At 430^@"C", an equilibrium mixture consists of 0.020 moles of "O"_2, 0.040 moles of "NO", and 0.96 moles of "NO"_2. Calculate Kp for the reaction, given the total pressure is "0.20 atm" ?

$2 {\text{NO"(g) + "O"_2(g) rightleftharpoons 2"NO}}_{2} \left(g\right)$. I know that the answer is $1.5 \cdot {10}^{5}$ atm with significant figures, but I am not sure how to get this. Help is much appreciated!

May 22, 2017

$1.5 \cdot {10}^{5} {\text{atm}}^{- 1}$

Explanation:

You know that

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

At ${430}^{\circ} \text{C}$, the reaction vessel contains

• ${\text{0.020 moles O}}_{2}$
• $\text{0.040 moles NO}$
• ${\text{0.96 moles NO}}_{2}$

Even without doing any calculations, you should be able to predict that

${K}_{p} \text{ >> } 1$

because the equilibrium mixture contains significantly more product than reactants, which implies that the equilibrium lies to the right, i.e. the forward reaction is favored at this temperature.

Now, at equilibrium, the total pressure inside the reaction vessel is equal to $\text{0.20 atm}$.

As you know, the partial pressures of the three gases will depend on their respective mole ratios, as described by Dalton's Law of Partial Pressures.

For nitrogen oxide, you will have

${P}_{\text{NO" = chi_ "NO" * P_"total}}$

Here

chi_ "NO" = (0.040 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) = 0.040/1.02

represents the mole fraction of nitrogen oxide in the reaction vessel at equilibrium.

Therefore,

${P}_{\text{NO" = 0.040/1.02 * "0.20 atm}}$

Use the same approach to calculate the partial pressure of oxygen gas and nitrogen dioxide. For oxygen gas, you will have

P_ ("O"_ 2) = (0.020 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) * "0.20 atm"

P_ ("O"_ 2)= 0.020/1.02 * "0.20 atm"

Similarly, for nitrogen dioxide, you will have

P_ ("NO"_ 2) = (0.96 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) * "0.20 atm"

P_ ("NO"_ 2) = 0.96/1.02 * "0.20 atm"

The equilibrium constant for this reaction can be written using the partial pressures of the three gases

${K}_{p} = \left({P}_{{\text{NO"_ 2)^2)/( P_ "NO"^2 * P_ ("O}}_{2}}\right)$

Plug in your values to find

K_p = ( (0.96/1.02 * 0.20)^2 color(red)(cancel(color(black)("atm"^2))))/( (0.040/1.20 * 0.20)^2 color(red)(cancel(color(black)("atm"^2))) * 0.020/1.02 * "0.20 atm")

${K}_{p} = \frac{{0.96}^{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(\frac{0.20}{1.02}\right)}^{2}}}} \cdot 1.02}{{0.040}^{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(\frac{0.20}{1.02}\right)}^{2}}}} \cdot 0.020 \cdot \text{0.20 atm}}$

K_p = (0.96^2 * 1.02)/(0.040^2 * 0.020 * "0.20 atm") = color(darkgreen)(ul(color(black)(1.5 * 10^5color(white)(.)"atm"^(-1))))

The answer is rounded to two sig figs.