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?

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)#