Question

Bromine monochloride, $BrCl$ decomposes into bromine and chlorine and reaches the equilibrium: $2BrC{l}_{(g})\rightleftharpoons B{r}_2(g)+C{l}_2(g)$ For which $K_c-32$ at $500K.$ If initially pure $BrCl$ is present at a concentration of $3.3x10^{-3}mol{L}^{-1}$, what is its molar concentration in the mixture at equilibrium?

• A. $3\times 10^{-4}mol{L}^{-1}$
• B. $4\times 10^{4}mol{L}^{-1}$
• C. $5\times 10^{-2}mol{L}^{-1}$
• D. $7\times 10^{-5}mol{L}^{-1}$

Answer 1

Answer: (C)

$\begin{array}{cccccc}& 2BrCl\left(g\right)& \text{rightleftharpoons}& B{r}_{2g}& +& C{I}_{2g}\\ Initial& 3.30\times 10^{-3}mol{L}^{-1}& & 0& & 0\\ Ateq.& \left(3.30\times 10^{-3}-x\right)& & \frac{x}{2}& & \frac{x}{2}\end{array}$
$K_c=\frac{\left(x2\right)\left(x2\right)}{\left(3.30\times 10^{-3}-x^2\right)}=32\left(Given\right)$
$\therefore \frac{x^2}{4{\left(3.30\times 10^{-3}-x\right)}^2}=32$

or, $\frac{x}{2\left(3.30\times 10^{-3}-x\right)}=\sqrt{32}=5.66$

or, $x=11.32\left(3.30\times 10^{-3}-x\right)$

or, $12.32x=11.32\times 3.30\times 10^{-3}$ or, $x=3.0\times 10^{-3}$

$\therefore$ At eq. $[BrCl]=\left(3.30\times 10^{-3}-3.0\times 10^{-3}\right)$

$=0.30\times 10^{-3}=3.0\times 10^{-4}mol{L}^{-1}$