Can Thermal Energy Break a Covalent Bond?

[Repetit]

Hey!

I want to estimate the temperature needed to break a covalent bond of strength 1.70 eV. Whould it be correct to compare the thermal energy $$k_B T$$ to the bond strength and find the temperature needed to make the thermal energy larger than the bond strength? I get a temperature of about 11000K, which seems a little high to me.

 

[Gokul43201]

Hey!

I want to estimate the temperature needed to break a covalent bond of strength 1.70 eV. Whould it be correct to compare the thermal energy $$k_B T$$ to the bond strength and find the temperature needed to make the thermal energy larger than the bond strength? I get a temperature of about 11000K, which seems a little high to me.

1. 1.7eV should be closer to 20,000K (room temperature is about 25meV).

2. Given enough time, you can break ONE covalent bond of 1.7eV strength even at room temperature. Assume the molecular temperatures follow a Boltzmann distribution; if the mean temperature is 300K, what fraction of molecules have a temperature of 20,000K? So, clearly, this is a question of kinetics.

3. In the limit of an infinite supply of the reactant molecule or continuously removed product (conditions far removed from equilibrium), the time taken to make the required quantity of product (ie: break the required number of covalent bonds) is given by the (likely first order) rate constant.

4. Note for instance, that thermal decomposition of water is usually carried out (industrially) at temperatures which are about an order of magnitude smaller than the bond energy; under quasi-equilibrium conditions the yield is typically about a percent.
http://adsabs.harvard.edu/abs/1983IJHE...8..675L

 

[Repetit]

Thanks for the reply! Your are right, the temperature is closer to 20,000K, I made a dumb mistake when calculating the value.

I believe things are not as simple as I wanted them to be, because time and other factors play an important role as well.