Question

For the reaction below, `K_c=1.8xx10^-6` at 184°C. What is the value of `K_p` for the following reaction (also below) at 184°C?

1. `2"NO"_2"(g)"\rightleftharpoons2"NO (g)"+"O"_2`
2. `"NO (g)"+1/2"O"_2"(g)"\rightleftharpoons"NO"_2"(g)"`

I'm assuming this needs use of formula `K_c=K_p(RT)^(\Deltan)`, so this is what I have so far:

`K_p=(1.8xx10^-6)(0.0821(184+273))^(\Deltan)` etc.
But what is `\Deltan`??

Answer 1

Well, if you notice, are any of these substances NOT gases? The `Deltan` can only be referring to gases, which should remind you of the ideal gas law. (It's no coincidence that you see `R`, `T`, and `n` in the same equation.)

`K_p(2) ~~ 122`

---

You assume that `K_c = 1.8 xx 10^(-6)` for

`2"NO"_2(g) rightleftharpoons 2"NO"(g) + "O"_2(g)`

and you want `K_p` for

`"NO"(g) + 1/2"O"_2(g) rightleftharpoons "NO"_2(g)`

Well, you first need the `K_p` for the first reaction... and as we know,

`K_p = K_c(RT)^(Deltan_"gas")`

But where does this come from? It uses the ideal gas law to rewrite `K_c` in terms of `"mol/L"` into `K_p` in terms of `"atm"` (hence the usage of `R = "0.0821 L"cdot"atm/mol"cdot"K"`).

`[A] = n_A/V_A = P_A/(RT)`

Hence, if

`K_c = ([C]^c[D]^d)/([A]^a[B]^b)`,

then since `P_i = (n_iRT)/V_i`,

`color(green)(K_p) = (P_C^cP_D^d)/(P_A^aP_B^b) = (((n_CRT)/V_C)^c((n_DRT)/V_D)^d)/(((n_ART)/V_A)^a((n_BRT)/V_B)^b)`

`= (((n_C)/V_C)^c((n_D)/V_D)^d)/(((n_A)/V_A)^a((n_B)/V_B)^b)(RT)^(d+c-(a+b))`

`= ([C]^c[D]^d)/([A]^a[B]^b)(RT)^(n_"products" - n_"reactants")`

`= color(green)(K_c(RT)^(Deltan_"gas"))`

So `Deltan` is just the mols of products minus reactants.

`K_p (1) = K_c(RT)^(Deltan_"gas")`

`= (1.8 xx 10^(-6) (cancel(("mol/L")^2) cdot cancel"mol/L")/cancel(("mol/L")^2))(0.0821 cancel"L"cdot"atm/"cancel"mol"cdotcancel"K" cdot (184+273.15 cancel"K"))^(2+1 - 2)`

`= 6.76 xx 10^(-5)` in implied units of `"atm"`.

But are we done? I hope not. This is reaction `(1)`, but we want reaction `(2)`. Recall:

• Reversed reactions have flipped equilibrium constants.
• Scaling reactions by constant coefficients raises the entire equilibrium constant to that coefficient.

Hence, if `K = 5` for

`A + B rightleftharpoons C + D`,

then for an arbitrary constant `q`,

`qC + qD rightleftharpoons qA + qB`,

we have the new equilibrium constant:

`K' = (1/K)^(q) = K^(-q) = 5^(-q)`

In your case, you just have `q = 1/2`, so:

`color(blue)(K_p(2)) = (1/(K_p(1)))^(1//2) = {K_p(1)}^(-1//2)`

`= (6.76 xx 10^(-5))^(-1//2)`

`~~ color(blue)(122)`