2 Moles Of Pcl5 Were Heated In A Closed Vessel Of 2 Litre Capacity. At Equilibrium, 40% Of PCl5 Is Dissociated Into PCl3 And Cl2. What Is The Value Of Equilibrium Constant?

As given in the question, the chemical equation is

[latex]PCl_{5}(g) \leftrightharpoons PCl_{3}(g) + Cl_{2}(g)[/latex]

With The degree of dissociation, α = 40 % = 0.4.

At equilibrium, the molar concentration of the component of the mixture is:

[latex][PCl_{5}](g) = \frac{2(1-\alpha )}{V}\\\Rightarrow \frac{1(1-0.4)}{2}\\\Rightarrow 0.6mol/L[/latex] [latex][PCl_{3}](g) = \frac{2\alpha}{V}\\\Rightarrow \frac{2 * 0.4}{2}\\\Rightarrow 0.4mol/L[/latex] [latex][Cl_{2}](g) = \frac{2\alpha}{V}\\\Rightarrow \frac{2 * 0.4}{2}\\\Rightarrow 0.4mol/L[/latex]

Therefore, [latex]Eq^{m} constant, K_{C} = \frac{[PCl_{3}](g) + [Cl_{2}](g)}{[PCl_{5}](g)}\\\Rightarrow \frac{0.4 * 0.4}{0.6}\\\Rightarrow 0.27mol/L[/latex]

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