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

                2BrCl_{(g)}\rightleftharpoons Br_{2}_{(g)}+Cl_{2}_{(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?

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

                    2BrCl_{(g)}\rightleftharpoons Br_{2}_{(g)}+Cl_{2}_{(g)} 

Initial Conc.     3.3\times10^{-3}          0            0

at equilibrium   3.3\times10^{-3}- 2x    x          x

Now, we can write,

K_c =\frac{[BR_2][Cl_2]}{[BrCl_2]^2}
32=\frac{x^2}{(3.3\times 10^{-3}-2x)^2}
5.66=\frac{x}{(3.3\times 10^{-3}-2x)}
x+11.32x=18.678 \times 10^{-3}

By solving the above equation we get, 

       x = 1.51 \times 10^{-3}

Hence, at equilibrium [BrCl_2] = 3.3\times 10^{-3} - (2\times 1.51\times 10^{-3})
                                                      = 0.3\times 10^{-3} M

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