Free radical 一問
2007-02-13 5:43 am
既然光 (hv) 可以break Cl to Cl single bond 去做 radical, 點解係正常狀態下Chlorine是diatomic molecule 而不是一堆radical 呢...?
回答 (2)
2007-02-14 5:56 pm
✔ 最佳答案
因為可以打破Cl-Cl single bond的不是普通的光.Covalent Bond是很強的bonding, 要打破Covalent Bond需要不少的能量, 並非普通光可以做到. 需要的是高能量的光, 如紫外線(UV). 而大氣層中的臭氧層(Ozone layer)已把大量波長短, 能量高的紫外線隔掉, 只餘小量低能量的到達地面. 所以我們生活環境中的Chlorine不會這麼容易變成free radicals.
而且, free radical本身是不穩定的, 所以才會form Covalent Bond達到穩定的狀態. 即使有chlorine free radicals出現, 也會很快和其他物質發生反應(如變回Cl2. 而在臭氧層上的Cl free radicals會和臭氧(Ozone, O3)發生反應, 是臭氧層被破壞的原因), 不會長時間有一大堆的free radical存在.
2007-02-15 8:01 pm
In the presence of uv light, only a small portion (not all) of Cl2 molecules would form Cl• free radicals. If there are no other reactive species, two Cl• free radicals will combine together to form Cl2 molecules. In the same way, a small portion of Cl2 molecules is destroyed and formed repeatedly. Therefore, Chorine contains mainly Cl2 molecules.
Normal sunlight (even diffused sunlight) has sufficient energy to break a small portion of Cl2 molecules into Cl• free radicals. For example, in diffused sunlight, CH4 reacts with Cl2 to give a mixture of products containing CH3Cl, CH2Cl2, etc. Furthermore, in direct strong sunlight, CH4 reacts explosively with Cl2 to give C and HCl. In the above reactions, the first step is that Cl2 molecules break into Cl• free radicals in the presence of light.
Normal sunlight (even diffused sunlight) has sufficient energy to break a small portion of Cl2 molecules into Cl• free radicals. For example, in diffused sunlight, CH4 reacts with Cl2 to give a mixture of products containing CH3Cl, CH2Cl2, etc. Furthermore, in direct strong sunlight, CH4 reacts explosively with Cl2 to give C and HCl. In the above reactions, the first step is that Cl2 molecules break into Cl• free radicals in the presence of light.
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