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.
