Question

Phosphorous pentachloride dissociates as follows:
$P{Cl}_5\left(g\right)\rightleftharpoons P{Cl}_3\left(g\right)+{Cl}_2\left(g\right)$
If total pressure at equilibrium is $P$ and the degree of dissociation of $P{Cl}_5$ is $x$, the partial pressure of $P{Cl}_3$ 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

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

Phosphorous pentachloride dissociates as follows:
$P{Cl}_5\left(g\right)\rightleftharpoons P{Cl}_3\left(g\right)+{Cl}_2\left(g\right)$
If total pressure at equilibrium is $P$ and the degree of dissociation of $P{Cl}_5$ is $x$, the partial pressure of $P{Cl}_3$ will be $\left(\frac{x}{x+1}\right)P$
Let $P_i$ be the inital pressure of $PC{l}_5$. The equilibrium pressure of $PC{l}_5$ is
$P_i(1-x)$.
$P_i\left(x\right)$ is the change in the pressure of $PC{l}_5$ to reach equilibrium.
The equilibrium pressure of $PC{l}_3$ is
$P_i\left(x\right)$.
The equilibrium pressure of $C{l}_2$ is
$P_i\left(x\right)$.
Total pressure at equilibrium is P. It is $P_i(1-x)+P_i\left(x\right)+P_i\left(x\right)=P_i(1+x)$
$P=P_i(1+x)$
$P_i=\frac{P}{(1+x)}$
The equilibrium pressure of $PC{l}_3$ is
$P_i\left(x\right)=\frac{P}{(1+x)}\times x=\left(\frac{x}{x+1}\right)P$.