At "340 K" and "1 atm", "N"_2"O"_4 is 66% dissociated into "NO"_2. What volume would "10 g" of "N"_2"O"_4 occupy under these conditions?

Oct 6, 2017

$V = \text{5 L}$

Explanation:

Start by writing the balanced chemical equation that describes this dissociation equilibrium

${\text{N"_ 2"O"_ (4(g)) rightleftharpoons 2"NO}}_{2 \left(g\right)}$

Now, you know that at a temperature of $\text{340 K}$ and a pressure of $\text{1 atm}$, 66% of the dinitrogen tetroxide dissociates to produce nitrogen dioxide.

This means that for every $100$ moles of dinitrogen tetroxide present in the sample, $66$ moles will dissociate to produce nitrogen dioxide. Notice that for every mole of dinitrogen tetroxide that dissociates, the reaction produces $2$ moles of nitrogen dioxide.

This means that for every $100$ moles of dinitrogen tetroxide present in the sample, the reaction will produce

overbrace(66 color(red)(cancel(color(black)("moles N"_2"O"_4))))^(color(blue)("what dissociates from 100 moles, i.e. 66%")) * "2 moles NO"_2/(1color(red)(cancel(color(black)("mole N"_2"O"_4)))) = "132 moles NO"_2

Use the molar mass of dinitrogen tetroxide to calculate the number of moles present in your sample

10 color(red)(cancel(color(black)("g"))) * ("1 mole N"_2"O"_4)/(92.011 color(red)(cancel(color(black)("g")))) = "0.1087 moles N"_2"O"_4

This means that the reaction vessel will contain

0.1087 color(red)(cancel(color(black)("moles N"_2"O"_4))) * "132 moles NO"_2/(100color(red)(cancel(color(black)("moles N"_2"O"_4)))) = "0.1435 moles NO"_2

At equilibrium, the reaction vessel will contain $0.1435$ moles of nitrogen dioxide and

0.1087 color(red)(cancel(color(black)("moles N"_2"O"_4))) * overbrace(("34 moles N"_2"O"_4)/(100color(red)(cancel(color(black)("moles N"_2"O"_4)))))^(color(blue)("what does not dissociate, i.e. 34%")) = "0.03696 moles N"_2"O"_4

The total number of moles of gas present in the reaction vessel will be

$\text{0.1435 moles " + " 0.03696 moles = 0.1805 moles}$

Now all you have to do is to use the ideal gas law equation to find the volume of the gas mixture.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Rearrange to solve for $V$

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

Plug in your values to find

$V = \left(0.1805 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 340color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm}}}}\right)$

$V = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{5 L}}}}$

The answer must be rounded to one significant figure, the number of sig figs you have for the mass of the dinitrogen tetroxide and the pressure of the mixture.