Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium $2 \mathrm{BrCl}(\mathrm{g}) \rightleftharpoons \mathrm{Br}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$ for which $K_c=32$ at $500 \mathrm{~K}$. If initially pure $\mathrm{BrCl}$ is present at a concentration of $3.3 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$, what is its molar concentration in the mixture at equilibrium ?

Question

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium $2 \mathrm{BrCl}(\mathrm{g}) \rightleftharpoons \mathrm{Br}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$ for which $K_c=32$ at $500 \mathrm{~K}$. If initially pure $\mathrm{BrCl}$ is present at a concentration of $3.3 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$, what is its molar concentration in the mixture at equilibrium ?,

MARKS App · Accepted Answer

$\mathrm{K}_{\mathrm{c}}=\frac{(\mathrm{x} / 2)(x / 2)}{\left(3.30 \times 10^{-3}-x\right)^2}=32$ (Given) $\therefore \frac{x / 2}{\left(3.30 \times 10^{-3}-x\right)}=\sqrt{32}$ or $x / 2=5.66 \times\left(3.30 \times 10^{-3}-x\right)$ $x=11.32\left(3.30 \times 10^{-3}-x\right)$ or $12.32 x=11.32\left(3.30 \times 10^{-3}\right)$ or $x=3.03 \times 10^{-3}$ at eqn. $[\mathrm{BrCl}]=\left(3.30 \times 10^{-3}-3.0 \times 10^{-3}\right)$ $=0.30 \times 10^{-3}=3.03 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$

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