If a reaction of sulfur dioxide with dioxygen gas to form sulfur trioxide starts with #"5 mols SO"_2# and #"5 mols O"_2#, and #60%# of the #"SO"_2# was consumed in the reaction, what is the total mols of gas at equilibrium?
1 Answer
Write the reaction:
#2"SO"_2(g) + "O"_2(g) -> 2"SO"_3(g)#
For this reaction, you know that the starting mols are:
#n_("SO"_2,i) = "5 mols SO"_2#
#n_("O"_2,i) = "5 mols O"_2#
We construct the ICE table to show the changes in mols at constant temperature and total pressure:
#2"SO"_2(g) + "O"_2(g) -> 2"SO"_3(g)#
#"I"" "5" "" "" "" "5" "" "" "0#
#"C"" "-2x" "" "-x" "" "+2x#
#"E"" "5 - 2x" "5-x" "" "2x#
Remember to include the stoichiometric coefficients in the change in concentration, i.e.
In this case, we know that
#5 - 2x = alpha xx 5#
#= (1 - 0.60)(5) = 2#
#5 - 2 = 2x#
#=> x = "1.5 mols gas"#
Therefore, at equilibrium, the mols of gas in the vessel are given by:
#color(blue)(n_(eq,t ot)) = (5 - 2x) + (5 - x) + (2x)#
#= 10 - x#
#= 10 - 1.5#
#=# #color(blue)("8.5 mols gas total")#