Light As An Activation Energy

[aj47]

Sorry long question but this has been bothering me for a while..

 

How is it that in some reactions light can provide the activation energy requred to break the neccesary bonds when in terms of temperature the energy is ridcoulously high.

 

e.g. chlorination of methane

Cl2 ==> Cl + Cl

CH4 + 2Cl => CH3 + HCl etc etc

 

In this reaction light is enough split the chlorine into free radicals while in its absence the heat required is over 400C.

 

I would imagine only 1 mole of photons would be required to break 1 mole of CL2 molecules so you could calculate the activation energy using E=hf right.

I can't be bothered to do the calculations but becasue of planks constant its going to be a small amount of energy compared to 400C.

 

Does anyone know why this is and what is the relationship between molecules that can gain energy from light, is it valancy of electrons or bond locations??

Help would be much appreciated.

[BhavinB]

There are inefficiencies and probabilities (related to chlorine concentrations) which play a role here. 1 mole of light will in no practical way beak one mole of chlorine gas.

 

As to answer your question though...both the valence electron states and the nuclear vibrational and rotational states play a big role in how light is absorbed by a molecule.

[aj47]

I understand that very few photons are going to be absorbed but if they don't then they won't count towards the activation energy, or are you saying a bond needs to constantly bomarded by light to break?

[jdurg]

Typicaly speaking, when 'light' is referred to they generally mean the shorter wavelength sections of light such as the blues and violets and even into the ultaviolets which actually have a good deal of energy.

 

As an example, if you mix equal moles of hydrogen and chlorine gas and then expose it red light, nothing will happen. Turn the light off and replace it with a blue filter and the reaction goes immediately.

[insane_alien]

1 mole of photons would be involved in the reaction yes. but the conversion would be only a small fraction say around 1% of the photons would actually strike a molecule and of those maybe 2% would strike it at the right angle. so my guesstimate is around 0.02% probably wrong though

[aj47]

but still, will the activation energy in light be proportial to the activation energy in heat?

[ajw2255]

No it wouldnt, becasue the light acts as a catalysis so the two activation energies wouldnt be equal. I dont know of a way to measure it but as previously stated, the light absorbacne would be small and the reaction involves a fair few intermediates, so its not that straight forward.

[aj47]

Ok thanks

[swansont]

I can't be bothered to do the calculations but becasue of planks constant its going to be a small amount of energy compared to 400C.

 

If you actually had bothered to do the calculation - even an order-of-magnitude estimate - you'd find your intuition was wrong. h may be small' date=' but so is k. Thermal energy is given by kT, and k ~10[sup']-4[/sup]eV/K. Visible light is of order 1 eV of energy, which is significantly more (at 400 C). So even accounting for the Maxwell-Boltzmann distribution of energies in a thermal sample, there is sufficient energy available from photons to provide the activation energy.

[aj47]

Would you say light energy always works by being converted into heat energy?

Becasue by definition temperature is the extent of the vibration of atoms in a substance which is also the effect of light has.

[swansont]

Would you say light energy always works by being converted into heat energy?

Becasue by definition temperature is the extent of the vibration of atoms in a substance which is also the effect of light has.

 

I don't think that's how photon activation works. I think it puts one molecule in an excited state through photon absorption, rather than being caused by collisional excitation. Or the photon strikes while the two reactants are interacting.

[aj47]

ah ok, thanks