Question

$PC{l}_5$ dissociates as $PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right).$ 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. $\left(\frac{x}{1+x}\right)\times P$
• B. $\left(\frac{x}{1+x}\right)\times P$
• C. $\left(\frac{2x}{1+x}\right)\times P$
• D. $x\times P$

Answer 1

The correct option is B $\left(\frac{x}{1+x}\right)\times P$

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
$(1-x)xx$
Total mole at equilibrium $=1-x+x+x=1+x$
$P_{PC{l}_3}=\left(\frac{x}{1+x}\right)\times P$