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Kp calculation help :3

Hi, im stuck on this question and not sure how to answer it. Any ideas? The attached image is what information youre given in the previous question. The question asks you to:

Find the partial pressure of x and y in the equilibrium mixture. I thought to do

22 = 6^2 / (y^3 x)
Y^3 x = 6^2 / 22
?
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Reply 1
Partial pressure = mole fraction * total pressure

Mole fraction if Y is 3 * mole fraction of X

So partial pressure of Y is 3 * partial pressure of X

So you have

p(X) + p(Y) + p(Z) = 22
p(X) + 3p(X) = 16
p(X) = 16/4 = 4.0 MPa

Now find p(Y)

p(Y) = 16-4 = 12.0 MPa
Reply 2
Original post by asinghj
Partial pressure = mole fraction * total pressure

Mole fraction if Y is 3 * mole fraction of X

So partial pressure of Y is 3 * partial pressure of X

So you have

p(X) + p(Y) + p(Z) = 22
p(X) + 3p(X) = 16
p(X) = 16/4 = 4.0 MPa

Now find p(Y)

p(Y) = 16-4 = 12.0 MPa


Sorry im confused wouldnt z be to the power of 2 and y to the power of 3?
Reply 3
Original post by kiiten
Sorry im confused wouldnt z be to the power of 2 and y to the power of 3?


If you have to calculate Kp then yes. But they aren't to any power here because the sum of individual partial pressures = total pressure. And that implies that p(Y) + p(X) + p(Z) = P(total)
Reply 4
Original post by asinghj
If you have to calculate Kp then yes. But they aren't to any power here because the sum of individual partial pressures = total pressure. And that implies that p(Y) + p(X) + p(Z) = P(total)


Yes, thank you i think i misunderstood but i understand now :biggrin:
Reply 5
Original post by asinghj
If you have to calculate Kp then yes. But they aren't to any power here because the sum of individual partial pressures = total pressure. And that implies that p(Y) + p(X) + p(Z) = P(total)


Hey if you dont mind please could you (or anyone else feel free to answer) help me on this calculation question. Ive attached the question (part d.) and my working (what i got right). Thanks :smile:

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Reply 6
Do you happen to know the final answer? Because im not completly sure, but i think it would be 35.5% to 3sf
Reply 7
Original post by asinghj
Do you happen to know the final answer? Because im not completly sure, but i think it would be 35.5% to 3sf


Yeah thats right but how did you get there?

- completely different question: when you make aspirin why is it cooled in ice? To allow crystals to form and what else??
(edited 7 years ago)
Reply 8
Original post by kiiten
Yeah thats right but how did you get there?

- completely different question: when you make aspirin why is it cooled in ice? To allow crystals to form and what else??


Ok, so what I did was to write the equation of the reaction happening:

CH3(CH2)14COOH + NaOH --> CH3(CH2)14COONa + H2O

You know the number of moles of NaOH, and the ratio of carboxylic acid to NaOH is 1:1 so number of moles of carboxylic acid = number of moles of NaOH

Now the ratio of carboxylic acid to fat is 1:3 so you divide the number of moles of carboxylic acid by 3 and then divide that by number of moles of fat that you worked out and then times by 100.


As for the other question do you cool aspirin in ice? Wow didn't know that. 😂😂😂
Original post by kiiten


- completely different question: when you make aspirin why is it cooled in ice? To allow crystals to form and what else??


That's it, to allow crystals to form.
I literally cannot think of anything else hat could come from lowering temperature, since i'm currently doing this practical.
Reply 10
Original post by asinghj
Ok, so what I did was to write the equation of the reaction happening:

CH3(CH2)14COOH + NaOH --> CH3(CH2)14COONa + H2O

You know the number of moles of NaOH, and the ratio of carboxylic acid to NaOH is 1:1 so number of moles of carboxylic acid = number of moles of NaOH

Now the ratio of carboxylic acid to fat is 1:3 so you divide the number of moles of carboxylic acid by 3 and then divide that by number of moles of fat that you worked out and then times by 100.


As for the other question do you cool aspirin in ice? Wow didn't know that. 😂😂😂



Thanks :biggrin: i kinda get what youre on about but could i have some numbers to clarify please?

Haha looll - yeah thats how you prepare aspirin

Original post by will'o'wisp
That's it, to allow crystals to form.
I literally cannot think of anything else hat could come from lowering temperature, since i'm currently doing this practical.


Yeah ive already done the prac - this was an exam ques but my teacher put an arrow (idk what its called) as if to say something is missing :s-smilie:
Original post by kiiten



Yeah ive already done the prac - this was an exam ques but my teacher put an arrow (idk what its called) as if to say something is missing :s-smilie:


Maybe, you missed a key word?
Reply 12
Original post by will'o'wisp
Maybe, you missed a key word?


Yeah but i cant find the mark scheme D: - i feel awkward asking my teacher because i just spoke to her this morning about a question.... :s-smilie:

asdfgdfg - ill just email her. I wont see her for a few days so it should be fine :smile:
Original post by kiiten
Yeah but i cant find the mark scheme D: - i feel awkward asking my teacher because i just spoke to her this morning about a question.... :s-smilie:

asdfgdfg - ill just email her. I wont see her for a few days so it should be fine :smile:


Develop the "i dun give nu ****s what anyone else thinks" attitude, you can ten ask away as many questions as you want without fear of embarrassment or others looking at you for asking such basic questions, i have it but haven't mastered it....
Reply 14
Original post by will'o'wisp
Develop the "i dun give nu ****s what anyone else thinks" attitude, you can ten ask away as many questions as you want without fear of embarrassment or others looking at you for asking such basic questions, i have it but haven't mastered it....


:yes::clap2: I will work towards it! :biggrin:
Original post by kiiten
Thanks :biggrin: i kinda get what youre on about but could i have some numbers to clarify please?

Haha looll - yeah thats how you prepare aspirin


So number of moles of NaOH = 24.5*0.15/1000 = was 3.675*10^-3 (as you correctly calculated)

Number of moles of fat in the sample = 3.675*10^-3/3 = 1.225*10^-3

Also number of moles of fat = 3.4491...*10^-3 (as you calculated)

% = (1.225*10^-3 / 3.4491...*10^-3) * 100
% = 35.5% (3sf)

And about aspirin, thanks for the heads up, I'm going to be doing that practical later on, maybe in January/February time
Reply 16
Original post by asinghj
So number of moles of NaOH = 24.5*0.15/1000 = was 3.675*10^-3 (as you correctly calculated)

Number of moles of fat in the sample = 3.675*10^-3/3 = 1.225*10^-3

Also number of moles of fat = 3.4491...*10^-3 (as you calculated)

% = (1.225*10^-3 / 3.4491...*10^-3) * 100
% = 35.5% (3sf)

And about aspirin, thanks for the heads up, I'm going to be doing that practical later on, maybe in January/February time


Yeah i got that - just wanted to clarify the moles of NaOH and fat bc theyre quite similar (in case i mixed them up).

Np :smile:, if youre unsure of anything you can ask me about the prac - not that youll need help lol

Thanks for your help :biggrin:
Original post by kiiten
Yeah i got that - just wanted to clarify the moles of NaOH and fat bc theyre quite similar (in case i mixed them up).

Np :smile:, if youre unsure of anything you can ask me about the prac - not that youll need help lol

Thanks for your help :biggrin:


We're not starting organic chem until January at least so I'll be sure to tag you when I get to that stage

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