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

I_(2(aq)) + I_((aq))^(-) hArr I_(3(aq))^(-).We started with I mole of

1.	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
Answer» 1.33 2.66 2 3 Solution :`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`