Answer : The initial concentration of HI and concentration of
at equilibrium is, 0.27 M and 0.386 M respectively.
Solution : Given,
Initial concentration of
and
= 0.11 M
Concentration of
and
at equilibrium = 0.052 M
Let the initial concentration of HI be, C
The given equilibrium reaction is,

Initially 0.11 0.11 C
At equilibrium (0.11-x) (0.11-x) (C+2x)
As we are given that:
Concentration of
and
at equilibrium = 0.052 M = (0.11-x)
0.11 - x = 0.052
x = 0.11 - 0.052
x = 0.058 M
The expression of
will be,
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)

By solving the terms, we get:
C = 0.27 M
Thus, initial concentration of HI = C = 0.27 M
Thus, the concentration of
at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M