[AhBeng]

Assume we have a reaction with 1 mole of gas on LHS, and 2 moles of gas on RHS. Eg. H2O(g) + C(s) --> H2(g) + CO(g).


Q. What happens to position of equilibrium when pressure is increased (ie. volume is decreased)?

A. Position of equilibirum shifts to the left (side with less moles of gas).


Q. What happens to position of equilibrium when an inert gas (eg. helium) is added at constant volume (eg. in a steel vessel)?

A. Adding helium to the equilibrium mixture at constant volume increases the total gas pressure and decreases the mole fractions of all 3 gases, but the partial pressure of each gas given by the product of its (reduced) mole fraction and the (increased) total pressure does not change. Hence, position of equilibrium does not change.


Q. What happens to position of equilibrium when an inert gas (eg. helium) is added at constant pressure, but not constant volume (wierd yeah, but hypothetically speaking, if it's possible)?

A. I honestly do not know. It seems to be an impossible setup, or inherently flawed. But hypothetically speaking, if adding an inert gas does not affect pressure (because volume is allowed to increase accordingly), then intuitively, there should be no effect on reactant and product molecules, and hence should have no effect on the position of equilibrium.

On the other hand, if volume increases, and thus molarity of all 3 gases decreases. As such, by the equilibrium constant or reaction quotient expression, Qc is now < Kc, and the position of reaction should shift to the right.

What am I not seeing correctly here? Which (if any) of the above is correct? Or is this question a moot point because the setup is impossible or inherently flawed? Please advise! Thanks in advance for any replies!

[Yggdrasil]

Assume we have a reaction with 1 mole of gas on LHS, and 2 moles of gas on RHS. Eg. H2O(g) + C(s) --> H2(g) + CO(g).


Q. What happens to position of equilibrium when pressure is increased (ie. volume is decreased)?

A. Position of equilibirum shifts to the left (side with less moles of gas).

Correct.

Q. What happens to position of equilibrium when an inert gas (eg. helium) is added at constant volume (eg. in a steel vessel)?

A. Adding helium to the equilibrium mixture at constant volume increases the total gas pressure and decreases the mole fractions of all 3 gases, but the partial pressure of each gas given by the product of its (reduced) mole fraction and the (increased) total pressure does not change. Hence, position of equilibrium does not change.

Correct.  Another way to think about this is that the concentrations of the gas do not change.

Q. What happens to position of equilibrium when an inert gas (eg. helium) is added at constant pressure, but not constant volume (wierd yeah, but hypothetically speaking, if it's possible)?

A. I honestly do not know. It seems to be an impossible setup, or inherently flawed. But hypothetically speaking, if adding an inert gas does not affect pressure (because volume is allowed to increase accordingly), then intuitively, there should be no effect on reactant and product molecules, and hence should have no effect on the position of equilibrium.

On the other hand, if volume increases, and thus molarity of all 3 gases decreases. As such, by the equilibrium constant or reaction quotient expression, Qc is now < Kc, and the position of reaction should shift to the right.

What am I not seeing correctly here? Which (if any) of the above is correct? Or is this question a moot point because the setup is impossible or inherently flawed? Please advise! Thanks in advance for any replies!

Think back to question two.  Is the important effect the effect on total pressure or the effect on the partial pressure of the reactants and products?

[AhBeng]

Thank you for replying, Yggdrasil.

Think back to question two.  Is the important effect the effect on total pressure or the effect on the partial pressure of the reactants and products?

The important effect is on the partial pressure (mole fraction x total pressure) of reactants versus products, and since the mole fractions of the gases are now reduced (on addition of helium), their partial pressures are also reduced (since total pressure remains the same).

Another way to think about it, is that with increased volume, the concentration/molarity of the reactant and product gases, all decrease.

Since on the RHS of equation (or numerator of Qc/Kc expression) has 2 gaseous terms (each unimolar), while the LHS of equation (or denominator of Qc/Kc expression) has only 1 (unimolar) gaseous term, in effect, the decrease in RHS/numerator is greater than the decrease in LHS/denominator, which translates into Qc < Kc, and the position of equilibrium (Kc) now is shifted to the right of Qc, so the net reaction is the forward reaction to produce more moles of gas.

So to summarize, the effect on total pressure is not important, it is volume, and hence concentration/molarity, which is important, in determining positional shifts of equilibrium.

Is this correct, Yggdrasil? Thanks again!