Question

At a certain temperature, 4.0 mol `NH_3` is introduced into a 2.0 L container, and the `NH_3` partially dissociates to `2NH_3(g)\rightleftharpoonsN_2(g)+3H_2(g)`. At equilibrium, 2.0 mol `NH_3` remains. What is the value of `K_c`?

A clearer version:

At a certain temperature, 4.0 mol `NH_3` is introduced into a 2.0 L container, and the `NH_3` partially dissociates to `2NH_3(g)\rightleftharpoonsN_2(g)+3H_2(g)`.

At equilibrium, 2.0 mol `NH_3` remains. What is the value of `K_c`?

Note: I started an ICE table but I don't know what to put in the `N_2` and `3H_2` columns.

Answer 1

Remember to include the coefficients in the change in concentratio, as well as the exponents. I get `K_c = 1.69`.

If `K_p = K_c(RT)^(Deltan_"gas")`, what would `K_p` be if this reaction has been occurring at `"298 K"`? `R = "0.08206 L"cdot"atm/mol"cdot"K"`. Highlight below to see.

`color(white)("From the reaction, "Deltan_"gas" = (1+3) - (2) = 2,)`
`color(white)("since there are 3 mols H"_2, "1 mol N"_2, "and 2 mols NH"_3.)`

`color(white)("So,")`

`color(white)(K_p = 1.69 cdot ("0.08206 L"cdot"atm/mol"cdot"K" cdot "298 K")^((1+3)-(2)))`

`color(white)(~~ 1010" in implied units of atm")`

---

Let's first find the concentrations, because `K_c` is in terms of concentrations.

`"4.0 mols NH"_3/("2.0 L") = "2.0 M"`

`"2.0 mols NH"_3/("2.0 L") = "1.0 M"`

The ICE table uses `"NH"_3` having the initial concentration of `"2.0 M"` and final of `"1.0 M"`. Therefore, we should be able to find `x` already, but let's assume we don't know `x` yet.

`color(red)(2)"NH"_3(g) rightleftharpoons "N"_2(g) + color(red)(3)"H"_2(g)`

`"I"" "2.0" "" "" "" "0" "" "" "0`
`"C"" "-color(red)(2)x" "" "+x" "" "+color(red)(3)x`
`"E"" "2.0-color(red)(2)x" "" "x" "" "color(red)(3)x`

Remember that the coefficients go into the change in concentration and exponents.

Now, the `K_c` would be:

`K_c = (x(color(red)(3)x)^color(red)(3))/(2.0 - color(red)(2)x)^color(red)(2)`

`= (27x^4)/(2.0 - 2x)^2`

But like mentioned, we know `x`. This is because

`2.0 - 2x = "1.0 M"`.

Therefore:

`x = "0.5 M"`.

As a result,

`color(blue)(K_c) = (27(0.5)^4)/(2.0 - 2(0.5))^2`

`= color(blue)(1.69)`