Question

$a$ moles of ${PCl}_5$ are heated in a closed container to equilibrate $P{Cl}_{5\left(g\right)}\rightleftharpoons P{Cl}_{5\left(g\right)}+{Cl}_{2\left(g\right)}$ at a pressure of $P$ atm. If $x$ moles of ${PCl}_5$ dissociate at equilibrium, then:

• A. $\frac{x}{a}=\frac{K_P}{K_P+P}$
• B. $\frac{x}{a}={\left(\frac{K_P+P}{K_P}\right)}^{1/2}$
• C. $\frac{x}{a}={\left(\frac{K_P}{P}\right)}^{1/2}$
• D. $\frac{x}{a}={\left(\frac{K_P}{K_P+P}\right)}^{1/2}$

Answer 1

The correct option is A $\frac{x}{a}=\frac{K_P}{K_P+P}$

${PCl}_{5\left(g\right)}\rightleftharpoons {PCl}_{3\left(g\right)}+{Cl}_{2\left(g\right)}$

At $a$ $0$ $0$

$t=0$

At $a-x$ $x$ $x$ total pressure $:$ $P$

$K_P=\frac{P^{{Cl}_2}\times P^{{PCl}_3}}{P^{{PCl}_5}}=\frac{\frac{x}{a+x}P\frac{x}{a+x}P}{\frac{a-x}{a+x}P}\Rightarrow \left(P\right)\frac{x^2}{a^2-x^2}=K_P$

$\frac{a^2-x^2}{x^2}=\frac{P}{K_P}$

$\frac{a^2}{x^2}=1+\frac{P}{K_P}$

$\Rightarrow \frac{x}{a}=\sqrt{\frac{K_P}{K_P+P}}$