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Topic: A doosy of a Gas Law Question  (Read 13585 times)

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TheOldBean

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A doosy of a Gas Law Question
« on: April 30, 2005, 12:52:04 AM »
Hello, my name is Dave and I'm a First Year Uni Student and I have an unusual text book question here which I was wondering if you brains out there could help me out with. Most appreciated!

An atmospheric chemist studying the reactionsof the pollutant SO2 places a mixture of SO2 and O2 in a 2.00-L container at 900. K and an initial pressure of 1.95 atm. When the reaction occurs, gaseous SO3 forms, and the pressure eventually falls to 1.65 atm. How many moles of SO3 form?

Thanks for your time!

Dave

EDIT: Sorry! Should say that the answer is 1.63 x 10-2 mol. If that helps!
EDIT x 2: I've taken a few steps myself. Considering that P is proportional to T, I assumed that when the pressure decreased by a certain percentage, the temperature would decrease by the same percentage. I'm not sure what else to do! This isn't like an assessment question, but it IS one that I would like to present to a small group, how to do it.
« Last Edit: April 30, 2005, 03:18:45 AM by The Old Bean »

charco

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Re:A doosy of a Gas Law Question
« Reply #1 on: April 30, 2005, 09:22:06 AM »
2SO2 + O2 --> 2SO3

under the conditions used all are gases therefore three moles becomes two moles

initial conditions: P= 1.95 atmos; V=2; T= 900K

PV=nRT thus n= PV/RT

n= 1.95 x 2/ 0.0821 x 900

n= 0.05278 moles

after reaction

n= PV/RT = 1.65 x 2 /0.0821 x 900

n= 0.04466

therefore change in the number of moles = 0.05278 - 0.04466 = 0.00812

this change in the number of moles represents one half of the number of moles of SO3 that are formed (see equation)

thus moles of SO3 = 0.00812 x 2 = 1.62 x 10^-2 moles


GCT

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Re:A doosy of a Gas Law Question
« Reply #2 on: April 30, 2005, 08:49:29 PM »
To clarify for the op a bit further.

For each mol of reaction there is a net change of moles of 1/3.  For instance, if 4 moles of sulfur dioxide were to react with 2 mols of oxygen gas, then 4 moles sulfur trioxide would have been formed.  The net change in moles is 4-6=-2, that's a loss of 2 mols for every 6 moles of gas which undergoes the reaction.

Thus .00812, if your calculations are correct, represents the net change in moles.  You would need to multiply this quanty by 3, to get the total moles of gas which underwent the reaction, than multiply by 4/6, overall multiplication factor is 2

TheOldBean

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Re:A doosy of a Gas Law Question
« Reply #3 on: April 30, 2005, 11:20:08 PM »
Thanks everyone for that! Just for the explaining of such:-

Charco:-
* Why did you use 0.0821 for R rather than 8.31?
* Still a little confused about why you multiplied it by two, I know there are two moles of SO3 formed, but I don't get how it's a half of SO3.

GCT:-

* I've finally understood where you got the 1/3, but I'm still wondering why you multiplied it by 3 then by 4/6.

Thanks!

Dave

charco

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Re:A doosy of a Gas Law Question
« Reply #4 on: May 01, 2005, 06:05:36 AM »
The gas constant is derived from PV =nRT and consequently depends on the value used for pressure which may be in ...
1. atmospheres
2. mmHg
3. Pa
4. KPa
5. Nm-2

each of these throws up a different value for the universal gas constant.
I just used the appropriate constant for pressure in atmospheres as the question gave the pressure in atmospheres.

As to why I multiplied by 2.

The maths gave the number of moles difference between reactants and products which given the stoichiometry was in a ratio of 1:2 compared to the actual moles of product.

i.e

2SO2 + O2 ---> 2SO3

the stoichiometry shows that for a net difference in moles of 1 there will be 2 moles of SO2 formed, therefore for a net difference of 'n' moles there will be 2n moles of SO2 formed.

Hence moles of SO2 = 2 x difference in moles between reactants and products

GCT

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Re:A doosy of a Gas Law Question
« Reply #5 on: May 03, 2005, 10:03:10 AM »
Thanks everyone for that! Just for the explaining of such:-

Charco:-
* Why did you use 0.0821 for R rather than 8.31?
* Still a little confused about why you multiplied it by two, I know there are two moles of SO3 formed, but I don't get how it's a half of SO3.

GCT:-

* I've finally understood where you got the 1/3, but I'm still wondering why you multiplied it by 3 then by 4/6.

Thanks!

Dave

I guess the most simplest way to approach this is to understand from the equation that for every 1 mole of reaction 2 moles of sulfur trioxide is formed.  You can use this conversion factor when you apply factor labeling.

Offline Donaldson Tan

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Re:A doosy of a Gas Law Question
« Reply #6 on: May 05, 2005, 04:12:02 PM »
* Why did you use 0.0821 for R rather than 8.31?

R = 8.314 J/mol.K = 0.0820578 L.atm/K.mol
if P is pascal, use the former. if P is atm, use the later.
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