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?

Explanation

Let x moles of `BrCI` decompose in order to attain the equilibrium . The initial molar concentration and the molar ocncentration at equilibrium point of different species may be represented as follows :

Applying Law of chemical equilibrium , `K_(c) =[[Br_(2)][CI_(2)]]/[[BrCI]^(2)] " or " 32 =((x//2) xx (x//2))/((0.0033 -x)^(2))`
On taking the square root, `5.656=(x//2)/((0.0033 -x))`
`(x)/((0.0033-x)) =11.31 " or " 12.31x = 0.037 , x (0.037)/(12.31) =0.003`
`:.` Molar concentration of `BeCI` at equilibrium point `=0.0033 -0.003`
`=0.0003 mol L^(-1) =3XX 10^(-4) mol L^(-1).`