Why methane burn at high temperature?

  • Thread starter chewchun
  • Start date
  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
901
3
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.
 
  • #3
24
0
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....
 
  • #4
901
3
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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