Login

Welcome, Guest. Please login or register.

November 01, 2025, 12:59:52 pm

Author Topic: partial pressures  (Read 1426 times)  Share 

0 Members and 1 Guest are viewing this topic.

lorelai

  • Victorian
  • Trailblazer
  • *
  • Posts: 27
  • Respect: +15
partial pressures
« on: October 17, 2011, 05:46:24 pm »
0
Could someone explain to me exactly what partial pressure is, and its relationship to the equilibrium constant and equilibrium systems?
This question was prompted by number three in the short answer of TSFX 2009.
2010: History of Revolutions 44 | Literature 36
2011: English 44 | Chemistry 38 | Mathematical Methods 38 | Specialist Mathematics 31 | Music Style & Composition 40
ATAR: 97.6

ARMYMAN0010

  • Victorian
  • Adventurer
  • *
  • Posts: 21
  • Respect: 0
Re: partial pressures
« Reply #1 on: October 17, 2011, 09:06:24 pm »
0
I'm not entirely sure of the theory on how this works, but I know for certain it's true. Say an equilibrium system consisted of A, B and C, and it was defined as A + B -> C (with an equilibrium arrow of course!). If the volume of the system was halved ( pressure doubled), the system would shift to the side which partially alleviates pressure-ie, side with less particles- C. Cool, that makes sense. What if the inert gas H2 were added? You would expect pressure to increase, and the same thing to occur. Well, apparently not. From what I can gather on wikipedia, the partial pressure- that is, the pressure of each individual molecule- will not be altered by the addition of an inert gas. So, equilibrium systems will NOT be altered by the addition of gasses not involved in the equilibrium reaction. The equilibrium constant wouldn't be altered either way- this value is constant at a constant temperature. However, the concentration fraction would change in the first example I gave.
hope that helps.. if somebody more knowledgeable than me ( I'm looking at you, uni chem people!) could explain what partial pressure is for both of us that would be cool too :P

RobM8

  • Guest
Re: partial pressures
« Reply #2 on: October 17, 2011, 11:59:34 pm »
0
My understanding of partial pressure:

Ideal Gas Law (pV=nRT) would predict that p doesn't change when you add an inert gas (that doesn't react with anything already in the container) and all other variables are kept the same and hence the equilibrium constant maintains the same value.

p is proportional to n as R T and V are constant/kept the same.
As n doesn't change when you add a non-reactive gas
Then p wouldn't change for that individual gas --> partial pressure is unaffected by addition of inert gas

In the context of equilibria that would mean that the addition (or removal) of an inert gas to (from) an equilibrium system has no effect on the position of equilibrium as individual concentrations (n/V) are kept constant.

I am no Niels Bohr but I would deduce that the net/effective pressure of the system would increase.

And to any keen chemists who stumble across this, the addition of an inert gas wouldn't increase the rate of a reaction right?

Dr.Lecter

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 261
  • Respect: +16
Re: partial pressures
« Reply #3 on: October 18, 2011, 12:12:53 am »
0

And to any keen chemists who stumble across this, the addition of an inert gas wouldn't increase the rate of a reaction right?

Addition of inert gas may increase pressure and hence more particle collisions
A census taker once tried to test me. I ate his liver with some fava beans and a nice Chianti.

Mao

  • CH41RMN
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 9181
  • Respect: +390
  • School: Kambrya College
  • School Grad Year: 2008
Re: partial pressures
« Reply #4 on: October 18, 2011, 02:01:22 am »
+1

And to any keen chemists who stumble across this, the addition of an inert gas wouldn't increase the rate of a reaction right?

Addition of inert gas may increase pressure and hence more particle collisions

No.

Addition of inert gas will increase the collisions between inert gas and the reactants, but since such collision doesn't yield any fruitful reaction, it has no net contribution to the reaction rate. Nothing (significant) changes.
Editor for ATARNotes Chemistry study guides.

VCE 2008 | Monash BSc (Chem., Appl. Math.) 2009-2011 | UoM BScHon (Chem.) 2012 | UoM PhD (Chem.) 2013-2015

Dr.Lecter

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 261
  • Respect: +16
Re: partial pressures
« Reply #5 on: October 18, 2011, 02:52:57 am »
0

And to any keen chemists who stumble across this, the addition of an inert gas wouldn't increase the rate of a reaction right?

Addition of inert gas may increase pressure and hence more particle collisions

No.

Addition of inert gas will increase the collisions between inert gas and the reactants, but since such collision doesn't yield any fruitful reaction, it has no net contribution to the reaction rate. Nothing (significant) changes.

I didn't say it contributes to reaction rate
A census taker once tried to test me. I ate his liver with some fava beans and a nice Chianti.

Mao

  • CH41RMN
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 9181
  • Respect: +390
  • School: Kambrya College
  • School Grad Year: 2008
Re: partial pressures
« Reply #6 on: October 18, 2011, 03:41:12 am »
+1
I didn't say it contributes to reaction rate

I'm going to just ignore this.



I am no Niels Bohr but I would deduce that the net/effective pressure of the system would increase.

Yes. The 'total pressure' of the system would increase.

Partial pressure is the contribution of a particular species towards the total pressure. If n, v and T are fixed, then this contribution remains constant, regardless of how much other gases you add in (provided n doesn't change due to any reactions, of course). We use partial pressure because it is a good measure of the rate of collision between gaseous species, and since equilibrium is all about equal forwards and backwards rates, it is the best quantity which we can use to describe gas equilibria.

You can relate addition of inert gas to the gas equation in two ways (assuming fixed container and constant temperature):

1. pp.V=n.R.T, where pp=partial pressure. In this case, adding more inert gas does not add more reagent gas, its partial pressure does not change, and the rate of reaction is not altered.

2. P.V=N.R.T, where P=total pressure, and N=total number of moles of all gaseous species (assuming they're all ideal gases). Adding more inert gas increases the total number of moles N, and thus the total pressure must increase. However, the total pressure doesn't tell us anything about the rate of reaction.

The total pressure is the sum of all partial pressures. This comes from the linear relationship between p and n in pV=nRT.
Editor for ATARNotes Chemistry study guides.

VCE 2008 | Monash BSc (Chem., Appl. Math.) 2009-2011 | UoM BScHon (Chem.) 2012 | UoM PhD (Chem.) 2013-2015