If 1.0 mole of `I_(2)` is introduced into 1.0 litre flask at 1000 K, at quilibrium `(K_(e )=10^(-6))`, which one is correct ?

Options

A. `[I_(2)(g)] gt [1^(-1)(g)]`
B. `[I_(2)(g)] lt [1^(-)(g)]`
C. `[I_(2)(g)]=[I^(-)(g)]`
D. `[I_(2)(g)]=(1)/(2)[I^(-)(g)]`

Answer

A

Explanation

`underset(I-x)(I_(2)) hArr underset(2x)(2I^(-))`
`K_(e )=((2x)^(2))/((1-x))=10^(-6)`
So In. shows that `(1-x)gt 2x therefore [I_(2)(g)] gt [I^(-)(g)]`