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

Question

#"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.

Naomi Aronson · Accepted 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 #"mols"#. The idea here is then: #A)#  #DeltaG_(rxn)^@ = sum_P nu_P DeltaG_(f,P)^@ - sum_R nu_R DeltaG_(f,R)^@#, where: #=> 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"")#

Axel Alvarado · Answer

undefined I need the specific data related to the reaction, such as initial concentrations, equilibrium concentrations, or equilibrium constant (if provided), to assist you in finding the moles at equilibrium.