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Kp for PCl5 dissociation in a vessel with an inert gas, from the total pressure at equilibrium

The same question, asked 2 times: 14 Jun 2022*, 29 Jun 2022. Every wording below, oldest first.

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  1. Question 1

    14 Jun 2022

    143
    PCl5PCl_{5} dissociates as
    PCl5( g)⇌PCl3( g)+Cl2( g)PCl_{5}(\text{ }g) \rightleftharpoons PCl_{3}(\text{ }g) +Cl_{2}(\text{ }g)
    5 moles of PCl5PCl_{5} are placed in a 200 litre vessel which contains 2 moles of N2N_{2} and is maintained at 600 K600\text{ }K. The equilibrium pressure is 2.46 atm2.46\text{ }atm. The equilibrium constant KpK_{p} for the dissociation of PCl5PCl_{5} is ____\_\_\_\_ ×10−3\times10^{- 3}. (nearest integer) (Given: R=0.082 L atm K−1 mol−1R = 0.082\text{ }L\text{ }atm{\text{ }K}^{- 1}{\text{ }mol}^{- 1} : Assume ideal gas behaviour)

  2. Question 2

    29 Jun 2022

    124
    4.0 moles of argon and 5.0 moles of PCl5PCl_{5} are introduced into an evacuated flask of 100 litre capacity at 610 K610\text{ }K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm6.0\text{ }atm. The KpK_{p} for the reaction is ______ [Given: R=0.082 L atm K−1mol−1R = 0.082\text{ }L\text{ }atm\text{ }K^{-1}mol^{-1}]

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