Q.

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 :

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1.33

2.66

2.0

3.0

answer is A.

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Detailed Solution

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

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