`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 :

Options

A. 1.33
B. 2.66
C. 2
D. 3

Answer

A

Explanation

`I_(2) + I^(-) rarr I_(2)^(-)`
Excess `AgNO_(3)` gives 0.25 mol of yellow ppt.
`(AgNO_(3) + I^(-) rarr AgI )`
`implies 0.5-x = 0.25 implies x = 0.25`
`implies K_(c) = [I_(3)^(-)]/([I_(2)][I^(-)]) = (x//V)/((1-x/V)(0.5-x/V)) = (0.25/1)/((0.75/1)(0.25/1)) = 1.33 (V = 1.0 L)`