What is the equilibrium concentration of Cl2in a 100 liter vessel containing o.235 mole of PCI5 and 0.174 mole of PCI3?

The compound #PCI_5# decomposes into #Cl_2# and #PCI_3#. The equilibrium of #PCI_5(g) rightleftharpoons Cl_2 (g) + PCI_3(g)# has a #K_(eq)# of #2.24 x 10^-2# at 327 #"^o#C.

1 Answer
Aug 10, 2017

Answer:

#["Cl"_2] = 0.00303# #M#

Explanation:

We're asked to find the equilibrium concentration of #"Cl"_2# in a given reaction.

Let's first write the chemical equation for this reaction:

#"PCl"_5(g) rightleftharpoons "PCl"_3(g) + "Cl"_2(g)# #" "#(which is given to us)

The equilirium constant expression for this reaction is thus

#K_"eq" = (["PCl"_3]["Cl"_2])/(["PCl"_5"]) = 2.24xx10^-2#

We're given the number of moles of two species:

  • #0.235# #"mol PCl"_5#

  • #0.174# #"mol PCl"_3#

And since the vessel has a volume of #100# #"L"#, the molar concentrations of each are

  • #["PCl"_5] = (0.235color(white)(l)"mol")/(100color(white)(l)"L") = color(red)(ul(0.00235color(white)(l)M#

  • #["PCl"_3] = (0.174color(white)(l)"mol")/(100color(white)(l)"L") = color(green)(ul(0.00174color(white)(l)M#

Plugging these values into the equilibrium constant expression equation:

#K_"eq" = ((color(green)(0.00174color(white)(l)M))["Cl"_2])/(color(red)(0.00235color(white)(l)M)) = 2.24xx10^-2#

Solving for #["Cl"_2]# yields

#color(blue)(ulbar(|stackrel(" ")(" "["Cl"_2] = 0.00303color(white)(l)M" ")|)#