How Much N2O4 Dissociates Upon Vaporization?

In summary: Hi Greg. I'm pretty sure that the products of N2O4 dissociation are 2 NO2, and I had the sense that the OP implied that he was told assume this. But, you're right about what you are saying. If there's any O2 formed, this changes the total number of moles formed.ChetIn summary, the problem involves the partial dissociation of nitrogen tetroxide into nitrogen oxide at ordinary temperature. Using the ideal gas law, the total number of moles of gas in the flask at 27 C (300K) was calculated to be 0.0149 moles. It was determined that 0.00555 moles of nitrogen tetrox
  • #1
SPhy
25
0
I managed to get the correct answer, however I don't know if my logic was sound.

Homework Statement



At ordinary temperature nitrogen tetroxide is partially dissociated (broken up) into nitrogen oxide.

Into an evacuated flask of 250 cm^3 volume, 0.86g of liquid N2O4 at 0 C is introduced. When the temperature in the bulb has risen to 27 C the liquid has all vaporized and the pressure is 1120mm of mercury. What percent of the nitrogen tetroxide has dissociated in this process?

Homework Equations



PV=nRT

The Attempt at a Solution


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First thing, I decided to convert .86 grams of N2O4 to moles. Found it to be 0.00935 moles

Then, using PV=nRT, determined the total number of "gas" moles in the flask at 27 C (300K). Found this to be 0.0149 moles.

0.0149 - 0.00935 = 0.00555 moles added ?
(0.00555/0.00935) x 100% = 59.3%

Initially this all made sense to me, but now I go back and read the question, it says all my liquid was vaporized (consequently expanded into the flask). So shouldn't my %dissociation be 100%?

In addition, if my method was correct, what does the 0.00555 actually mean? Is that the amount of N2O4 that didn't/did turn into gas?

Any help appreciated.
 
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  • #2
Presumably there will be an equilibrium of N2O4 as a gas and NO as a gas.

Edit: @SPhy : On second thought, there must be more species in the flask than just those two alone. What are the possible products when N2O4 dissociates? (equations must balance).
 
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  • #3
gneill said:
Presumably there will be an equilibrium of N2O4 as a gas and NO as a gas.

Edit: @SPhy : On second thought, there must be more species in the flask than just those two alone. What are the possible products when N2O4 dissociates? (equations must balance).

For the equation to balance we must require 2NO2 .

So the number of moles of gas that occupy the flask contain both gases? Thus the .00555 value is actually the disassociated amount?
 
  • #4
The problem statement indicates that one of the products is NO. So, how many NO's can you get from one N2O4. What's left over?
 
  • #5
If it dissociates to 2NO2 molecules, then when a molecule of N2O4 dissociates, one molecule disappears and two new molecules are formed, so the net is one extra molecule for every molecule of N2O4 that dissociates. So the 0.00555 moles represents the number of N2O4 moles that dissociated (out of the original 0.00935 moles). In the end, the container holds 0.0038 moles of N2O4 and 0.0111 moles of NO2.

Chet
 
  • #6
Chestermiller said:
If it dissociates to 2NO2 molecules, then when a molecule of N2O4 dissociates, one molecule disappears and two new molecules are formed, so the net is one extra molecule for every molecule of N2O4 that dissociates. So the 0.00555 moles represents the number of N2O4 moles that dissociated (out of the original 0.00935 moles). In the end, the container holds 0.0038 moles of N2O4 and 0.0111 moles of NO2.

Chet

Ahhh! Got it, thanks!
 
  • #7
Chestermiller said:
If it dissociates to 2NO2 molecules, then when a molecule of N2O4 dissociates, one molecule disappears and two new molecules are formed, so the net is one extra molecule for every molecule of N2O4 that dissociates. So the 0.00555 moles represents the number of N2O4 moles that dissociated (out of the original 0.00935 moles). In the end, the container holds 0.0038 moles of N2O4 and 0.0111 moles of NO2.

Chet
Hi Chet. The difficulty I see with the problem statement is that they specifically name Nitrogen Oxide (NO) as a product. That leaves one to wonder what happens to the extra oxygens. At room temp they might remain as individual O's or combine to O2.
 
  • #8
gneill said:
Hi Chet. The difficulty I see with the problem statement is that they specifically name Nitrogen Oxide (NO) as a product. That leaves one to wonder what happens to the extra oxygens. At room temp they might remain as individual O's or combine to O2.
Hi Greg. I'm pretty sure that the products of N2O4 dissociation are 2 NO2, and I had the sense that the OP implied that he was told assume this. But, you're right about what you are saying. If there's any O2 formed, this changes the total number of moles formed.

Chet
 

FAQ: How Much N2O4 Dissociates Upon Vaporization?

What is the ideal gas law?

The ideal gas law, also known as the Pv=nRT law, is a fundamental equation in thermodynamics that describes the relationship between pressure (P), volume (V), amount of substance (n), and temperature (T) of an ideal gas. It is expressed as PV = nRT, where R is the universal gas constant.

What does each variable in the ideal gas law represent?

P represents pressure, V represents volume, n represents the amount of substance (in moles), R represents the universal gas constant, and T represents temperature (in Kelvin).

What is the significance of the ideal gas law in thermodynamics?

The ideal gas law is important in thermodynamics because it allows us to predict the behavior of ideal gases under different conditions. It also helps us understand the relationships between different thermodynamic properties, such as pressure, volume, and temperature.

What are the assumptions of the ideal gas law?

The ideal gas law is based on several assumptions, including that the gas particles are infinitely small and have no intermolecular forces, that they are in constant random motion, and that they do not interact with each other or their container. These assumptions are only valid for ideal gases at low pressures and high temperatures.

Can the ideal gas law be applied to real gases?

The ideal gas law is a simplification of the behavior of real gases, which do not always behave ideally. However, at low pressures and high temperatures, real gases often behave similarly to ideal gases and the ideal gas law can be applied with reasonable accuracy. At higher pressures and lower temperatures, the equations of state for real gases must be used instead.

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