JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Equilibrium

PCl_{5}

dissociates as

PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g)

5 moles of

PCl_{5}

are placed in a 200 litre vessel which contains 2 moles of

N_{2}

and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant

K_{p}

for the dissociation of

PCl_{5}

is _____

\times 1 0^{- 3}

. (nearest integer)
(Given:

R = 0.082 L atm K^{- 1} mol^{- 1}

: Assume ideal gas behaviour)

Answer: 1107

Solution

At equilibrium:

PCl_{5}

dissociates by

x

mol. Total moles =

5 - x + x + x + 2 = 7 + x

. Using

P V = n R T

2.46 \times 200 = (7 + x) \times 0.082 \times 600

, solving gives

x = 3

. At equilibrium: 2 mol

PCl_{5}

, 3 mol each of

PCl_{3}

and

Cl_{2}

, total 10 mol.

K_{p} = \frac{n _{PCl_{3}} \cdot n _{Cl_{2}}}{n _{PCl_{5}}} \times \frac{P _{total}}{n _{total}} = \frac{3 \times 3}{2} \times \frac{2.46}{10} = 1.107 = 1107 \times 1 0^{- 3}

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Integer answer