$P C l_{5} ⇌ P C l_{3} + C l_{2}$ in the reversible reaction, the moles of $P C l_{5}$ , $P C l_{3}$ and $C l_{2}$ are a, b and c respectively at equilibrium and total pressure is P, then value of $K_{p}$ is:

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n mole of PCl3 and n mole Cl2 are allowed to react at constant temperature T to have a total equilibrium pressure P, as :PCl3(g)+Cl2(g)⇌PCl5(g). If y mole of PCl3 are formed at equilibrium, find kp for the given reaction.

The correct option is B (2n−y)y(n−y)2.PPCl3+Cl2⇌PCl5Total Moles =n−y+n−y+y=2n−yPPCl5=yP2n−yKp=yP2n−y×(2n−y)2(n−y)2P2=y(2n−y)(n−y)2P

n mole of $P C l_{3}$ and n mole $C l_{2}$ are allowed to react at constant temperature T to have a total equilibrium pressure $P$ , as : $P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$ . If y mole of $P C l_{3}$ are formed at equilibrium, find $k_{p}$ for the given reaction.

The correct option is B $\frac{(2 n - y) y}{(n - y)^{2} . P}$
$P C l_{3} + C l_{2} ⇌ P C l_{5}$
Total Moles = $n - y + n - y + y$
$= 2 n - y$
$P_{P C l_{5}} = \frac{y P}{2 n - y}$
$K_{p} = \frac{y P}{2 n - y} \times \frac{(2 n - y)^{2}}{(n - y)^{2} P^{2}}$
$= \frac{y (2 n - y)}{(n - y)^{2} P}$