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??

1 Answer
Jul 17, 2018

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)