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

` {:(,Fe O(s) , +, CO(g) ,hArr, Fe(s) + CO_2(g) ),( " Initial pressure " , 3.30 xx 10^(3) ,, -,, - ),(" Conc. at eq . ( mol "L^(-1) , (3.30 xx 10^(-3) x) , , (x)/(2),,(x)/(2)):}`
` " " K_ c= ( [Br_2(g) ][Cl_2(g) ])/( [BrCl(g) ]^(2)) `
or ` " " 32= ((x)/(2) xx(x)/(2))/((3.30 xx 10 ^(-3) -x) ^(2))`
The square root on both sides , we get
` " " 5.66 = ((x)/(2))/( 3.30 xx 10 ^(-3) -x) = (x)/( 2(3.30xx10 ^(-3)-x))`
or ` " " 0.337 - 11.32 x = x `
or ` " " ( 1+ 11 .32 ) x = 0.037`
` or " " x= ( 0.037 )/( 1+11.32 )= 3.0 xx 10 ^(-3)`
` therefore ` Equilibrium concentration of BrCl `= 3.30 xx 10 ^(-3) `
` " " = 3.30 xx 10 ^(-3) - 3.0 xx 10 ^(-3)`
`" " = 3.0 xx 10 ^(-4) " mol" L^(-1)`