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Question: Given, \({{K}_{c}}\) for \(PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)\), is 0.04 ...

Given, Kc{{K}_{c}} for PCl5(g)⇌PCl3(g)+Cl2(g)PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g), is 0.04 at 2500C^{0}C. How many moles of PCl5PC{{l}_{5}} must be added to a 3 litre flask to obtain a Cl2C{{l}_{2}} concentration of 0.15M?

Explanation

Solution

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 : PCl5(g)⇌PCl3(g)+Cl2(g)PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)
Temperature = 2500C^{0}C
Volume of the flask = 3 litre
Concentration of Cl2C{{l}_{2}}= 1.15M
Now we know that PCl5PC{{l}_{5}}dissociates according to the given equation in the question:
PCl5(g)⇌PCl3(g)+Cl2(g)PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)

a00Initial concentration
a(1−x)a(1-x)axaxFinal concentration

Where x is the degree of dissociation
The concentration of Cl2C{{l}_{2}}= 0.1= ax
The value of equilibrium constant i.e. Kc{{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
Kc{{K}_{c}}=[PCl3][Cl2][PCl5]\dfrac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}
Kc=[ax][ax][a(1−x)]{{K}_{c}}=\dfrac{[ax][ax]}{[a(1-x)]}= 0.041
ax2=0.0414a{{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 PCl5PC{{l}_{5}} added is = 0.215 moles.

Note:
The magnitude Kc{{K}_{c}}indicates the extent to which the reaction will proceed.
If the value of K is a larger number (Kc{{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 (Kc{{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 Kc{{K}_{c}} = 1 than the system contains equal amount of reactant and product.