Can O2 combine with H2 to form water without activation energy?

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Hydrogen gas (H2) and oxygen gas (O2) can theoretically react to form water (H2O) without external intervention, but this occurrence is extremely unlikely at room temperature. Although some molecules may possess enough kinetic energy to overcome the activation energy barrier, the probability is minimal, resulting in a reaction rate that would take billions of years. As temperature decreases, the likelihood of reaction diminishes further due to a reduction in the number of molecules with sufficient kinetic energy. The reaction rate is mathematically described by the equation e^(-Ea/kT), indicating that lower temperatures lead to significantly slower reaction rates. Activation energy remains a necessary requirement for the reaction to occur, confirming that spontaneous reaction without it is not feasible.
kevin_tee
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Let say that I have hydrogen gas and oxygen gas mix together, is there any chance that some H2 and O2 will react to form water without doing anything to it? I know that there needs to be activation energy to start the reaction, but are there any chance of reaction happening without activation energy? Thank you
 
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Even at room temperature, some small fraction of the gas molecules has high enough kinetic energy to react with each other (look up Boltzmann distribution). However, the fraction is very small and the reaction would probably take literally billions of years at room temp.
 
Thank you, so in lower temperature the less probability of H2 and O2 combining is lower because the gas molecule with high kinetic is less than high temperature, did I understand it correctly?
 
Yes, the reaction rate is proportional to ##e^{-\frac{E_{a}}{kT}}##, where ##E_{a}## is the activation energy and ##k## is the Boltzmann constant. Because of the behavior of the exponential function, the reaction rate very rapidly becomes slower when temperature is decreased.
 
hilbert2 said:
Yes, the reaction rate is proportional to ##e^{-\frac{E_{a}}{kT}}##, where ##E_{a}## is the activation energy and ##k## is the Boltzmann constant. Because of the behavior of the exponential function, the reaction rate very rapidly becomes slower when temperature is decreased.

Thank you, now I understand
 
kevin_tee said:
I know that there needs to be activation energy to start the reaction, but are there any chance of reaction happening without activation energy? Thank you

No, you still need activation energy.

2~H_{2}(g)+O_{2}(g)~\xrightarrow{\Delta}~2~H_{2}O(g)~\ \ \ \ \ \ \ \ \ \ \Delta H^{\circ}=-483.6~kJ~mol^{-1}
 
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