`X_(2) + X^(-) hArr X_(3)^(-)` (x = iodine) This reaction is set up in aqueous medium. We start with 1 mol of `X_(2)` and 0.5 mol of `X^(-)` in 1L flask. After equilibrium is reached, excess of `AgNO_(3)` gave 0.25 mol of yellow ppt. equilibrium constant is

Options

A. 1.33
B. 2.66
C. `2.00`
D. `3.00`

Answer

A

Explanation

`{:(X_(2), +,X^(-), hArr, X_(3)^(-)), (1,, 0.5,,0),((1-x),,(0.5-x),,x):}`
`(0.5-x)`= unreacted `X^(-)`
`0.5 -x = 0.25 , x = 0.25 `
`K_(c ) = ([X_(3)^(-)])/([X_(2)][X^(-)])=(0.25)/(0.75 xx 0.25) = 1.33`