# Question #7316a

##### 1 Answer

#### Explanation:

**!! EXTREMELY LONG ANSWER !!**

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*

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

Right from the start, the very,very, very small value of **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_"total" xx P_"total" " "# and#" " "P"_(O_2) = V_(O_2)/V_"total" xx P_"total"#

Let's take a sample of

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

Similarly,

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

#"N"_text(2(g]) " "+" " "O"_text(2(g]) " "rightleftharpoons" " color(red)(2)"NO"_text((g])#

By definition,

#K_p = (("NO")^color(red)(2))/(("N"_2) * ("O"_2)) = (color(red)(2)x)^2/((0.78-x) * (0.21 - x))#

#K_p = (4x^2)/((0.78-x)(0.21-x)) = 4.10 * 10^(-31)#

Since

This means that you have

#K_p = (4x^2)/(0.78 * 0.21) = 4.10 * 10^(-31)#

This means that

#x = sqrt((0.78 * 0.21 * 4.10 * 10^(-31))/4) = 1.3 * 10^(-16)#

The equilibrium partial pressure of nitric oxide will thus be

#P_(NO) = 2 * x = 2.6 * 10^(-16)"atm"#

The volume of nitric gas in the sample will be

#V_(NO) = P_(NO)/P_"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

#"1 cm"""^3 = 10^(-3)"L"#

#PV = nRT implies n = (PV)/(RT)#

#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 = n/V = (1.91 * 10^(-34)"moles")/("100 cm"""^3) = color(green)(1.9 * 10^(-36)"mol/cm"""^3)#

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) = (2.49 * 10^(-5)"moles")/("100 cm"""^3) = color(green)(2.5 * 10^(-7)"mol/cm"""^3)#

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 = (1.8 * 10^(-6) "moles")/("100 cm"""^3) = color(green)(1.80 * 10^(-8)"mol/cm"""^3)#