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Standard XII

Chemistry

Reaction with Metals

Bromine monoc...

Question

Bromine monochloride, $B r C l$ , decomposes into bromine and chlorine and reaches equilibrium as $2 B r C l_{(g)} ⇌ B r_{2 (g)} + C l_{2 (g)}$ ; for which $K_{c} = 32$ at $500 K$ .

If initially, pure $B r C l$ is present at a concentration of $3.30 \times 10^{- 3} m o l l i t r e^{- 1}$ , what is its molar concentration in the mixture at equilibrium?

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Solution

The initial concentration of $B r C l, B r_{2}$ and $C l_{2}$ are $3.30 \times 10^{- 3} M, 0 M$ and $0 M$ respectively.
Let x M be the amount of BrCl dissociated.
The equilibrium concentrations are $(3.30 \times 10^{- 3} \times 10^{- 3} - x), x / 2, x / 2$ respectively.
The equilibrium constant expression is
$K_{c} = \frac{[B r_{2}] [C l_{2}]}{[B r C l]^{2}} = \frac{x / 2 \times x / 2}{(3.30 \times 10^{- 3} - x)^{2}} = 32$
$x = 3.00 \times 10^{- 3}$
Hence, the equilibrium concentration of $B r C l$ is $3.30 \times 10^{- 3} - 3.00 \times 10^{- 3} = 0.30 \times 10^{- 3} = 3.0 \times 10^{- 4} M$ .

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Bromine monochloride, BrCl, decomposes into bromine and chlorine and reaches equilibrium as 2BrCl(g)⇌Br2(g)+Cl2(g); for which Kc=32 at 500 K. If initially, pure BrCl is present at a concentration of 3.30×10−3mol litre−1, what is its molar concentration in the mixture at equilibrium?