# Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

Question:

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

$2 \mathrm{BrCl}(\mathrm{g}) \longleftrightarrow \mathrm{Br}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})$

for which $K_{c}=32$ at $500 \mathrm{~K}$. If initially pure $\mathrm{BrCl}$ is present at a concentration of $3.3 \times 10^{-3} \mathrm{molL}^{-1}$, what is its molar concentration in the mixture at equilibrium?

Solution:

Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

Now, we can write,

$\frac{\left[\mathrm{Br}_{2}\right]\left[\mathrm{Cl}_{2}\right]}{[\mathrm{BrCl}]^{2}}=K_{\mathrm{C}}$

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

$\Rightarrow \frac{x}{3.3 \times 10^{-3}-2 x}=5.66$

$\Rightarrow x=18.678 \times 10^{-3}-11.32 x$

$\Rightarrow 12.32 x=18.678 \times 10^{-3}$

$\Rightarrow x=1.5 \times 10^{-3}$

Therefore, at equilibrium,

$[\mathrm{BrCl}]=3.3 \times 10^{-3}-\left(2 \times 1.5 \times 10^{-3}\right)$

$=3.3 \times 10^{-3}-3.0 \times 10^{-3}$

$=0.3 \times 10^{-3}$

$=3.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$