Question #6fe0b

1 Answer
Nathan L.
Jul 19, 2017

#4.5# #"mol PCl"_3#

Explanation:

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

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

And so the equilibrium constant expression is

#K_c = (["PCl"_3]["Cl"_2])/(["PCl"_5])#

Let's make a list of our initial concentrations (no volume is given, but this doesn't matter as we're dealing just with quantity):

Initial:

  • #"PCl"_5#: #3# #"mol"#

  • #"PCl"_3#: #3# #"mol"#

  • #"Cl"_2#: #2# #"mol"#

Using the coefficients of the chemical equation (which are all #1:1# molar ratios), we can determine the change in concentration:

Change:

  • #"PCl"_5#: #-x#

  • #"PCl"_3#: #+x#

  • #"Cl"_2#: #+x#

[I'll make the assumption that the reaction quotient #Q_c < K_c#; i.e. it proceeds to the right (which it does, because there's less #"PCl"_5# once equilibrium is established.)]

Since at equilibrium, there are #1.5# #"mol PCl"_5#, the change in concentration #x# must be

#3# #"mol"# #- 1.5# #"mol"# #= 1.5# #"mol"#

Which means the final concentrations are

  • #"PCl"_5#: #3# #"mol"# #- 1.5# #"mol"# #= 1.5# #"mol"#

  • #"PCl"_3#: #3# #"mol"# #+ 1.5# #"mol"# #= color(red)(4.5# #color(red)("mol"#

  • #"Cl"_2#: #2# #"mol"# #+ 1.5# #"mol"# #= 3.5# #"mol"#

Thus, at equilibrium, there will be #color(red)(4.5# #color(red)("mol PCl"_3#.

(You can even experiment the situation with different volumes to prove that you'll get this same value every time, no matter the volume.)

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