Ncert Exercises & Solutions

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

2BrCl (g) Br2 (g) + Cl2 (g)

for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of
3.3 × 10–3 mol L–1, what is its molar concentration in the mixture at equilibrium?

Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

$2 {BrCl}_{(g)} \leftrightarrow {Br}_{2 (g)} + {Cl}_{2 (g)}$
$Initial conc . 3.3 \times 10^{- 3} 0 0$
$At equilibrium 3.3 \times 10^{- 3} - 2 x x x$

Now, we can write,

$\frac{[{Br}_{2}] [{Cl}_{2}]}{{[BrCl]}^{2}} = K_{c}$
$\Rightarrow \frac{x \times x}{{(3.3 \times 10^{- 3} - 2 x)}^{2}} = 32$
$\Rightarrow x = 18.678 \times 10^{- 3} - 11.32 x$
$\Rightarrow 12.32 x = 18.678 \times 10^{- 3}$
$\Rightarrow x = 1.5 \times 10^{- 3}$

Therefore, at equilibrium,

$[BrCl] = 3.3 \times 10^{- 3} - (2 \times 1.5 \times 10^{- 3})$
$= 3.3 \times 10^{- 3} - 3.0 \times 10^{- 3}$
$= 0.3 \times 10^{- 3}$
$= 3.0 \times 10^{- 4} {molL}^{- 1} \begin{aligned} \end{aligned}$

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