[Colahate]

The question is :
0.05 moles of SO2 and 0.1 moles of O2 gases are introduced into a flask of volume 1 Liter at a constant temperature 325 K. The following equilibrium is established:
SO2(g)+1/2 O2(g) ----> SO3(g)
At equilibrium, the number of moles of SO3 is 0.01 mole and the total pressure of the mixture of 2 gases is 2 atm.(R=0.082 atm.L.k-1.mol-1)

My question is: when I calculate total number of moles using equilibrium I get
SO3 at equilibrium = x= 0.01 moles
So
SO2 at equilibrium = 0.05-x=0.04 moles
And
O2 at equilibrium = 0.1-x/2=0.095 moles

Their sum is ntotal= 0.01+0.04+0.095=0.145 moles

However, when I use pv=nrt, given the total pressure at equilibrium is 2 atm
n=pv/rt= 2*(1L)/0.082*325= 0.075 moles

Why is that? Why is pv=nrt wrong ?

Thanks in advance!

[Hunter2]

Something wrong with the pressure, because pV =nRT with your datas gives for p =0,145 mol * 0.082 atm.L.k-1.mol-1* 325K/1l = 3,86 atm.

The 2 atm is given for SO₂ and O₂: 

total pressure of the mixture of 2 gases is 2 atm

Later you have 3 gases and the pressure will be higher.

[Colahate]

Thanks for the quick reply
Sorry that was a typo. It is the pressure of the 3 gases..

[Hunter2]

I think your calculations are correct. If it is for the 3 gas mixture then the pressure is still to low.  Even if we think we mix 0,05 mol SO₂ and 0,1 mol O₂ and look the first seconds before the reaction takes place we have 0,15 mol what guides also to to 3,99 atm. Later we have 3 gases and this should be lower?

[Colahate]

I thought there was something wrong with the given exercise, but it is from a book that has been used for about 16 years in my country... So i think there is something wrong with my thinking?

[mikasaur]

Can you write out the question word for word from the text? Are they just asking you to solve for the total number of moles at equilibrium?

[Colahate]

They are asking for partial pressures at equilibrium, yet thats not my problem. I knew how to solve it, and it turned out to be correct, but I'm still not convinced why the total number of moles differs. When I used equilibrium to solve for the partial pressures( finding mole fraction and multiplying by total pressure at equilibrium)  i got the correct answer. However, calculating using pv=nrt gives totally different answers. Thats my question.. Thanks for your replies!

[mikasaur]

Hmm, well it could just be a poorly-thought-out question. At some temperature the mole fractions given by the stoichiometry of the reaction will give you 2 atm of total pressure for the three gases, but it could be that the authors of the question didn't bother to find out what that temperature would be!

As Hunter2 was saying... before the reaction even begins we have 4.00 atm of pressure. If the reaction went to completion we'd go from .15 moles of gas to .125 moles of gas (all of the SO₂ would be converted to SO₃ and 0.025 moles of O₂ would be consumed). So the reduction in pressure would be minimal.

I think you're right. And 2 atm just doesn't make sense.

[Colahate]

Thank you so much Hunter2 and mikasaur!
And yes thats exactly my point!
Finally, i found people who understood what i meant..