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laras_creed
Group Title
Chlorofluorocarbons (CFC's) are organic compounds that have been implicated in ozone depletion. When the CFC known as Freon 12 (CCl2F2) is exposed to UV radiation (wavelength in the range of 10 nm to 400 nm, a bit shorter than visible light), a chlorine atom breaks off from the rest of the molecule. Prove that this is possible by calculating the maximum wavelength capable of breaking the CCl bond.
Average Bond Energy (KJ/mol):
HH: 435
FF: 155
ClCl: 242
CC: 347
C=C: 610
CC triple bound: 836
 2 years ago
 2 years ago
laras_creed Group Title
Chlorofluorocarbons (CFC's) are organic compounds that have been implicated in ozone depletion. When the CFC known as Freon 12 (CCl2F2) is exposed to UV radiation (wavelength in the range of 10 nm to 400 nm, a bit shorter than visible light), a chlorine atom breaks off from the rest of the molecule. Prove that this is possible by calculating the maximum wavelength capable of breaking the CCl bond. Average Bond Energy (KJ/mol): HH: 435 FF: 155 ClCl: 242 CC: 347 C=C: 610 CC triple bound: 836
 2 years ago
 2 years ago

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emblem101 Group TitleBest ResponseYou've already chosen the best response.0
You have the enthalpies of the CC bond and the ClCl bond, both of which are homonuclear. To find the heteronuclear bond enthalpy, CCl, you have you use the formula Professor Cima showed in the lecture. It is the geometric mean of the two homonuclear enthalpies + 96.3(difference of the two electronegativities)^2 ext, use E = ħc / λ where E = energy per photon ħ = plancks constant = 6.626x10^34 J s c = speed of light = 3.00x10^8 m/s λ = wavelength.. E = ħc / λ rearranging λ = ħc / E for CCl... λ = (6.626x10^34 J s) x (3.00x10^8 m/s) / (5.63x10^19 J) = 3.53x10^7 m = 353x10^9 m = 353 nm
 2 years ago

laras_creed Group TitleBest ResponseYou've already chosen the best response.0
Thank you for your help. I have seen this answer on other sites. What i cant get is the 5.63x10^19 J. I keep getting 325 as E(total). ???
 2 years ago
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