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One mole of $\mathrm{PCl}_{5}$ is heated in a closed $2 \mathrm{dm}^{3}$ vessel. At equilibrium $40 \% \mathrm{PCl}_{5}$ is dissociated. Calculate the equilibrium constant.
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The correct answer is:
$0.133$
$\quad \mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}$
Initial conc. $\frac{1}{2}=0.5 \quad 0 \quad 0$
Equili. conc. $\frac{1-0.4}{2}=\frac{0.6}{2} \quad \frac{0.4}{2} \quad \frac{0.4}{2}$
$K_{C}=\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}=\frac{\frac{0.4}{2} \times \frac{0.4}{2}}{\frac{0.6}{2}}=\frac{(0.4)^{2}}{2 \times 0.6}=0.133$
Initial conc. $\frac{1}{2}=0.5 \quad 0 \quad 0$
Equili. conc. $\frac{1-0.4}{2}=\frac{0.6}{2} \quad \frac{0.4}{2} \quad \frac{0.4}{2}$
$K_{C}=\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}=\frac{\frac{0.4}{2} \times \frac{0.4}{2}}{\frac{0.6}{2}}=\frac{(0.4)^{2}}{2 \times 0.6}=0.133$
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