Question

For the following reaction in equilibrium
$P{Cl}_5\left(g\right)\rightleftharpoons P{Cl}_3\left(g\right)+{Cl}_2\left(g\right)$
Vapour density is found to be $100$ when $1mole$ of $P{Cl}_5$ is taken in a $10litre$ flask at $27$. Calculate the equilibrium pressure. Also calculate percentage dissociation of $P{Cl}_5$.

Answer 1

Given reaction is $PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$

equilibrium pressure

$K_c=\left[PC{l}_3\right]\left[C{l}_2/PC{l}_3\right]=1.66\times {10}^{-3}$

now, $kc=kp(RT{)}^n$

and change in moles $=2-1=1$

$\therefore kp=kc/RT=\frac{1.66\times {10}^{-3}}{24.6}=6.7\times {10}^{-2}atm$

$\%$ dissociation of $PC{l}_s$:

molecular mass $=2\times v$ density $=2\times 100=200$

ans thearibical mass $u208$

dissociation $\left(x\right)=\frac{208-200}{200}=0.04$

$\%$ dissociation will be $0.04\times 100=4\%$