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Topic: Pressure and equilibrium of decomposition.  (Read 5655 times)

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Offline Moose100

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Pressure and equilibrium of decomposition.
« on: September 09, 2015, 12:50:46 AM »
SO3 decomposes on heating:
 2SO2(g)  ::equil:: 2SO2(g) + O2(g)
 A sample of pure SO3 was heated to a high temperature T. At equilibrium, the mole ratio of SO2 to
SO3 was 0.152 and the total pressure was 2.73 atm. If the piston was pushed in to cut the volume in half what is the new pressure?


Please *delete me*!  UI tried using the ratio to get the pressures on the left. 0.414 and 2.32 respectively.Also tried multiplying 2.73 by 2 but don't think this is the way.

Please help !! Thanks a million.
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Offline billnotgatez

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Re: Pressure and equilibrium of decomposition.
« Reply #1 on: September 09, 2015, 01:54:49 AM »
Quote from: Moose100 on September 09, 2015, 12:50:46 AM
SO3 decomposes on heating:
 2SO2(g)  ::equil:: 2SO2(g) + O2(g)
 A sample of pure SO3 was heated to a high temperature T. At equilibrium, the mole ratio of SO2 to
SO3 was 0.152 and the total pressure was 2.73 atm. If the piston was pushed in to cut the volume in half what is the new pressure?


Please *delete me*!  UI tried using the ratio to get the pressures on the left. 0.414 and 2.32 respectively.Also tried multiplying 2.73 by 2 but don't think this is the way.

Please help !! Thanks a million.


Do you have a typo in the formula you posted?
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Online Borek

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Re: Pressure and equilibrium of decomposition.
« Reply #2 on: September 09, 2015, 02:52:12 AM »
Trivial typo.

Many ways to skin that cat. Start by calculating K, you will need that. Then you can try to assume initial volume to be V and express initial number of moles of SO3 in the initial mixture as a function of V. Next try to derive expression for the final pressure (new volume being V/2, and using K to find the new equilibrium) - you will find that V canceled out.
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Offline Moose100

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Re: Pressure and equilibrium of decomposition.
« Reply #3 on: September 09, 2015, 09:50:29 AM »
Yes I did do the volume as V/2. Thanks I will be back later!
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Offline Moose100

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Re: Pressure and equilibrium of decomposition.
« Reply #4 on: September 09, 2015, 05:43:55 PM »
Those concentrations I got ARE correct right?
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Online Borek

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Re: Pressure and equilibrium of decomposition.
« Reply #5 on: September 10, 2015, 03:59:12 AM »
Quote from: Moose100 on September 09, 2015, 05:43:55 PM
Those concentrations I got ARE correct right?

Hard to tell not seeing them.
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Offline Moose100

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Re: Pressure and equilibrium of decomposition.
« Reply #6 on: September 10, 2015, 08:35:52 AM »
I'm talking about the ones I posted earlier. I will put the new ones later. I'm thinking if those are wrong the new ones will be?
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Offline mjc123

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Re: Pressure and equilibrium of decomposition.
« Reply #7 on: September 10, 2015, 08:49:08 AM »
If you mean 0.414 and 2.32 atm, that is wrong. 0.414 = 0.152*2.73, but 0.152 is the ratio of SO2 to SO3, not the ratio of SO2 to total pressure.
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Offline Moose100

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Re: Pressure and equilibrium of decomposition.
« Reply #8 on: September 10, 2015, 01:53:37 PM »
Right that is the ratio and I multiplied it by the beginning pressure to get 0.414 for S02 and 2.32 for Oxygen. I use these to get the constant. I wanted to ask before I put it down if this was the correct approach. Sorry if im being tedious. :-\
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Offline Enthalpy

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Re: Pressure and equilibrium of decomposition.
« Reply #9 on: September 10, 2015, 01:54:54 PM »
And is there any indication about what the temperature does when the piston is moved?
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Offline Moose100

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Re: Pressure and equilibrium of decomposition.
« Reply #10 on: September 10, 2015, 02:45:19 PM »
Not at all they just say its really high.
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Online Borek

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Re: Pressure and equilibrium of decomposition.
« Reply #11 on: September 10, 2015, 06:18:46 PM »
Quote from: Moose100 on September 10, 2015, 01:53:37 PM
Right that is the ratio and I multiplied it by the beginning pressure

And you were already told it is wrong - question clearly says what is the ratio of SO2/SO3, not what is the ratio ratio of SO2/everything.
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Offline Moose100

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Re: Pressure and equilibrium of decomposition.
« Reply #12 on: September 10, 2015, 07:43:01 PM »
OOps sorry. I will take a closer look here.
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Offline Moose100

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Re: Pressure and equilibrium of decomposition.
« Reply #13 on: September 13, 2015, 08:27:27 PM »
Equilibrium conc of:

SO3: 2.73-2x
SO2: 2x
O2  : x

2x/(2.73-2x)=0.152

x=0.180

Eventually I get to the point where K= 0.00415

Am I in the ballpark here?




« Last Edit: September 13, 2015, 09:37:34 PM by Moose100 »
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Offline Moose100

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Re: Pressure and equilibrium of decomposition.
« Reply #14 on: September 14, 2015, 07:10:49 PM »
Ptotal=pO2+pSO3+pSO2

pSO2/pSO3=0.152

pO2/pSO2= 0.5

Ptotal= 0.5pSO2 +pSO2/0.152 + pSO2

Is this closer? I feel like everything I try is wrong.
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