Sciencemadness Discussion Board - Chemical Kinetics - Good Problem

Chemical Kinetics - Good Problem

Claisen - 6-1-2011 at 12:14

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.

vulture - 6-1-2011 at 13:06

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]

Magpie - 6-1-2011 at 16:41

I worked this problem with a result of P(total)=379.5mmHg. Anyone else give it a try?

Claisen - 7-1-2011 at 02:42

@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.

Magpie - 7-1-2011 at 09:19

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.

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.

Claisen - 7-1-2011 at 12:22

Quote: Originally posted by Magpie

Yes, I had some math errors. This time I checked my work.
I get P(tot) = 532.4mmHg.

Wrong again

The answer given is 401.58 mmHg

Quote: Originally posted by Magpie

assuming there is a constant vapor pressure component of 32.5mmHg due to the presence of liquid S.

How is that assumption valid? What if S changes its state? How do you find that out?

[Edited on 7-1-2011 by Claisen]

Magpie - 7-1-2011 at 13:14

Quote: Originally posted by Claisen

Quote: Originally posted by Magpie

Yes, I had some math errors. This time I checked my work.
I get P(tot) = 532.4mmHg.

Wrong again

The answer given is 401.58 mmHg

OK, I will present my calculations and we can discuss where I went wrong.

Quote: Originally posted by Claisen

Quote: Originally posted by Magpie

assuming there is a constant vapor pressure component of 32.5mmHg due to the presence of liquid S.

How is that assumption valid? What if S changes its state? How do you find that out?

That S is a liquid is given in the problem statement, ie: S(l)

Claisen - 7-1-2011 at 14:29

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.

Magpie - 7-1-2011 at 14:56

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]

Claisen - 7-1-2011 at 15:26

I have no idea what is wrong in the solution. Perhaps other members might help.

Claisen - 8-1-2011 at 07:29

Anyone else? Please help me.

vulture - 8-1-2011 at 09:42

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

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]

Claisen - 8-1-2011 at 11:10

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.

kmno4 - 9-1-2011 at 07:42

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 ?

Claisen - 9-1-2011 at 09:56

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]

spirocycle - 9-1-2011 at 10:24

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?

Magpie - 9-1-2011 at 10:41

Quote: Originally posted by spirocycle

you keep sayin that p is due to R and Q, but what about S?

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

spirocycle - 9-1-2011 at 11:21

gotcha

Claisen - 9-1-2011 at 13:58

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.

vulture - 9-1-2011 at 14:09

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.

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.

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]

Magpie - 9-1-2011 at 14:18

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]

Claisen - 10-1-2011 at 00:44

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 vulture

You assume that S is liquid at all times but for t= 30 you find it's a gas??

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.

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

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]

Magpie - 10-1-2011 at 07:17

Quote: Originally posted by Claisen

I am just a high school student.

Yikes! What kind of a high school gives problems like this?

Claisen - 10-1-2011 at 09:20

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

kmno4 - 10-1-2011 at 14:43

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

Claisen - 10-1-2011 at 16:09

Yeah you are right, its not so tough, just moderate. There are tough ones too

. Most of the questions which they ask require very critical thinking.

Magpie - 10-1-2011 at 16:42

Yeah, that problem was really easy. That's why we worked on it for days and had so much help.