0
$\begingroup$

I heard that spontaneous reaction happens if $\Delta G=\Delta H-T\Delta S$ is negative.

For combustion of methane, according to Chemguide:

  • $\Delta H=\pu{-891.1 kJ K^{-1} mol^{-1}}$
  • $\Delta S=\pu{-0.2422 kJ K^{-1} mol^{-1}}$

Thus, I would deduce that at normal temperature, $\Delta G<0$, combustion of methane is spontaneous, while at very high temperatures, it will stop.

But, from another reference, I deduce the opposite conclusion: if we consider the diagram below (taken from https://fr.wikipedia.org/wiki/R%C3%A9action_chimique),

Energy diagram for methane

I would think that in order to be able to go to the intermediate state, one needs to provide energy, thus that it could not be spontaneous for this reason. One needs to provide energy, for example by heating, that is increasing temperature.

How is that possible? Any comment?

$\endgroup$
11
  • 3
    $\begingroup$ Practically every spontaneous reaction requires activation energy. This comes from thermal motion of molecules, thermal, mechanical or electric shock, light, breaking crystals or from already ongoing reaction. Or, spontaneous reaction is not ongoing, if needed activation energy is too high. We then say reactants are thermodynamically unstable, but kinetically stable. Typical example for that are diamonds. $\endgroup$
    – Poutnik
    Oct 6 '20 at 22:04
  • 1
    $\begingroup$ Thermodynamics and kinetics are two different things. $\endgroup$
    – Jon Custer
    Oct 6 '20 at 23:21
  • 4
    $\begingroup$ Does this answer your question? Difference between thermodynamic and kinetic stability $\endgroup$
    – Mithoron
    Oct 7 '20 at 1:16
  • 2
    $\begingroup$ A catalyst can reduce the activation energy, but that "bump" still must be passed to "roll downhill", even though the reaction is thermodynamically favorable. For example, methanol won't ignite at room temperature, but a Pt catalyst can set it on fire. Look up "catalytic lighter". $\endgroup$ Oct 7 '20 at 2:48
  • 1
    $\begingroup$ Thank you all. Wikipedia saus something similar : "Because spontaneous processes are characterized by a decrease in the system's free energy, they do not need to be driven by an outside source of energy.". But so the word spontaneous is a bit clumsy $\endgroup$ Oct 7 '20 at 6:07
3
$\begingroup$

"Spontaneous" means different things in different contexts

Your penultimate paragraph captures a key idea. The explanation for why this is right requires a recognition of the context of the term "spontaneous".

The context of the statement at the start of the question $\Delta G=\Delta H-T\Delta S$ is negative is thermodynamic stability. But this is somewhat at variance with the more natural use of the term which implies "things happen without being pushed". This idea is closer to the idea of kinetic stability in chemistry.

You correctly identify the need to add energy to get the reaction past the transition state. Even if the reaction overall releases energy (thermodynamically spontaneous) the reaction won't just happen if there is a huge barrier to getting over the transition state. There are big kinetic barriers that stop the reaction "just happening". Oxygen and gasoline will react to release energy but this doesn't happen without the push given by the spark plug in the engine of a car.

That barrier is so low in some cases a compound will react with nearly everything with little excuse (chlorine trifluoride will set fire to asbestos). That is a spontaneous reaction in any context. Luckily, few thermodynamically spontaneous reactions are also kinetically spontaneous or humans would catch fire in air.

So when you see the term "spontaneous" ask what is the context: thermodynamic or kinetic? And don't confuse them.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks a lot for explanations $\endgroup$ Oct 7 '20 at 16:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.