Question

'a' moles of $PC{l}_5$ are heated in a closed container till equilibrium is established.
$PC{l}_5\left(g\right)\leftrightharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$ is maintianed at a pressure of P atm. If 'b' moles of $PC{l}_5$ dissociates at equilibrium, then:

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

Answer 1

The correct option is B $\frac{a}{b}=(\frac{K_p+P}{K_p}{)}^{\frac{1}{2}}$

$PC{l}_5\leftrightharpoons PC{l}_3+C{l}_2$
$initialmolesa00$
$atequilibriuma-bbb$



Total no. of moles at equilibrium,
= a - b + b + b = a + b
Let total pressure = P
Therefore,
Partial pressure of $PC{l}_5=\left(\frac{a-b}{a+b}\right)P$
Partial pressure of $PC{l}_3=\left(\frac{b}{a+b}\right)P$
Partial pressure of $C{l}_2=\left(\frac{b}{a+b}\right)P$
So
$K_p=\frac{PC{l}_3\times C{l}_2}{PC{l}_5}$
$\Rightarrow K_p=\frac{\left(\frac{b}{a+b}\right)P\times \left(\frac{b}{a+b}\right)P}{\left(\frac{a-b}{a+b}\right)P}\Rightarrow K_p=\frac{b^{2}P}{a^2-b^2}\Rightarrow \frac{a^2-b^2}{b^2}=\frac{P}{K_p}\Rightarrow \frac{a^2}{b^2}-1=\frac{P}{K_p}\Rightarrow \frac{a^2}{b^2}=\frac{P}{K_p}+1\Rightarrow (\frac{a}{b}{)}^2=\frac{P+K_p}{K_p}\Rightarrow \frac{a}{b}=(\frac{P+K_p}{K_p}{)}^{\frac{1}{2}}$