Question

A mixture of 0.439 M H 2 , 0.317 M I 2 , and 0.877 M HI is enclosed in a vessel and heated to 430 °C. H 2 ( g ) + I 2 ( g ) − ⇀ ↽ − 2 HI ( g ) K c = 54.3 at 430 ∘ C Calculate the equilibrium concentrations of each gas at 430 ∘ C .

Answer 1

Answer : The concentration of
`H_2,I_2\text{ and }HI` at equilibrium is, 0.244 M, 0.122 M and 1.267 M respectively.

Explanation :

The given chemical reaction is:

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

Initial conc. 0.439 0.317 0.877

At eqm. (0.439-x) (0.317-x) (0.877+2x)

As we are given:

`K_c=54.3`

The expression for equilibrium constant is:

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

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

`54.3=((0.877+2x)^2)/((0.439-x)* (0.317-x))`

x = 0.195 and x = 0.690

We are neglecting the value of x = 0.690 because equilibrium concentration can not be more than initial concentration.

Thus, the value of x = 0.195 M

The concentration of
`H_2` at equilibrium = (0.439-x) = (0.439-0.195) = 0.244 M

The concentration of
`I_2` at equilibrium = (0.317-x) = (0.317-0.195) = 0.122 M

The concentration of
`HI` at equilibrium = (0.877+2x) = (0.877+2\times 0.195) = 1.267 M