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 . (It's no coincidence that you see $R$, $T$, and $n$ in the same equation.)

$K_p(2) ~~ 122$

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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 .
• 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:

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

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

$~~ color(blue)(122)$