In a $500 \mathrm{~mL}$ flask, the degree of dissociation of $\mathrm{PCl}_5$ at equilibrium is $40 \%$ and…

In a $500 \mathrm{~mL}$ flask, the degree of dissociation of $\mathrm{PCl}_5$ at equilibrium is $40 \%$ and the initial amount is 5 moles. The value of equilibrium constant in $\mathrm{mol} \mathrm{L}^{-1}$ for the decomposition of $\mathrm{PCl}_5$ is
  1. $2.33$
  2. $2.66$
  3. $5.32$
  4. $4.66$

Solution

$\begin{aligned} & \mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3+\mathrm{Cl}_2 \text { initial moles } \\ & \left.5(1)^5 \alpha\right) \quad 5 \alpha \quad 5 \alpha \text { moles at equilibrium } \\ & \frac{5(1-\alpha)}{0.5} \quad \frac{5 \alpha}{0.5} \quad \frac{5 \alpha}{0.5} \text { conc at equilibrium } \\ & \alpha=40 \% \\ & =0.4 \\ & K_c=\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]} \\ & =\frac{\left(\frac{5 \times 0.4}{0.5}\right)\left(\frac{5 \times 0.4}{0.5}\right)}{\left(\frac{5 \times 0.6}{0.5}\right)}=\frac{16}{6} \\ & =2.66 \mathrm{~mol} / \mathrm{L} \\ & \end{aligned}$

Asked in: AP EAMCET 2008

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