Question

Calculate the concentration at equilibrium?

`H_2+I_2 rightleftharpoons 2HI`
`Keq = 49.5`

If 5.0M HI is initially placed into a container, what will be the equilibrium [HI]?

Answer 1

Your numbers given are wrong. To calculate the conc. of HI we need to know the initial concentration of `H_2` and `I_2`. Assuming we have 5.0M of `H_2` and `I_2`; conc. of HI = 1M

Explanation:

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

`49.5 = [HI]^2/(5.00*5.00)`

`49.5*25 = [HI]^2`

`sqrt1237.5 = [HI]`

[HI] = 1M

Answer 2

I got `"3.9 M"`.

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Given that `K = 49.5` for this reaction

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

we have `K' = 1/K = 0.0202` for this reaction:

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

`"I"" "5.0" "" "" "0" "" "" "0`
`"C"" "-2x" "" "+x" "+x`
`"E"" "5.0-2x" "x" "" "x`

So the mass action expression for this reaction is:

`K' = (["H"_2]_(eq)["I"_2]_(eq))/(["HI"]_(eq)^2)`

`= 0.0202 = x^2/(5.0 - 2x)^2`

This is readily solvable:

`sqrt(0.0202) = x/(5.0 - 2x)`

`5.0sqrt(0.0202) - 2sqrt(0.0202)x = x`

`5.0sqrt(0.0202) = (2sqrt(0.0202) + 1)x`

`x = (5.0sqrt(0.0202))/(2sqrt(0.0202) + 1)`

`=` `"0.553 M"`

That means the equilibrium `["HI"]` is:

`color(blue)(["HI"]_(eq)) = "5.0 M" - 2("0.553 M")`

`=` `color(blue)("3.9 M")`