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If we would have benzene in a closed barrel for quite a while at room temperature, some vapor would form fast due to its volatility, right? But let's say that when the barrel was closed, some air got trapped inside. So it's inside at any time. Is there any chance that the benzene would react with the components in the air at room temperature? Or would we simply have benzene vapor and air? Or would we have benzene, some products of reaction and air? I made a sketch for a better understanding of my question.

enter image description here

So which one is the real one? And if it is the second one, would the reaction be similar to combustion, or?Is temperature a major player in this? And if so, how strongly would it affect the reaction?

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  • $\begingroup$ Note that the air (really it's the oxygen that may slowly react with the benzene) will also dissolve into the liquid benzene. In this case, for the reaction $\ce{C6H6 + O2 -> products}$, the concentration of one reactant (benzene) would be greatly increased. Depending on the solubility of $\ce{O2}$ in benzene, the reaction would probably mainly take place in the liquid phase (or maybe at the liquid/vapor interface). Still, regardless of where or how the reaction might take place, the key to your question is the kinetic issues discussed in the answer by Chester Miller. $\endgroup$ – airhuff Feb 24 '17 at 19:24
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Combustion is occurring, but the kinetics are very very very slow at room temperature. So, on a time scale of days, weeks, and months, the system will be closer to the first picture. On a time scale of decades, centuries, and millenia, the system will be closer to the 2nd picture.

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    $\begingroup$ to add a little to you nice answer, it is the activation energy of the reaction benzene/oxygen that is so large that the rate of reaction is so slow. The same is true of many organics molecules, those that make up you and me for example :) $\endgroup$ – porphyrin Feb 24 '17 at 14:31
  • $\begingroup$ Thank you both for your answers! So if one would add enough thermal energy to go over the activation energy of the reaction, then there will be a full combustion process, right? $\endgroup$ – Physther Feb 24 '17 at 19:33
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    $\begingroup$ Sure. Just add a spark. $\endgroup$ – Chet Miller Feb 24 '17 at 19:35

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