Question

Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Determine the initial and equilibrium concentration of HI if initial concentrations of H2 and I2 are both 0.11 M and their equilibrium concentrations are both 0.052 M at 430°C.

Answer 1

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)

0.11 - x = 0.052

x = 0.11 - 0.052

x = 0.058 M

The expression of
`K_c` will be,

`K_c=([HI]^2)/([H_2][I_2])`

`54.3=((C+2(0.058))^2)/((0.052)* (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