Given the following data, how do I find the mols at equilibrium for this reaction?

#n_(BrCl(g),eq) = "0.81 mols"#
#n_(Br_2(g),eq) = "0.29 mols"#
#n_(Cl_2(g),eq) = "0.29 mols"#

DISCLAIMER: LONG ANSWER!

Well, if we assume a rigid container, we can ignore the volume of the vessel and just use #"mols"#. The idea here is then:

1. Find #DeltaG_(rxn)^@#, i.e. at #25^@ "C"# and #"1 atm"# using standard Gibbs' free energies of formation.
2. Use the equilibrium condition to find #K#.
3. Use #K# to find #n_(Br_2(g),eq)# and #n_(Cl_2(g),eq)#.

#A)#

#DeltaG_(rxn)^@ = sum_P nu_P DeltaG_(f,P)^@ - sum_R nu_R DeltaG_(f,R)^@#,

where:

• #P# and #R# indicate products and reactants, respectively.
• #nu# is the stoichiometric coefficient.
• #DeltaG_f^@# is the standard change in Gibbs' free energy (of formation), due to forming each substance from its elements in their elemental state (that is, at #25^@ "C"# and #"1 atm"#).
• #DeltaG_f^@# for elements in their elemental state is thus #0#.

#=> ul(DeltaG_(rxn)^@) = overbrace((1/2 cdot "3.11 kJ/mol" + 1/2 cdot "0 kJ/mol"))^"Products" - overbrace((1 cdot -"0.98 kJ/mol"))^"Reactants"#

#=# #ul("2.535 kJ/mol")#

Normally, we could calculate #DeltaG_(rxn)# at nonequilibrium conditions with:

#DeltaG_(rxn) = DeltaG_(rxn)^@ + RTlnQ#,

where #RTlnQ# accounts for the shift away from standard conditions and #Q# would be the reaction quotient.

At chemical equilibrium, the reaction has no tendency to move either direction, so #DeltaG = 0# and #Q -= K#. Thus,

#DeltaG_(rxn)^@ = -RTlnK#,

and the equilibrium constant (by dividing by #-RT# and exponentiating both sides) is:

#ulK = "exp"(-DeltaG_(rxn)^@//RT)#

#= e^(-"2.535 kJ/mol"//("0.008314472 kJ/mol"cdot"K" cdot "298.15 K"))#

#= ul0.3597#

At this point, we can now determine the mols present of #"BrCl"(g)# at equilibrium. Construct an ICE table using #"mols"#:

#"BrCl"(g) rightleftharpoons 1/2"Br"_2(g) + 1/2"Cl"_2(g)#

#"I"" "1.40" "" "" "0" "" "" "" "0#
#"C"" "-x" "" "+x//2" "" "+x//2#
#"E"" "1.40 - x" "x//2" "" "" "x//2#

#K = ((x//2)^cancel(1//2))^cancel(2)/(1.40 - x) = (x//2)/(1.40 - x) = 0.3597#

This #K# is not small, but solving this is not that bad. Eventually we obtain the physical answer as:

#x = |Deltan_(BrCl(g))| = "0.5858 mols"#

#= 2n_(Br_2(g),eq) = 2n_(Cl_2(g),eq)#

And that means...

#color(blue)(n_(BrCl(g),eq)) = 1.40 - 0.5858 = ulcolor(blue)("0.81 mols")#

Everything follows from here. Now the rest is easy.

#B)#

Refer to the ICE table above to realize that:

#color(blue)(n_(Br_2(g),eq)) = x/2 ~~ ulcolor(blue)("0.29 mols")#

#C)#

Refer to the ICE table above to realize that:

#color(blue)(n_(Cl_2(g),eq)) = x/2 ~~ ulcolor(blue)("0.29 mols")#

And as a check, is #K# still correct?

#K = ((0.5858//2)^(1//2))^2/(1.40 - 0.5858) ~~ 0.3597# #color(blue)(sqrt"")#