Question

`H_2` (g) + `F_2` (g) `\rightleftharpoons2HF` (g), find final values of `[H_2]_(eq)`, `[F_2]_(eq)`, and `[HF]_(eq)`?

`k=1.15xx10^2`
If `3.000` mol of `H_2`, `F-2` and `HF` are added to a `1.500` L flask...

(find final values)

Answer 1

`["H"_2]_(eq) = ["F"_2]_(eq) = "0.471 M"`
`["HF"]_(eq) = "5.06 M"`

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I assume you have given the equilibrium constant `K_c` (not `k`, which is a rate constant). Also, we used `K_c` because you gave concentrations.

The reaction was:

`"H"_2(g) + "F"_2(g) rightleftharpoons 2"HF"(g)`

In a `"1.500 L"` flask, the volume is shared, so the initial concentrations are:

`["H"_2]_i = "3.000 mols"/"1.500 L" = "2.000 M"`

`["F"_2]_i = "3.000 mols F"_2/"1.500 L" = "2.000 M"`

`["HF"]_i = "3.000 mols HF"/"1.500 L" = "2.000 M"`

So, the ICE table uses concentrations to give:

`"H"_2(g) " "+" " "F"_2(g) " "rightleftharpoons" " 2"HF"(g)`

`"I"" "2.000" "" "" "2.000" "" "" "2.000`
`"C"" "-x" "" "" "-x" "" "" "+2x`
`"E"" "2.000-x" "2.000-x" "2.000+2x`

...remembering that coefficients go into the change in concentration.

Here we assumed that the reaction goes forward. If you calculate `Q_c`, you would find that here it equals `1`:

`Q_c = ("2.000 M HF")^2/("2.000 M H"_2 cdot "2.000 M F"_2) = 1`

(Since `Q_c < K_c`, the reaction wants to shift towards the products, so our assumption is good.)

Therefore, we can set up the `K_c` expression:

`1.15 xx 10^2 = (2.000 + 2x)^2/((2.000 - x)(2.000 - x))`

This is nice in that it is a perfect square. That is one situation where no approximations need to be made and it is quite solvable without the quadratic formula.

`sqrt(1.15 xx 10^2) = 10.72 = (2.000 + 2x)/(2.000 - x)`

We only take `K_c > 0` because equilibrium constants can never be negative. Proceed to solve for `x`:

`10.72(2.000 - x) = 2.000 + 2x`

`21.45 - 10.72x = 2.000 + 2x`

`19.45 = 12.72x`

`x = "1.53 M"`

Note that we lost a solution of `x = "2.69 M"`, but that's OK because it is nonphysical; `2.000 - x` would have given a negative concentration for the reactants.

Anyways, we now get the equilibrium concentrations to be:

`color(blue)(["H"_2]_(eq) = ["F"_2]_(eq) = 2.000 - x = "0.471 M")`

`color(blue)(["HF"]_(eq) = 2.000 + 2x = "5.06 M")`

And just to check,

`K_c = (5.06^2)/((0.471)(0.471)) = 115.41 ~~ 115` `color(blue)(sqrt"")`