Now I have a breaker which is containing $\ce{CH4}$ of some amount at standard conditions and then I put it on a burner.

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In Thermodynamics we are taught about how to find the bond enthalpy of poly-atomic atoms. They say that such molecules (here in this case methane) all four $C-H$ bonds are identical. However, the energies required to break the individual $C-H$ bond in each successive step differs $$\begin{alignat}{2} \ce{CH4(g) &-> CH3(g) + H(g)}\qquad &&\Delta_\text{bond}H^\circ=+427\ \mathrm{kJ\cdot mol^{-1}}\\ \ce{CH3(g) &-> CH2(g) + H(g)}\qquad &&\Delta_\text{bond}H^\circ=+439\ \mathrm{kJ\cdot mol^{-1}}\\ \ce{CH4(g) &-> CH(g) + H(g)}\qquad &&\Delta_\text{bond}H^\circ=+452\ \mathrm{kJ\cdot mol^{-1}}\\ \ce{CH(g) &-> C(g) + H(g)}\qquad &&\Delta_\text{bond}H^\circ=+347\ \mathrm{kJ\cdot mol^{-1}} \end{alignat}$$

and now their bond enthalpy is the mean of all those numbers.

now going back to the diagram we heat up the beaker containing methane then the energy that is absorbed by any one of the $\ce{C-H}$ bond should be equal to the energy absorbed by the other $\ce{C-H}$ bond because all the bond in a molecule should absorb equal amount of energy at the same time because at time $t$ the temperature is same all over the beaker and all bonds are equal. so it should be like that while calculating the bond dissociation energy of a poly-atomic molecule (take the example of methane) calculate it like $4\times 427\ \mathrm{kJ\cdot mol^{-1}}$. (is this a common sense?)

all bonds Should broke up at once!

but we know that this does not happen, which means that all the bonds in a molecule does not absorb heat of equal amount at the same time or are they unequal? And fairly speaking it is painful to me to visualize this thing maybe because I am not able to understand this reality.

How do I comprehend this thing?

In Short

  • all bonds of $CH_4$ are equal

  • The temperature of beaker is same all over its volume

  • energy that is absorbed by any one of the $\ce{C-H}$ bond should be equal to the energy absorbed by the other $\ce{C-H}$ bond
  • bonds should absorb equal energy at time t
  • the should all break at once at same time t

this do not happen so how to understand this thing?


You are making several assumptions that assume equilibrium conditions. In this sense all the bonds are equal: they are on average over time.

However, at a molecular level (one molecule at a time) this is not true especially if enough energy is being added to break some bonds. The energy distribution for each bond will be Boltzmann-Like so some bonds will have enough energy to break. Once broken, of course, the other bonds no longer are CH4 bonds but are CH3 bonds of some sort. Hence we would not expect the next dissociation energy to be the same as the first.

The critical idea here is that the thermodynamic quantities of bond dissociation are averages and at a molecular level the actual energy is distributed according to statistical rules: some atoms or some of their bonds have more energy than others. The reactions occur not when the average energy input is equal to the dissociation energy, but when a small number of bonds at the extremes of the distribution have enough energy.

This is a very important idea (called statistical thermodynamics) in understanding chemical reactions.



The C-H bonds in $\ce{CH4(g)}$ are not the same as the C-H bonds in $\ce{CH3(g)}$. The $\ce{CH3(g)}$ "molecule" is electrically neutral but it is a radical with an unpaired electron in one orbit. In fact the physical orientation of the molecule changes to a planar configuration.

enter image description here

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    $\begingroup$ You can use \ce{} to format chemical expressions. See this meta post or, if you have lots of time, the mhchem package documentation. $\endgroup$ – orthocresol Oct 26 '15 at 17:41
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    $\begingroup$ okay, i got that, that bonds of $CH_4$ not same as that of $CH_3$, i pointed out that at time t all the bonds should break up at once but it is not that, so i want to know that why this doesn't happen $\endgroup$ – Sanmveg saini Oct 26 '15 at 17:46
  • $\begingroup$ If guess you're missing the point that C(g) doesn't exist at STP. A gaseous carbon atom would only exist in a very rarefied gas. The same goes for the rest of the "weird" molecules. // A chemical reaction only happens when two "things" interact "physically." How would you pluck 4 atoms of H from methane at the same time (four directions at once)? $\endgroup$ – MaxW Oct 26 '15 at 20:03
  • $\begingroup$ To look at it another way, let's imagine two very high velocity methane molecules (thus each molecule has a high "temperature") colliding in deep space. I can calculate the kinetic energy necessary to break all 8 bonds and yield 10 free atoms. That chemical reaction just wouldn't happen. You'd get some weird assortment of polyatomic species depending on the orientations of methane molecules as they collided. $\endgroup$ – MaxW Oct 26 '15 at 20:44

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