Equilibrium



How meq of hypo =15.7*1/10

Here, initial moles of HI = given mass/ molar mass = 0.96/ 128 gm = 0.0075 moles = 7.5 x 103 moles
So, the reaction is, 2HI <=> H2 + I2Here, no. of moles of reactants and products are same so, Kp = Kc
Hence, Initial conc.0.0075mol...0 ...0
Equbm. Conc. .... (0.0075 x) x/2 x/2Further, the same amount of I2 reacts with Na2S2O3,
So, I2 + 2Na2S2O3 > 2NaI + Na2S4O6
It may be written as :
I2 + 2S2O32 > S4O62 + 2I
Here, we can see that, meq. of I2 = meq. of Na2S4O6So, we can write as (w/E) x 1000 = 15.7 x 1/10
Or, (w/E) for I2 =1.57 x 103So, moles of I2 formed at equilibrium
= 1.57 x 103/ 2
or, x/2 = 1.57 x 103/ 2
So, x = 1.57 x 103Hence, degree of dissociation of HI = moles dissociated/ initial moles of HI = 1.57 x 103/ 7.5 x 103 = 0.209
Also,%age of dissociation = 20.9 %

@AbhishekSingh
What the defination of meq?

@YashKapoor
Normality into volume gives meq