Ok, so I learned about the equilibrium constant. Now, I've seen that the equilibrium constant of burning is extremely small $(K \ll 1)$. here, I have a question. you see, $K$ is still NOT 0, which means that the forward reactions happen at least a tiny bit. Then, shouldn't we see some parts of anything burning at least a little bit?
Answer
The equilibrium constant for combustion of organic matter in air with oxygen is not small, but extremely large ($K_\mathrm{eq} \gg 1$), as is expected from a reaction that is simultaneously very exothermic and (usually) increases entropy due to the formation of more gaseous molecules than the input oxygen.
The major reason carbon-based life can exist at ambient temperature in an oxygen atmosphere is purely kinetic, not thermodynamic. You, the tree outside and everything else made of carbon is right now undergoing continuous combustion. However, in the absence of catalyst, this process is too slow below a couple hundred degrees Celsius for it to be self-sustaining. More technically, combustion of organic matter is a highly exergonic process, but the activation energy is high. The meagre amount of heat generated by the handful of molecules reacting is too quickly diluted into the surroundings, and the reaction does not accelerate and spiral out of control (a fire, as described by the eternal Feynman).
Very luckily for us, Life figured out this vast untapped source of chemical energy held back by kinetics approximately three billion years ago and developed a whole metabolic process to extract this energy in a stepwise fashion using catalysis, which we call aerobic respiration. Without it, multicellular organisms could well never had evolved.
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