Phosphorus pentachloride dissociates as follows in a closed reaction vessel, PCl5(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, PCl5(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/x+1))P (B) ((2x/1-x))P (C) ((x/x-1))P (D) ((x/1-x))P

Tardigrade Admin · Accepted Answer

underset text(1-x)PCl5(g) <=> underset textxPCl3(g) + underset textxCl2(g) Total number of moles at equilibrium = ( 1 - x) + x + x = 1 + x pPCl3=((x/1+x)) × p

Q. Phosphorus pentachloride dissociates as follows in a closed reaction vessel,
$PC l_{5 (g)} ⇌ PC l_{3 (g)} + C l_{2 (g)}$
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

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

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

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

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

$(1-x) PCl X_{5} X_{(g)}$ $x PCl X_{3} X_{(g)} +$ $x Cl X_{2} X_{(g)}$
Total number of moles at equilibrium
$= (1 - x) + x + x = 1 + x$
$p_{PC l_{3}} = (\frac{x}{1 + x}) \times p$

Q. Phosphorus pentachloride dissociates as follows in a closed reaction vessel, PCl5(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

(x+1x​)P

(1−x2x​)P

(x−1x​)P

(1−xx​)P