# 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.

**Complete answer:**

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.