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I(2)(aq)+I^(-)(aq)hArr(aq). We started w...

`I_(2)(aq)+I^(-)(aq)hArr(aq).` We started with 1 mole of `I_(2)` and 0.5 mole of `I_(-)` in one litre flask.After equilibrium is reached , excess of `AgNO_(3)` gave 0.25 mole of yellow precipitate. Equilibium constant is :

A

`1.33`

B

`2.66`

C

`2.00`

D

`3.00`

Text Solution

Verified by Experts

The correct Answer is:
A
`I_(2)+I^(Θ)hArr_(3)^(Θ)`
`{:("moles at" t=0,1,0.5,0),("moles at" t=t_(Eq),1-x,0.5-x,x):}`
Excess `AgNO_(3)` gave `0.25` mol of yellow precipitate.
`(AgNO_(3) + I^(ɵ)rarr underset("yellow ppt.")(AgI))`
`rArr 0.5-x=0.25 rArr x=0.25`
`rArr K_(c )=([I_(3)^(ɵ)])/([I_(2)][I^(ɵ)])=(x//V)/(((1-x)/V)((0.5-x)/V))`
`(0.25//1)/((0.75/1)(0.25/1))=1.33 , (V=1.0 L)`
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