Question 7316a

Sep 27, 2015

["NO"] = 1.9 * 10^(-36)"mol/cm"""^3

Explanation:

The way I see it, you can use a sample volume to figure out the partial pressures of nitrogen gas and oxygen gas.

You can then use these partial pressures and the equilibrium constant ${K}_{p}$ to get the partial pressure of nitric oxide.

Finally, you could use the ideal gas law to find the number of moles of nitric oxide you get at that pressure, temperature, and volume.

At $\text{298 K}$, you have ${K}_{p} = 4.10 \cdot {10}^{- 31}$.

Right from the start, the very,very, very small value of ${K}_{p}$ tells you that the volume of nitric oxide will be insignificant, since the equilibrium lies far to the left.

For ideal gases, you know that the percent each gas has in the total volume is proportional to the percent the partial pressure of each gas has in the total pressure.

This means that you have

${P}_{{N}_{2}} = {V}_{{N}_{2}} / {V}_{\text{total" xx P_"total" " }}$ and $\text{ " "P"_(O_2) = V_(O_2)/V_"total" xx P_"total}$

Let's take a sample of ${\text{100 cm}}^{3}$ of air. The partial pressures of nitrogen gas and oxygen gas will thus be

P_(O_2) = (21color(red)(cancel(color(black)("cm"""^3))))/(100color(red)(cancel(color(black)("cm"""^3)))) * "1.0 atm" = "0.21 atm"

Similarly, ${P}_{{N}_{2}} = \text{0.78 atm}$.

Now you can use an ICE Table with the partial pressures of the gases

${\text{N"_text(2(g]) " "+" " "O"_text(2(g]) " "rightleftharpoons" " color(red)(2)"NO}}_{\textrm{\left(g\right]}}$

color(purple)("I")" " " " "0.78" " " " " " " " 0.21" " " " " " " " "0
color(purple)("C")" " ""(-x)" " " " " "(-x)" " " " " " (+color(red)(2)x)
color(purple)("E")" "(0.78-x)" "(0.21-x)" " " " "(color(red)(2)x)

By definition, ${K}_{p}$ will be

${K}_{p} = \left(\left({\text{NO")^color(red)(2))/(("N"_2) * ("O}}_{2}\right)\right) = {\left(\textcolor{red}{2} x\right)}^{2} / \left(\left(0.78 - x\right) \cdot \left(0.21 - x\right)\right)$

${K}_{p} = \frac{4 {x}^{2}}{\left(0.78 - x\right) \left(0.21 - x\right)} = 4.10 \cdot {10}^{- 31}$

Since ${K}_{p}$ is so small, you can approximate $\left(0.78 - x\right)$ and $\left(0.21 - x\right)$ with $0.78$ and $0.21$, respectively.

This means that you have

${K}_{p} = \frac{4 {x}^{2}}{0.78 \cdot 0.21} = 4.10 \cdot {10}^{- 31}$

This means that $x$ will be equal to

$x = \sqrt{\frac{0.78 \cdot 0.21 \cdot 4.10 \cdot {10}^{- 31}}{4}} = 1.3 \cdot {10}^{- 16}$

The equilibrium partial pressure of nitric oxide will thus be

${P}_{N O} = 2 \cdot x = 2.6 \cdot {10}^{- 16} \text{atm}$

The volume of nitric gas in the sample will be

${V}_{N O} = {P}_{N O} / {P}_{\text{total" xx V_"total}}$

V_(NO) = (2.16 * 10^(-16)color(red)(cancel(color(black)("atm"))))/(1.0color(red)(cancel(color(black)("atm")))) * "100 cm"""^3 = 2.16 * 10^(-14)"cm"""^3

Now you can (finally) use the ideal gas law equation to get the number of moles of nitric acid. Do not forget to convert the volumes from ${\text{cm}}^{3}$ to liters by using the conversion factor

$\text{1 cm"""^3 = 10^(-3)"L}$

$P V = n R T \implies n = \frac{P V}{R T}$

n_(NO) = (2.16 * 10^(-16)color(red)(cancel(color(black)("atm"))) * 2.16 * 10^(-14) * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 298color(red)(cancel(color(black)("K")))) = 1.91 * 10^(-34)"moles NO"

The concentration of nitric oxide will thus be

$C = \frac{n}{V} = \left(1.91 \cdot {10}^{- 34} {\text{moles")/("100 cm"""^3) = color(green)(1.9 * 10^(-36)"mol/cm}}^{3}\right)$

To the same for nitrogen gas and oxygen gas to get their concentrations.

n_(N_2) =(0.78color(red)(cancel(color(black)("atm"))) * 0.78 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 298color(red)(cancel(color(black)("K")))) = 2.5 * 10^(-5) "moles N"""_2

The concentration will thus be

${C}_{{N}_{2}} = \left(2.49 \cdot {10}^{- 5} {\text{moles")/("100 cm"""^3) = color(green)(2.5 * 10^(-7)"mol/cm}}^{3}\right)$

and

n_(O_2) =(0.21color(red)(cancel(color(black)("atm"))) * 0.21 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 298color(red)(cancel(color(black)("K")))) = 1.8 * 10^(-6) "moles O"""_2#

The concentration will thus be

$C = \left(1.8 \cdot {10}^{- 6} {\text{moles")/("100 cm"""^3) = color(green)(1.80 * 10^(-8)"mol/cm}}^{3}\right)$