Question

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?

Answer 1

`V = "5 L"`

Explanation:

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

`"N"_ 2"O"_ (4(g)) rightleftharpoons 2"NO"_ (2(g))`

Now, you know that at a temperature of `"340 K"` and a pressure of `"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

`"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.