At $340 K$ $"340 K"$ and $1 atm$ $"1 atm"$ , $N_{2} O_{4}$ $"N"_2"O"_4$ is $66 %$ $66%$ dissociated into ${NO}_{2}$ $"NO"_2$ . What volume would $10 g$ $"10 g"$ of $N_{2} O_{4}$ $"N"_2"O"_4$ occupy under these conditions?

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At $340 K$ $"340 K"$ and $1 atm$ $"1 atm"$ , $N_{2} O_{4}$ $"N"_2"O"_4$ is $66 %$ $66%$ dissociated into ${NO}_{2}$ $"NO"_2$ . What volume would $10 g$ $"10 g"$ of $N_{2} O_{4}$ $"N"_2"O"_4$ occupy under these conditions?

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Answer 1 · 2017-10-06T00:53:31

$V = 5 L$ $V = "5 L"$

Explanation:

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

$N_{2} O_{4 (g)} ⇌ 2 {NO}_{2 (g)}$ $"N"_ 2"O"_ (4(g)) rightleftharpoons 2"NO"_ (2(g))$

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

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

This means that for every $100$ $100$ 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

$"0.1435 moles " + " 0.03696 moles = 0.1805 moles"$

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

$color(blue)(ul(color(black)(PV = nRT)))$

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 $0.0821("atm L")/("mol K")$

$T$ is the absolute temperature of the gas

Rearrange to solve for $V$

$PV = nRT implies V = (nRT)/P$

Plug in your values to find

$V = (0.1805 color(red)(cancel(color(black)("moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 340color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm"))))$

$V = color(darkgreen)(ul(color(black)("5 L")))$

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.

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At $340 K$ $"340 K"$ and $1 atm$ $"1 atm"$ , $N_{2} O_{4}$ $"N"_2"O"_4$ is $66 %$ $66%$ dissociated into ${NO}_{2}$ $"NO"_2$ . What volume would $10 g$ $"10 g"$ of $N_{2} O_{4}$ $"N"_2"O"_4$ occupy under these conditions?

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Explanation:

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At "340 K"340 K"340 K" and "1 atm"1 atm"1 atm", "N"_2"O"_4N2O4"N"_2"O"_4 is 66%66%66% dissociated into "NO"_2NO2"NO"_2. What volume would "10 g"10 g"10 g" of "N"_2"O"_4N2O4"N"_2"O"_4 occupy under these conditions?

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At $340 K$ $"340 K"$ and $1 atm$ $"1 atm"$ , $N_{2} O_{4}$ $"N"_2"O"_4$ is $66 %$ $66%$ dissociated into ${NO}_{2}$ $"NO"_2$ . What volume would $10 g$ $"10 g"$ of $N_{2} O_{4}$ $"N"_2"O"_4$ occupy under these conditions?