4 . 0 moles of argon and 5 . 0 moles of PCl _ 5 are introduced into...

${PCl}_{5} = 5$ mole

$Ar = 4$ mole

${P}_{Total} = \dfrac{9 \times 0 . 082 \times 610}{100} = 4 . 5atm$

${P}_{{PCl}_{5}} = \dfrac{5 \times 4 . 5}{9} = 2 . 5; {P}_{Ar} = \dfrac{4 \times 4 . 5}{9} = 2$

${PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2}$\begin{cases}2 . 5 - P & P & P\end{cases}$$</p><p>${P}_{total} = 2 . 5 - P + P + P + {P}_{Ar} = 6$</p><p>$P = 1 . 5$</p><p>${K}_{p} = \dfrac{1 . 5 \times 1 . 5}{1} = 2 . 25$