Question

Bromine monochloride, $BrCl$ decomposes into bromine and chlorine and reaches the equilibrium: $2BrCl_{(g})\rightleftharpoons Br_2 {(g)} + Cl_2{(g)}$ For which $K_c - 32 $ at $500 K.$ If initially pure $BrCl$ is present at a concentration of $ 3.3 x 1 0^{-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)

$\begin{matrix}&2BrCl\left(g\right)&\text{rightleftharpoons}&Br_{2g}&+&CI_{2g}\\ Initial&3.30\times10^{-3} mol L^{-1}&&0&&0\\ Ateq.&\left(3.30\times10^{-3}-x\right)&&\frac{x}{2}&&\frac{x}{2}\end{matrix}$
$K_{c} =\frac{\left(x 2\right)\left(x 2\right)}{\left(3.30\times10^{-3}-x^{2}\right)} =32 \left(Given\right) $
$\therefore \quad \frac{x^{2}}{4\left(3.30 \times 10^{-3}-x\right)^{2}}=32$

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

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

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