Question

What are the reactions involved for ozone layer depletion in the stratosphere?

Answer 1

Hint: To answer this question, we should know that ozone depletion is the gradual thinning of Earth’s ozone layer in the upper atmosphere caused by the release of chemical compounds containing gaseous compounds from industry and other human activities.

Step by step solution:
As we know that ozone layer or ozone shield is a region of Earth's stratosphere that absorbs most of the Sun's ultraviolet radiation. It contains a high concentration of ozone (${{O}_{3}}$) in relation to other parts of the atmosphere. We should note this important thing that the ozone layer contains less than 10 parts per million of ozone, while the average ozone concentration in Earth's atmosphere as a whole is about 0.3 parts per million. We should know that the ozone layer absorbs 97 to 99% of the Sun's medium-frequency ultraviolet light (from about 200 nm to 315 nm wavelength), which otherwise would potentially damage exposed life forms near the surface.
We should know that the ozone layer can be depleted by free radical catalysts, including nitric oxide (NO), nitrous oxide (${{N}_{2}}O$), hydroxyl (OH), atomic chlorine (Cl), and atomic bromine (Br). The concentrations of chlorine and bromine increased markedly in recent decades because of the release of large quantities of man-made compounds, especially chlorofluorocarbons (CFCs) and bromo-fluorocarbons.
We should know that in the stratosphere, there is a constant conversion between different molecules of oxygen. The ozone layer is created when ultraviolet rays react with oxygen molecules (${{O}_{2}}$) to create ozone (${{O}_{3}}$) and atomic oxygen (O). This process is called the Chapman cycle.
At first oxygen molecule is photolyzed by solar radiation, creating two oxygen radicals:
$h\nu +{{O}_{2}}\to 2\overset{\bullet }{\mathop{O}}\,$
Then in second step, oxygen radicals then react with molecular oxygen to produce ozone:
${{O}_{2}}+\overset{\bullet }{\mathop{O}}\,\to {{O}_{3}}$
After this we should understand the chemistry of ozone depletion. We should know that CFCs molecules are made up of chlorine, fluorine and carbon atoms. These are extremely stable and this allows CFC's to slowly make their way into the stratosphere. This prolonged life in the atmosphere allows them to reach great altitudes where photons are more energetic. The following reaction displays how Cl atoms have an ozone destroying cycle. Once released CFCs mix with atmospheric gases and reach the stratosphere, where they are decomposed by UV radiations.

& C{{F}_{2}}C{{l}_{2}}\to \overset{\bullet }{\mathop{Cl}}\,+\overset{\bullet }{\mathop{C}}\,{{F}_{2}}Cl \\\ & \overset{\bullet }{\mathop{Cl}}\,+O_3 \to Cl\overset{\bullet }{\mathop{O}}\,+{{O}_{2}} \\\ & Cl\overset{\bullet }{\mathop{O}}\,+\overset{\bullet }{\mathop{O}}\,\to \overset{\bullet }{\mathop{Cl}}\,+{{O}_{2}} \\\ \end{aligned}$$ We should know that chlorine is able to destroy so much of the ozone because it acts as a catalyst. Chlorine starts the breakdown of ozone and combines with freed oxygen to create two oxygen molecules. After each reaction, chlorine begins the destructive cycle again with another ozone molecule. One chlorine atom can thereby destroy thousands of ozone molecules. Because ozone molecules are being broken down they are unable to absorb any ultraviolet light so we experience more intense UV radiation at the earth’s surface. Note: Now, we should know about the condition of Antarctica. We know that Antarctica has the coldest winter temperatures on earth, often reaching -110 F. These chilling temperatures result in the formation of polar stratospheric clouds (PSC's) which are a mixer of frozen $${{H}_{2}}O$$ and $$HN{{O}_{3}}$$. Due to their extremely cold temperatures, PSC's form an electrostatic attraction with CFC molecules as well as other halogenated compounds. As spring comes to Antarctica, the PSC's melt in the stratosphere and release all of the halogenated compounds that were previously absorbed to the cloud. In the antarctic summer, high energy photons are able to photolyze the halogenated compounds, freeing halogen radicals that then catalytically destroy $${{O}_{3}}$$.