Problem 59 - Chemistry and Chemical Reactivity

Question

Balance the following equations: (a) for the synthesis of urea, a common fertilizer \(\mathrm{CO}_{2}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (b) for the reactions used to make uranium(VI) fluoride for the enrichment of natural uranium \(\mathrm{UO}_{2}(\mathrm{s})+\mathrm{HF}(\mathrm{aq}) \rightarrow \mathrm{UF}_{4}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\) \(\mathrm{UF}_{4}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{g}) \rightarrow \mathrm{UF}_{6}(\mathrm{s})\) (c) for the reaction to make titanium(IV) chloride, which is then converted to titanium metal \(\mathrm{TiO}_{2}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{TiCl}_{4}(\ell)+\mathrm{CO}(\mathrm{g})\) \(\operatorname{TiCl}_{4}(\ell)+\operatorname{Mg}(\mathrm{s}) \rightarrow \operatorname{Ti}(\mathrm{s})+\operatorname{Mg} \mathrm{Cl}_{2}(\mathrm{s})\)

Step-by-Step Solution

Answer

The balanced equations are: (a) \( \mathrm{CO}_{2} + 2 \mathrm{NH}_{3} \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2} + \mathrm{H}_{2} \mathrm{O} \) (b) \( \mathrm{UO}_{2} + 4 \mathrm{HF} \rightarrow \mathrm{UF}_{4} + 2 \mathrm{H}_{2} \mathrm{O} \) and \( \mathrm{UF}_{4} + \mathrm{F}_{2} \rightarrow \mathrm{UF}_{6} \) (c) \( \mathrm{TiO}_{2} + 2 \mathrm{Cl}_{2} + 2 \mathrm{C} \rightarrow \mathrm{TiCl}_{4} + 2 \mathrm{CO} \) and \( \mathrm{TiCl}_{4} + 2 \mathrm{Mg} \rightarrow \mathrm{Ti} + 2 \mathrm{MgCl}_{2} \).

Step 1: Balance Equation (a)

For the equation \( \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell) \), we start by counting atoms for each element.- Carbon (C): 1 in \(\mathrm{CO}_2\) ➔ 1 in \(\mathrm{NH}_{2} \mathrm{CONH}_{2}\)- Oxygen (O): 2 in \(\mathrm{CO}_2\) ➔ 1 in \(\mathrm{H}_2\mathrm{O}\) and 1 in \(\mathrm{NH}_{2} \mathrm{CONH}_{2}\)- Nitrogen (N): 2 in \(\mathrm{NH}_3\) ➔ 2 in \(\mathrm{NH}_{2} \mathrm{CONH}_{2}\)- Hydrogen (H): 3 from \(\mathrm{NH}_3\) affects the compound formula.The balanced equation is:\[ \mathrm{CO}_{2}(\mathrm{g}) + 2 \mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{s}) + \mathrm{H}_{2} \mathrm{O}(\ell) \]

Step 2: Balance Equation (b) Part 1

For the first reaction in equation (b), \( \mathrm{UO}_{2}(\mathrm{s})+\mathrm{HF}(\mathrm{aq}) \rightarrow \mathrm{UF}_{4}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell) \), track each element:- Uranium (U): 1 in \(\mathrm{UO}_2\) ➔ 1 in \(\mathrm{UF}_4\)- Oxygen (O): 2 in \(\mathrm{UO}_2\) ➔ 2 in 2 \(\mathrm{H}_2 \mathrm{O}\)- Hydrogen (H) and Fluorine (F): Adjust for four \(\mathrm{F}\) atoms in \(\mathrm{UF}_4\) using \(\mathrm{HF}\).The balanced equation is:\[ \mathrm{UO}_{2}(\mathrm{s}) + 4 \mathrm{HF}(\mathrm{aq}) \rightarrow \mathrm{UF}_{4}(\mathrm{s}) + 2 \mathrm{H}_{2} \mathrm{O}(\ell) \]

Step 3: Balance Equation (b) Part 2

For the second reaction in equation (b), \( \mathrm{UF}_{4}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{g}) \rightarrow \mathrm{UF}_{6}(\mathrm{s}) \), consider:- Uranium (U): 1 in \(\mathrm{UF}_4\) ➔ 1 in \(\mathrm{UF}_6\)- Fluorine (F): 4 in \(\mathrm{UF}_4\) and 2 from \(\mathrm{F}_2\) ➔ 6 in \(\mathrm{UF}_6\)The equation is already balanced:\[ \mathrm{UF}_{4}(\mathrm{s}) + \mathrm{F}_{2}(\mathrm{g}) \rightarrow \mathrm{UF}_{6}(\mathrm{s}) \]

Step 4: Balance Equation (c) Part 1

For the first reaction in equation (c), \( \mathrm{TiO}_{2}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{TiCl}_{4}(\ell)+\mathrm{CO}(\mathrm{g}) \), balance:- Titanium (Ti): 1 in \(\mathrm{TiO}_{2}\) ➔ 1 in \(\mathrm{TiCl}_4\)- Oxygen (O): 2 in \(\mathrm{TiO}_{2}\) ➔ forms 2 \(\mathrm{CO}\)- Carbon (C): 1 carbon used forms 2 \(\mathrm{CO}\)- Chlorine (Cl): 2 \(\mathrm{Cl}_2\) ➔ 4 in \(\mathrm{TiCl}_4\)The balanced equation:\[ \mathrm{TiO}_{2}(\mathrm{s}) + 2 \mathrm{Cl}_{2}(\mathrm{g}) + 2 \mathrm{C}(\mathrm{s}) \rightarrow \mathrm{TiCl}_{4}(\ell) + 2 \mathrm{CO}(\mathrm{g}) \]

Step 5: Balance Equation (c) Part 2

In the second reaction for equation (c), \( \mathrm{TiCl}_{4}(\ell)+\mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Ti}(\mathrm{s})+\mathrm{Mg} \mathrm{Cl}_{2}(\mathrm{s}) \), ensure:- Titanium (Ti): 1 in \(\mathrm{TiCl}_4\) ➔ 1 in \(\mathrm{Ti}\)- Chlorine (Cl): 4 in \(\mathrm{TiCl}_4\) balanced with 2 \(\mathrm{MgCl}_2\) output- Magnesium (Mg): Adjust with 2 \(\mathrm{Mg}\)The balanced equation:\[ \mathrm{TiCl}_{4}(\ell) + 2 \mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Ti}(\mathrm{s}) + 2 \mathrm{Mg} \mathrm{Cl}_{2}(\mathrm{s}) \]

Key Concepts

Stoichiometry

Stoichiometry is like a recipe for chemical reactions. It tells you how much of each ingredient you need to start with and how much product you can expect at the end. To understand stoichiometry, imagine baking a cake without a recipe. You might end up with too much flour or not enough sugar. In chemistry, stoichiometry is the balanced equation that ensures the right proportions of reactants, like ingredients in a recipe. A balanced chemical equation represents this relationship and ensures the law of conservation of mass is observed.
When balancing equations, count the number of atoms of each element on both sides of the reaction until they match. For example, in the synthesis of urea \[ \mathrm{CO}_{2}(\mathrm{g}) + 2 \mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{s}) + \mathrm{H}_{2} \mathrm{O}(\ell) \] we ensure there is 1 carbon, 4 hydrogen, 1 oxygen, and 2 nitrogen atoms on both sides. Stoichiometry is what guides us to use "2" moles of ammonia for each mole of carbon dioxide to keep the balance.

• Count atoms for each element on both sides of the equation.
• Adjust coefficients to get the same number of each type of atom on both sides.
• Check your work by recounting atoms.

Synthesis Reaction

A synthesis reaction involves combining two or more simple substances to form a more complex product. It's like putting together pieces of a puzzle to create a complete picture. In chemistry, these reactions are expressed as A + B \( \rightarrow \) AB. For example, in the synthesis of urea: \[ \mathrm{CO}_{2}(\mathrm{g}) + 2 \mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{s}) + \mathrm{H}_{2} \mathrm{O}(\ell) \] We're essentially taking simple molecules of carbon dioxide and ammonia and forming a more complex compound, urea, plus water as a by-product.

• Typically involve two or more reactants and one product.
• Often release heat; hence referred to as exothermic reactions.
• Can be seen in biological, industrial, and environmental processes.

These reactions are foundational for creating new compounds from simpler ones and are crucial in industries for producing fertilizers, medicines, and many other products.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances, called products. These transformations occur due to interactions at the molecular or atomic level.
It's important to note that during chemical reactions, the masses of products and reactants are conserved, as ordered by the law of conservation of mass. Each type of reaction has distinct characteristics. For instance, the reaction to make titanium(IV) chloride involves complex stoichiometry and balancing: \[ \mathrm{TiO}_{2}(\mathrm{s}) + 2 \mathrm{Cl}_{2}(\mathrm{g}) + 2 \mathrm{C}(\mathrm{s}) \rightarrow \mathrm{TiCl}_{4}(\ell) + 2 \mathrm{CO}(\mathrm{g}) \] This involves transformations of titanium dioxide, chlorine, and carbon into titanium chloride and carbon monoxide. Such reactions are central to materials science, energy production, and environmental technology. They underline how new materials are created and how energy can be stored and released.

• Involves breaking old bonds and forming new ones.
• May involve matter and energy changes.
• Comprise four main types: synthesis, decomposition, single replacement, and double replacement reactions.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between substances. One substance gets oxidized (loses electrons), and another gets reduced (gains electrons). These reactions are essential for a multitude of biological and industrial processes.
For instance, the reaction of uranium compounds with fluoride is a redox reaction where uranium switches from one oxidation state to another: \[ \mathrm{UF}_{4}(\mathrm{s}) + \mathrm{F}_{2}(\mathrm{g}) \rightarrow \mathrm{UF}_{6}(\mathrm{s}) \] Here, uranium is reduced as it forms a higher state of uranium hexafluoride by gaining fluorine atoms. Understanding redox reactions helps in fields like energy storage systems, such as in batteries, and also highlights their importance in biochemical cycles.

• Involves identification of oxidizing and reducing agents.
• Essential for energy production processes like respiration and combustion.
• Usually tracked by changes in oxidation numbers of the elements involved.