`I_(2)+I^(ɵ)hArrI_(3)^(ɵ)` This reaction is set-up in aqueous medium. We start with `1` mol of `I_(2)` and `0.5` mol of `I^(ɵ)` in `1 L` flask. After equilibrium reached, excess of `AgNO_(3)` gave `0.25` mol of yellow precipitate. Equilibrium constant isA. 1.33B. 2.66C. 2D. 3

`I_(2)+I^(ɵ)hArrI_(3)^(ɵ)` This reaction is set-up in aqueous medium

1.	`I_(2)+I^(ɵ)hArrI_(3)^(ɵ)` This reaction is set-up in aqueous medium. We start with `1` mol of `I_(2)` and `0.5` mol of `I^(ɵ)` in `1 L` flask. After equilibrium reached, excess of `AgNO_(3)` gave `0.25` mol of yellow precipitate. Equilibrium constant isA. 1.33B. 2.66C. 2D. 3
Answer» Correct Answer - A `{:(,I_(2),+,I^(-),hArr,I_(3)^(-)),("Initial",1M,,0.5M,,-),("At eqm,. " ,(1-x)M,,0.5M,,x "mol"):}` Moles of `AgI=` Moles of `I^(-)` in solution `:. [I^(-)]_("eqn")=0.25M` `0.5-x=0.2:. x=0.25` `K=(x)/((1-x)(0.5-x))` `=(0.25)/(0.75xx0.25)=1.33`

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`I_(2)+I^(ɵ)hArrI_(3)^(ɵ)` This reaction is set-up in aqueous medium. We start with `1` mol of `I_(2)` and `0.5` mol of `I^(ɵ)` in `1 L` flask. After equilibrium reached, excess of `AgNO_(3)` gave `0.25` mol of yellow precipitate. Equilibrium constant isA. 1.33B. 2.66C. 2D. 3

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