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

Step-by-step 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=([HI]^2)/([H_2]* [I_2])`

We are given:

`K_c=52`

Putting values in above equation, we get:

`52=((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