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Claisen
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Chemical Kinetics - Good Problem
I need help in understanding a concept.
Consider a first order decomposition reaction
2P(g) -----> 4Q(g) + R(g) + S(l)
taking place at a temperature T. After 30 minutes from the start of the decomposition in a closed vessel, the total pressure developed is found to be
317 mmHg and after a long time, the total pressure observed was 617 mmHg. Calculate the toal pressure in the vessel after 75 minutes. (Given : V.P. of
S(l) at temperature T=32.5 mmHg)
Attempt:
Let the initial pressure of P be Pi.
At t=30 minutes, let 2x be the decrease in pressure of P.
There will be a pressure of 4x,x, ? due to Q,R,S respectively.
I have a problem in finding out the pressure of S. How is the total pressure in the vessel related to the state of S? i.e. How do I know whether S
will remain a liquid or change into gas? Someone suggested me that if the pressure due to S is say 'y', it will remain a liquid if y > 32.5mmHg.
I don't understand the concept behind it. Can somebody explain?
Thanks in advance.
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vulture
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The temperature is kept constant during the experiment at T, the starting pressure is 317mmHg and the end pressure is 617mmHg. This means that during
the whole of the experiment, S will be liquid, since its vapor pressure is lower than the pressure in the vessel.
If we assume that decomposition is complete, we can also deduce that and the end of the experiment the pressure of
Q + R = 617mmHg - 32.5mmHg.
Does that help?
[Edited on 6-1-2011 by vulture]
One shouldn't accept or resort to the mutilation of science to appease the mentally impaired.
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Magpie
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I worked this problem with a result of P(total)=379.5mmHg. Anyone else give it a try?
The single most important condition for a successful synthesis is good mixing - Nicodem
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Claisen
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@vulture
For the element S to remain as liquid, I think we need to consider its pressure during the reaction not the total pressure of the vessel. There will
be an increase in partial pressure due to S during the course of the reaction, and its state might depend on its partial pressure. But I don't have
any reason for it. That is problematic.
@Magpie
Sorry your answer is wrong. Try again. If you want I will tell you the correct answer but that will take away the beauty of the problem.
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Magpie
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Quote: Originally posted by Claisen  |
@Magpie
Sorry your answer is wrong. Try again. If you want I will tell you the correct answer but that will take away the beauty of the problem.
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Yes, I had some math errors. This time I checked my work.
I get P(tot) = 532.4mmHg.
As vulture says, I'm assuming there is a constant vapor pressure component of 32.5mmHg due to the presence of liquid S.
The single most important condition for a successful synthesis is good mixing - Nicodem
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Claisen
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Wrong again 
The answer given is 401.58 mmHg
How is that assumption valid? What if S changes its state? How do you find that out?
[Edited on 7-1-2011 by Claisen]
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Magpie
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OK, I will present my calculations and we can discuss where I went wrong.
That S is a liquid is given in the problem statement, ie: S(l)
The single most important condition for a successful synthesis is good mixing - Nicodem
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Claisen
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No its not about the calculations. I doubt your concept/approach to the problem.
Yes it is given that S is a liquid at Temperature T in the chemical reaction. As the reaction progresses, more of S is formed. T is constant. V (of S)
changes. P (of S) must change in accordance with Boyle's Law. This change in pressure may cause a change in its state. I think that's why the vapour
pressure is given. I am not sure about it. I already tried the problem keeping the pressure of S as 32.5 mmHg but did not get the correct answer. If
you get the correct answer by your method then I am wrong.
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Magpie
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Here's my solution. I did find another math error, but the answer still does not agree. So, where am I making error(s)?
Assumptions
2P(g) = 4Q(g) + R(g) + S(l)
At time = 0 min P=2 moles ( an arbitrary assumption)
At time = 30 min
... R=x moles
...Q=4x moles
....S=x moles
....P=2-2x moles
p(tot) = 317mmHg; p (due to P, Q, and R) = 317-32.5 = 284.5mmHg
At time = 75 min
....R=y moles
...Q=4y moles
...S=y moles
...P=2-2y moles
p(tot) -32.5mmHg =p (due to P, Q, and R)
At time = infinity
....R=1 mole
....Q=4 moles
...S= 1 mole
...P=0 moles
p(tot) = 617mmHg; p (due to R and Q) = 617-32.5 = 584.5mmHg
Let V=the volume due to gases P, Q, and R. This a constant.
Temperature, T, is also constant.
Assuming ideal gas behavior: pV=nRT; p/n = RT/V = a
constant.
volume changes due to S(l) are negligible
Solution
At time = infinity
p/n = 584.5/5 = 116.9
At time = 30 minutes, p/n = 284.5/[(2-2x) + x + 4x] = 284.5/(2+3x)
Then 284.5/(2+3x) = 116.9; x=0.1446 moles
For a 1st order decomposition of P, at time = 30 min:
(2-2x)/2 = e ^[-k(30)]; ln[(2-2x)/2] = -30k;
ln0.8554 = -30k; -0.1562 = -30k; k = 0.0052
At time = 75 minutes
(2-2y)/2 = e^[(-0.0052)(75)]; y= 0.3229
[p(tot) -32.5]/[(2-2y) +y +4y] = 116.9
p(tot) -32.5 = 116.9(2+3y) = 347.0
p(tot) = 379.5mmHg
[Edited on 8-1-2011 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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Claisen
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I have no idea what is wrong in the solution. Perhaps other members might help.
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Claisen
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Anyone else? Please help me.
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vulture
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The vapor pressure of S is given and it's always lower than the total pressure in the reaction, which increases constantly because the volume is
fixed.
| Quote: |
For the element S to remain as liquid, I think we need to consider its pressure during the reaction not the total pressure of the vessel
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The pressure during the reaction is the total pressure because you have a closed vessel.
A substance can only exist as gas when the total pressure is equal or lower than its vapor pressure. When vapor pressure equals the ambient pressure
the liquid boils.
[Edited on 8-1-2011 by vulture]
[Edited on 8-1-2011 by vulture]
One shouldn't accept or resort to the mutilation of science to appease the mentally impaired.
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Claisen
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Ok.
Why don't we get the correct answer then?
P.S. : The answer provided cannot be wrong because it is as given by a national question paper setter.
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kmno4
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| Quote: Originally posted by Claisen | Ok.
Why don't we get the correct answer then?
P.S. : The answer provided cannot be wrong because it is as given by a national question paper setter. |
Just for fun I tried to solve it and I got almost the same result as Magpie got: 379,6 mmHg
Can you prove that it is not correct ?
ps.
2P(g) -----> 4Q(g) + R(g) + S(l)
is equivalent to:
2P(g) -----> 5Q(g) + S(l) or 5R(g) + S(l)
why complicate it ?
Have you rewritten it correctly ?
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Claisen
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Thanks for looking into this thread kmno4.
The correct answer provided is 401.58 mmHg. Hence 379.6 mmHg is incorrect. Wait for some days more when I some how find how to get 401.58 mmHg and
prove it for you 
Yes that makes it a little simpler.
[Edited on 9-1-2011 by Claisen]
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spirocycle
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| Quote: Originally posted by Magpie |
At time = 30 min
... R=x moles
...Q=4x moles
....S=x moles
....P=2-2x moles
p(tot) = 317mmHg; p (due to P, Q, and R) = 317-32.5 = 284.5mmHg
At time = 75 min
....R=y moles
...Q=4y moles
...S=y moles
...P=2-2y moles
p(tot) -32.5mmHg =p (due to P, Q, and R)
At time = infinity
....R=1 mole
....Q=4 moles
...S= 1 mole
...P=0 moles
p(tot) = 617mmHg; p (due to R and Q) = 617-32.5 = 584.5mmHg
[Edited on 8-1-2011 by Magpie] |
you keep sayin that p is due to R and Q, but what about S?
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Magpie
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No, p is due to P, R, and Q.
This p is useful in the calculations. But it is not equal to the total pressure p(tot) which does include the vapor pressure of S.
p(tot) = p + 32.5mmHg
The single most important condition for a successful synthesis is good mixing - Nicodem
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spirocycle
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gotcha
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Claisen
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Eureka!!
I got the correct answer at last with someone else's help.
This question is very conceptual. The state of S doesnot depend on the total pressure in the vessel as Vulture said, infact it depends on its own
partial pressure.
If the pressure due to S at any time is greater than 32.5 mmHg, then the additional pressure causes condensation and S remains as liquid as the
saturation pressure is 32.5 mmHg.
If partial pressure of S is less than 32.5 mmHg, then it exists in some other state. It is a gas in our case.
Make an assumption that at t=infinity, S exists as a liquid and take its pressure as 32.5 mmHg. This assumption comes out to be true (by
calculations). Similarly make such assumption for other times given and then solve the problem.
I got that at t= infinity, S is a liquid
at t = 30 min, S is a gas
at t = 75 min, S is a liquid.
I hope magpie, vulture and kmno4 understood their mistake.
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vulture
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Ehm, I have some fundamental problems with this solution.
| Quote: |
Make an assumption that at t=infinity, S exists as a liquid and take its pressure as 32.5 mmHg. |
Which is exactly what I said...
| Quote: |
This assumption comes out to be true (by calculations). Similarly make such assumption for other times given and then solve the problem.
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You assume that S is liquid at all times but for t= 30 you find it's a gas??
| Quote: |
The state of S doesnot depend on the total pressure in the vessel as Vulture said, infact it depends on its own partial pressure.
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My physical chemistry is a bit rusty, but this doesn't make sense. Take a vessel with water and air. If the total pressure inside the vessel is higher
than the vapor pressure of water, the water will simply not boil.
[Edited on 9-1-2011 by vulture]
One shouldn't accept or resort to the mutilation of science to appease the mentally impaired.
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Magpie
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| Quote: Originally posted by Claisen |
Make an assumption that at t=infinity, S exists as a liquid and take its pressure as 32.5 mmHg.
==============
I got that at t= infinity, S is a liquid
at t = 30 min, S is a gas
at t = 75 min, S is a liquid.
I hope magpie, vulture and kmno4 understood their mistake. |
Without redoing the calculations I can see where the assumption that S is a gas at 30 minutes will give a different answer.
What I don't see is the basis for making that assumption. How are we supposed to know that there is not enough partial pressure (of
S) for it to be a liquid at 30 minutes? You could just as well assume that there is enough partial pressure.
But I'm glad you have resolved this and I found the exercise very interesting. It also refreshed some rusty parts of my education. 
===================
I think I've seen the light. This solution requires some trial & error, ie:
Use the nomenclature from my above posted calculations, only this time let p = the total pressure. Assume, again that the number of original moles of
P = 2.
1. At time = infinity: Assume that S is a gas.
p = 617mmHg and the total moles = 1+4+1 =6.
The partial pressure of S = (1/6)(617mmHg) = 102.83. This exceeds its vapor pressure at T (32.5mmHg) so this assumption is wrong and S is partially a
liquid at infinite time.
Then moles of S(g) = 32.5/617 = 0.0527
p/n=RT/V=617/5.0527 = 122.1, a constant at any time.
2. At time =30 minutes: Assume S is a gas.
p/n= 122.1 = 317mmHg/[(2-2x) +x +4x +x)]
x=0.1491 moles; partial pressure of S = [0.1491/(2+4x)](317)
=18.2mmHg. This is less than the vapor pressure of S at T, so S is a gas at time = 30 minutes.
For a 1st order decomposition, P/P(0) = e^(-kt)
where P(0) = original moles of P at time t = 0.
(2-2x)/2 = e^[-k(30)] ; k = 0.0054
3. At time = 75 minutes: moles P remaining =2-2y
(2-2y)/2 = e^[-(0.0054)(75)]; y =0.3330moles
Again assume S is a gas. Then p/(2+4y) =122.1
p =122.1[(2+4(0.3330)] = 406.8mmHg
Checking the partial pressure of S
= (0.3330/[2+4(0.3330)])406.8mmHg = 40.6mmHg which is greater than the vapor pressure of S at temperature T, so this assumption is not true and S is a
partially a liquid at t= 75 minutes.
So the partial pressure of S=32.5mmHg.
moles S(g) = (32.5/p)n
p/n = 122.1
moles S(g) = 32.5/(p/n) = 32.5/122.1 = 0.2662
p = [(2+3y+0.2662)/0.2662]32.5mmHg = 397.5mmHg
This is quite close to the answer of 401.5mmHg. This small difference can be affected by rounding errors made during the calculations.
[Edited on 10-1-2011 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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Claisen
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Hello everyone,
I wrote a reply for it yesterday in notepad but was not able to post it. Anyway I see that Magpie posted his solution which is correct.
| Quote: Originally posted by Magpie |
What I don't see is the basis for making that assumption. How are we supposed to know that there is not enough partial pressure (of
S) for it to be a liquid at 30 minutes? You could just as well assume that there is enough partial pressure.
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I would like to clear this problem.
............2P-------> 5Q + S(l)
t=0......Po............0............0
t=30....Po-x..........5x/2......x/2 or 32.5
t=inf....0...............5Po/2.....Po/2 or 32.5
At t=30min, S will exist as a gas if x/2 < 32.5 ( I already stated why it is so)
Similarly, at t=inf, S will exist as a gas if Po/2 < 32.5
Let us assume that S exists as a liquid at t=inf
so,
5Po/2 + 32.5 = 617
from here Po = 233.8 mmHg
Po/2 = 116.9 mmHg which is greater than 32.5 mmHg. Hence our assumption is correct.
Now let us assume that S exists as a liquid at t=30mins
so,
Po-x + 5x/2 + 32.5 = 317
Substitute Po=233.8
we get x as 33.8 mmHg so that x/2 = 16.9 mmHg which is less than 32.5 mmHg. Hence our assumption is incorrect. S exists as a gas.
| Quote: Originally posted by vulture | My physical chemistry is a bit rusty, but this doesn't make sense. Take a vessel with water and air. If the total pressure inside the vessel is higher
than the vapor pressure of water, the water will simply not boil.
[Edited on 9-1-2011 by vulture] |
I am just a high school student. I can't explain you in much depth. This is what I think is your problem.
| Quote: | | While below the boiling point a liquid evaporates from its surface, at the boiling point vapor bubbles come from the bulk of the liquid. For this to
be possible, the vapor pressure must be sufficiently high to win the atmospheric pressure |
| Quote: | | If the temperature in a system remains constant (an isothermal system), vapor at saturation pressure and temperature will begin to condense into its
liquid phase as the system pressure is increased. Similarly, a liquid at saturation pressure and temperature will tend to flash into its vapor phase
as system pressure is decreased. |
Here the author wrote 'system' for one element as far as I think. For more than 1 element, partial pressures come into play rather than total
pressure. That is why partial pressure is important.
Source- http://en.wikipedia.org/wiki/Boiling
http://en.wikipedia.org/wiki/Boiling_point
[Edited on 10-1-2011 by Claisen]
[Edited on 10-1-2011 by Claisen]
[Edited on 10-1-2011 by Claisen]
[Edited on 10-1-2011 by Claisen]
[Edited on 10-1-2011 by Claisen]
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Magpie
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Yikes! What kind of a high school gives problems like this? 
The single most important condition for a successful synthesis is good mixing - Nicodem
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Claisen
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I am preparing for the world's toughest Engineering Entrance exam, as it is said. Its called Indian Institute of Technology - Joint Entrance
Examination. Such questions are normal
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kmno4
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This was not difficult problem. At the begining I suspected that for 30 min p+32,5 can be not true. For a moment I even wanted to calculate it in a
correct way but I was thinking that: ... no, no it cannot be so "complicated" ( =laziness)
It turns out that liquid phase appears after ~50 minutes .......
My pressure for 75 minutes is 402,1 mmHg.
Not difficult but also not banal
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