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

`{:(,2BrCl(g) hArr , Br_2(g)+,Cl_2(g)),("Initial conc.", 3.30xx10^(-3),,),("At equi.", (3.30xx10^(-3)-x),x//2,x//2):}` where x is the amount of BrCl dissociated
`K_c=([Br_2][Cl_2])/[BrCl]^2 =(x//2 xx x//2)/(3.30xx10^(-3)-x)^2 =32`
Solving
`x=3.00xx10^(-3)`
`therefore` At equilibrium [BrCl]
`=3.30xx10^(-3)-3.00xx10^(-3)`
`=0.30xx10^(-3)`
or `=3.0xx10^(-4) "mol L"^(-1)`