JEE Main · 2022 · Shift-IIhardCEQ-055

4.0 moles of argon and 5.0 moles of PCl5 are introduced into an evacuated flask of 100 litre capacity at 610 K. The…

Chemical Equilibrium · Class 11 · JEE Main Previous Year Question

Question

4.0 moles of argon and 5.0 moles of PCl5\mathrm{PCl_5} are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The KpK_p for the reaction is:

(R=0.082 L atm K−1 mol−1)(R = 0.082\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}})

Options
  1. a

    2.25

    ✓
  2. b

    6.24

  3. c

    12.13

  4. d

    15.24

Correct Answera

2.25

Detailed Solution

Step 1 — Total moles at equilibrium: ntotal=PVRT=6.0×1000.082×610=60050.02≈12 moln_{total} = \frac{PV}{RT} = \frac{6.0 \times 100}{0.082 \times 610} = \frac{600}{50.02} \approx 12\ \mathrm{mol}

Step 2 — Moles of PCl₅ system: Argon = 4 mol (inert) ⇒\Rightarrow PCl₅ system moles = 12−4=812 - 4 = 8 mol

Reaction: PCl5⇌PCl3+Cl2\mathrm{PCl_5 \rightleftharpoons PCl_3 + Cl_2}

Let xx = moles dissociated:

  • PCl₅: 5−x5-x, PCl₃: xx, Cl₂: xx ⇒\Rightarrow total = 5+x=8⇒x=35+x = 8 \Rightarrow x = 3

Step 3 — Partial pressures: PPCl5=212×6=1.0 atmP_{\mathrm{PCl_5}} = \frac{2}{12} \times 6 = 1.0\ \mathrm{atm}

PPCl3=PCl2=312×6=1.5 atmP_{\mathrm{PCl_3}} = P_{\mathrm{Cl_2}} = \frac{3}{12} \times 6 = 1.5\ \mathrm{atm}

Step 4 — KpK_p: Kp=PPCl3⋅PCl2PPCl5=1.5×1.51.0=2.25 atmK_p = \frac{P_{\mathrm{PCl_3}} \cdot P_{\mathrm{Cl_2}}}{P_{\mathrm{PCl_5}}} = \frac{1.5 \times 1.5}{1.0} = 2.25\ \mathrm{atm}

Answer: Option (1) — 2.25

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