P}}_{{{\text{O}}_{\text{2}}}}} \\\ \end{aligned}$
Then total pressure of the reaction is written as,
$\begin{aligned} & \text{ }{{\text{P}}_{\text{Total}}}\text{ = }{{\text{P}}_{\text{HN}{{\text{O}}_{\text{3}}}}}+4{{\text{P}}_{{{\text{O}}_{\text{2}}}}}+{{\text{P}}_{{{\text{O}}_{\text{2}}}}}+2{{\text{P}}_{{{\text{O}}_{\text{2}}}}} \\\ & \Rightarrow \text{ }{{\text{P}}_{\text{Total}}}\text{ = }{{\text{P}}_{\text{HN}{{\text{O}}_{\text{3}}}}}+7{{\text{P}}_{{{\text{O}}_{\text{2}}}}} \\\ \end{aligned}$
The total pressure of the reaction is given as 30 atm and the partial pressure of $\text{ HN}{{\text{O}}_{\text{3 }}}$is equal to 2 atm.Lets substitute these values in total vapour pressure .we have,
$\begin{aligned} & \text{ }\Rightarrow \text{ }{{\text{P}}_{\text{Total}}}\text{ = }{{\text{P}}_{\text{HN}{{\text{O}}_{\text{3}}}}}+7{{\text{P}}_{{{\text{O}}_{\text{2}}}}} \\\ & \Rightarrow {{\text{P}}_{{{\text{O}}_{\text{2}}}}}=\dfrac{30-2}{7}=\dfrac{28}{7}=4\text{ atm } \\\ \end{aligned}$
Therefore, partial pressure of oxygen gas is 4 atm.
Now let’s substitute value of pressure of oxygen, nitric acid, water in equation (1).we have,
$\begin{aligned} & \text{ }{{\text{K}}_{\text{P }}}=\text{ }\dfrac{\text{P}_{\text{N}{{\text{O}}_{\text{2}}}}^{4}\times {{\text{P}}_{{{\text{H}}_{\text{2}}}\text{O}}}\times {{\text{P}}_{{{\text{O}}_{\text{2}}}}}}{\text{P}_{\text{HN}{{\text{O}}_{3}}}^{4}}\text{ =}\dfrac{{{\left( 4\times {{\text{P}}_{{{\text{O}}_{\text{2}}}}} \right)}^{4}}\times \left( 2\times {{\text{P}}_{{{\text{O}}_{\text{2}}}}} \right)\times {{\text{P}}_{{{\text{O}}_{\text{2}}}}}}{{{\left( 4 \right)}^{4}}} \\\ & \Rightarrow {{\text{K}}_{\text{P }}}=\dfrac{{{\left( 4\times 4 \right)}^{4}}\times \left( 2\times 4 \right)\times 4}{{{\left( 2 \right)}^{4}}}={{2}^{20}}\, \\\ \end{aligned}$
The equilibrium constant at pressure $\text{ }{{\text{K}}_{\text{P }}}$ and concentration $\text{ }{{\text{K}}_{\text{C }}}$ are related by the following relation.
$\text{ }{{\text{K}}_{\text{P }}}=\text{ }{{\text{K}}_{\text{C }}}{{\left( \text{RT} \right)}^{\Delta n}}\text{ }$
The $\text{ }\Delta \text{n }$ for reaction is calculated by subtracting the number of moles of products and reactant as follows,
$\begin{aligned} & \text{ }\Delta \text{n }=\text{ }\left( 4+2+1 \right)-\left( 4 \right) \\\ & \Rightarrow \Delta \text{n}=3 \\\ \end{aligned}$
At last, let’s determine the equilibrium constant$\text{ }{{\text{K}}_{\text{C }}}$.
$\begin{aligned} & \text{ }{{\text{K}}_{\text{C }}}=\text{ }\dfrac{{{\text{K}}_{\text{P }}}}{{{\left( \text{RT} \right)}^{\Delta n}}}\text{ } \\\ & \Rightarrow {{\text{K}}_{\text{C }}}=\dfrac{{{\left( 2 \right)}^{20}}}{{{\left( 0.08\times 400 \right)}^{3}}}=\dfrac{{{\left( 2 \right)}^{20}}}{{{\left( 32 \right)}^{3}}} \\\ & \therefore {{\text{K}}_{\text{C }}}=32 \\\ \end{aligned}$
Therefore, the equilibrium constant is 32.

Note: Note that,
i) When $\,\Delta \text{n}=0\text{ }$ the value of $\text{ }{{\text{K}}_{\text{C }}}=\text{ }{{\text{K}}_{\text{P }}}$ .
ii) When $\,\Delta \text{n}>0\text{ }$or $\,\Delta \text{n}<0\text{ }$the value of $\text{ }{{\text{K}}_{\text{C }}}\ne \text{ }{{\text{K}}_{\text{P }}}$ .
The values of $\text{ }{{\text{K}}_{\text{C }}}$ and $\text{ }{{\text{K}}_{\text{P }}}$ are independent of its units in which the partial pressure and molar concentration of various species are expressed.