I2(aq)+I-(aq)⇌I3-(aq) We started with 1 mole of l2 and 0.5 mole of I- in one liter flask.  After equilibrium is reached, excess of Ag NO 3 gave 0.25 mole of yellow precipitate. Equilibrium constant is :

I2(aq.) + I-(aq.) ⇌ I3-(aq.) t  =0 ; 1 mole   0.5 mole           0teq. ;   1-x          0.5-x                 xAgNO3 reacts with I-and forms precipitate.   i.e. , AgNO3+I-⟶AgI(s)+NO3-(aq.) ⇒ (0.5-x)=0.25 ⇒ x=0.25 kc=I3-I2I-=0.250.75×0.25=1.33