So, regarding a generic reaction (for the sake of argument, let us assume that it is completely gas phase), we have Le Chatelier's Principle. Let us use this example:

$$\ce{2SO2(g) + O2(g) <=> 2SO3(g)} \quad\quad \Delta H < 0$$

Now, we have the questions: oxygen gas is added, heat is added, a noble gas is added, etc.

For the addition of oxygen gas, we obviously shift right, and for heat, we shift left.

For a noble gas, I would expect no shift to happen. Indeed, this is what the SAT's seem to say, and I completely agree. However, my teacher seems to have the conception that when one adds a noble gas to a reaction such as above, you decreases molecular space, and the reaction does shift/speed up.

I have a test tomorrow, and if she gives me a question like this, I am planning to write that there is no shift or rate change. Now, if she marks this wrong, I need a formal argument to support my claim.

Am I correct in saying that with the addition of a noble gas, there is no change in either equilibrium shift or rate change? Why?

  • $\begingroup$ Do you mean "ideal gas" (= a gas that obeys the equation of state $pV = nRT$), or "noble gas" (= He, Ne, Ar, Kr, Xe)? $\endgroup$ Commented Jun 9, 2017 at 16:17
  • $\begingroup$ Noble gas, for example, many problems use Ar. $\endgroup$
    – Equinox
    Commented Jun 9, 2017 at 20:10
  • $\begingroup$ Oh, ok, I see. I guess the point is that the gas does not react i.e. it is inert. $\endgroup$ Commented Jun 9, 2017 at 20:22
  • $\begingroup$ @Equinox Congradulations on your Yearling badge! $\endgroup$
    – A.K.
    Commented Apr 18, 2019 at 19:32
  • $\begingroup$ For the record, SO3 is described by Wikipedia as "Colorless to white crystalline solid which will fume in air.[1] Colorless liquid and gas.[2]" And, "Melting point 16.9 °C (62.4 °F; 290.0 K)" and "Boiling point 45 °C (113 °F; 318 K)". So why does this equilibrium not specify a temperature as the SO3 could well be NOT in a gaseous state? Temperature permitting, presence could be as a particle suspension in air, so increasing pressure is expected to benefit SO3 formation. $\endgroup$
    – AJKOER
    Commented Mar 7, 2020 at 13:53

1 Answer 1


See, in an equilibrium reaction in gas phase at constant temperature, noble gases can be added in two ways―

  1. At constant volume in the system
  2. At constant pressure in the system.

Effect of adding noble gases at constant volume

At constant temperature and volume, if noble gas is added, total number of molecules,i.e., amount of the element increases (or you can say, 'number of moles' increases, though it's not correct, have you ever heard of number of kilograms?) in the system. As a result, also total pressure of the system increases, but there is no change in partial pressure of the reactants and products (because total pressure increases and mole fraction decreases). So, there is no effect on equilibrium by adding noble gases in a system at constant temperature and volume.

Effect of adding noble gases at constant pressure

At constant temperature and pressure, addition of noble gas to the system leads to increase in volume (cause, amount of the element increases). As a result, partial pressure of the reactants and products decreases (because total pressure constant but mole fraction decreases) as well as the sum of partial pressure of the reactants and products also decreases. Here, according to Le Chatelier's Principle, the equilibrium would be shifted in that direction where number of molecules or volume increases.

  • $\begingroup$ @chail10...So the backward reaction is favoured. Interesting! $\endgroup$ Commented Jun 9, 2017 at 9:55
  • $\begingroup$ @Eashaan Godbole In this case, yes! As total number of gaseous molecules in the left (3) > total no. of gaseous molecules in the right (2) $\endgroup$
    – chail10
    Commented Jun 9, 2017 at 10:02

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