Question

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium
$2BrCl\left(g\right)\to B{r}_2\left(g\right)+C{l}_2\left(g\right)$
For which $K_c=32$ at 500K. If initially pure BrCl is present at a concentration of $3.3\ast {10}^{-3}mol{L}^{-}$, what is it's molar concentration in the mixture of equilibrium?

Answer 1

$\begin{array}{cccc}2{BrCl}_2\left(g\right)& \to & {Ar}_2\left(g\right)& +{Cl}_2\left(g\right)\\ 3.3\times {10}^{-3}m& & 0& 0\\ 3.3\times {10}^{-3}-2\alpha & & \alpha & \alpha \end{array}$

$\therefore K_c=\frac{\alpha \times \alpha }{(3.3\times {10}^{-3}-2\alpha {)}^2}$

$\Rightarrow \frac{{\alpha }^2}{(3.3\times {10}^{-3}-2\alpha {)}^2}=32$

$\Rightarrow \frac{\alpha }{(3.3\times {10}^{-3}-2\alpha {)}^2}=4\sqrt{2}$

$\Rightarrow \alpha =18.67\times {10}^{-3}-8\sqrt{2}\alpha$

$\Rightarrow (1+8\sqrt{2})\alpha =18.67\times {10}^{-3}$

$\Rightarrow \alpha =1.5162\times {10}^{-3}$

$\therefore$ Molar concentration of $Brcl$ at equation

$=3.3\times {10}^{-3}-2\alpha$

$=3.3\times {10}^{-3}-2\times 1.562\times {10}^{-3}$

$=0.267\times {10}^{-3}M$