Is there a general effect on the rate of reaction of a dynamic equilibrium when an inert gas is introduced at a constant volume? I know that the position of equilibrium won't change, but much like a catalyst which increases the rate of both forward and reverse reactions, I'm wondering if the introduction of an inert gas will have a similar effect.

My preliminary thoughts were that the inert gas would 'block' successful collisions by the reactants and products, thereby decreasing both forward and reverse rates. However, these 'blocked' collisions could simply be the result of more collisions taking place since there are more molecules in the same volume so the rate doesn't actually change.

My other thought was that certain inert gases might increase the rate of reaction by acting as a heterogenous catalyst – a surface for the reaction to take place.

Clearly, my thoughts haven't led to a definite answer so I was wondering if anyone could clear up exactly what happens to the rate of reaction when an inert gas is introduced. Thanks!


2 Answers 2


At otherwise the same partial pressure of the reacting gas, an inert gas slows down heterogenous reactions of a gas with condensed phases.

It is caused by local additional decreasing of partial pressure of the reacting gas at the boundary layer, where it gets partly depleted and is not being resupplied fast enough from the bulk volume.

The effect is the most obvious if products stay in the condensed phase. Like

$$\ce{O2(g) + A(s) ->[O2 0.21 atm vs air] B(s)}$$

Pure oxygen would have boundary pressure equal to bulk pressure. But with nitrogen and fast enough reaction, oxygen boundary partial pressure may drop significantly -- having near pure nitrogen there -- and so would the reaction rate.

This inert gas effect is in breathing or fuel burning context empirically confirmed by better tolerance of lower oxygen partial pressure due altitude, compared to lowered oxygen percentage at normal pressure.

The typical threshold range for burning organic/fossil fuel is $\pu{16-18 \%}$ of oxygen in air, with the partial pressure $\ce{0.16-0.18 atm}$ . But in mountains, locals burn wood at much lower oxygen pressure, due lower air pressure.

For homogenous reactions, the effect is more subtle, but an extra inert gas generally slows down reactions, by the following effects:

  • Its molecules take way part of energy released by the reactions, in macro scale via heat capacity and on molecular level. It decreases the probability of reactant collisions leading to the reaction.
  • It affects diffusion of limiting reactant, especially if the inert molecules are slow or reactants are not in the stoichiometric ratio.
  • If some reactions steps involves some intermittent reactants in excited state, it may cancel their excitation before they can react.

There is possibly more effects.

"Unimolecular" reactions are already well described by Porphyrin.

  • $\begingroup$ Do you know any information about what happens for non-hetereogenous reactions? In other words, just gases reacting like in the Haber process. $\endgroup$ Commented Apr 12, 2023 at 7:56
  • $\begingroup$ @PenandPaper But that IS a heterogenous reaction, gases reacting on surface of solid catalyst. If we added to H2 and N2 e.g. Ar, there wouldb be locally decreased partial pressures of N2 and H2 on catalyst surface. I mean even more than just due formed NH3. $\endgroup$
    – Poutnik
    Commented Apr 12, 2023 at 8:02
  • $\begingroup$ Could you also just explain why the inert gas at the boundary layer stops the reacting gases from being replenished. Wouldn't this happen without the inert gas since the reagents are reacting anyway? $\endgroup$ Commented Apr 13, 2023 at 0:52
  • $\begingroup$ @PenandPaper see the question update. $\endgroup$
    – Poutnik
    Commented Apr 13, 2023 at 5:35

In the gas phase in so called 'Unimolecular Reactions' an inert gas does play a vital part by collisionally exciting a potentially reactive molecule. The scheme is

$$\ce{A + M } \overset{k_2}{ \underset{k_{-2}}{\rightleftharpoons}} A^* + M \\ \ce{A^* \overset{k_1}\to product}$$

where $M$ is the inert gas and $A^*$ is vibrationally excited or 'energised' $A$. The rate law depends on $[M]$ and $[A]$ and can be found by applying steady state conditions to $[A^*]$. Methyl isocyanide $\ce{CH3NC}$ to acetonitrile $\ce{CH3CN}$ isomerisation is an example of such a reaction. The simple scheme (above) is due to Lindemann, a more advanced one is called RRKM theory.

Additional comment.

The rate of product formation with steady state on $A^*$is

$$\displaystyle \frac{dP}{dt}=\frac{k_1k_2[A][M]}{k_1+k_{-2}[M]}$$

so that the rate is small and bi-molecular at low pressures of $[M]$ i.e. $(k_2[A][M])$ but becomes larger and unimolecular at high pressure when $k_{-2}[M]\gg k_1$ and is $k_1k_2[A]/k_{-2}$. Often a plot of $\displaystyle \frac{1}{[A]}\frac{dP}{dt}$ vs $[A]$ is made better to illustrate these effects.

  • $\begingroup$ So the inert has in this case increases the rate? $\endgroup$ Commented Apr 13, 2023 at 0:50
  • $\begingroup$ its not quite as simple as that,I've added a comment in my answer due to lack of space here. $\endgroup$
    – porphyrin
    Commented Apr 13, 2023 at 7:44

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