JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Equilibrium

4.0 moles of argon and 5.0 moles of

PCl X_{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

K_{p}

for the reaction is [Given:

R = 0.082 L atm K^{- 1} mo l^{- 1}

]

(A)2.25✓ Correct
(B)6.24
(C)12.13
(D)15.24

Answer: (A)

Solution

Initial total pressure (before equilibrium) =

\frac{n R T}{V} = \frac{9 \times 0.082 \times 610}{100} = 4.5

atm. Initial partial pressures:

P_{PCl X_{5}} = \frac{5}{9} \times 4.5 = 2.5

atm,

P_{Ar} = 2.0

atm. For the reaction

PCl X_{5} PCl X_{3} + Cl X_{2}

, let P atm of

PCl X_{5}

dissociate. At equilibrium: total pressure =

(2.5 - P) + P + P + 2.0 = 6.0

atm, giving P = 1.5 atm. Therefore,

K_{p} = \frac{P _{PCl X_{3}} \times P _{Cl X_{2}}}{P _{PCl X_{5}}} = \frac{1.5 \times 1.5}{1.0} = 2.25

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Equilibrium

Solution

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