Sulfur dioxide’s atmospheric lifetime is determined by two processes: Oxidation in the gas phase (by the...

Sulfur dioxide’s atmospheric lifetime is determined by two processes: Oxidation in the gas phase (by the OH radical) and oxidation in the liquid phase (e.g. in cloud droplets). If the associated lifetimes were four days and five days, respectively, what is the total lifetime? How could that lifetime be shortened despite keeping the concentration of the OH radicals constant?

Expert Solution

Sulfur dioxide’s atmospheric lifetime is determined by two processes:

Oxidation in the gas phase (by the OH radical) and

Oxidation in the liquid phase (e.g. in cloud droplets)

But if the associated lifetimes were four days and five days, respectively As Hydroxyl (OH) radical concentrations not only decrease with latitude & altitude since both the concentrations of sunlight & water vapour intensity decrease as you move towards the poles.

OH radical is very reactive as it reacts as soon as it formed in the atmosphere and it is the most important oxidant in the atmosphere but its value become zero at night therefore, total lifetime will be half.i.e. when Oxidation in the gas phase (by the OH radical) is four then total lifetime will be Two days.

if the OH radical becomes constant then the lifetime be shortened because of the reactive and oxidative nature of OH radical which will reduce the lifetime of sulfur dioxide.

queen_honey_blossom answered 2 years ago

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