Kinetics of the Photochemical Reaction Between Hydrogen and Chlorine
H2 + Cl2 ---h𝜈→ 2HCl
The possible mechanism of the above reaction is-
- Cl2 + h𝜈 ---k1→ 2Cl
- Cl + H2 ---k2→ HCl + H
- H + Cl2 ---k3→ HCl + Cl
- Cl + Cl ---k4→ Cl2
where, k1, k2, k3 and k4 are rate constants.
If oxygen is present, the chain is terminated due to the formation of HO2 intermediate.
H + O2 → HO2
Hence quantum yield is low in the presence of oxygen.
The Cl atoms are formed in steps- i and iii and disappear in steps ii and iv. Hence, applying SSA we get-
k1Iabs + k3[H][Cl2] = k2[Cl][H2] + k4[Cl]2 -----Equation-1
The H atoms are formed in steps ii and disappears in step iii. Hence applying SSA we get-
k2[Cl][H2] = k3[H][Cl2] -----Equation-2
Fropm equation-1 and equation-2 we have-
k1Iabs = k4[Cl]2
or, [Cl] = (k1Iabs / k4)1/2
Therefore, the overall rate of formation of HCl-
d[HCl]/dt = k2[Cl][H2] + k3[H][Cl2]
putting the value of k3[H][Cl2] from equation-2 we get-
or, d[HCl]/dt = k2[Cl][H2] + k2[Cl][H2]
or, d[HCl]/dt = 2k2[Cl][H2]
Now putting the value of [Cl] from equation-3 we get-
d[HCl]/dt = 2k2 (k1Iabs / k4)1/2 [H2]
or, d[HCl]/dt = 2k2 (k1/k4)1/2 (Iabs)1/2[H2]
Hence, the rate of formation of HCl is directly proportional to the square root of the intensity of light(Iabs).
The second step is highly endothermic for H2 + Br2 reaction. Hence reaction is extreamly slow but corresponding reaction for H2 + Cl2 is exothermic. Hence, it occurs instantaneously. Therefor, later reaction has extreamly high quantum yield or photoefficiency.
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