Question

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
$2BrCl\leftrightarrow B{r}_2\left(g\right)+C{l}_2\left(g\right)$
Fro which $K_c=32$ at 500 K.
If initially pur BrCl is present at a concentration of $3.3\times {10}^{-3}mo{l}^{-1}$, what is its molar concentration in the mixture at equilibrium?

Answer 1

Let the amount of bromine and chlorice formed at equilibrium be x. The given reaction is:
$2BrC{l}_{\left(g\right)}\leftrightarrow B{r}_{2\left(g\right)}+C{l}_{2\left(g\right)}$
$Initialconc.3.3\times {10}^{-3}00$
$Atequilibrium3.3\times {10}^{-3}-2xxx$
Now, we can write,
$\frac{\left[B{r}_2\right]\left[C{l}_2\right]}{[BrCl{]}^2}=K_c$
$\Rightarrow \frac{x\times x}{(3.3\times {10}^{-3}-2x{)}^2}=$$\frac{x^2}{(3.3\times {10}^{-3}-2x{)}^2}=32$
$\Rightarrow \frac{x}{3.3\times {10}^{-3}-2x}=5.66$ ($\because \sqrt{32}=5.66$)
$\Rightarrow x=18.678\times {10}^{-3}-11.32x$
$\Rightarrow 12.32x=18.678\times {10}^{-3}$
$\Rightarrow x=1.5\times {10}^{-3}$
Therefore, at equilibrium,
$\left[BrCl\right]=3.3\times {10}^{-3}-(2\times 1.5\times {10}^{-3})$
$=3.3\times {10}^{-3}-3.3\times {10}^{-3}$
$=0.3\times {10}^{-3}$
$=3.0\times {10}^{-4}mol{L}^{-1}$