Bromine monochloride,$BrCl$ decomposes into bromine and chlorine and reaches the equilibrium:
$2 BrCl_{(g)} \longleftrightarrow Br_{2(g)} + Cl_{2(g)}$
for which $K_c = 32$ at $500 \ K$.
If initially pure $BrCl$ is present at a concentration of $3.3 \times 10^{-3} \ mol \ L^{-1}$,what is its molar concentration in the mixture at equilibrium?

(D) Let the amount of $Br_2$ and $Cl_2$ formed at equilibrium be $x \ mol \ L^{-1}$.
The reaction is: $2 BrCl_{(g)} \longleftrightarrow Br_{2(g)} + Cl_{2(g)}$
Initial concentration: $3.3 \times 10^{-3} \ M$,$0$,$0$
Equilibrium concentration: $(3.3 \times 10^{-3} - 2x)$,$x$,$x$
$K_c = \frac{[Br_2][Cl_2]}{[BrCl]^2} = 32$
$\frac{x^2}{(3.3 \times 10^{-3} - 2x)^2} = 32$
Taking the square root on both sides: $\frac{x}{3.3 \times 10^{-3} - 2x} = \sqrt{32} \approx 5.657$
$x = 5.657 \times (3.3 \times 10^{-3} - 2x)$
$x = 18.668 \times 10^{-3} - 11.314x$
$12.314x = 18.668 \times 10^{-3}$
$x \approx 1.516 \times 10^{-3} \ mol \ L^{-1}$
At equilibrium,$[BrCl] = 3.3 \times 10^{-3} - 2x$
$[BrCl] = 3.3 \times 10^{-3} - 2(1.516 \times 10^{-3})$
$[BrCl] = 3.3 \times 10^{-3} - 3.032 \times 10^{-3} = 0.268 \times 10^{-3} \ mol \ L^{-1} \approx 2.7 \times 10^{-4} \ mol \ L^{-1}$