Understanding the bond dissociation energy of poly-atomic molecules Now I have a breaker which is containing $\ce{CH4}$ of some amount at standard conditions and then I put it on a burner.



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
$$\begin{alignat}{2}
\ce{CH4(g) &amp;-&gt; CH3(g) + H(g)}\qquad &amp;&amp;\Delta_\text{bond}H^\circ=+427\ \mathrm{kJ\cdot mol^{-1}}\\
\ce{CH3(g) &amp;-&gt; CH2(g) + H(g)}\qquad &amp;&amp;\Delta_\text{bond}H^\circ=+439\ \mathrm{kJ\cdot mol^{-1}}\\
\ce{CH4(g) &amp;-&gt; CH(g) + H(g)}\qquad &amp;&amp;\Delta_\text{bond}H^\circ=+452\ \mathrm{kJ\cdot mol^{-1}}\\
\ce{CH(g) &amp;-&gt; C(g) + H(g)}\qquad &amp;&amp;\Delta_\text{bond}H^\circ=+347\ \mathrm{kJ\cdot mol^{-1}}
\end{alignat}$$ 

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?
Whoa!!!

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.