# 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?

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#2"NO"_2"(g)"\rightleftharpoons2"NO (g)"+"O"_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# ??

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

I'm assuming this needs use of formula

#K_p=(1.8xx10^-6)(0.0821(184+273))^(\Deltan)# etc.

But what is#\Deltan# ??

##### 1 Answer

Well, if you notice, are any of these substances NOT gases? The

#K_p(2) ~~ 122#

You assume that

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

and you want

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

Well, you first need the

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

But where does this come from? It uses the **ideal gas law** to rewrite

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

Hence, if

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

then since

#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

#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

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

Hence, if

#A + B rightleftharpoons C + D# ,

then for an arbitrary constant

#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

#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)#