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Prove alpha=sqrt((K(p)/(P+K(p)))) for ...

Prove `alpha=sqrt((K_(p)/(P+K_(p))))` for
`PCl_(5) hArr PCl_(3)+Cl_(2)`
where `alpha` is the degree of dissociation at temperature when equilibrium constant is `K_(p)`.

Text Solution

Verified by Experts


`P_(PCI_(5)) =(P(1-x))/((1+x)) : P_(PCI _(3)) =(P(x))/((1+x)) , P_(CI_(2)) =(P(x))/((1+x))`
Applying Law of Chemical equilibrium.
`K_(p) =(P_(PCI_(3))xxPCI_(2))/(P_(PCI_(5)))=((P(x))/((1+x))xx(P(x))/((1+x)))/((P(1+x))/(1+x))`
`K_(p) =(Px^(2))/(1+x^(2)) , x^(2)P =K_(P) -K_(p)x^(2) , x^(2) (P +K_(p)) =K_(p)`
`" or "" " x^(2) =(k_(p))/((P +K_(p))) :. x =((K_(p))/(P+K_(p)))^(1//2)`
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In the reaction, PCl_(5)hArrPCl_(3)+Cl_(2) , the amounts of PCl_(5) , PCl_(3) and Cl_(2) , 2 moles each at equilibrium and the total pressure is 3 atm . Find the equilibrium constant K_(p) .

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Knowledge Check

  • For the reaction PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) , the equation connecting the degree of dissociation (alpha) of PCl_(5)(g) with the equilibrium constant K_(p) is

    A
    `alpha=(K_(P)//P)/(4+K_(P)//P)`
    B
    `alpha=[(K_(p)//P)/(K_(p)//P+1)]^(1//2)`
    C
    `alpha=[(K_(P)//P)/(K_(P)//P+1)]`
    D
    `alpha=[(K_(P)//P)/(4+K_(P)//P)]^(1//2)`

  • In the dissociation of PCl_(5) as PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) If the degree of dissociation is alpha at equilibrium pressure P, then the equilibrium constant for the reaction is

    A
    `K_(p)=alpha^(2)/(1+alpha^(2)P)`
    B
    `K_(p)=(alpha^(2)P^(2))/(1-alpha^(2))`
    C
    `K_(p)=P^(2)/(1-alpha^(2))`
    D
    `K_(p)=(alpha^(2)P)/(1-alpha^(2))`

  • In the dissociation of PCl_5 PCl_(5(g)) Leftrightarrow PCl_(3(g)) +CL_(2(g)) if as if the degree of dissociation is alpha at equilibrium pressure P, then the equilibrium constant for the reaction is

    A
    `K_p=(alpha^2)/(1+alpha^2P)`
    B
    `K_(p)=(alpha^2 P^2)/(1-alpha^2)`
    C
    `K_(p)=(P^2)/(1-alpha^2)`
    D
    `K_(p)=(alpha^2P)/(1-alpha^2)`

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