How Much N2O4 Dissociates Upon Vaporization?

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Homework Help Overview

The discussion revolves around the dissociation of nitrogen tetroxide (N2O4) into nitrogen dioxide (NO2) upon vaporization in a controlled environment. The original poster presents a scenario involving the introduction of a specific mass of N2O4 into a flask, where temperature and pressure changes are observed, leading to questions about the extent of dissociation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the moles of N2O4 and the resulting gases, questioning the implications of the vaporization process and the products formed during dissociation. There is a focus on understanding the balance of the chemical equation and the interpretation of the calculated moles of gas.

Discussion Status

The discussion is active with participants examining different interpretations of the dissociation products and the implications of the calculations. Some guidance is offered regarding the relationship between the moles of N2O4 and the resulting gases, but there is no explicit consensus on the final interpretation of the dissociation products.

Contextual Notes

Participants note the potential confusion arising from the problem statement, particularly regarding the identification of products as nitrogen oxide (NO) versus nitrogen dioxide (NO2), which raises questions about the completeness of the dissociation reaction and the resulting species in the flask.

SPhy
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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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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).
 
Last edited:
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?
 
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?
 
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
 
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!
 
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.
 
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
 

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