Question #5b826

• #["Br"_2]_(eq) = "0.0108 M"#

• #["Cl"_2]_(eq) = "0.0108 M"#

• #["BrCl"]_(eq) = "0.0284 M"#

What we're really doing here is that we're given an initial state with #["BrCl"]_i = "0.050 M"#, and we're given #K_(eq)#, which is defined for a final state we define as equilibrium, when the rate of the forward reaction is equal to that of the reverse reaction.

Concentrations are a matter of "what did we start with?", and "what did we end with?", and not "how did we get there?".

We would construct an ICE table to describe the change:

#color(orange)(2) "BrCl"color(red)((g)) " "" "rightleftharpoons" " "Br"_2color(red)((g)) " "+" " "Cl"_2color(red)((g))#

#"I"" ""0.050 M"" "" "" "" "" ""0 M"" "" "" "" ""0 M"#
#"C"" "-color(orange)(2)x" "" "" "" "" "+x" "" "" "+x#
#"E"" "(0.050 - color(orange)(2)x) "M"" "" "x" M"" "" "" "x" M"#

where #x# represents #["Br"_2]_(eq)#, #["Cl"_2]_(eq)#, etc.

Notice how this reaction does NOT occur in aqueous solution... I will assume low pressure, so that bromine is a gas... Anyways, that gives an equilibrium state (note the stoichiometry in #"BrCl"#!):

#K_(eq) = 0.145 = (["Br"_2]["Cl"_2])/(["BrCl"]^2)#

#= x^2/(0.050 - 2x)^2#

And we are lucky to have a perfect square...

#sqrt(0.145) = x/(0.050 - 2x)#

#sqrt(0.145)(0.050 - 2x) = x#

#sqrt(0.145)cdot0.050 - 2sqrt(0.145)x = x#

#sqrt(0.145)cdot0.050 = x + 2sqrt(0.145)x#

#x = (sqrt(0.145)cdot0.050)/(1 + 2sqrt(0.145)) = "0.01081 M"#

Thus, we have:

• #color(blue)(["Br"_2]_(eq) = "0.0108 M")#

• #color(blue)(["Cl"_2]_(eq) = "0.0108 M")#

• #color(blue)(["BrCl"]_(eq) = "0.0284 M")#

Verify that this still gives #K_(eq) ~~ 0.145#.