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?

Question

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?

Naomi Aronson · Accepted Answer

#"8.5 mols of gas"# Write the response: #2"SO"_2(g) + "O"_2(g) -> 2"SO"_3(g)# You are aware of the initial mols for this reaction as follows: #n_("SO"_2,i) = "5 mols SO"_2# #n_("O"_2,i) = "5 mols O"_2# To display the variations in mols at constant temperature and total pressure, we build the ICE table as follows: #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. #-2x# for a reactant written with a coefficient of #2#, #+x# for a product with a coefficient of #1#, etc. In this case, we know that #60%# of the #"SO"_2# was consumed, so the fraction of dissociation #alpha# is given as #alpha = 1 - 0.60#: #5 - 2x = alpha xx 5# #= (1 - 0.60)(5) = 2# #5 - 2 = 2x# #=> x = "1.5 mols gas"# As a result, at equilibrium, the gas in the vessel's mols is determined by: #color(blue)(n_(eq,t ot)) = (5 - 2x) + (5 - x) + (2x)# #= 10 - x# #= 10 - 1.5# #=# #color(blue)("8.5 mols gas total")#

Owen Ambrose · Answer

undefined 2 mols.