I already know that nitrogen's triple bond requires a lot of energy to be broken, so the activation energy is very high, but the book also lists the endothermic energy change as a reason.

  1. Does the positive enthalpy change have something to do with the reaction not being spontaneous i.e. due to positive free energy?

  2. Is there a link between $E_a$ and free energy when deciding whether a reaction will occur?


1 Answer 1


Yes, you are right, the reason primarily looking at it from a theoretical point of view is the high bond dissociation energy of $\ce{N#N}$

Let's look at this aspect from every point of view

  1. Theoretical- High bond dissociation energy of $\ce{N#N}$

  2. Kinetics- We have the Arrhenius equation given below, which states the relationship between the rate constant and activation energy of the reaction.

$$k = A \mathrm e^{-E_\mathrm a/(R T)}$$ So essentially, this relation shows that if activation energy is higher, then the rate is slower. However, if we increase the temperature sufficiently, then the reaction does start to happen significantly. (Hence, oxygen and nitrogen do react when lightning strikes)

  1. Thermodynamics- We have the very famous relation of thermodynamics

$$\Delta G= \Delta H-T \Delta S$$ So now lets again look at each term here

  • $H$- Enthalpy of the reaction, it is very high, as we discussed before, the bond dissociation energy of $\ce{N#N}$ is very high.
  • $S$- Entropy change won't be any significant as the reactants and products both are gases, so this is not an entropy-driven reaction.

Thus , in every way possible we can conclude that nitrogen and oxygen do not normally react at room temperature.

  • $\begingroup$ Thanks a lot! You clearer a lot of things up. One last thing. From what I understand, even if the reaction was exothermic somehow and was thus thermodynamically favourable, it still wouldn’t occur due to the high activation energy barrier. So should the enthalpy change be listed at all as a reason for the reaction not occurring? $\endgroup$ Commented Jan 27, 2020 at 10:37
  • $\begingroup$ chemistry.stackexchange.com/questions/105580/… $\endgroup$ Commented Jan 27, 2020 at 14:24
  • $\begingroup$ It is endothermic not exothermic $\endgroup$ Commented Jan 27, 2020 at 14:24
  • $\begingroup$ I know it is. That was not my question $\endgroup$ Commented Jan 27, 2020 at 17:35
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    $\begingroup$ Ohh I got your question, we'll see what you are talking about is two different things. Activation is a kinetically factor, and enthalpy is thermodynamic factor. They are related. But the realtion between kinetics and thermodynamics depends on other factors too. Hence, in every reaction you would hear phrases like it is kinetically favourable but not thermodynamically so, although activatiom and enthalpy are related, don't use only one of them to conclud your reaction $\endgroup$ Commented Jan 28, 2020 at 10:59

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