How many mols of #"CO"_2# were added at constant temperature and volume to a #"1-L"# container containing #"0.400 mols"# of #"CO"_2# and of #"H"_2#, and #"0.200 mols"# of #"CO"# and of #"H"_2"O"(g)# (#K_c = 4#) to cause #"CO"_2# to drop to #"0.300 mols"#?
1 Answer
I got
This is just asking you to:
- Determine the equilibrium constant for an initial state.
- Use this previous equilibrium configuration as your new initial state, where the amount of
#"CO"_2# you now start with is plus an unknown amount,#y# - The reaction will go backwards to reach its second equilibrium configuration, and thus the change in mols,
#x# , will have the reversed sign compared to the forward reaction.
Since the volume of the container is shared, we can use just the mols of everything instead of the molarity. The first equilibrium constant will stay the same, and thus the starting equilibrium configuration has
#K_c = (["CO"_2]["H"_2])/(["CO"]["H"_2"O"]) = (0.400^2)/(0.200^2) = 4#
We want to add
Your initial ICE table should look like this (after reaching the first equilibrium configuration already, and adding the
#"CO"(g) " "+" " "H"_2"O"(g) " "rightleftharpoons" " "CO"_2(g) + "H"_2(g)#
#"I"" "0.200" "" "" "0.200" "" "" "" "0.400 + y" "" "0.400#
#"C"" "+x" "" "" "+x" "" "" "" "" "" "-x" "" "-x#
#"E"" "0.200 + x" "0.200 + x" "0.400 + y - x" "0.400 - x#
Now, we want the final mols of
#n_(CO) = 0.200 + x = 0.300#
Thus,
#((0.300 + y)(0.300))/((0.300)(0.300)) = 4#
Solve for the amount of
#(4(0.300^2))/(0.300) = 0.300 + y#
#=> color(blue)(y) = (4(0.300^2))/(0.300) - 0.300 = color(blue)("0.900 mols CO"_2 " added")#