The chemical equation is
"PCl"_5 ⇌ "PCl"_3 + "Cl"_2
"Amount of PCl"_5color(white)(l) "dissociated" = "0.40 × 5 mol" = "2.0 mol"
∴ The equilibrium moles of each component are:
"PCl"_5 = "5 mol - 2.0 mol" = "3.0 mol"
"PCl"_3 = "2.0 mol"
"Cl"_2color(white)(ll) = "2.0 mol"
The volume of the container is "5 L", so the equilibrium concentrations of each component are
["PCl"_5] = "3.0 mol"/"5 L" = "0.60 mol/L"
["PCl"_3] = "2.0 mol"/"5 L" = "0.40 mol/L"
["Cl"_2] = "2.0 mol"/"5 L" = "0.40 mol/L"
We can set up part of an ICE table to solve this problem.
color(white)(mmmmmll)"PCl"_5 ⇌ "PCl"_3+ "Cl"_2
"E/mol·L"^"-1": color(white)(l)"0.60color(white)(mm)0.40color(white)(mll)0.40
The equilibrium constant expression is
K_text(c) = (["PCl"_3]["Cl"_2])/(["PCl"_5])
∴ K_text(c) = (0.40 × 0.40)/0.60 = 0.27
Note: The answer can have only one significant figure, because that is all you gave for the moles of "PCl"_5 and for the volume of the container.
However, I did the calculation to two significant figures for you.
