Question

For the reaction, $PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$, the degree of dissociation 'x' is related to pressure 'P(x<<1)

• A. $x\propto \frac{1}{P}$
• B. $x\propto \frac{1}{\sqrt{P}}$
• C. $x\propto \frac{1}{P^2}$
• D. $x\propto \frac{1}{P^2}$

Answer 1

The correct option is B

$x\propto \frac{1}{\sqrt{P}}$

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
$\underset{mole}{Intial}100$
$Ateql{b}^{m}1-xxx$

Total moles $n_{total}=(1-x)+x+x=(1+x)$

Partial pressure of $PC{l}_5\left(g\right)$, $PC{l}_3\left(g\right)$ & $C{l}_2\left(g\right)$ are

$P_{PC{l}_5}=\frac{n_{PC{l}_3}}{n_{total}}\times P=\frac{(1-x)}{(1+x)}\times P$;

$P_{PC{l}_3}=\frac{n_{PC{l}_3}}{n_{total}}\times P=\frac{x}{(1+x)}\times P;P_{C{l}_2}=\frac{n_{C{L}_2}}{n_{total}}\times P=\frac{x}{(1+x)}\times P$

$K_P=\frac{P_{PC{l}_2}\times P_{C{l}_2}}{P_{PC{l}_3}}=\frac{(\frac{x}{(1+x)}\times P)(\frac{x}{(1+x)}\times P)}{\frac{(1-x)}{(1+x)}\times P}=\frac{x^2\times P}{(1-x^2)}$

$\because (1-x^2)\cong 1;K_P=x^2\times P;x=\sqrt{\frac{K_P}{P}};x\propto \frac{1}{\sqrt{P}}$