Question

$PC{l}_5$ dissociates as $PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$. If $PC{l}_5$ is 10% dissociated at 1 atm, then the percentage of dissociation at 4 atm is:

• A. 2.5%
• B. 4%
• C. 5%
• D. 10%

Answer 1

The correct option is C 5%

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
$P_1$
$P_1(1-\alpha )P_1\alpha P_1\alpha$
$0.9P_{1}0.1P_{1}0.1P_1$
Now,
$K_P=\frac{P_{PC{l}_3}\times P_{C{l}_2}}{P_{PC{l}_5}}=\frac{(0.1P_1{)}^2}{0.9P_1}=\frac{P_1}{90}$
Again,

$1.1P_1=1atm$ -- Total pressure
$\Rightarrow P_1=\frac{1}{1.1}$

So, $K_P=\frac{1}{99}$
For new condition,
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
$P_2$
$P_2(1-\alpha )P_2\alpha P_2\alpha$
$\Rightarrow P_2(1+\alpha )=4$ -- Total pressure

$K_P=\frac{P_2{\alpha }^2}{1-\alpha }=\frac{1}{99}$
Note: $1-{\alpha }^2\approx 1$ since ${\alpha }^2<<1$

${\alpha }^2=\frac{1}{396}\Rightarrow \alpha =0.05$