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Loss of atmospheric ozone has led to an ozone "hole" over Antarctica. The loss occurs in part through three consecutive steps: (1) Chlorine atoms react with ozone \(\left(\mathrm{O}_{3}\right)\) to form chlorine monoxide and molecular oxygen. (2) Chlorine monoxide forms \(\mathrm{ClOOCl}\). (3) \(\mathrm{ClOOCl}\) absorbs sunlight and breaks into chlorine atoms and molecular oxygen. (a) Write a balanced equation for each step. (b) Write an overall balanced equation for the sequence.

Short Answer

Expert verified
Step 1: \( \text{Cl} + \text{O}_3 \rightarrow \text{ClO} + \text{O}_2 \)Step 2: \( 2 \text{ClO} \rightarrow \text{ClOOCl} \)Step 3: \( \text{ClOOCl} \rightarrow 2 \text{Cl} + \text{O}_2 \)Overall: \( \text{O}_3 \rightarrow 2 \text{O}_2 \)

Step by step solution

01

- Write the equation for Step 1

In the first step, chlorine atoms react with ozone to form chlorine monoxide and molecular oxygen. The reaction is: \[ \text{Cl} + \text{O}_3 \rightarrow \text{ClO} + \text{O}_2 \]
02

- Write the equation for Step 2

In the second step, chlorine monoxide reacts to form \text{ClOOCl}. The reaction is: \[ 2 \text{ClO} \rightarrow \text{ClOOCl} \]
03

- Write the equation for Step 3

In the third step, \text{ClOOCl} absorbs sunlight and breaks into chlorine atoms and molecular oxygen. The reaction is: \[ \text{ClOOCl} \rightarrow 2 \text{Cl} + \text{O}_2 \]
04

- Write the overall balanced equation

To find the overall balanced equation, combine the equations from steps 1, 2, and 3. Notice that \text{ClO} and \text{ClOOCl} are intermediates and cancel out in the overall reaction. Add the equations together: \[ \text{Cl} + \text{O}_3 \rightarrow \text{ClO} + \text{O}_2 \] \[ 2 \text{ClO} \rightarrow \text{ClOOCl} \] \[ \text{ClOOCl} \rightarrow 2 \text{Cl} + \text{O}_2 \] Combining them, we get: \[ \text{Cl} + \text{O}_3 \rightarrow \text{Cl} + 2 \text{O}_2 \] The chlorine atoms cancel out on both sides, leaving the overall balanced equation: \[ \text{O}_3 \rightarrow \text{O}_2 + \text{O}_2 \] or simply: \[ \text{O}_3 \rightarrow 2 \text{O}_2 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balanced Chemical Equations
Balanced chemical equations are crucial for understanding and representing chemical reactions. They ensure the same number of each type of atom on both sides of the equation, following the Law of Conservation of Mass. In the context of ozone depletion, each step involves individual reactions that contribute to the overall process. For example, in the first reaction:
\[ \text{Cl} + \text{O}_3 \rightarrow \text{ClO} + \text{O}_2 \]
We start with one chlorine atom (Cl) and one ozone molecule (\( \text{O}_3 \)). The products are chlorine monoxide (\( \text{ClO} \)) and molecular oxygen (\( \text{O}_2 \)). This reaction is balanced because the number and types of atoms on each side are equal.
Balancing each step individually is fundamental before combining them into an overall balanced equation. By ensuring balance in each step, especially when dealing with intermediates and final products, we can understand the chemical changes occurring in complex processes like ozone depletion.
Chlorine-Catalyzed Ozone Depletion
Chlorine-catalyzed ozone depletion involves a series of chemical reactions primarily driven by the presence of chlorine atoms in the atmosphere. Chlorine atoms, often originating from chlorofluorocarbons (CFCs), play a significant role in the breakdown of ozone (\( \text{O}_3 \)).
Here is how the process works, step-by-step:
  • In Step 1, a chlorine atom reacts with an ozone molecule to produce chlorine monoxide (\( \text{ClO} \)) and molecular oxygen (\( \text{O}_2 \)).
  • In Step 2, two \( \text{ClO} \) molecules combine to form \( \text{ClOOCl} \) through a intermediary reaction.
  • In Step 3, \( \text{ClOOCl} \) absorbs sunlight and decomposes, releasing two chlorine atoms and one more molecule of oxygen (\( \text{O}_2 \)).

The chlorine atoms act as a catalyst because they are not consumed in the overall reaction, allowing them to continuously break down ozone molecules. The net result of this series of reactions is the conversion of ozone into molecular oxygen, as shown in the overall balanced equation:
\[ \text{O}_3 \rightarrow 2 \text{O}_2 \]
This cycle can continue, leading to significant ozone depletion in the stratosphere, particularly over regions like Antarctica.
Ozone Hole Over Antarctica
The ozone hole over Antarctica is a dramatic example of the consequences of chlorine-catalyzed ozone depletion. The phenomenon is particularly severe over Antarctica due to specific atmospheric conditions.
Here are key points to understand why the ozone hole forms:
  • Extremely cold temperatures in the Antarctic stratosphere during the winter lead to the formation of polar stratospheric clouds. These clouds provide a surface for the chemical reactions that release chlorine atoms from compounds like CFCs.
  • With the return of sunlight in the spring, the reactions accelerate, increasing chlorine atom concentration, which then breaks down ozone.
  • The recurring destruction of ozone creates a significant 'hole' in the ozone layer, resulting in reduced ozone levels above the continent.

The decrease in ozone allows more harmful ultraviolet (UV) radiation to reach the Earth's surface, contributing to health risks like skin cancer and cataracts, as well as environmental damage.

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Most popular questions from this chapter

Nitrogen (N), phosphorus (P), and potassium (K) are the main nutrients in plant fertilizers. By industry convention, the numbers on a label refer to the mass percents of \(\mathrm{N}, \mathrm{P}_{2} \mathrm{O}_{5},\) and \(\mathrm{K}_{2} \mathrm{O},\) in that order. Calculate the N/P/K ratio of a \(30 / 10 / 10\) fertilizer in terms of moles of each element, and express it as \(x / y / 1.0\).

Hydroxyapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH}),\) is the main mineral component of dental enamel, dentin, and bone. Coating the compound on metallic implants (such as titanium alloys and stainless steels) helps the body accept the implant. When placed in bone voids, the powder encourages natural bone to grow into the void. Hydroxyapatite is prepared by adding aqueous phosphoric acid to a dilute slurry of calcium hydroxide. (a) Write a balanced equation for this preparation. (b) What mass (g) of hydroxyapatite could form from \(100 . \mathrm{g}\) of \(85 \%\) phosphoric acid and \(100 . \mathrm{g}\) of calcium hydroxide?

What is the molecular formula of each compound? (a) Empirical formula \(\mathrm{CH}_{2}(\mathscr{A}=42.08 \mathrm{~g} / \mathrm{mol})\) (b) Empirical formula \(\mathrm{NH}_{2}(\mathscr{M}=32.05 \mathrm{~g} / \mathrm{mol})\) (c) Empirical formula \(\mathrm{NO}_{2}(\mathscr{M}=92.02 \mathrm{~g} / \mathrm{mol})\) (d) Empirical formula CHN \((\mathscr{M}=135.14 \mathrm{~g} / \mathrm{mol})\)

Convert the following into balanced equations: (a) When lead(II) nitrate solution is added to potassium iodide solution, solid lead(II) iodide forms and potassium nitrate solution remains. (b) Liquid disilicon hexachloride reacts with water to form solid silicon dioxide, hydrogen chloride gas, and hydrogen gas. (c) When nitrogen dioxide is bubbled into water, a solution of nitric acid forms and gaseous nitrogen monoxide is released.

Metal hydrides react with water to form hydrogen gas and the metal hydroxide. For example, $$ \operatorname{SrH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \operatorname{Sr}(\mathrm{OH})_{2}(s)+2 \mathrm{H}_{2}(g) $$ You wish to calculate the mass (g) of hydrogen gas that can be prepared from \(5.70 \mathrm{~g}\) of \(\mathrm{SrH}_{2}\) and \(4.75 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) (a) What amount (mol) of \(\mathrm{H}_{2}\) can be produced from the given mass of \(\mathrm{SrH}_{2} ?\) (b) What amount (mol) of \(\mathrm{H}_{2}\) can be produced from the given mass of \(\mathrm{H}_{2} \mathrm{O} ?\) (c) Which is the limiting reactant? (d) How many grams of \(\mathrm{H}_{2}\) can be produced?

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