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?

Answer

`3 xx 10^(-4) "molL"^(-1)`

Explanation

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)`