Question

In a $500ml$ flask, the degree of dissociation of $PC{{l}_{5}}$ at equilibrium is $40%$ and the initial amount is $5$ moles. The value of equilibrium constant in $mol{{L}^{-1}}$ for the decomposition of $PC{{l}_{5}}$ is
A. $2.33$
B. $2.66$
C. $5.32$
D. $4.66$

Answer 1

Hint: Equilibrium constant K is the ratio of concentration of products and reactants once the reaction has reached equilibrium. It is important in the context of chemical reactions as it tells us about how reversible the reaction is.

Formula used: ${{K}_{eq}}=\dfrac{\left[ P \right]}{\left[ R \right]}$
Where, ${{K}_{eq}}$ is the dissociation constant
$P$ is the concentration of product
$R$ is the concentration of reactant

Complete step by step answer:
The ratio between the product of molar concentrations of the products to that of molar concentration of reactants with each concentration term raised to a power equal to stoichiometric coefficient in the balanced chemical reaction at a given temperature is called an equilibrium constant.
Here, it is given that the degree of dissociation $\left( \alpha \right)=40%$
$\alpha =\dfrac{D.M}{I.M}$
Where, $\alpha $ is the degree of dissociation
$D.M$ is the dissociated moles
$I.M$ is the initial moles
Here, $D.M=40$ and $I.M=100$
Now, substituting the value in the given formula we get,
$\alpha =\dfrac{40}{100}$
$\alpha =0.4$
To calculate the value of ${{K}_{c}}$ , that is, concentration at equilibrium, the formula can be written as:
${{K}_{C}}=\dfrac{\left[ P \right]}{\left[ R \right]}$
${{K}_{C}}$ is the concentration at equilibrium
$\left[ P \right]$ is the concentration of product
$\left[ R \right]$ is the concentration of reactant
The decomposition of $PC{{l}_{5}}$ is given as:
$PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}$

	$PC{{l}_{5}}$	$PC{{l}_{3}}$	$C{{l}_{2}}$
Initial concentration	$5$	$0$	$0$
Final concentration	$5(1-\alpha )$	$5\alpha $	$5\alpha $
Concentration at equilibrium	$\dfrac{5(1-\alpha )}{0.5}$	$\dfrac{5\alpha }{0.5}$	$\dfrac{5\alpha }{0.5}$ .

Since the degree of dissociation is $40%$ so the moles at equilibrium:
${{K}_{C}}=\dfrac{\left[ PC{{l}_{3}} \right]\left[ C{{l}_{2}} \right]}{\left[ PC{{l}_{5}} \right]}$
Concentration of $PC{{l}_{3}}=\dfrac{5\alpha }{0.5}$
 $\Rightarrow \dfrac{5\times 0.4}{0.5}$
Concentration of $C{{l}_{2}}=\dfrac{5\alpha }{0.5}$
 $\Rightarrow \dfrac{5\times 0.4}{0.5}$
Concentration of $PC{{l}_{5}}=\dfrac{5(1-\alpha )}{0.5}$
 $\Rightarrow \dfrac{5\times (1-0.4)}{0.5}=\dfrac{5\times 0.6}{0.5}$
Now, substituting the value of $\alpha $ in concentration at equilibrium and further substituting them in the above formula, we get,
${{K}_{C}}=\dfrac{\left( \dfrac{5\times 0.4}{0.5} \right)\left( \dfrac{5\times 0.4}{0.5} \right)}{\left( \dfrac{5\times 0.6}{0.5} \right)}$
On further solving we get.
${{K}_{C}}=2.66$ $Mol/l$

So, the correct answer is Option B.

Note: When the reaction is in equilibrium, ratio of product of concentration of products to that of reactants is called as reaction quotient and is denoted by $Q$ and when the reaction is in equilibrium it is called as equilibrium constant.
The expression of equilibrium constant for the reversed reaction is written as the reciprocal of that for the forward reaction.
Degree of dissociation is defined as the ratio of dissociated moles to the initial number of moles.