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?

Related questions
See all questions in Dynamic Equilibrium
Impact of this question
2645 views around the world
You can reuse this answer
Creative Commons License