Question

Phosphorus pentachloride dissociates as follows in a closed reaction vessel, $PCl_{5\left(g\right)} {\rightleftharpoons} PCl_{3\left(g\right)}+Cl_{2\left(g\right)}$ 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

• 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)

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

Answer 2

$\underset{\text{(1-x)}}{ {PCl5_{(g)}}} {<=>}$ $\underset{\text{x}}{ {PCl3_{(g)}}}\,+$ $\underset{\text{x}}{ {Cl2_{(g)}}}$ Total number of moles at equilibrium $= ( 1 - x) + x + x = 1 + x$ $p_{PCl_3}=\left(\frac{x}{1+x}\right) \times p$