Question

What is the concentration when equilibrium is reached?

`2HI(g)` `harr` `H_2`(g) + `I_2`(g) K=0.0180

If 0.068 mol `HI(g)` is sealed in a 1.0L flask, what is the concentration of `HI(g)` when equilibrium is reached?

Answer 1

The equilibrium concentration of `"HI"` is 0.054 mol/L.

Explanation:

The balanced equation is

`"2HI(g)" ⇌ "H"_2(g) + "I"_2(g)`

Now, we can set up an ICE table.

`color(white)(mmmmmmm)"2HI(g)" ⇌ color(white)(l)"H"_2(g) + color(white)(l)"I"_2(g)`
`"I/mol·L"^"-1": color(white)(mm)0.068 color(white)(mmmll)0color(white)(mmmll)0`
`"C/mol·L"^"-1": color(white)(mm)"-2"xcolor(white)(mmmll)"+"xcolor(white)(mmll)"+"x`
`"E/mol·L"^"-1": color(white)(m)"0.068-2"xcolor(white)(mmll)xcolor(white)(mmmll)x`

`K_text(c) = (["H"_2]["I"_2])/(["HI"]^2) = x^2/("0.0068-2"x)^2 = 0.0180`

`x/("0.068-2"x) = 0.134`

`x = 0.134("0.068-2"x) =0.00912 - 0.268x`

`1.268x = "0.009 12"`

`x = "0.007 19"`

`"[HI]" = ("0.068-2"x)color(white)(l) "mol/L" = "(0.068 -2×0.007 19) mol/L" = "(0.068-0.0144) mol/L" = "0.054 mol/L"`

Check:

`K = "0.007 19"^2/0.054^2 = 0.018"`

It checks!

Answer 2

`x = "0.00719 M"`. I'll leave it to you to figure out how this leads to the equilibrium concentrations by reading below.

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Here we have nonzero amounts of `"HI"(g)` initially, so it must react (through dissociation) to form `"H"_2(g)` and `"I"_2(g)`.

`2"HI"(g) rightleftharpoons "H"_2(g) + "I"_2(g)`

`"I"" "0.068" "" "" "0" "" "0`
`"C"" "-2x" "" "+x" "+x`
`"E"" "0.068-2x" "x" "" "x`

The equilibrium expression is then:

`K_c = 0.0180 = x^2/(0.068 - 2x)^2`

This is a perfect square, so...

`sqrt(0.0180) = x/(0.068 - 2x)`

We take the positive square root, as it must lead to some `K_c' = sqrt(K_c)`, for the same reaction with half the coefficients.

`sqrt(0.0180)(0.068) - 2sqrt(0.0180)x = x`

`x = (sqrt(0.0180)(0.068))/(1 + 2sqrt(0.0180))`

The physical solution is then `x = "0.00719 M"`. Therefore, the equilibrium concentrations are:

`x = color(blue)([H_2] = [I_2] = "0.00719 M")`

`0.068 - 2x = color(blue)([HI] = "0.0536 M")`

And this can be verified:

`K_c = ([H_2][I_2])/([HI]^2) = (0.00719)^2/(0.0536)^2 ~~ 0.0180` `color(blue)(sqrt"")`