In the dissociation of PCl 5 as: PCl 5( g ) → PCl 3( g )+ Cl 2( g ), If the degree of dissociation is α at equilibrium pressure "P", then the equilibrium constant for the reaction is:

Question

In the dissociation of PCl 5 as: PCl 5( g ) → PCl 3( g )+ Cl 2( g ), If the degree of dissociation is α at equilibrium pressure "P", then the equilibrium constant for the reaction is: (A) Kp=(α2/1+α2 p) (B) Kp=(α2 p2/1-α2) (C) Kp=(α p2/1-α2) (D) Kp=(α2 p/1-α2)

Tardigrade Admin · Accepted Answer

underset textequilibrium; textInitial: underset1 - αPCl 5 leftharpoons undersetαPCl 3+ undersetαCl 2 Total moles: 1-α+α+α=1+α Total pressure = P ∴ PPCl5=((1-α) P/1+α), PPCl3=(α P/1+α), PC l2=(α P/1+α) Kp=(PP Cl3, × PC l2/PP C l5)=(α2 P2/(1+α)(1-α)(P))=(α2 P/1-α2)

Q. In the dissociation of $PC l_{5}$ as:
$PC l_{5} (g) \to PC l_{3} (g) + C l_{2} (g)$ , If the degree of dissociation is $α$ at equilibrium pressure "P", then the equilibrium constant for the reaction is:

$K_{p} = \frac{α ^{2}}{1 + α ^{2} p}$

$K_{p} = \frac{α ^{2} p ^{2}}{1 - α ^{2}}$

$K_{p} = \frac{α p ^{2}}{1 - α ^{2}}$

$K_{p} = \frac{α ^{2} p}{1 - α ^{2}}$

Q. In the dissociation of PCl5​ as: PCl5​(g)→PCl3​(g)+Cl2​(g), If the degree of dissociation is α at equilibrium pressure "P", then the equilibrium constant for the reaction is:

Kp​=1+α2pα2​

Kp​=1−α2α2p2​

Kp​=1−α2αp2​

Kp​=1−α2α2p​

equilibrium; Initial :​1−αPCl5​​⇌αPCl3​​+αCl2​​ Total moles: 1−α+α+α=1+α Total pressure =P ∴PPCl5​​=1+α(1−α)P​,PPCl3​​=1+ααP​,PCl2​​=1+ααP​ Kp​=PPCl5​​PPCl3​​,×PCl2​​​=(1+α)(1−α)(P)α2P2​=1−α2α2P​

Q. In the dissociation of $PC l_{5}$ as:
$PC l_{5} (g) \to PC l_{3} (g) + C l_{2} (g)$ , If the degree of dissociation is $α$ at equilibrium pressure "P", then the equilibrium constant for the reaction is:

$K_{p} = \frac{α ^{2}}{1 + α ^{2} p}$

$K_{p} = \frac{α ^{2} p ^{2}}{1 - α ^{2}}$

$K_{p} = \frac{α p ^{2}}{1 - α ^{2}}$

$K_{p} = \frac{α ^{2} p}{1 - α ^{2}}$