Question

Finding initial concentration by using Kc and equilibrium concentration?!

At a certain temp the Kc for the Equillibrium reaction `H_2+ I_2 rightleftharpoons 2HI` is 51,50.
An unknown amount of HI was placed in a 1 dm^3 container. At equilibrium 0,218 mol of Hydrogen gas (H2) was present.

i)How much HI was originally placed in the container?
ii) How much I2 and HI were present at equilibrium?

Answer 1

Here's what I get.

Explanation:

This looks like the appropriate time to set up an ICE table.

`color(white)(mmmmmmm)"H"_2 + "I"_2 ⇌ "2HI"`
`"I/mol·dm"^3: color(white)(mll)0color(white)(mll)0color(white)(mmm)y`
`"C/mol·dm"^3: color(white)(ll)"+"xcolor(white)(m)"+"xcolor(white)(mm)"-2"x`
`"E/mol·dm"^3: color(white)(l)"0,218"color(white)(ll)xcolor(white)(mml)y"-"2x`

The equilibrium concentration of `"H"_2` is `"0,218 mol/dm"^3`, so `x" = 0,218"`.

Our ICE table now becomes

`color(white)(mmmmmmml)"H"_2color(white)(m) +color(white)(m) "I"_2color(white)(mm) ⇌color(white)(ml) "2HI"`
`"I/mol·dm"^3: color(white)(mm)0color(white)(mmmm)0color(white)(mmmmmm)y`
`"C/mol·dm"^3: color(white)(ll)"+0,218"color(white)(m)"+0,218"color(white)(mmm)"-0,436"`
`"E/mol·dm"^3: color(white)(mll)"0,218"color(white)(mm)"0,218"color(white)(mml)y-"0,436"`

We can now insert these numbers into the equilibrium constant expression.

`K_text(eq) = "[HI]"^2/(["H"_2]["I"_2]) = (y" - 0,436")^2/("0,218" × "0,218") = "51,50"`

`(y" - 0,436")^2= "0,218"^2 × "51,50" ="2,447"`

`y" - 0,436" = 1.564"`

`y = "1,564+ 0,436" = "2,000"`

`["I"_2] = "0,218 mol·dm"^"-3"`

`["HI"] =(y" - 0,436") color(white)(l)"mol·dm"^"-3" = ("2,000 - 0,436")color(white)(l)"mol·dm"^"-3" = "1,56 mol·dm"^"-3"`

The volume of the container was `"1 dm"^3`, so

`"i)"color(white)(m)"the initial amount of HI was 2,00 mol."`
`"ii)"color(white)(ll) "the equilibrium amount of I"_2 color(white)(l)"was 0,218 mol".`
`color(white)(mll)"the equilibrium amount of HI was 1,56 mol."`