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Q.

4.0 moles of argon and 5.0 moles of $P C l_{5}$ 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: $R = 0.082 Latm K^{- 1} {mol}^{- 1}$ ]

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a

6.24

b

12.13

c

15.24

d

2.25

answer is A.

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Detailed Solution

$\begin{array}{l} {PCl}_{5} = 5 mole \\ Ar = 4 mole \\ P_{Total} = \frac{9 \times 0 .82 \times 610}{100} = 4 .5 atm \\ P_{{PCl}_{5}} = \frac{5 \times 4 .5}{9} = 2 .5; P_{Ar} = \frac{4 \times 4 .5}{9} = 2 \\ {PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2} \\ 2 .5 - P P \\ P_{total} = 2 .5 - P + P + P + P_{Ar} = 6 \\ P = 1 .5 \\ K_{p} = \frac{1 .5 \times 1 .5}{1} = 2 .25 \end{array}$

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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 system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The Kp for the reaction is [Given: R=0.082Latm⁡K−1mol−1]