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?
#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)#