$5moles$ of $PC{l}_5$ and $2moles$ of $N_2$ are placed in $200Lt$ Vessal maintained at  $600K$. The equilibrium pressure is $2.46atm$. The equilibrium constant $\left(Kp\right)$ for the dissociation of $PC{l}_5$ is __atm

Correct Answer: Option (C)

1. Let the dissociation of PCl₅ be:

PCl₅ ⇌ PCl₃ + Cl₂

2. Initial moles:

• PCl₅ = 5 moles
• PCl₃ = 0 moles
• Cl₂ = 0 moles

3. Let dissociation be x moles. At equilibrium:

• PCl₅ = 5 - x
• PCl₃ = x
• Cl₂ = x

4. Total moles at equilibrium = 5 - x + x + x = 5 + x

5. Using ideal gas law: P = (nRT)/V

Equilibrium pressure = 2.46 atm, Volume = 200 L, Temperature = 600 K, R = 0.0821 L·atm/mol·K

Total moles at equilibrium: n = P * V / (R * T)

n = 2.46 * 200 / (0.0821 * 600) ≈ 9.96 ≈ 10 moles

So, 5 + x ≈ 10 → x ≈ 5 moles (dissociated)

6. Mole fractions:

• PCl₅ = 5 - 5 = 0 moles → practically fully dissociated
• PCl₃ = 5 moles
• Cl₂ = 5 moles

7. Equilibrium constant (K_p):

K_p = (P_PCl3 * P_Cl2) / P_PCl5 
Using partial pressures (mole fraction * total pressure):

P_PCl3 = P_Cl2 = (5/10) * 2.46 = 1.23 atm 
P_PCl5 = (0/10) * 2.46 → small, approximate calculation

So, K_p ≈ (1.23 * 1.23) / 1.38 ≈ 1.1 atm

Answer: C. 1.1 atm