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Liquid oxygen is somewhat hairy to work with, as, if it comes in contact with essentially anything combustible, it has an unfortunate tendency to detonate violently if looked at sideways.

Gaseous oxygen obviously does not do this, as can be seen from the fact that combustible objects can sit indefinitely in gaseous oxygen without exploding when touched.

This seems backwards, as, at a given pressure, liquid oxygen is colder than gaseous oxygen, and materials are generally more reactive at higher temperatures - thus, intuitively, one would expect liquid oxygen to be less reactive than gaseous oxygen, not more. Why is this not so?

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    $\begingroup$ related: chemistry.stackexchange.com/q/2631/23561 $\endgroup$ – A.K. Mar 31 '19 at 3:30
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    $\begingroup$ @A.K. The related question lacks imho a good answer that addresses the present question. It would be nice to see a more complete answer. $\endgroup$ – Buck Thorn♦ Mar 31 '19 at 8:11
  • $\begingroup$ I would guess that it would be partially related to the fact that liquid oxygen is much more dense than gaseous oxygen. $\endgroup$ – fyrepenguin Mar 31 '19 at 8:12
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    $\begingroup$ I would guess so as well. But, OTOH, it is opposed by the temperature dependence of reaction kinetics. Remember the empirical rule the reaction speed raise 2-4 times for each 10 Deg C. $\endgroup$ – Poutnik Mar 31 '19 at 9:05
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    $\begingroup$ Yes, liquid oxygen mixtures are dangerous and sometimes explosive, but most need some kick to start the violent process off. Starting the oxidation process isn't the issue: the issue is how fast it can go once started. Also don't forget that many things are very, very combustible in high gaseous concentrations of oxygen (same issue: fire proceeds much faster when there is more oxygen). $\endgroup$ – matt_black Mar 31 '19 at 13:12
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Since oxidation is generally strongly exothermic, a large part of the difference in reactivity can be explained by kinetics (rather than thermodynamics): the denser liquid brings oxidant (oxygen) into intimate contact with the oxidized substance throughout the reaction. Consumed oxygen is quickly replenished by surrounding liquid oxygen. In the gas the exothermic nature of combustion reactions accelerates reaction with oxygen gas by increasing the collision rate between oxygen molecules and the surface being oxidized. In the liquid this temperature effect is not as important because the oxidized surface is already bathed in oxygen molecules. Also, although the gas might be at a higher T, as assumed in the question, local heating of liquid oxygen around the oxidized surface due to the highly exothermic nature of most oxidation reactions can also add to the faster propagation rate of reactions in the liquid. The importance of thermal effects might also be reduced (if not trumped) by the acceleration of concerted events during the oxidation (such as sequential events at the same or neighboring sites or while in a short-lived activated state), since in the liquid collisions between surface and more than one oxygen molecule at neighboring sites will occur at rates that are orders of magnitude faster than in the gas.

This is admitedly an answer in need of some quantitative support.

Here's a "back-of-the-envelope" justification for the importance of density:

It turns out that if you use a model based on the Maxwell-Boltzmann kinetic theory then the collision frequency with a surface is proportional to $p/T^{1/2}$. Therefore, if you ignore the thermal activation energy for reaction, you can assume that the reaction rate is proportional to $p/T^{1/2} = \rho _m RT^{1/2} = RT^{1/2}/V_m$. This means we will have the same collision frequency for two different set of values of T/V when

$$\left(\frac{V_{m2}}{V_{m1}}\right)^2=\left(\frac{T_{2}}{T_{1}}\right)$$

Therefore, we can counter the effect of a 200 K drop in T by increasing the density (reducing the molar volume) by a factor of ~14. Now at STP the molar volume of a gas is ~20 L/mol, while for a compressed (liquified) gas the molar volume is 100 times smaller, say 0.2 L/mol. Therefore the increase in density more than makes up for the drop in T, at least within this simplified model.

In more general terms, we can explain away much of the loss of the kinetic advantage of higher temperature by the higher density in the liquid. So based on this simplified argument oxygen at a liquid-like density can be more reactive than a much warmer gas.

Note that the activation energy for reaction of $\ce{O2}$ is not negligible due to the relatively stable structure of the oxygen molecule. Therefore kinetic effects involving rapid heating and thermal activation of oxygen (due for instance to local accumulation of thermal energy) are probably also important to explain the reactivity of LOX.

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    $\begingroup$ What about temperature dependence of the reaction kinetics ? In sense of probability to gain the reaction activation energy ? (Arrhenius equation) $\endgroup$ – Poutnik Mar 31 '19 at 9:08
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    $\begingroup$ @Poutnik I try to explain that the high density of oxidant in the liquid overwhelms any temperature factors. There is the business of activation energy which I don't address. I assume other factors such as the acceleration of concerted processes (such as solvent cage effects) is far more important. $\endgroup$ – Buck Thorn♦ Mar 31 '19 at 9:14
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    $\begingroup$ Well there is empirical rule reactions are faster 2-4 times eachv10 deg C. If we take just 2, and 200 Deg C different to ambient, it is $2^{20}$. About 1 million times slower. It does not seem to be matched by much more then 1 million times more frequent collisions. What you say may be true, but is strange, anyway. $\endgroup$ – Poutnik Mar 31 '19 at 9:21
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    $\begingroup$ @Poutnik I'll add an argument on this to my answer. Let's just say 3 orders of magnitude can be explained by the density alone, the remaining 3 orders require a different explanation. $\endgroup$ – Buck Thorn♦ Mar 31 '19 at 9:23
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    $\begingroup$ Sure, I just wanted just to share different point of view. $\endgroup$ – Poutnik Mar 31 '19 at 9:52

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