What is the equilibrium concentration of each of the substance in the equilibrium when the initial concentration of `ICl` was `0.78 M`?
`2ICl(g) hArr I_(2)(g)+Cl_(2)(g), K_(c)=0.14`

Answer

`[I_(2)]= [Cl_(2)] = 0.167M. [ICl] = 0.0446 M`

Explanation

The given reaction is
`2ICl(g) hArr I_(2)(g) + Cl_(2)(g)`
`{:("Initial conc.",0.78 M,0,,0At,),("equilibrium",(0.78-2x)M,,xM,,xM):}`
Now, we c an write, `([I_(2)][Cl_(2)])/([Icl]^(2)) = K_(c)`
`rArr (x xx x )/((0.78-2x)^(2)) = 0.14`
`rArr (x^(2))/((0.78 -2x)^(2)) = 0.14`
`rArr (x)/(0.78 - 2x) = 0.374`
`rArr x = (0.292 - 0.748x)`
`rArr 1.748x = 0.292`
`rArr x = 0.167`
Hence the equilibrium
`[lCl] =`
`[l2] = 0.167 M[lCl] = [0.78 - 2 xx 0.167]M = 0.44M
`6 M`