Question

$a$ moles of $P{Cl}_5$ is heated in a closed container to equilibriate $P{Cl}_{5\left(g\right)}\rightleftharpoons P{Cl}_{3\left(g\right)}+{Cl}_{2\left(g\right)}$ at a pressure of $P$ atm. If $x$ moles of $P{Cl}_5$ 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 D $\frac{x}{a}={\left(\frac{K_p}{K_p+P}\right)}^{1/2}$

$\underset{1-\alpha }{P{Cl}_{5\left(g\right)}}\rightleftharpoons \underset{\alpha }{P{Cl}_{3\left(g\right)}}+\underset{\alpha }{{Cl}_{2\left(g\right)}}$

$\alpha =\frac{x}{a}$ (degree of dissociation)

Total moles $=1-\alpha +\alpha +\alpha +\alpha =1+\alpha$

$K_p=\frac{P_{P{Cl}_5}\times P_{{Cl}_2}}{P_{P{Cl}_5}}=\frac{[\frac{\alpha }{1+\alpha }.p][\frac{\alpha }{1+\alpha }.p]}{\frac{1-\alpha }{1+\alpha }.p}$

$=\frac{{\alpha }^{2}p}{1-{\alpha }^2}\Rightarrow \alpha ={\left(\frac{K_p}{K_p+P}\right)}^{1/2}$

Hence, option D is correct.