Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, PCl,5(g) leftharpoons PCl3(g) + Cl2(g) If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be:

Question

Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, PCl,5(g) leftharpoons PCl3(g) + Cl2(g) If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be: (A) ((x/1-x))P (B) ((x/x+1))P (C) ((2x/1-x))P (D) ((x/x-1))P

Tardigrade Admin · Accepted Answer

underset overset11-xPCl5(g) leftharpoons underset overset0xPCl3(g) + underset overset0xCl2(g) ∴ Kc= (x2/1-x)[(Peq/1+x)] Also Ppcl3=(Peq.x/(1+x))

Q. Phosphorus pentachloride dissociates as follows, in a closed reaction vessel,
$PCl,_{5 (g)} ⇌ PC l_{3 (g)} + C l_{2 (g)}$
If total pressure at equilibrium of the reaction mixture is $P$ and degree of dissociation of $PC l_{5}$ is $x$ , the partial pressure of $PC l_{3}$ will be:

$(\frac{x}{1 - x}) P$

$(\frac{x}{x + 1}) P$

$(\frac{2 x}{1 - x}) P$

$(\frac{x}{x - 1}) P$

$1 - x 1 PC l_{5 (g)} ⇌ x 0 PC l_{3 (g)} + x 0 C l_{2 (g)}$
$∴ K_{c} = \frac{x ^{2}}{1 - x} [\frac{P _{e q}}{1 + x}]$
Also $Pp c l_{3} = \frac{P _{e q} . x}{( 1 + x )}$

Q. Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, PCl,5(g)​⇌PCl3(g)​+Cl2(g)​ If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5​ is x, the partial pressure of PCl3​ will be:

(1−xx​)P

(x+1x​)P

(1−x2x​)P

(x−1x​)P