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

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

#DeltaG_f^@ ("Br"_2(g)) = "3.11 kJ/mol"#

#DeltaG_f^@("BrCl"(g)) = -"0.98 kJ/mol"#

#A)# Find the mols of #"BrCl"(g)# at equilibrium if the volume of the container is fixed at #"1.0 L"# .

#B)# Find the mols of #"Br"_2(g)# at equilibrium.

#C)# Find the mols of #"Cl"_2(g)# at equilibrium.

##### 1 Answer

#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

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

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

#P# and#R# indicateproductsandreactants, respectively.#nu# is thestoichiometric coefficient.#DeltaG_f^@# is thestandard 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 hasno tendencyto 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 anICE tableusing#"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.

Refer to the ICE table above to realize that:

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

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