Question

Gaseous hydrogen iodide is placed in a closed container at 425°C, where it partially decomposes to hydrogen and iodine: 2HI(g)⇌H₂(g)+I₂(g) At equilibrium it is found that [HI]= 3.51×10⁻³ M, [H₂]= 4.76×10⁻⁴ M, and [I₂]= 4.76×10⁻⁴ M.

What is the value of
`K_c` at this temperature? Express the equilibrium constant to three significant digits.

Answer 1

Answer : The value of
`K_c` at this temperature is 0.0184

Explanation : Given,

Concentration of
`HI` at equilibrium =
`3.51* 10^(-3)M`

Concentration of
`H_2` at equilibrium =
`4.76* 10^(-4)M`

Concentration of
`I_2` at equilibrium =
`4.76* 10^(-4)M`

The given equilibrium reaction is,

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

The expression of
`Kc` will be,

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

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

`K_c=((4.76* 10^(-4))* (4.76* 10^(-4)))/((3.51* 10^(-3))^2)`

`K_c=0.0184`

Therefore, the value of
`K_c` at this temperature is 0.0184