The value of `K_(c)` for the reaction `2HI(g)hArrH_(2)+I_(2)(g)` is `1xx10^(-4)`. At a given time, t he composition of reaction mixture is
`[HI]=2xx10^(-5) mol`,`[H_(2)]=1xx10^(-5)` mol and `[l_(2)]=1xx10^(-5)` mol
In which direction will the reaction proceed ?

Explanation

For the reactio, `2HI(g)hArr(g)+l_(2)(g)`
Reaction quotient `=([H_(2)][l_(2)])/([HI]^(2))=((1xx10^(-5) mol ) xx(1xx10^(-5) mol))/((2xx10^(-5)mol )^(2))=(1)/(4)`
As the value of Q (0.25) is more than the `K_(c)` value `(1xx10^(-4))`, the reaction will proceed in the reverse diection.