Inert gases and gaseous equilibrium mixtures

Written by Adrian
On February 02, 2015
Categories: AP | AP TOPIC 13 | Big Idea 6

What happens when an inert gas is added to a gaseous equilibrium mixture at constant volume?

Before we answer that question, it might be worth defining what we mean by “an inert gas”. In this context we mean a gas that is not involved in the equilibrium reaction, or one that reacts with any of the components of the equilibrium reaction. Now, it is true that the gases that are usually added in the context of these questions do come from group 18, e.g., are often gases like He, Ne or Ar, but being a member of group 18 (the inert/noble gases), is not a requirement to be considered ‘inert’ in this context.

OK, let’s get back to the answer to the original question. It’s a simple, ‘nothing‘.

19127341_sThere are a number of different ways to think about why this is true, and each explanation exhibits varying degrees of accuracy, depth and level of understanding, but I think that they all have some merit depending on the situation, and what you are trying to convey. Here are a few ways to think about this;

Q and K expressions only contain the gases in the chemical equilibrium reaction, so if the inert gas is not part of the chemical reaction and does not change anything that is in the equilibrium mixture, then it cannot be part of the Q or K expression, and therefore it does not effect anything in the Q or K expression. As such, Q does not change, and there is no need for the equilibrium to shift to restore its value to that of K.

If the volume is constant (which it always will be if the equilibrium mixture is contained in a rigid vessel), and the number of moles of the reactants and products in the equilibrium mixture do not change because they do not interact with the inert gas, then the concentrations (mol/L) and partial pressures (atm) of the reactants and products do not change. As such, Q does not change, i.e., the equilibrium will not need to shift in order to get back to the original K value.

Most misconceptions occur because of the following misunderstandings;

Addition of an inert gas at constant volume DOES increase the total pressure in the system, and as a result, people tend to want to say that the equilibrium WILL shift. They want to say that the equilibrium will shift, because the total pressure has increased, and that they think that therefore the gaseous reaction will shift to the side with the least number of moles of gas, in order to reduce the pressure. This is NOT correct! The only reason that an equilibrium position shifts to the side with the fewer moles of gas when volume is decreased (and therefore pressure is increased), is that the partial pressures of all of the gases present are all increased. As a result, Q changes, and therefore the equilibrium must shift to make Q = K once more. In the case of adding an inert gas, the partial pressures of the gases in the equilibrium system do not change, and therefore Q does not change and no shift is required.

In addition, because people remember that partial pressures can be calculated by taking the mole fraction of each gas in the equilibrium mixture and multiplying by the total pressure, they think that since total pressure has increased, then partial pressures will have increased, and that Q will change, causing a shift in equilibrium position to maintain the original value of K. However, in this case one can think of ‘total pressure’ as being only the total of the pressures of the gases in the equilibrium system, and NOT any ‘total pressure’ that includes any gases that are NOT in the equilibrium system. You may or may not like this idea, mainly because you may want to include the inert gas in the total pressure (which honestly is strictly the correct way to think about things), but if you do want to do that, then consider the following;

If one were to include the inert gas in the ‘total pressure’, then one would also need to include the inert gas in the mole fraction part of the calculation too. The calculation below shows how this ultimately makes no difference to the partial pressures of the gases in the equilibrium system.


Consider a mixture of three gases A, B and C, in equilibrium with one another, with a total pressure of 10 atm.

A(g) + B(g) ⇌ C(g)

Gas A 1.0 mols
Gas B 1.0 mols
Gas C 2.0 mols

Total moles of gas = 4.0 mols
Total pressure = 10. atm

Partial Pressure A = (1/4)*10. = 2.5 atm
Partial Pressure B = (1/4)*10. = 2.5 atm
Partial Pressure C = (2/4)*10. = 5.0 atm

Q = (5.0)/(2.5*2.5)

If one then adds 2.0 moles of inert gas, it exerts another 5.0 atm of pressure, and we can include the inert gas in the calculations. If we do, we find;

Total moles = 6.0 mols
Total pressure = 15 atm

Partial Pressure A = (1/6)*15 = 2.5
Partial Pressure B = (1/6)*15 = 2.5
Partial Pressure C = (2/6)*15 = 5.0
Partial Pressure Inert = (2/6)*15 = 5.0

Q = (5.0)/(2.5*2.5)

i.e., the partial pressures that are entered into the Q expression have not changed, and therefore Q does not change, and therefore there is no shift in equilibrium.

1 Comment

  1. Pamela Hill

    Thank you so much for sharing this. It really helped me understand this topic.

    Reply

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