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 $PC{l}_5$ is $x$, the partial pressure of $PC{l}_3$ will be:

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

Answer 1

The correct option is C $\left[\frac{x}{x+1}\right]P$

$PC{l}_5\to {PCl}_3\left(g\right)+{Cl}_2\left(g\right)$

$(1-x)xx$

Total pressure $=1-x+x+x=1+x$

${PCl}_3=\left(\frac{x}{1+x}\right)\times P$