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:

$K=\frac{[HI]^2}{[H_2][I_2]}$

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

$54.3=\frac{(0.877+2x)^2}{(0.439-x)\times (0.317-x)}$

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