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,

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

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

$50.3=\frac{(0.100+2x)^2}{(0.100-x)\times (0.100-x)}$

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