Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, $\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$ If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of $\mathrm{PCl}_{5}$ is x , the partial pressure of $\mathrm{PCl}_{3}$ will be

(A) $\left(\frac{x}{x+1}\right) P$

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

(C) $\left(\frac{x}{x-1}\right) P$

(D) $\left(\frac{x}{1-x}\right) P$

JEE Main · 2006 · 30 Apr 2006 · Offline · Equilibrium

Phosphorus pentachloride dissociates as follows, in a closed reaction vessel,

PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g)

If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of

PCl_{5}

is x , the partial pressure of

PCl_{3}

will be

Answer: (A)

Begin with one mole of

PC l_{5}

dissociating to a degree

x

. At equilibrium there are

(1 - x)

moles of

PC l_{5}

x

PC l_{3}

and

x

C l_{2}

, a total of

(1 + x)

moles. The partial pressure of a gas is its mole fraction times the total pressure, so for

PC l_{3}

it is

\frac{x}{1 + x} P

, which may be written

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

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option