Problem 37 - Chemistry: The Central Science in SI Units, Global Edition

Question

Write balanced equations for each of the following reactions. (a) When mercury(II) oxide is heated, it decomposes to form \(\mathrm{O}_{2}\) and mercury metal. ( \(\mathbf{b}\) ) When copper(II) nitrate is heated strongly, it decomposes to form copper(II) oxide, nitrogen dioxide, and oxygen. (c) Lead(II) sulfide, \(\mathrm{PbS}(s)\) reacts with ozone to form \(\mathrm{PbSO}_{4}(s)\) and \(\mathrm{O}_{2}(g) .\) (d) When heated in air, \(\mathrm{ZnS}(s)\) is converted to \(\mathrm{ZnO}\). (e) Potassium peroxide reacts with \(\mathrm{CO}_{2}(g)\) to give potassium carbonate and \(\mathrm{O}_{2}\). (f) Oxygen is converted to ozone in the upper atmosphere.

Step-by-Step Solution

Answer

(a) 2 HgO -> 2 Hg + \(O_2\) (b) 2 Cu(NO3)2 -> 2 CuO + 4 NO2 + \(O_2\) (c) PbS + 2 O3 -> PbSO4 + \(O_2\) (d) ZnS + \(O_2\) -> 2 ZnO (e) K2O2 + CO2 -> K2CO3 + \(O_2\) (f) 3 \(O_2\) -> 2 O3

Step 1: (a) Mercury(II) oxide decomposition

Mercury(II) oxide decomposes to form oxygen gas and mercury metal. Write the unbalanced equation as: HgO -> Hg + \(O_2\) To balance the equation, simply add a coefficient of 2 in front of HgO: 2 HgO -> 2 Hg + \(O_2\)

Step 2: (b) Copper(II) nitrate decomposition

Copper(II) nitrate decomposes to form copper(II) oxide, nitrogen dioxide, and oxygen gas. Write the unbalanced equation as: Cu(NO3)2 -> CuO + NO2 + \(O_2\) To balance the equation, use the coefficients 2, 1, 4, and 1, respectively: 2 Cu(NO3)2 -> 2 CuO + 4 NO2 + \(O_2\)

Step 3: (c) Lead(II) sulfide reacting with ozone

Lead(II) sulfide reacts with ozone to form lead(II) sulfate and oxygen gas. Write the unbalanced equation as: PbS + O3 -> PbSO4 + \(O_2\) To balance the equation, use a coefficient of 2 for ozone: PbS + 2 O3 -> PbSO4 + \(O_2\)

Step 4: (d) Zinc sulfide heated in air

When heated in air, zinc sulfide is converted to zinc oxide. Let's express this reaction as: ZnS + \(O_2\) -> ZnO To balance the equation, use a coefficient of 2 for both oxygen and zinc oxide: ZnS + \(O_2\) -> 2 ZnO

Step 5: (e) Potassium peroxide reacting with carbon dioxide

Potassium peroxide reacts with carbon dioxide to give potassium carbonate and oxygen gas. Write the unbalanced equation as: K2O2 + CO2 -> K2CO3 + \(O_2\) This equation is already balanced, so no need to add coefficients.

Step 6: (f) Oxygen to ozone conversion

Oxygen is converted to ozone in the upper atmosphere. Write the unbalanced equation as: \(O_2\) -> O3 To balance the equation, add a coefficient of 3 in front of ozone and a coefficient of 2 in front of oxygen: 3 \(O_2\) -> 2 O3

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry. It ensures the law of conservation of mass is obeyed. This law states that matter cannot be created or destroyed in a chemical reaction. Therefore, the number of atoms of each element must be the same on both sides of the equation. Here's how you can balance an equation:

• Write down the unbalanced equation.
• Count the atoms of each element on both sides.
• Add coefficients in front of compounds to balance the atoms.
• Repeat the process for each element until the equation is balanced.

Remember, coefficients are whole numbers placed in front of compounds or elements. It's important to never change the subscripts in chemical formulas, as this changes the substances involved. Balancing equations can sometimes be trial and error, especially with complex reactions. Patience and practice are key.

Decomposition Reactions

Decomposition reactions involve breaking a compound into simpler substances. These reactions usually require energy input in the form of heat, light, or electricity. A classic example is the decomposition of mercury(II) oxide into mercury and oxygen—a crucial discovery by Joseph Priestley that led to identifying oxygen. Consider this equation: 2 HgO (s) → 2 Hg (l) + O y2 (g) In this reaction, mercury(II) oxide decomposes into mercury and oxygen gas when heated. Decomposition reactions often have the general format of: AB → A + B Where a binary compound like AB breaks down into simpler products A and B upon heating or other input energy sources. Common examples include the breakdown of peroxides, nitrates, and sulfates.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, also known as redox reactions, are chemical reactions involving the transfer of electrons between two substances. This transfer results in changes in oxidation states of the reacting species. These reactions are crucial in chemical processes ranging from respiration to combustion. Let's look at how oxidation and reduction occur:

• Oxidation: Loss of electrons or gain of oxygen. Increase in oxidation state.
• Reduction: Gain of electrons or loss of oxygen. Decrease in oxidation state.

An exemplary redox reaction is seen when lead(II) sulfide reacts with ozone: PbS + 2 O3 → PbSO4 + O2 Here, lead(II) sulfide is oxidized while ozone is reduced, illustrating the exchange of electrons and changes in oxidation states. Identifying oxidation and reduction in reactions helps in balancing redox equations and understanding electron flow.

Atmospheric Chemistry

Atmospheric chemistry focuses on the chemical composition and reactions occurring in the Earth's atmosphere. It involves understanding both natural processes and human impacts on the atmosphere, such as pollution. A significant aspect of atmospheric chemistry is the formation of ozone high in the atmosphere: 3 O2 → 2 O3 This conversion of oxygen to ozone primarily occurs in the stratosphere, where ultraviolet light catalyzes the reaction. Ozone plays a vital role in protecting life on Earth by absorbing the majority of the sun's harmful ultraviolet radiation. Moreover, atmospheric chemistry covers topics like the greenhouse effect, air quality, and acid rain. It's vital to study these to predict climate changes and devise measures for pollution control. Understanding these processes helps in making informed environmental policies and promotes a sustainable future.