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Question: In a 500 mL flask, the degree of dissociation of \(PCl_5\) at equilibrium is 40% and the initial amo...

In a 500 mL flask, the degree of dissociation of PCl5PCl_5 at equilibrium is 40% and the initial amount is 5 moles. The value of equilibrium constant in molL−1molL^{-1} for the decomposition of PCl5PCl_5 is:
A.2.33
B.2.66
C.5.32
D.4.66

Explanation

Solution

Calculate the final amount of PCl5PCl_5 and the products formed by its dissociation using the degree of dissociation and then calculate molarity for them. Using that calculate the equilibrium constant in molL−1molL^{-1}.

Formula Used: We can use the below formula to calculate equilibrium constant.
k= [PCl3][Cl2][PCl5]k = {\text{ }}\dfrac{{[{\text{PC}}{{\text{l}}_3}][{\text{C}}{{\text{l}}_2}]}}{{[{\text{PC}}{{\text{l}}_5}]}}

Complete step by step answer: Let’s start with writing the things that are given to us in this question. We are provided with 500mL flask, a degree of dissociation of 40% which will be 0.4 and the initial amount of PCl5PCl_5 is 5 moles. We have to find the equilibrium constant.
We know
PCl5→PCl3 + Cl2{\text{PC}}{{\text{l}}_{\text{5}}} \to {\text{PC}}{{\text{l}}_{\text{3}}}{\text{ + C}}{{\text{l}}_{\text{2}}}
The initial amount of PCl5PCl_5 is 5 moles, hence the final amount of PCl5PCl_5 will be the degree of dissociation times the initial amount which gives the final amount of PCl5PCl_5 is equal to 3 moles. Hence,
PCl5→PCl3 + Cl2{\text{PC}}{{\text{l}}_{\text{5}}} \to {\text{PC}}{{\text{l}}_{\text{3}}}{\text{ + C}}{{\text{l}}_{\text{2}}}
5 0 0{\text{5 0 0}}
3 2 2{\text{3 2 2}}
Applying the formula of Molality which is
Molality = No. of Moles(mol)Volume(L)\dfrac{{{\text{No}}{\text{. of Moles(mol)}}}}{{{\text{Volume(L)}}}}
This gives the Molarity of PCl5PCl_5 is 6M, PCl3 is 4M and Cl2 is 4M as the volume given is 500mL.
The formula of equilibrium constant K is given as
k= [PCl3][Cl2][PCl5]k = {\text{ }}\dfrac{{[{\text{PC}}{{\text{l}}_3}][{\text{C}}{{\text{l}}_2}]}}{{[{\text{PC}}{{\text{l}}_5}]}}
Putting the values we get
⇒\Rightarrow k= 4×46=166k = {\text{ }}\dfrac{{4 \times 4}}{6} = \dfrac{{16}}{6}
⇒\Rightarrow k=2.66molL−1k = 2.66mol{L^{ - 1}}
Hence the answer of this question is option B.

Note: We should remember that the equilibrium constant is used to determine the rate of the reaction i.e the rate at which the reactants are disappearing and the products are being formed. The magnitude of equilibrium constant gives us an idea about the relative amount of reactant and the product. Equilibrium constant only applicable for gas which depends on the temperature of the system and not dependent on quantity of substances, volume, pressure, catalyst used in the reaction.