Question

Consider the gas reaction: H2 (g)2(g) 2HI (g) The equilibrium constant at 731 K is 50.3. Equal amounts of all three gases (0.100 M) are introduced in a container, calculate the concentration of each gas after the system reaches equilibrium. Express your results with the right number of significant figures.

Answer 1

Answer : The concentration of
`H_2,I_2` and
`HI` at equilibrium 0.033 M, 0.033 M and 0.234 M respectively.

Solution : Given,

`K_c=50.3`

Concentration = 0.100 M

The given equilibrium reaction is,

`H_2(g)+I_2(g)\rightleftharpoons 2HI(g)`

Initially conc. 0.100 0.100 0.100

At equilibrium (0.100-x) (0.100-x) (0.100+2x)

The expression of
`K_c` will be,

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

Now put all the given values in this expression, we get:

`50.3=((0.100+2x)^2)/((0.100-x)* (0.100-x))`

By solving the term x, we get

`x=0.067\text{ and }0.158`

From the values of 'x' we conclude that, x = 0.158 can not more than initial concentration. So, the value of 'x' which is equal to 0.158 is not consider.

Thus, the concentration of
`H_2` and
`I_2` at equilibrium = (0.100-x) = 0.100 - 0.067 = 0.033 M

The concentration of
`HI` at equilibrium = (0.100+2x) = 0.100 + 2(0.067) = 0.234 M