Chemistry - Chemical Equilibrium Question with Solution | TestHub

If $x$ mol of PCl₅ dissociates, the equilibrium amounts are $(4-x)$ mol PCl₅, $x$ mol each of PCl₃ and Cl₂, plus 2 mol argon. The total amount is $(6+x)$ mol. The ideal-gas law at $$\begin{gathered} T=800 \\ , \\ mathrmK \end{gathered}$$ gives:

$$6+x=\frac{PV}{RT}=\frac{6.568\times 80}{0.0821\times 800}=8$$

Therefore, $x=2$.

Thus, each reactive species has a concentration of $$\begin{gathered} 2/80=0.025 \\ , \\ mathrmM \end{gathered}$$.

The equilibrium constant $K_c$ is calculated as:

$$K_c=\frac{[\mathrm{PC}{\mathrm{l}}_3][\mathrm{C}{\mathrm{l}}_2]}{[\mathrm{PC}{\mathrm{l}}_5]}=\frac{(0.025)(0.025)}{(0.025)}=0.025$$

And $K_p$ is:

$$\begin{gathered} K_p=K_{c}RT=0.025\times 0.0821\times 800=1.642 \\ , \\ mathrmatm \end{gathered}$$

The total number of moles is 8 mol, not 6 mol.