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PCl(5),PCl(3) and Cl(2) are at equilibri...

`PCl_(5),PCl_(3)` and `Cl_(2)` are at equilibrium at `500K` and having concentration `1.59M PCl_(3), 1.59M CL_(2)` and `1.41 MPCl_(5)`. Calculate `K_(c)` for the reaction `PCl_(5)hArrPCl_(3)+Cl_(2)`

Text Solution

Verified by Experts

The equilibrium constant `K_(c)` for the above reaction can be written as
`K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=((1.59)^(2))/((1.41))=1.79`
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PCl_(5) , PCl_(3) and Cl_(2) are at equilibrium at 500K and having concentration 1.59M PCl_(3) , 1.59M Cl_(2) and 1.41 M PCl_(5) . Calcualte K_(c) for the reaction PCl_(5)hArrPCl_(3)+Cl_(2)

PCl_(5),PCl_(3) and Care at equlibrium at 500 K and having concentration 1.59 M PCl_(3), 1.59 M Cl_(2) and 1.41 M PCl_(5) Calculate K_(C) for the reaction. PCl_(5) hArr PCl_(3)+Cl_(3)

Knowledge Check

  • Initially 0.8 mole of PCl_(5) and 0.2 mole of PCl_(3) are mixed in one litre vessel . At equilibrium 0.4 mole of PCl_(3) is present . The value of K_(C) for the reaction PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)) is

    A
    0.13 `molL^(-1)`
    B
    `0.05 molL^(-1)`
    C
    0.065 `molL^(-1)`
    D
    `0.1 molL^(-1)`

  • PCl _(5) hArr PCl _(3) + PCl_(3) If the equilibrium cnstant (K _(c)) for the above reaction at 500 K is 1.79 and the equilibrium concentration of PCl _(5) and PCl _(3) are 1.41 M and 1.59M, respectively, then the concentration of Cl _(2) is approximately.

    A
    `1.26M`
    B
    `3.59M`
    C
    `0.59M`
    D
    `1.59M `

  • For PCl_(5 (g)) hArr PCl_(3 (g)) + Cl_(2 (g)) at equilibrium , K_(P) = P//3 . Then degree of dissociation of PCl_(5) at that temperature is

    A
    `0.75`
    B
    `0.25`
    C
    `0.9`
    D
    `0.5`

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