Why methane burn at high temperature?

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Methane combustion requires a high activation energy of around 1200K, which is attributed to the need for sufficient energy to initiate the reaction between methane and oxygen. Although the reaction is thermodynamically favorable at all temperatures, as indicated by the negative Gibbs free energy change (dG), it does not occur spontaneously without overcoming this activation energy barrier. The reaction produces water and carbon dioxide, which have more degrees of freedom, contributing to a positive entropy change (dS). To initiate combustion, a small energy input, such as from a spark plug, can create a localized high energy density, triggering a chain reaction. Alternatively, increasing the overall temperature can also facilitate the reaction. The complexity of methane's combustion mechanism, despite its simple molecular structure, adds to the challenges in understanding its activation energy.
chewchun
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Methane burns at 1200k without any spark or ignition.
But why is this so?Why so high temperature?
Isn't it that when there is the right proportion of methane and oxygen, combustion will occur with a small energy needed?
Or is it because of high activation energy?
 
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methane reacting with oxygen is thermodynamically favorable at essentially all temperatures.

That's because according to the constant temperature (which is a very nice approximation) free energy equation, dG = dH - TdS, dG will ALWAYS be negative: the binding energy of methane + oxygen is higher (less negative) than the binding energy of the products water and carbon dioxide, so dH is negative, and dS is always positive because the equation

CH4 + 2 O2 -> CO2 + 2 H2O has molecules with more degrees of freedom on the right hand side than on the left hand side.

To make it quantitative, you look up numbers on a thermodynamic data chart.

However, thermodynamically favorable =/= actually happens because to actually happen, the activation energy must be overcome. There's 2 ways to do it. One is to apply a small amount of energy in a tiny place so that the energy density is very high, such as with a spark plug, and a reaction occurs at that point, and the energy released by the reaction causes a chain reaction. Or you can just raise the temperature of everything.
 
chill_factor said:
methane reacting with oxygen is thermodynamically favorable at essentially all temperatures.

That's because according to the constant temperature (which is a very nice approximation) free energy equation, dG = dH - TdS, dG will ALWAYS be negative: the binding energy of methane + oxygen is higher (less negative) than the binding energy of the products water and carbon dioxide, so dH is negative, and dS is always positive because the equation

CH4 + 2 O2 -> CO2 + 2 H2O has molecules with more degrees of freedom on the right hand side than on the left hand side.

To make it quantitative, you look up numbers on a thermodynamic data chart.

However, thermodynamically favorable =/= actually happens because to actually happen, the activation energy must be overcome. There's 2 ways to do it. One is to apply a small amount of energy in a tiny place so that the energy density is very high, such as with a spark plug, and a reaction occurs at that point, and the energy released by the reaction causes a chain reaction. Or you can just raise the temperature of everything.

Is there any reason for the high activation energy of methane combustion?
To me,such a simple molecule seems to be pretty complex upon what you have said...
 
yes there is a reason but it is complicated. i think i remember being taught this in my physical chemistry class but i did not learn it well and forgot.
 
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