N mole of PCl-3 and n mole Cl-2 are allowed to react constant temperature T to have a total equilibrium pressure P- as -PCl-3-g-Cl-2-g- rightleftharpoons PCl-5-g- If y mole of PCl-3 are formed equilibrium- k-p the given reaction- dfrac-n-y-2-2n-y-P-y-dfrac-2n-y-y-n-y-2-P-dfrac-y-n-y-2-2n-y-P-dfrac-n-y-2-P-2n-y-y-

Question

Toppr Admin · Accepted Answer

PCl3-Cl2-x21CC-PCl5Total Moles -n-x2212-y-n-x2212-y-y-2n-x2212-yPPCl5-yP2n-x2212-yKp-yP2n-x2212-y-xD7-2n-x2212-y-2-n-x2212-y-2P2-y-2n-x2212-y-n-x2212-y-2P

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. (n−y)2(2n−y)Py(2n−y)y(n−y)2.Py(n−y)2(2n−y)P(n−y)2.P(2n−y)y

PCl3+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

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