at eqbm conc. of I- is 0.25 moles/lit as AgNO3 reacts with 0.25 moles of I- to form AgI
So,
I2 + I- --> I3-
t=0 1 0.5 0
t=teq 0.75 0.25 0.25
So Keq = Kc = (0.25)/(0.25*0.75)
= 1.33 lit/moles
I2(aq)+I-(aq) → I3-(aq)
We started with 1 mole of I2 and 0.5 mole of I- in one litre flask.After equlibrium is reached,excess of AgNO3 gave 0.25 mole of yellow ppt.Find the Equilibrium constant of the rxn.
Note--- → represents reversible rxn..
at eqbm conc. of I- is 0.25 moles/lit as AgNO3 reacts with 0.25 moles of I- to form AgI
So,
I2 + I- --> I3-
t=0 1 0.5 0
t=teq 0.75 0.25 0.25
So Keq = Kc = (0.25)/(0.25*0.75)
= 1.33 lit/moles