Question

$PC{l}_5$ dissociates as follows in a closed reaction vessel.

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$

If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of $PC{l}_3$ is x, the partial pressure of $PC{l}_5$ will be :

• A. $\left(\frac{x}{x+1}\right)P$
• B. $\left(\frac{2x}{1-x}\right)P$
• C. $\left(\frac{x}{x-1}\right)P$
• D. $\left(\frac{x}{1-x}\right)P$

Answer 1

Answer: (A)

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

The total pressure is P and the degree of dissociation is x.

	$PC{l}_5$$⟶$	$PC{l}_3$	+$C{l}_2$
Initial moles	1	0	0
Change	x	x	x
Moles at equilibrium	1-x	x	x

Total number of moles $=1-x+x+x=1+x$.

The partial pressure of $PC{l}_3=\left(\frac{x}{1+x}\right)P$.