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# Chemical Equilbria Watch

1. I've been trying to solve this one problem for hours and I have no idea how to go about it so I was wondering if anyone had any ideas on how it could be solved

Question: On warming, N2O4(g) dissociates into NO2(g).

Measurement of the molar mass of an equilibrium mixture at a certain temperature gave a value of 80.2 gmol-1. Calculate the degree of dissociation at this temperature.

Any suggestions would be incredibly helpful
2. For a gas apparent molar mass is the weighted average of the real molar masses of all gases present. For example "molar mass" of air is around 29 g/mol.

Does it help?
3. (Original post by Borek)
For a gas apparent molar mass is the weighted average of the real molar masses of all gases present. For example "molar mass" of air is around 29 g/mol.

Does it help?
Still slightly baffled unfortunately
4. (Original post by oscartales)
Still slightly baffled unfortunately
Are you familiar with calculating atomic masses from two isotopes? Say 35Cl and 37Cl to get the Ar of 35.5? It's very similar to that.
5. (Original post by EierVonSatan)
Are you familiar with calculating atomic masses from two isotopes? Say 35Cl and 37Cl to get the Ar of 35.5? It's very similar to that.
Oh no I'm quite all right with that step It's what comes after that's making me confused because the question doesn't give you a value for Kp so I'm not sure how it would be possible to find the degree of dissociation
6. (Original post by oscartales)
Oh no I'm quite all right with that step It's what comes after that's making me confused because the question doesn't give you a value for Kp so I'm not sure how it would be possible to find the degree of dissociation
Then you've done the hard work, you've calculated the percentage of each species present So you know how much is dissociated i.e. the degree of dissociation.
7. (Original post by EierVonSatan)
Then you've done the hard work, you've calculated the percentage of each species present So you know how much is dissociated i.e. the degree of dissociation.
Hmm I keep getting the wrong answer in the end according to my answer book Would it be possible to guide me through the first part if it's possible?
8. (Original post by oscartales)
Hmm I keep getting the wrong answer in the end according to my answer book Would it be possible to guide me through the first part if it's possible?
Better if you go through what you've got and what the book's answer is? This is so we can see where you might be going wrong, instead
9. (Original post by EierVonSatan)
Better if you go through what you've got and what the book's answer is? This is so we can see where you might be going wrong, instead
Sure

Judging from the question, I assumed it would require us to find the molar mass present for each at equilibrium from the molar mass of the mixture (which was given - 80.2 gmol-1).

Therefore 80.2 gmol-1 would be ntotal. ntotal = nN2O4 + 2nNO2.

Let the moles of N2O4 at equilbrium be: 1-x and let the mass of NO2 be 2x.

I'm actually getting to this part and I'm realising that I don't know the mass of the compound so I have no idea how I got to an answer in the first place haha

I understand that if you use the total molar mass of the mixture then for the other components you must use: mass/no of. moles but everything's confusing me now that I'm writing it out

The answer given is 0.147 (for the degree of dissociation)
10. (Original post by oscartales)
Sure

Judging from the question, I assumed it would require us to find the molar mass present for each at equilibrium from the molar mass of the mixture (which was given - 80.2 gmol-1).

Therefore 80.2 gmol-1 would be ntotal. ntotal = nN2O4 + 2nNO2.

Let the moles of N2O4 at equilbrium be: 1-x and let the mass of NO2 be 2x.

I'm actually getting to this part and I'm realising that I don't know the mass of the compound so I have no idea how I got to an answer in the first place haha

I understand that if you use the total molar mass of the mixture then for the other components you must use: mass/no of. moles but everything's confusing me now that I'm writing it out
Not to worry Before dissociation we have a number of N2O4 molecules floating around - call this N. Some if it dissociates into NO2 and the proportions of each are given by 80.2 = xN2O4 + yNO2, since x and y must equal 1 we can work these out:

80.2 = xN2O4 + (1-x)NO2
80.2 = 96x + (1-x)46x
some algebra later...
x = 0.744... and y = 0.256...

or the equivalent percentages

Now we just have to relate x and y back to N, using the stoichiometry of the reaction (2:1): N = x + 0.5y = 0.872

Finally the amount that is dissociated is number of molecules dissociated divided by the total starting amount = 0.5y/N which equals...

The answer given is 0.147 (for the degree of dissociation)
11. (Original post by EierVonSatan)
Not to worry Before dissociation we have a number of N2O4 molecules floating around - call this N. Some if it dissociates into NO2 and the proportions of each are given by 80.2 = xN2O4 + yNO2, since x and y must equal 1 we can work these out:

80.2 = xN2O4 + (1-x)NO2
80.2 = 96x + (1-x)46x
some algebra later...
x = 0.744... and y = 0.256...

or the equivalent percentages

Now we just have to relate x and y back to N, using the stoichiometry of the reaction (2:1): N = x + 0.5y = 0.872

Finally the amount that is dissociated is number of molecules dissociated divided by the total starting amount = 0.5y/N which equals...

Ah thank you so much for your help and sorry for the hassle But it's definitely much more clear now so thank you
12. (Original post by oscartales)
Ah thank you so much for your help and sorry for the hassle But it's definitely much more clear now so thank you
Welcome, and no hassle at all Interesting question

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