Q. Four moles of $PCl_{5}$ are heated in a closed $4 \, dm^{3}$ container to reach equilibrium at 400 K. at equilibrium 50% of $PCl_{5}$ is dissociated. What is the value of $\text{K}_{\text{c}}$ in $\text{mol/L}$ for the dissociation of $PCl_{5}$ into $PCl_{3}$ and $Cl_{2}$ at 400 K is

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Solution:

$ \, \, PCl_{5}\rightleftharpoonsPCl_{3}+Cl_{2}$
Initial conc. $\frac{4}{4}$ 0 0
Equili. conc. $\frac{2}{4} \, \frac{2}{4} \, \frac{2}{4}$
$K=\frac{\left[P C l_{3}\right] \left[C l_{2}\right]}{\left[P C l_{5}\right]}$
$ \, \, =\frac{2 \times 2 \times 4}{4 \times 4 \times 2}=\frac{1}{2}=0.5$