#3# moles of #"SO"_3# gas are taken in an #"8-L"# container at #800^@"C"#. At equilibrium, #0.6# moles of #"O"_2# are formed. What is the value of #K_c# for #2"SO"_2(g) + "O"_2(g) rightleftharpoons 2"SO"_3(g)# ?
1 Answer
Explanation:
The first thing to notice here is that the problem wants you to figure out the equilibrium constant,
#2"SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons 2"SO"_ (3(g))#
but that you start with sulfur trioxide as the only gas present in the reaction vessel. This implies that the reverse reaction, i.e. the reaction that consumes sulfur trioxide and produces sulfur dioxide and oxygen gas, is favored here.
#2"SO"_ (3(g)) rightleftharpoons 2"SO"_ (2(g)) + "O"_ (2(g))#
The equilibrium constant for this reaction is
#K_c^' = 1/K_c#
Now, you know that you start with
#["SO"_ 3]_ 0 = "3 moles"/"8 L" = 3/8color(white)(.)"M"#
Now, the balanced chemical equation tells you that in order for the reaction to produce
At equilibrium, you know that you have
#["O"_ 2] = "0.6 moles"/"8 L" = 3/40color(white)(.)"M"#
You can say that in order for the reaction to produce
#["SO"_ 2]_ "produced" = 2 xx ["O"_ 2]_ "consumed"#
#["SO"_ 2]_ "produced" = 2 xx 3/40color(white)(.)"M"#
#["SO"_ 2 ]_ "produced" = 3/20color(white)(.)"M"#
and consume
#["SO"_ 3]_ "consumed" = ["O"_ 2]_ "produced"#
#["SO"_ 3]_ "consumed" = 3/40color(white)(.)"M"#
This means that the equilibrium concentration of sulfur trioxide will be
#["SO"_ 3] = ["SO"_ 3]_ 0 - ["SO"_ 3]_ "consumed"#
#["SO"_ 3] = 3/8color(white)(.)"M" - 3/40color(white)(.)"M"#
#["SO"_ 3] = 3/10color(white)(.)"M"#
By definition, the equilibrium constant for the initial equilibrium when the reverse reaction is favored is equal to
#K_c^' = (["SO"_2]^2 * ["O"_2])/(["SO"_3]^2)#
This implies that the equilibrium constant for the initial equilibrium when the forward reaction is favored will be
#K_c = 1/K_c^'#
#K_c = (["SO"_3]^2)/(["SO"_2]^2 * ["O"_2])#
Plug in the values you have for the equilibrium concentrations of the three gases to find--I'll leave the expression without added units!
#K_c = ( (3/10)^2)/( (3/20)^2 * 3/40) = 160/3 = 53.3#
Rounded to one significant figure, the answer will be
#color(darkgreen)(ul(color(black)(K_c = 50)))#
So, this tells you that at
#2"SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons 2"SO"_ (3(g))#
has
#K_c = 50#
which implies that, at this temperature, the forward reaction is favored, i.e. the equilibrium will lie to the right.