The reaction, PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is started in a five-litre container by taking one mole of PCl₅. If 0.3 mole of PCl5 is there at equilibrium, concentration of PCl₃ and K_C will respectively be:

The correct answer is: 0.14 , 49/150

Step 1: Write the balanced equilibrium reaction and note the given data.

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

Given:
• Volume of the container = 5 L
• Initial moles of PCl₅ = 1 mol
• Moles of PCl₅ at equilibrium = 0.3 mol

Step 2: Determine the moles that decomposed.

Since only PCl₅ was initially present and it partially dissociates:

Moles of PCl₅ decomposed = Initial moles − Equilibrium moles = 1 − 0.3 = 0.7 mol

Step 3: Set up the ICE table (in moles).

From the stoichiometry of the reaction (1 : 1 : 1), for every 1 mole of PCl₅ that decomposes, 1 mole of PCl₃ and 1 mole of Cl₂ are formed.

Step 4: Calculate the equilibrium concentrations.

Concentration = Moles ÷ Volume (in litres)

[PCl₅] = 0.3 / 5 = 0.06 M

[PCl₃] = 0.7 / 5 = 0.14 M

[Cl₂] = 0.7 / 5 = 0.14 M

Step 5: Calculate K_C.

The equilibrium constant expression for this reaction is:

$K_C=\frac{\left[{\mathrm{PCl}}_3\right]\left[{\mathrm{Cl}}_2\right]}{\left[{\mathrm{PCl}}_5\right]}$

Substituting the values:

$K_C=\frac{0.14\times 0.14}{0.06}=\frac{0.0196}{0.06}$

Converting to a fraction:

$K_C=\frac{196}{600}=\frac{49}{150}$

Final Answer:
The concentration of PCl₃ at equilibrium is 0.14 M and K_C = 49/150.