According to the Wikipedia page on Chemical Equations:

Symbols are used to differentiate between different types of reactions. To denote the type of reaction:
"$=$" symbol is used to denote a stoichiometric relation.
"$\rightarrow$" symbol is used to denote a net forward reaction.
"$\rightleftarrows$" symbol is used to denote a reaction in both directions.
"$\rightleftharpoons$" symbol is used to denote an equilibrium.

How is a "reaction in both directions" and an "equilibrium" any different? Aren't they supposed to be identical?

  • 2
    $\begingroup$ Unfortunately this is the same definition as in the goldbook and no other explanations were given. $\endgroup$ Nov 29, 2014 at 17:14

2 Answers 2


As far as I know, they are used indistinguishably and interchangeably. One or another could be chosen to emphasize whether you are talking about chemical equilibrium or just trying to show that a reaction can take place in both directions depending on the conditions (which can be understood as completely the same...).

I saw the $⇄$ symbol used to specify some conditions for the reaction to take place in one direction and other kind of conditions for the reaction to occur in the other direction, and $⇌$ if conditions are the same for both reactions

For example:

enter image description here

First case is not an equilibrium, but it's true that the reaction can go both directions using the appropriated reagents in each case. But that could not be expressed using $⇌$.

The second reaction is an equilibrium, both reactions are happening continuously under the same conditions. It also could be expressed using $⇄$.

That is the only difference I can tell. Like Martin pointed out, IUPAC does not give more hints about it, so the question cannot be answered for 100% sure.

  • $\begingroup$ This seems reasonable. You can have effects in both directions without them being in equilibrium and I think that this is what is being denoted here. $\endgroup$ Nov 30, 2014 at 16:32
  • $\begingroup$ What about $\ce{<<=>}$ and $\ce{<=>>}$? $\endgroup$
    – Shub
    Jan 29, 2023 at 7:25

A reaction in both directions may be equilibrating, but is not necessarily equilibrating (like if the system is being excited), and it certainly isn't necessarily at equilibrium.

For example, if one puts pure ammonia gas in a container, it will be far from equilibrium with respect to its decomposition products (nitrogen and hydrogen gas) but it will spontaneously proceed in the following direction: $2\mathrm{NH}_3 \rightarrow \mathrm{N}_2 + 3\mathrm{H}_2$. Some time after that, the rate of the reverse reaction becomes non-negligible; $\mathrm{N}_2 + 3\mathrm{H}_2 \rightarrow 2\mathrm{NH}_3$, so that both reactions simultaneously occurring can be summarized as: $2\mathrm{NH}_3 \rightleftarrows \mathrm{N}_2 + 3\mathrm{H}_2$. That doesn't mean the system's at equilibrium, just that the forward and reverse rates are non-negligible.

Finally, the equilibrium condition (so that the concentration of all species remains constant over time) applies, and the forward and reverse rates become equivalent. That's the distinction that's specified with the double harpoon arrows: $2\mathrm{NH}_3 \rightleftharpoons \mathrm{N}_2 + 3\mathrm{H}_2$.

I think the harpoon arrows are most often used to specify reaction rates, so the equilibrium condition, where the forward and reverse rates balance, is marked using them.


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