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?

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.