# 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?

## $2 {\text{NO"_2"(g)"\rightleftharpoons2"NO (g)"+"O}}_{2}$ $\text{NO (g)"+1/2"O"_2"(g)"\rightleftharpoons"NO"_2"(g)}$ I'm assuming this needs use of formula ${K}_{c} = {K}_{p} {\left(R T\right)}^{\setminus \Delta n}$, so this is what I have so far: ${K}_{p} = \left(1.8 \times {10}^{-} 6\right) {\left(0.0821 \left(184 + 273\right)\right)}^{\setminus \Delta n}$ etc. But what is $\setminus \Delta n$??

Jul 17, 2018

Well, if you notice, are any of these substances NOT gases? The $\Delta n$ 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.)

${K}_{p} \left(2\right) \approx 122$

You assume that ${K}_{c} = 1.8 \times {10}^{- 6}$ for

$2 {\text{NO"_2(g) rightleftharpoons 2"NO"(g) + "O}}_{2} \left(g\right)$

and you want ${K}_{p}$ for

${\text{NO"(g) + 1/2"O"_2(g) rightleftharpoons "NO}}_{2} \left(g\right)$

Well, you first need the ${K}_{p}$ for the first reaction... and as we know,

${K}_{p} = {K}_{c} {\left(R T\right)}^{\Delta {n}_{\text{gas}}}$

But where does this come from? It uses the ideal gas law to rewrite ${K}_{c}$ in terms of $\text{mol/L}$ into ${K}_{p}$ in terms of $\text{atm}$ (hence the usage of $R = \text{0.0821 L"cdot"atm/mol"cdot"K}$).

$\left[A\right] = {n}_{A} / {V}_{A} = {P}_{A} / \left(R T\right)$

Hence, if

${K}_{c} = \frac{{\left[C\right]}^{c} {\left[D\right]}^{d}}{{\left[A\right]}^{a} {\left[B\right]}^{b}}$,

then since ${P}_{i} = \frac{{n}_{i} R T}{V} _ i$,

$\textcolor{g r e e n}{{K}_{p}} = \frac{{P}_{C}^{c} {P}_{D}^{d}}{{P}_{A}^{a} {P}_{B}^{b}} = \frac{{\left(\frac{{n}_{C} R T}{V} _ C\right)}^{c} {\left(\frac{{n}_{D} R T}{V} _ D\right)}^{d}}{{\left(\frac{{n}_{A} R T}{V} _ A\right)}^{a} {\left(\frac{{n}_{B} R T}{V} _ B\right)}^{b}}$

$= \frac{{\left(\frac{{n}_{C}}{V} _ C\right)}^{c} {\left(\frac{{n}_{D}}{V} _ D\right)}^{d}}{{\left(\frac{{n}_{A}}{V} _ A\right)}^{a} {\left(\frac{{n}_{B}}{V} _ B\right)}^{b}} {\left(R T\right)}^{d + c - \left(a + b\right)}$

$= \frac{{\left[C\right]}^{c} {\left[D\right]}^{d}}{{\left[A\right]}^{a} {\left[B\right]}^{b}} {\left(R T\right)}^{{n}_{\text{products" - n_"reactants}}}$

$= \textcolor{g r e e n}{{K}_{c} {\left(R T\right)}^{\Delta {n}_{\text{gas}}}}$

So $\Delta n$ is just the mols of products minus reactants.

${K}_{p} \left(1\right) = {K}_{c} {\left(R T\right)}^{\Delta {n}_{\text{gas}}}$

= (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)

$= 6.76 \times {10}^{- 5}$ in implied units of $\text{atm}$.

But are we done? I hope not. This is reaction $\left(1\right)$, but we want reaction $\left(2\right)$. 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

$A + B r i g h t \le f t h a r p \infty n s C + D$,

then for an arbitrary constant $q$,

$q C + q D r i g h t \le f t h a r p \infty n s q A + q B$,

we have the new equilibrium constant:

$K ' = {\left(\frac{1}{K}\right)}^{q} = {K}^{- q} = {5}^{- q}$

In your case, you just have $q = \frac{1}{2}$, so:

$\textcolor{b l u e}{{K}_{p} \left(2\right)} = {\left(\frac{1}{{K}_{p} \left(1\right)}\right)}^{1 / 2} = {\left\{{K}_{p} \left(1\right)\right\}}^{- 1 / 2}$

$= {\left(6.76 \times {10}^{- 5}\right)}^{- 1 / 2}$

$\approx \textcolor{b l u e}{122}$