Bromine monochloride, `BrCl` decomposition into bromine and chlorine and reaches the equilibrium:
`2BrCl (g) hArr Br_(2)(g) + Cl_(2)(g)`
for which `K_(c) = 32` at `500 K`. If initially pure BrCl is present at a concentration of `3.3 xx 10^(-3) "mol" L^(-1)`. What is its molar concentration in the mixture at equilibrium ?

Explanation

`{:(,2BrCl(g),hArr,Br_(2)(g),+,Cl_(2)(g)),("Initial",3.30xx10^(-3),,,,),(,"mol " L^(-1),,,,),("At Eq",(3.30xx10^(-3)-x),,x//2,,x//2):}`

`K_(c )=((x//2)(x//2))/((3.30xx10^(-3)-x^(2)))=32` (Given)
`:. X^(2)/(4(3.30xx10^(-3)-x))=32`
or `x/(2(3.30xx10^(-3)-x))=sqrt(32)=5.66`
`x=11.32(3.30xx10^(-3)-x)`
or `12.32x=11.32xx3.30xx10^(-3)` or `x=3.0xx10^(-3)`
`:. At eqm`.
`[BrCl]=(3.30xx10^(-3)-3.0xx10^(-3))=0.30xx19^(-3)`
`=3.0xx10^(-4) mol L^(-1)`.