Question

Bromine monochloride, $BrCl$ decomposes into bromine and chlorine and reaches the equilibrium.
$2BrCl\left(g\right)\rightleftharpoons B{r}_2\left(g\right)+C{l}_2\left(g\right)$; $K_c=32$ 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

The correct option is A $3\times {10}^{-4}$

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

The equilibrium concentrations of $BrCl$, $B{r}_2\left(g\right)$ and $C{l}_2$ are $(3.3\times {10}^{-3}-2x)M,xM$ and $xM$ respectively.

The expression for the equilibrium constant is $K_C=\frac{\left[B{r}_2\right]\left[C{l}_2\right]}{\left[BrCl\right]}=\frac{x^2}{(3.3\times {10}^{-3}-2x{)}^2}$.

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

Thus, $\left[BrCl\right]=33\times {10}^{-4}-2x=33\times {10}^{-4}-30\times {10}^{-4}=3\times {10}^{-4}$.

Hence, the correct option is A.