Question

The equilibrium constant for the reaction H2+I2<->2HI is 54.8 at 425°. What are the concentrations for all gases present if there is 1.000mol of each gas in a one litre container?

Answer 1

The equilibrium concentrations are:

`["H"_2] = ["I"_2] = "0.681 mol/L"`
`["HI"] = "2.362 mol/L"`

Explanation:

The balanced equation is

`"H"_2 + "I"_2 ⇌ "2HI"`

Now, we can set up an ICE table.

`color(white)(mmmmmmm)"H"_2 +color(white)(mml) "I"_2 ⇌color(white)(mll) "2HI"`
`"I/mol·L"^"-1": color(white)(ml)1.000color(white)(mm)1.000color(white)(mml)1.000`
`"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmll)"-"xcolor(white)(mmml)"+2"xcolor`
`"E/mol·L"^"-1": color(white)(m)"1.000-"xcolor(white)(m)"1.000-"xcolor(white)(m)"1.000+2"x`

`K_text(c) = (["HI"]^2)/(["H"_2]["I"_2]) = (1.000 + 2x)^2/((1.000-x)(1.000-x)) = 54.8`

`(1.000 + 2x)/(1.000-x) = 7.403`

`1.000 + 2x = 7.403 - 7.403x`

`9.403x = 6.403`

`x = 6.403/9.403 = 0.681`

∴ `["H"_2] = ["I"_2] = "0.681 mol/L"`

`["HI"] = "(1.000 + 2×0.681) mol/L = 2.362 mol/L"`

Check:

`(1.000 + 2x)^2/(1.000 -x)^2 = 2.362^2/0.3191^2 = 5.579/ 0.1018= 54.8`

It checks!