Question

When $PC{l_5}$ heated in $2L$ capacity vessel at eq of $40\%$ $PC{l_5}$ dissociate into $PC{l_3}$ and $C{l_2}$. Value of ${K_c}$.

Answer 1

We can calculate the equilibrium constant of the reaction using the concentrations of $PC{l_5}$, $PC{l_3}$ and $C{l_2}$. We can calculate the value of equilibrium constant by multiplying the concentration of the products and dividing it by the concentration of reactants. We can calculate the concentration of the reactants and products, their moles and volume.

Complete answer:
Given data contains,
Volume of the equilibrium mixture is $2L$.
Degree of dissociation $\left( \alpha \right)$ is $40\% = \dfrac{{40}}{{100}} = 0.4$.
We can write the chemical reaction as,
$PC{l_5}\left( g \right) \rightleftharpoons PC{l_3}\left( g \right) + C{l_2}\left( g \right)$

We can write the ICE table for the given reaction as,
\begin{matrix}
&PC;{l_5}\left( g \right)\rightleftharpoons&PC;{l_3}\left( g \right) +&C;{l_2}\left( g \right)\\
\text{Initial}&{{\text{2mol}}}&{{\text{0mol}}}&{{\text{0mol}}}\\
\text{Change}&{2\alpha {\text{mol}}}&{2\alpha{\text{ mol}}}&{2\alpha {\text{mol}}}\\
\text{Equilibrium}&{2\left( {1 - \alpha } \right) \text{mol}}&{2\alpha \text{mol}}&{2\alpha \text{mol}}
\end{matrix}

We can calculate the molar concentrations of $PC{l_5}$, $PC{l_3}$ and $C{l_2}$ of the mixture using the values of degree of dissociation and volume.
We can calculate the molar concentration of $PC{l_5}$ as,
$\left[ {PC{l_5}\left( g \right)} \right] = \dfrac{{2\left( {1 - \alpha } \right)}}{V}$

Let us substitute the value of degree of dissociation and volume.
$\left[ {PC{l_5}\left( g \right)} \right] = \dfrac{{2\left( {1 - 0.4} \right)}}{2}$
$\left[ {PC{l_5}\left( g \right)} \right] = 0.6mol/L$
The molar concentration of $PC{l_5}$ is $0.6mol/L$.
We can calculate the molar concentration of $PC{l_3}$ as,
$\left[ {PC{l_3}\left( g \right)} \right] = \dfrac{{2\left( \alpha \right)}}{V}$

Let us substitute the value of degree of dissociation and volume.
$\left[ {PC{l_3}\left( g \right)} \right] = \dfrac{{2\left( {0.4} \right)}}{2}$
$\left[ {PC{l_3}\left( g \right)} \right] = 0.4mol/L$
The molar concentration of $PC{l_3}$ is $0.4mol/L$.
We can calculate the molar concentration of $C{l_2}$ as,
$\left[ {Cl{ _2}\left( g \right)} \right] = \dfrac{{2\left( \alpha \right)}}{V}$

Let us substitute the value of degree of dissociation and volume.
$\left[ {C{l_2}\left( g \right)} \right] = \dfrac{{2\left( {0.4} \right)}}{2}$
$\left[ {C{l_2}\left( g \right)} \right] = 0.4mol/L$
The molar concentration of $C{l_2}$ is $0.4mol/L$.

We know that the formula to calculate the equilibrium constant is written as,
${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{\left[ {{\text{Products}}} \right]}}{{\left[ {{\text{Reactants}}} \right]}}$
We can write the rate law for the given reaction as,
${K_c} = \dfrac{{\left[ {PC{l_3}\left( g \right)} \right]\left[ {C{l_2}\left( g \right)} \right]}}{{\left[ {PC{l_5}\left( g \right)} \right]}}$

Let us now substitute the value of molar concentrations of $PC{l_5}$, $PC{l_3}$ and $C{l_2}$.
We can now calculate the equilibrium constant as,
${K_c} = \dfrac{{\left[ {0.4} \right]\left[ {0.4} \right]}}{{\left[ {0.6} \right]}}$
${K_c} = 0.2666$
The value of equilibrium constant is $0.2666mol/L$.

Note:
-We can determine the composition of the system at equilibrium using the initial composition of the system and known values of equilibrium constant.
-The reciprocal of forward reaction gives the equilibrium expression of the reverse reaction.
-The different types of equilibrium constants are stability constants, binding constants, formation constants, dissociation constants, and association constants.
-Factors such as temperature, ionic strength, and solvents affect the value of equilibrium constant.