Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reaction:

PCl3(g) + Cl2(g) -> PCl5(g)

Keq = 2.01 at 500K. A 1.000-L reaction vessel is charged with 0.300 mol of PCl5 and allowed to equilibrate at this temperature. The equilibrium partial pressure of PCl5 is ______ atm.
a) 10.1
b) 0.386
c) 2.24
d) 2.48
e) 0.211

[tex]\textbf{Given:}\\\begin{itemize} \item $n = 0.300 \, \text{mol}$ (moles of PCl5) \item $V = 1.000 \, \text{L}$ (volume of the container) \item $T = 500 \, \text{K}$ (temperature) \item $R = 0.0821 \, \text{atm} \cdot \text{L} / \text{mol} \cdot \text{K}$\end{itemize}[/tex]

Plug into ideal gas law:

[tex]P_{\text{PCl5}} = \frac{n_{\text{PCl5}}RT}{V} \\\\ = \frac{(0.300 \, \text{mol})(0.0821 \, \text{atm} \cdot \text{L} / \text{mol} \cdot \text{K})(500 \, \text{K})}{1.000 \, \text{L}} \\\\ \approx \boxed{12.315 \, \text{atm}}[/tex]

[tex]\hrulefill[/tex]

Now with the pressure, construct an ICE table.

[tex]\boxed{\text{~~~ ~Species ~~ }} \boxed{PCl_5} \boxed{PCl_3} \boxed{Cl_2} \\\boxed{\text{~Initial (atm)~}} \boxed{12.3} \boxed{0} \boxed{0} \\\boxed{\text{Change (atm)}} \boxed{-x} \boxed{+x} \boxed{+x} \\\boxed{\text{Equilibrium (atm)}} \boxed{12.3 - x} \boxed{x} \boxed{x} \\[/tex]

Notice, the final is [tex]12.3 -x[/tex], so we will be using that in our setup.

Solving:

[tex]K_p = \frac{12.3-x}{x^2}[/tex]

[tex]2.01 = \frac{12.3-x}{x^2}[/tex]

[tex]2.01x^2+x-12.3 = 0[/tex]

[tex]\boxed{x=2.23}[/tex]

Now subtract the final pressure from the initial pressure of PCl5 in order to get the partial pressure.

[tex]12.3-2.23 \approx \boxed{10.1}[/tex]

[tex]\hrulefill[/tex]

The correct answer is a) 10.1