1.

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 ?

Answer»

`[I_(2)(g)] gt [1^(-1)(g)]`
`[I_(2)(g)] LT [1^(-)(g)]`
`[I_(2)(g)]=[I^(-)(g)]`
`[I_(2)(g)]=(1)/(2)[I^(-)(g)]`

Solution :`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)]`


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