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I(2(aq)) + I((aq))^(-) hArr I(3(aq))^(-)...

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

A

1.33

B

2.66

C

2

D

3

Text Solution

Verified by Experts

The correct Answer is:
A

`underset(1-alpha)underset(1)(I_(2(aq)))+ underset(0.5-alpha)underset(0.5)(I_((aq))^(-)) harr underset(alpha)underset(0)(I_(3(aq))^(-))`
At equilibrium
`(0.5-alpha)` mole of `I^(-)`
reacts with `AgNO_(3)` to gave 0.25 mole of AgI
`:. 0.5 - alpha=0.25, alpha=0.25`
`K_(c)=([I_(3)^(-)])/([I_(2)](I^(-)))=(2)/((1-alpha)(0.5-alpha))=(0.25)/(0.75 xx 0.25)=(4)/(3)=1.33`
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