At a certain temperature, #"PCl"_5(g)# decomposes. What are the concentrations of #"PCl"_5# and #"PCl"_3# in a #"3.70-L"# container that begins with #"0.287 mols"# #"PCl"_5(g)# that dissociates? #K_c = 1.80# at this temperature.

1 Answer
Truong-Son N.
Jun 26, 2017

#"0.00308 M"#
#"0.0745 M"#

I'll let you decide which is which, by reading the answer down below. ;)

Construct an ICE table and mass action expression in terms of concentration:

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

#"I"" "0.287/3.70" "" "" "0" "" "" "0#

#"C"" "-x" "" "" "+x" "" "+x#

#"E"" "0.287/3.70 - x" "x" "" "" "x#

And thus,

#K_c = x^2/(0.287/3.70 - x)#

#K_c# is not small enough for the small #x# approximation, so this requires the full quadratic formula.

#0.287/3.70K_c - K_cx = x^2#

#=> x^2 + K_cx - 0.287/3.70K_c#

#= x^2 + 1.80x - 0.1396 = 0#

Solve to obtain a physical value of #x = 0.07449# #"M"#. Therefore:

#[PCl_5] = 0.287/3.70 - 0.0745 = ul"0.00308 M"#

#[PCl_3] = ul"0.0745 M"#

Why would the small #x# approximation fail? What is the percent dissociation here?

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