My chemistry teacher was talking about how there are two requirements for a reaction to occur:

  1. The reactants must have enough energy; and
  2. they must also have the correct orientation for a reaction to occur.

What does it mean for molecules to have the "correct orientation", perhaps using the gas-phase reaction as an example?

$$\ce{2H2 + O2 -> 2H2O}$$

  • 6
    $\begingroup$ Generally two or or more atoms need to approach in a way that they aren't blocked by other atoms so they can contact directly and angle between axis of their contact and existing bonds is right for the specific reaction. $\endgroup$
    – Mithoron
    Jul 9, 2016 at 22:51
  • 1
    $\begingroup$ If two atoms approach and react then orientation is obviously not important, for molecules orientation is important as stated by @Mithoron. In solution molecules diffuse together then have to find the correct orientation to react. This they do by mutual translational and rotational diffusion (randomly turning around) until they are in the correct position to react. If they diffuse apart before they find the correct alignment, then no reaction occurs at this encounter. $\endgroup$
    – porphyrin
    Jul 10, 2016 at 7:50
  • 1
    $\begingroup$ Related for the specific reaction: Reaction mechanism of combustion of H2? It is, however, not a really good example of "correct orientation". $\endgroup$ Mar 28, 2017 at 9:42

1 Answer 1


What does it mean for molecules to have the "correct orientation", perhaps using the gas-phase reaction as an example? $$\ce{2H2 + O2 -> 2H2O}$$

This is a good example, but a complicated one. It is a chain reaction with three steps that have radical species as reactants and products (chain propagation), and one step that has radical species as reactants but not as products (chain termination). If you add up reactions 1 through 4, you will get the net reaction.

$$\ce{ O + H2 -> OH + H [1]}$$ $$\ce{ H2 + OH -> H2O + H [2]}$$ $$\ce{ O2 + H -> O + OH [3]}$$ $$\ce{ H + OH -> H2O [4]}$$

In order for the chain reaction to start, you need an initiation reaction starting with stable molecules making radicals. This is a slow reaction requiring a lot of energy, and there is some debate what it is. Here are two suggestions discussed in the literature.

$$\ce{ O2 + H2 -> OH + OH [5]}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ce{ O2 + H2 -> H2O + O [6]}$$

These two reactions require different orientations of the dioxygen and the dihydrogen molecules when they collide. This is because different bonds are broken and made when the different products are made. The picture below shows a "good" orientation for reaction [5] on the left and for reaction [6] on the right (The circles show the reactant molecules, and the dashed lines show the bonds that have to be made to get the products). enter image description here

With the orientation on the left, it is easier to make two OH radicals and more difficult to make a water (with an O atom remaining). With the orientation on the right, it is easier to make a water molecule (with an O atom remaining) and much more difficult to make two OH radicals. Depending what the products are, some orientations upon collision are productive (lead to product) and some orientations are non-productive (no reaction occurs).

There is a third suggestion for the initiation step in the literature, and it requires an orientation for the collision different from the two orientations shown above. Can you figure out the best orientation for reaction 7?

$$\ce{ H2 + O2 -> H + HO2 [7]}$$

Source for reaction mechanism: NASA report


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