# Kinetics of the Formation of HCl

## Kinetics of the Photochemical Reaction Between Hydrogen and Chlorine

H_{2}+ Cl

_{2}---h𝜈→ 2HCl

The possible mechanism of the above reaction is-

i. Cl

_{2}+ h𝜈 ---k

_{1}→ 2Cl

ii. Cl + H

_{2}---k

_{2}→ HCl + H

iii. H + Cl

_{2}---k

_{3}→ HCl + Cl

iv. Cl + Cl ---k

_{4}→ Cl

_{2}

where, k

_{1}, k

_{2}, k

_{3}and k

_{4}are rate constants.

If oxygen is present, the chain is terminated due to the formation of HO

_{2}intermediate.

H + O

_{2}→ HO

_{2}

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-

k

_{1}I

_{abs}+ k

_{3}[H][Cl

_{2}] = k

_{2}[Cl][H

_{2}] + k

_{4}[Cl]

^{2}-----Equation-1

The H atoms are formed in steps ii and disappears in step iii. Hence applying SSA we get-

k

_{2}[Cl][H

_{2}] = k

_{3}[H][Cl

_{2}] -----Equation-2

Fropm equation-1 and equation-2 we have-

k

_{1}I

_{abs}= k

_{4}[Cl]

^{2}

or, [Cl] = (k

_{1}I

_{abs}/ k

_{4})

^{1/2}

Therefore, the overall rate of formation of HCl-

d[HCl]/dt = k

_{2}[Cl][H

_{2}] + k

_{3}[H][Cl

_{2}]

putting the value of k

_{3}[H][Cl

_{2}] from equation-2 we get-

or, d[HCl]/dt = k

_{2}[Cl][H

_{2}] + k

_{2}[Cl][H

_{2}]

or, d[HCl]/dt = 2k

_{2}[Cl][H

_{2}]

Now putting the value of [Cl] from equation-3 we get-

d[HCl]/dt = 2k

_{2}(k

_{1}I

_{abs}/ k

_{4})

^{1/2}[H

_{2}]

or, d[HCl]/dt = 2k

_{2}(k

_{1}/k

_{4})

^{1/2}(I

_{abs})

^{1/2}[H

_{2}]

Hence, the rate of formation of HCl is directly proportional to the square root of the intensity of light(I

_{abs}).

The second step is highly endothermic for H

_{2}+ Br

_{2}reaction. Hence reaction is extreamly slow but corresponding reaction for H

_{2}+ Cl

_{2}is exothermic. Hence, it occurs instantaneously. Therefor, later reaction has extreamly high quantum yield or photoefficiency.

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