# Partial Pressure Problem?

## Sodium azide, NaN3(s), reacts with chlorine gas according to the equation: 2 NaN3(s) + Cl2(g) → 2 NaCl(s) + 3 N2(g) If 50.0 g of NaN3(s) is placed in a 125.0 L tank pressurized with 4.00x102 torr of Cl2(g) at 25.0C, calculate the partial pressure of nitrogen and chlorine after the reaction goes to completion and calculate the total pressure in the tank. Assume that chlorine is the only gas is present in the tank before the start of the reaction, that the temperature remains constant during reaction, and that the volume of the solid materials may be neglected.

May 29, 2016

Here's what I got.

#### Explanation:

The first thing to do here is use the ideal gas law equation to find the number of moles of chlorine gas present in the reaction vessel before the reaction takes place

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

Make sure to convert the pressure and the temperature to the units used in the expression of the ideal gas constant!

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

${n}_{C {l}_{2}} = \left(\frac{4 \cdot {10}^{2}}{760} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 125.0 color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25.0)color(red)(cancel(color(black)("K}}}}\right)$

${n}_{C {l}_{2}} = {\text{2.688 moles Cl}}_{2}$

Use the molar mass of sodium azide to determine how many moles you have in that $\text{50.0-g}$ sample

50.0 color(red)(cancel(color(black)("g"))) * "1 mole NaN"_3/(65.01color(red)(cancel(color(black)("g")))) = "0.7691 moles NaN"_3

Now, notice that the reaction consumes $\textcolor{red}{2}$ moles of sodium azide and produces $\textcolor{b l u e}{3}$ moles of nitrogen gas for every mole of chlorine gas that takes part in the reaction.

$\textcolor{red}{2} {\text{NaN"_ (3(s)) + "Cl"_ (2(g)) -> 2"NaCl"_ ((s)) + color(blue)(3)"N}}_{2 \left(g\right)}$

This tells you that sodium azide will act as a limiting reagent, since that many moles will only allow for

0.7691 color(red)(cancel(color(black)("moles NaN"_3))) * "1 mole Cl"_2/(color(red)(2)color(red)(cancel(color(black)("moles NaN"_3)))) = "0.3846 moles Cl"_2

to take part in the reaction. The number of moles of chlorine gas that do not take part in the reaction will be equal to

n_("Cl"_2color(white)(a)"unreacted") = "2.688 moles" - "0.3846 moles"

n_("Cl"_2color(white)(a)"unreacted") = "2.3034 moles Cl"_2

Moreover, the reaction will produce

0.7691 color(red)(cancel(color(black)("moles NaN"_3))) * (color(blue)(3)color(white)(a)"moles N"_2)/(color(red)(2)color(red)(cancel(color(black)("moles NaN"_3)))) = "1.154 moles N"_2

The total number of moles of gas that are present in the vessel after the reaction is complete will be

n_"total" = overbrace("2.3034 moles")^(color(purple)("unreacted Cl"_2)) + overbrace("1.154 moles")^(color(brown)("produced N"_2))

${n}_{\text{total" = "3.4574 moles gas}}$

Now, when temperature and volume are kept constant, pressure and number of moles of gas have a direct relationship.

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

This means that you have

${P}_{1} / {n}_{1} = {P}_{\text{total"/n_"total}}$

Here

${P}_{1}$ - the initial pressure in the reaction vessel
${n}_{1}$ - the initial number of moles of chlorine gas
${P}_{\text{total}}$ - the pressure in the vessel after the reaction is complete

Rearrange to solve for the ${P}_{\text{total}}$

${P}_{\text{total" = n_"total}} / {n}_{1} \cdot {P}_{1}$

Plug in your values to find

${P}_{\text{total" = (3.4574 color(red)(cancel(color(black)("moles"))))/(2.688color(red)(cancel(color(black)("moles")))) * 4.0 * 10^2"torr}}$

P_"total" = color(green)(|bar(ul(color(white)(a/a)color(black)(5.14 * 10^2"torr")color(white)(a/a)|)))

The partial pressure of chlorine and of nitrogen gas can be found using Dalton's Law of Partial Pressures, which states that the partial pressure of a component $i$ of a gaseous mixture is proportional to its mole fraction in the mixture, ${\chi}_{i}$

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{i} = {\chi}_{i} \times {P}_{\text{total}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The mole fraction of chlorine will be given by the number of moles of chlorine that did not take part in the reaction and the total number of moles of all gaseous products present in the vessel

${\chi}_{C {l}_{2}} = \left(2.3034 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(3.4574color(red)(cancel(color(black)("moles}}}}\right) = 0.6662$

You will thus have

P_(Cl_2) = 0.6662 * 5.14 * 10^2"torr" = color(green)(|bar(ul(color(white)(a/a)color(black)(3.42 * 10^2"torr")color(white)(a/a)|)))

Similarly, you will have

${\chi}_{{N}_{2}} = \left(1.154 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(3.4574color(red)(cancel(color(black)("moles}}}}\right) = 0.3338$

and

P_(N_2) = 0.3338 * 5.14 * 10^2"torr" = color(green)(|bar(ul(color(white)(a/a)color(black)(1.72 * 10^2"torr")color(white)(a/a)|)))

Notice that

${P}_{{N}_{2}} = {P}_{\text{total}} - {P}_{C {l}_{2}}$

${P}_{{N}_{2}} = 5.14 \cdot {10}^{2} \text{torr" - 3.42 * 10^2"torr" = 1.72 * 10^2"torr}$

which makes sense because the partial pressures of the two gases must add up to give the total pressure in the vessel.

Furthermore, the mole fractions of ${N}_{2}$ and $C {l}_{2}$ also add up as $0.6662 + 0.3338 = 1.0000$, as they should.

All the answers are rounded to three sig figs.