Question

$a$ moles of $PC{l}_5$ are heated in a closed container to equilibriate $PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$ at a pressure of $P$ atm. If $x$ moles of $PC{l}_5$ dissociate at equilibrium, then:

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

Answer 1

The correct option is C $\frac{x}{a}={\left(\frac{K_p}{K_p+P}\right)}^{\frac{1}{2}}$

$PC{l}_5\left(g\right)\leftrightharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$

Moles of $PC{l}_5$ dissociated $=x$

Moles of $PC{l}_5$ left $=(a-x)$

Moles of $PC{l}_3$ formed $=x$

Moles of $C{l}_2$ formed $=x$

Total moles $=(a+x)$

Degree of dissociastion $\alpha =\frac{x}{a}$

Pressure due to $PC{l}_5=\frac{(1-\alpha )}{(1+\alpha )}P$

Pressure due to $PC{l}_3=\frac{\alpha }{(1+\alpha )}P$

Pressure due to $C{l}_2=\frac{\alpha }{(1+\alpha )}P$

Now,

$K_p=\frac{X_{PC{l}_3}.X_{C{l}_2}}{X_{Pc{l}_5}}P=\frac{{\alpha }^2}{1-{\alpha }^2}P$

$K_p(1-{\alpha }^2)=K_p-K_p{\alpha }^2={\alpha }^{2}P$

$K_p=(K_p+P){\alpha }^2\Rightarrow \alpha =\sqrt{\frac{K_p}{K_p+P}}=\frac{x}{a}$

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