Question

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

$2BrCl\left(g\right)\rightleftharpoons B{r}_2\left(g\right)+{Cl}_2\left(g\right)$

for which $K=32$ at $500K$. 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?

• A. $3\times {10}^{-4}$
• B. $1\times {10}^{-4}$
• C. $1.5\times {10}^{-4}$
• D. $6\times {10}^{-4}$

Answer 1

Answer: (A)

$3\times {10}^{-4}$

Solution:- (A) $3\times {10}^{-4}$

$2{BrCl}_{\left(g\right)}\rightleftharpoons {{Br}_2}_{\left(g\right)}+{{Cl}_2}_{\left(g\right)}$

Initially:-

$\left[BrCl\right]=3.3\times {10}^{-3}$

$\left[{Br}_2\right]=0$

$\left[{Cl}_2\right]=0$

At equillibrium:-

$\left[BrCl\right]=3.3\times {10}^{-3}-2x$

$\left[{Br}_2\right]=x$

$\left[{Cl}_2\right]=x$

From the above reaction:-

$K_C=\frac{\left[{Br}_2\right]\left[{Cl}_2\right]}{{\left[BrCl\right]}^2}$

$\therefore 32=\frac{x.x}{{(3.3\times {10}^{-3}-2x)}^2}(\because K_C=32)$

$\Rightarrow 5.65=\frac{x}{3.3\times {10}^{-3}-2x}$

$\Rightarrow 12.3x=18.645\times {10}^{-3}$

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

Therefore, at equillibrium

$\left[BrCl\right]=3.3\times {10}^{-3}-2\times 1.5\times {10}^{-3}=3\times {10}^{-4}$

Hence the molar concentration of $BrCl$ in the mixture at equilibrium is $3\times {10}^{-4}$.