# Question 5b826

Aug 3, 2017
• ["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 ${K}_{e q}$, 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:

$\textcolor{\mathmr{and} a n \ge}{2} {\text{BrCl"color(red)((g)) " "" "rightleftharpoons" " "Br"_2color(red)((g)) " "+" " "Cl}}_{2} \textcolor{red}{\left(g\right)}$

$\text{I"" ""0.050 M"" "" "" "" "" ""0 M"" "" "" "" ""0 M}$
$\text{C"" "-color(orange)(2)x" "" "" "" "" "+x" "" "" } + x$
$\text{E"" "(0.050 - color(orange)(2)x) "M"" "" "x" M"" "" "" "x" M}$

where $x$ represents ${\left[{\text{Br}}_{2}\right]}_{e q}$, ${\left[{\text{Cl}}_{2}\right]}_{e q}$, 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 $\text{BrCl}$!):

${K}_{e q} = 0.145 = \left({\left[\text{Br"_2]["Cl"_2])/(["BrCl}\right]}^{2}\right)$

$= {x}^{2} / {\left(0.050 - 2 x\right)}^{2}$

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

$\sqrt{0.145} = \frac{x}{0.050 - 2 x}$

$\sqrt{0.145} \left(0.050 - 2 x\right) = x$

$\sqrt{0.145} \cdot 0.050 - 2 \sqrt{0.145} x = x$

$\sqrt{0.145} \cdot 0.050 = x + 2 \sqrt{0.145} x$

$x = \frac{\sqrt{0.145} \cdot 0.050}{1 + 2 \sqrt{0.145}} = \text{0.01081 M}$

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 ${K}_{e q} \approx 0.145$.