# Given, ${{K}_{c}}$ for $PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)$, is 0.04 at 250$^{0}C$. How many moles of $PC{{l}_{5}}$ must be added to a 3 litre flask to obtain a $C{{l}_{2}}$ concentration of 0.15M?

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Hint: Equilibrium constant is defined as that product of concentration of product to the product of concentration of reactants each raise to the power of their respective stoichiometric coefficient.

Given in the question:
Value of equilibrium constant for the reaction = 0.04
Reaction give is : $PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)$
Temperature = 250$^{0}C$
Volume of the flask = 3 litre
Concentration of $C{{l}_{2}}$= 1.15M
Now we know that $PC{{l}_{5}}$dissociates according to the given equation in the question:
$PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)$

 a 0 0 Initial concentration $a(1-x)$ ax ax Final concentration

Where x is the degree of dissociation
The concentration of $C{{l}_{2}}$= 0.1= ax
The value of equilibrium constant i.e. ${{K}_{c}}$ will be product of concentration of product to the product of concentration of reactants each raise to the power of their respective stoichiometric coefficient
${{K}_{c}}$=$\dfrac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}$
${{K}_{c}}=\dfrac{[ax][ax]}{[a(1-x)]}$= 0.041
$a{{x}^{2}}=0.0414$
And the value of ax= 0.1, the value of x will be= 0.414
And the value of a = 0.2415
Hence the number of moles of $PC{{l}_{5}}$ added is = 0.215 moles.

Note:
The magnitude ${{K}_{c}}$indicates the extent to which the reaction will proceed.
If the value of K is a larger number (${{K}_{c}}$ >>1)it means that products have a large value of concentration, which means that the concentration will increase when the reaction moves to the right side of the reaction.
If the value of K is a larger number (${{K}_{c}}$ <<1)it means that reactants have a large value of concentration, which means that the concentration will increase when the reaction moves to the left side of the and if the value of reaction ${{K}_{c}}$ = 1 than the system contains equal amount of reactant and product.