2 moles of \(PCl_5\) were heated in a closed vessel of a 2 litre capacity. At equilibrium 40% of \(PCl_5\) dissociated into \(PCl_3\) and \(Cl_2\). The value of the equilibrium constant is

0.267

The correct answer is option 1. 0.267.

Let us solve the problem step-by-step to determine the equilibrium constant \(K_c\).

The reaction will be as follows:

\( PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \)

Initial moles of \(PCl_5\): 2 moles

Volume of the vessel: 2 liters

Initial concentration of \(PCl_5\):

\( [PCl_5]_{\text{initial}} = \frac{2 \text{ moles}}{2 \text{ liters}} = 1 \text{ M} \)

40% of \(PCl_5\) dissociates.

Moles of \(PCl_5\) dissociated:

\( 2 \text{ moles} \times 0.4 = 0.8 \text{ moles} \)

At equilibrium:

Moles of \(PCl_5\) remaining: \(2 - 0.8 = 1.2 \text{ moles}\)

Moles of \(PCl_3\) formed: 0.8 moles

Moles of \(Cl_2\) formed: 0.8 moles

\( [PCl_5] = \frac{1.2 \text{ moles}}{2 \text{ liters}} = 0.6 \text{ M} \)

\( [PCl_3] = \frac{0.8 \text{ moles}}{2 \text{ liters}} = 0.4 \text{ M} \)

\( [Cl_2] = \frac{0.8 \text{ moles}}{2 \text{ liters}} = 0.4 \text{ M} \)

The equilibrium expression for the dissociation is:

\(K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \)

Substitute the equilibrium concentrations into the expression:

\( K_c = \frac{(0.4)(0.4)}{0.6} \)

\( K_c = \frac{0.16}{0.6} \)

\( K_c = \frac{16}{60}\)

\( K_c = \frac{4}{15} \)

\( K_c \approx 0.267 \)

The value of the equilibrium constant \(K_c\) is \(0.267\).