At moderately high temperatures, SbCl5 decomposes into SbCl3 and Cl2 as follows: SbCl5(g) <---> SbCl3(g) + Cl2(g) ?

#K_P = "0.335 atm"#, although you may want to report your answer without units. I wrote the units because this is nonstandard temperature.

The initial state of the system contains #"65.4 g"# of #"SbCl"_5(g)# in a #"5.00 L"# container. As it approaches equilibrium, the fraction of dissociation is #0.358#.

So first, calculate the starting total pressure of #"SbCl"_5(g)# (i.e. it is the only gas present so far). Assume we are looking at ideal gases.

#P_(SbCl_5) V = n_(SbCl_5)RT#

where as usual, #P# is pressure in #"atm"#, #n# is mols, #V# is volume in #"L"#, #R = "0.08206 L"cdot"atm/mol"cdot"K"# is the universal gas constant, and #T# is temperature in #"K"#.

The initial pressure of antimony(V) chloride is then:

#=> P_(SbCl_5) = (n_(SbCl_5)RT)/V#

#= (65.4 cancel"g" xx "1 mol"/(299.02 cancel("g SbCl"_5)) cdot "0.08206 L"cdot"atm/mol"cdot"K" cdot (195 + "273.15 K"))/("5.00 L")#

#=# #ul"1.680 atm"#

Now we can construct the ICE table in terms of partial pressures.

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

#"I"" "1.680" "" "" "0" "" "" "" "0#
#"C"" "-x" "" "" "+x" "" "+x#
#"E"" "1.680-x" "x" "" "" "" "x#

Here we really have #P_(SbCl_3) = P_(Cl_2) = x# and #P_(SbCl_5) = 1.680 - x#.

That's how we tend to write it, but we are given that the fraction of dissociation is #0.358#, so that fraction of #"1.680 atm"# is lost from #"PCl"_5(g)#.

#x = 0.358 cdot 1.680 = "0.6016 atm"#

Therefore:

#color(blue)(K_P) = (P_(SbCl_3)P_(Cl_2))/(P_(SbCl_5))#

#= x^2/(1.680 - x) = ("0.6016 atm")^2/("1.680 atm" - "0.6016 atm")#

#=# #color(blue)("0.335 atm")#