# Bromine monochloride, (BrCl) decomposes into bromine and chlorine and reaches the equilibrium. 2BrCl_((g))hArrBr_(2(g))+Cl_(2(g)) For which K_(c)

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Bromine monochloride, (BrCl) decomposes into bromine and chlorine and reaches the equilibrium.
2BrCl_((g))hArrBr_(2(g))+Cl_(2(g))
For which K_(c)=32 at 500 K. If initially pure BrCl is present at a concentration of 3.30xx10^(-3) mol litre^(-1), what is its molar concentration in the mixture at equilibrium?

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Correct Answer - 3 xx 10^(-4) "molL"^(-1)
Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:
{:(,2BrCl_((g)),hArr,Br_(2(g)),+,Cl_(2(g))),("Initial conc.",3.3xx10^(-3),,0,,0),("At equilibrium",3.3xx10^(-3)-2x,,x,,x):}
Now, we can write,
([Br_(2)][Cl_(2)])/([BrCl]^(2)) = K_(c)
rArr (x xx x )/((3.3xx10^(-3)-2x)^(2))=32
rArr (x)/(3.3xx10^(-3)-2x)=5.66
rArr x = 18.678 xx 10^(-3) = 5.66
rArr x = 18.678 xx 10^(-3) - 11.32x
rArr 12.32 x = 18.678 xx 10^(-3)
rArr x = 1.5 xx 10^(-3)
Therefore, at equilibrium,
[BrCl] = 3.3 xx 10^(-3) - (2 xx 1.5 xx 10^(-3))
= 3.3 xx 10^(-3) - 3.0 xx 10^(-3)
= 0.3 xx 10^(-3)
= 3.0 xx 10^(-4) "molL"^(-1)`