[ice888]

Hi all, i've encountered this question below during my coursework...

In an experiment, x moles of A placed in a flask underwent a reaction to produce B.
A(g) <-> 2B(g)   [reversible reaction]
At equilibrium, y moles of A had reacted and the total pressure in the flask was P atm.

If z moles of inert C is added to the closed flask, how will the value of equilibrium constant and the position of equilibrium be affected?

I've worked my answer out for this and my answer differs from my tutor's regarding the position of equilibrium. I strongly think that the position of equilibrium will shift left as by Le Chatelier's Principle, the closed system will relieve the stress (due to the addition of C molecules, therefore the backward reaction is favoured.

However, my tutor claimed otherwise and said because C is inert, it doesn't affect the system's equilibrium position.

Please advise and convince me that I'm wrong.  

[sdekivit]

Hi all, i've encountered this question below during my coursework...

In an experiment, x moles of A placed in a flask underwent a reaction to produce B.
A(g) <-> 2B(g)   [reversible reaction]
At equilibrium, y moles of A had reacted and the total pressure in the flask was P atm.

If z moles of inert C is added to the closed flask, how will the value of equilibrium constant and the position of equilibrium be affected?

I've worked my answer out for this and my answer differs from my tutor's regarding the position of equilibrium. I strongly think that the position of equilibrium will shift left as by Le Chatelier's Principle, the closed system will relieve the stress (due to the addition of C molecules, therefore the backward reaction is favoured.

However, my tutor claimed otherwise and said because C is inert, it doesn't affect the system's equilibrium position.

Please advise and convince me that I'm wrong.  

the equilibrium will shift as the pressure of the system rises.  The equilibriumconstant will stay unchanged.

[Demotivator]

the equilibrium will shift as the pressure of the system rises.  

Wrong!
The addition of inert C does not alter the partial pressures of the reactants. No shift in equilibrium. If a shift occured, the K wouldn't be the same value.

If the pressure was altered through compression or expansion, the partial pressures of the reactants would change and a shift would occur.

[sdekivit]

Wrong!
The addition of inert C does not alter the partial pressures of the reactants. No shift in equilibrium. If a shift occured, the K wouldn't be the same value.

If the pressure was altered through compression or expansion, the partial pressures of the reactants would change and a shift would occur.

but according to Chatelier principle a shift will undo an effect on the equilibrium so that the system has the same K after equilibrium is reached again.

But as we have more C the mole fractions of the equilibriumparticipants changes more for B then for A.

[Demotivator]

but according to Chatelier principle a shift will undo an effect on the equilibrium so that the system has the same K after equilibrium is reached again.

But as we have more C the mole fractions of the equilibriumparticipants changes more for B then for A.

Yes, I know that that shifts maintain the same K. I mean in this particular case, a shift would change the K because there has been no change in the concentrations for a shift to occur in the first place.
As for mole fractions, that doesn't count. These are gases and the equilibrium equation refers to partial pressures of the active species only.

[Borek]

but according to Chatelier principle a shift will undo an effect on the equilibrium so that the system has the same K after equilibrium is reached again.

But as we have more C the mole fractions of the equilibriumparticipants changes more for B then for A.

What counts here are partiall presures, not the total system pressure.

A <-> 2B

K = p_B²/p_A

p_X - X partial pressure
n_X - number of moles of X

Before C addition:

p_B = P n_B/(n_B+n_A)
p_A = P n_A/(n_B+n_A)

after C addition

p'_B = P' n_B/(n_B+n_A+n_C)
p'_A = P' n_A/(n_B+n_A+n_C)

in general

p = n RT/V

so

P' = (n_B+n_A+n_C)/(n_B+n_A) P

substituting P' into p' equations:

p'_B = (n_B+n_A+n_C)/(n_B+n_A) P n_B/(n_B+n_A+n_C)
p'_A = (n_B+n_A+n_C)/(n_B+n_A) P n_A/(n_B+n_A+n_C)

canceling:

p'_B = P n_B/(n_B+n_A) = p_B
p'_A = P n_A/(n_B+n_A) = p_A

Partial presures were not changed, equilibrium is not moving.

[ice888]

Thank you !
Really some rigor to go through the mathematics, and i'm convinced, and also learnt some new stuffs too  
I've one part which i don't get it "What counts here are partial presures, not the total system pressure." Why doesn't the total system pressure matter? Isn't the partial pressure derived from the total system pressure?

So am i right to say now that if an inert gas is added to a close system, the respective partial pressure of the gas A, and B doesn't change? Then under what circumstances will the partial pressure change? Because i'd always thought that since P_A=P n_A/n_(A+B) then when i add n_c of inert gas into the system, the partial pressure of the gases will be redistributed such as P_A=P' n_A/n_(A+B+C) giving a different result since the mole ratio and total pressure have changed.

Again, thank you !

[Borek]

I've one part which i don't get it "What counts here are partial presures, not the total system pressure."

Partial pressures are used just like concentrations in the K quotient definition. Only partial pressures have any significance when you look at the equilibrium.

Why doesn't the total system pressure matter? Isn't the partial pressure derived from the total system pressure?

Total pressure matter only as long as its changes partial pressures. In this case adding an inert gas you change only partial pressure of this inert gas (that's because the volume of the system haven't changed).

So am i right to say now that if an inert gas is added to a close system, the respective partial pressure of the gas A, and B doesn't change?

If there is no volume change - yes (as long as we are in the area where ideal gas approximation works).

Then under what circumstances will the partial pressure change?

Think about it this way: gases are indepenedent. Adding inert gas doesn't change situation of the gas you are observing. If you don't change number of moles, or volume, or temperature - state of your gas doesn't change.

Because i'd always thought that since P_A=P n_A/n_(A+B) then when i add n_c of inert gas into the system, the partial pressure of the gases will be redistributed such as P_A=P' n_A/n_(A+B+C) giving a different result since the mole ratio and total pressure have changed.

Total pressure increase is just as large as needed for the partial pressures of A and B to remain unchanged - remember these things that canceled out?

[sdekivit]

but pressure will rise after adding an inert gas to an equilibrium of gasses:

http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/LeChatelier-Intro.html

pressure is the collision of molecules to a surface. When there are more molecules in a same volume there are more collision to the surface and thus the pressure rises.

But because the inert gas does not take place in the equilibriumreactions the inert gas won't affect the equilibrium, in this case. (think i assumed a constant pressure and thus a volumechange due to adding the inert gas, but it's a closed flask).

[Borek]

pressure is the collision of molecules to a surface. When there are more molecules in a same volume there are more collision to the surface and thus the pressure rises.

Don't mistake partial pressure with the total pressure.

As long as we are in the area where the gas can be treated as ideal gas, adding  inert gas won't change a thing.