Question

Hydrogen gas, iodine vapor, hydrogen iodine are mixed in a flask and heated to 696°C. H2(g) + I2(g) ⇋ 2 HI(g) Kc = 52 at 696°C If the initial concentrations of hydrogen gas and iodine vapor are both 0.044 mol/L and the concentration of hydrogen iodine is 0.177 mol/L what is the equilibrium concentration of hydrogen gas? Enter a number to 4 decimal places.

Answer 1

Answer: The equilibrium concentration of hydrogen gas is 0.0269 M

Explanation:

The chemical reaction follows the equation:

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

At t = 0          0.044M     0.044M              0.177M  

At $t=t_{eq}$    (0.044-x)M    (0.044-x)M      (0.177+x)M

The expression for $K_c$ for the given reaction follows:

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

We are given:

$K_c=52$

Putting values in above equation, we get:

$52=\frac{(0.177+x)^2}{(0.044-x)^2}$

$x=0.0171M$

Hence, the equilibrium concentration of hydrogen gas is (0.044-x) M =(0.044-0.0171) M= 0.0269 M