Chemistry - Chemical Equilibrium Question with Solution | TestHub

$$\begin{array}{c}\text{Moles}\text{of}{\text{Cl}}_{\text{2}}\text{initially}\text{=}\text{0.448/22.4}\text{=}\text{0.02}\text{mol}\text{Moles}\text{of}{\text{PCl}}_{\text{5}}\text{=}\text{2.085/208.5}\text{=}\\ \text{0.01}\text{mol}\\ \text{Let}\text{x}\text{mol}\text{of}{\text{PCl}}_{\text{5}}\text{dissociate:}{\text{PCl}}_{\text{5}}\text{=}\text{0.01}\text{-}\text{x,}{\text{PCl}}_{\text{3}}\text{=}\text{x,}{\text{Cl}}_{\text{2}}\text{=}\text{0.02}\text{+}\text{x}\\ \text{Take}\text{R}\text{=}\text{0.0821}\text{L}\text{atm/mol/K}\text{and}\text{T}\text{=}\text{473}\text{K,}\text{so}\text{RT/V}\text{=}\text{(0.0821)(473)/0.448}\text{=}\text{86.7}\\ \text{atm}\text{per}\text{mol}\\ {\text{K}}_{\text{P}}\text{=}\text{[x(0.02+x)/(0.01-x)](86.7)}\text{=}\text{5}\\ {\text{x}}^{\text{2}}\text{+}\text{0.0777x}\text{-}\text{5.77}\times {\text{10}}^{\text{-4}}\text{=}\text{0,}\text{so}\text{x}\text{=}\text{0.0068}\text{mol}\\ \text{Total}\text{moles}\text{=}\text{0.03}\text{+}\text{0.0068}\text{=}\text{0.0368}\text{mol}\\ \text{P}\text{=}\text{(0.0368)(86.7)}\text{=}\text{3.19}\text{atm}\\ \text{Nearest}\text{integer}\text{is}\text{3}\end{array}$$