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
Truong-Son N.
Aug 24, 2017

#"8.5 mols of gas"#

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. #-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"#

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")#

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