In a collision between two molecules, can the relative velocity of one with respect to the other be too large for a reaction to occur? For example, suppose that two molecules collide with (a) a perfect orientation to enable a reaction and (b) kinetic energy in excess of the activation energy, but their relative velocity is very large in magnitude. Will the two molecules simply bounce and part ways chemically unchanged?
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$\begingroup$ If molecules are colliding with good orientation and a little more than the activation energy, they will react. $\endgroup$– MauriceCommented Aug 16, 2022 at 19:07
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1$\begingroup$ ... but it may happen, if they cannot release the the reaction energy by other way, they revert the reaction back. Like 2H <=> H2* -> H2 + energy $\endgroup$– PoutnikCommented Aug 16, 2022 at 19:13
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$\begingroup$ @Poutnik - did you mean the last $H2$ to be $2H$? $\endgroup$– Jon CusterCommented Aug 16, 2022 at 19:40
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$\begingroup$ @JonCuster No, I did not. If H2* is not able to pass the energy, the bond breaks. If it is, H2 stays. Notice the first arrow is bidirectional. $\endgroup$– PoutnikCommented Aug 16, 2022 at 22:56
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$\begingroup$ Too high kinetic energy may lead to alternative, destructive reactions. O may react with O2 to form O3, but can with enough energy also form 3 O instead. $\endgroup$– PoutnikCommented Aug 17, 2022 at 11:46
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1 Answer
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One important thing to remember is that rarely is there ever only one possible reaction for two molecules to engage in; we often make distinctions between "kinetic" reactions (reactions that occur because their activation energies are low) and "thermodynamic" reactions (reactions that occur because the products have favorable energies). However, every reaction we'd reasonably do in a lab is technically a "kinetic" reaction, because extremely high-energy velocities will eventually surpass the activation barriers to extremely unstable products like carbanions and random molecular fragments. Basically, go fast enough and your molecules will just explode on contact. We like to think of gas-phase molecules as billiard balls bouncing around, and that can be a pretty good model! Especially considering that if you hit two billiard balls together fast enough, they will also just explode.
So we're asking "are there kinetic energies below the activation energies of competing reactions that will make collisions faster than the time-scale of bond formation?" And the answer is a very confident "no"—bond formation and rearrangement occurs on the femto- and picosecond time-scales (mere billionths of a second long) and there is simply no way to make them go fast enough such that they'd spend too little time in the regions where suitable electron-wavefunction overlap for a transition-state occurs, especially if you were hoping they don't just explode.
If you're interested in high-energy reactions, I recommend exploring some of the theoretical mechanisms behind fragmentations in mass spectrometry. Much of mass spectrometry leverages kinetic energy to access high-energy fragmentary reactions for structural analysis of compounds.
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$\begingroup$ What about molecules travelling at 99% speed of light passing close to each other? Close enough for chemical reactions to occur. $\endgroup$ Commented Aug 18, 2022 at 19:39
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1$\begingroup$ basically all our models for chemistry break down at relativistic speeds. the Schrodinger equation—which dictates all modeling for orbitals that engage in bonding—is not Lorentz invariant, and thus cannot be applied in cases of relativistic speeds. one could attempt solving for a simple molecule at relativistic speeds using the Dirac equation or QED, but these computations might be insoluble and would almost definitely not yield interactions relevant to chemistry of large enough molecules. interesting thought experiment though, I might play around with it. $\endgroup$ Commented Aug 18, 2022 at 21:27
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$\begingroup$ so we don't know the answer yet for relativistic speeds? $\endgroup$ Commented Aug 22, 2022 at 14:30
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1$\begingroup$ @suyashsingh234 i don’t mean to imply that—*i’m* uncertain, but this may have been investigated before. treatments with the Dirac equation are not unheard of; it’s roughly used to approximate the behavior of core electrons in large molecules, but i personally am unaware of treatments for valence/bonding electrons (i’m sure it’s been looked at) $\endgroup$ Commented Aug 22, 2022 at 16:26
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$\begingroup$ It takes time for molecules to rearrange, dispose of excess energy etc. This could be a reason why exothermic reactions become less favorable as temperature is increased and molecules decompose at high temperatures. The potential energy for bond formation must be released either by radiation or transfer to colder molecules. When this cannot happen, the bond does not form. $\endgroup$– jimchmstCommented Nov 17, 2022 at 19:38