What is #K_c# if #40%# of #"PCl"_5(g)# is not dissociated?

Dissociation reaction of #"PCl"_5(g)# can be expressed as

#"PCl"_5(g) rightleftharpoons"PCl"_3(g)+"Cl"_2(g)# ......(1)

Let us begin with #1#mole of #"PCl"_5#

ICE table is
I #" "1" mol "0 " mol "0" mol"#

C #" "-alpha" mol "alpha " mol "alpha" mol"#

E #" "(1-alpha)" mol "alpha " mol "alpha" mol"#

Where degree of dissociation #alpha=60%=0.6# (given that #40%# is not dissociated).

Volume of equilibrium mixture #V = 1 "L"#

Molar concentrations of the components of the mixture at equilibrium are

#["PCl"_5(g) ] =((1-alpha))/V=(x(1-0.6))/1= 0.4" mol·L"^-1#

#["PCl"_3(g) ]=(alpha)/V=0.6/1= 0.6" mol·L"^-1#

#["Cl"_2(g) ]=alpha/V=0.6/1=0.6" mol·L"^-1#

By definition #"K"_c=(["PCl"_3]["Cl"_2])/(["PCl"_5])# ........(2)
Inserting various mole concentrations in (2) , we get

#"K"_c=(0.6 xx0.6)/(0.4)#

#=>"K"_c=0.9#

I got #K_c = 0.9# (based on #"mol/L"# units).

The thermal dissociation of #"PCl"_5# is:

#"PCl"_5(g) rightleftharpoons "PCl"_3(g) + "Cl"_2(g)#

#"I"" "" "1" "" "" "" "0" "" "" "0#
#"C"" "-alpha" "" "" "+alpha" "" "+alpha#
#"E"" "1 - alpha" "" "" "alpha" "" "" "alpha#

where #alpha# indicates the fraction of #"PCl"_5# dissociated. Since #40%# was NOT dissociated, #alpha = 100% - 40% = 60%# dissociated.

The total mols of gas, assuming #"1 mol"# of #"PCl"_5# to begin with, is then:

#alpha + alpha + 1 - alpha = 1 + alpha#

I must assume that because you only have given relative values to us. Therefore, we cannot determine the actual #"mols"# of gas otherwise, nor the total pressure.

The expression for #K_c# is:

#K_c = (alpha^2)/(1 - alpha)#

Since we know that #alpha = 0.6#, we then get:

#color(blue)(K_c) = (0.6^2)/(1 - 0.6)#

#=# #color(blue)(0.9)#

That aside, I do want to show what would happen if we assumed #n# mols of #"PCl"_5# instead of #1#. We thus scale up the reaction by a factor of #n#, and we would be looking at:

#n"PCl"_5(g) rightleftharpoons n"PCl"_3(g) + n"Cl"_2(g)#

Remember that #alpha# is a fraction of dissociation, so it is multiplied by the amount #n#. We assume a #"1-L"# container to get:

#n"PCl"_5(g) rightleftharpoons n"PCl"_3(g) + n"Cl"_2(g)#

#"I"" "" "n" "" "" "" "0" "" "" "" "0#
#"C"" "-alphan" "" "" "+alphan" "" "+alphan#
#"E"" "(1 - alpha)n" "" "alphan" "" "" "alphan#

We then proceed to see how #K_c'# compares to the old #K_c#:

#K_c' = ((alphan)^(2n))/((1 - alpha)n)^n = ((alpha^2)/(1 - alpha))^n(n^2)^n/(n^n)#

which shows that #K_c' = n^n(K_c)^n#. And so, for #n# mols of gas, we have

#K_c' = n^n(0.9)^n#

which means this problem was ill-posed; depending on what quantity of #"PCl"_5(g)# was assumed, we could have gotten an infinity of solutions #K_c' = n^n(0.9)^n# for any number of mols of reactant gas.