Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M respectively.
Solution : Given,
Initial concentration of H_2 and I_2 = 0.11 M
Concentration of H_2 and I_2 at equilibrium = 0.052 M
Let the initial concentration of HI be, C
The given equilibrium reaction is,
H_2(g)+I_2(g)\rightleftharpoons 2HI(g)
Initially 0.11 0.11 C
At equilibrium (0.11-x) (0.11-x) (C+2x)
As we are given that:
Concentration of H_2 and I_2 at equilibrium = 0.052 M = (0.11-x)
The expression of K_c will be,
K_c=\frac{[HI]^2}{[H_2][I_2]}
54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}
By solving the terms, we get:
C = 0.27 M
Thus, initial concentration of HI = C = 0.27 M
Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M
