Question

Question #5b826

Answer 1

• `["Br"_2]_(eq) = "0.0108 M"`

• `["Cl"_2]_(eq) = "0.0108 M"`

• `["BrCl"]_(eq) = "0.0284 M"`

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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`.