Problem 87 - Chemistry: The Central Science in SI Units, Global Edition

Question

Write a balanced chemical reaction for the condensation reaction between \(\mathrm{H}_{3} \mathrm{PO}_{4}\), molecules to form \(\mathrm{H}_{6} \mathrm{P}_{4} \mathrm{O}_{13}\).

Step-by-Step Solution

Answer

The balanced chemical equation for the condensation reaction between H₃PO₄ molecules to form H₆P₄O₁₃ is: \(8H_{3}PO_{4} \rightarrow H_{6}P_{4}O_{13}\)

Step 1: Identify the reactants and products

In this condensation reaction, the reactant is phosphoric acid (H₃PO₄) and the product is condensed phosphoric acid (H₆P₄O₁₃).

Step 2: Write the unbalanced chemical equation

An unbalanced chemical equation is the one in which both sides of the equation have different numbers of atoms for each element. Write the unbalanced equation as follows: \(H_{3}PO_{4} \rightarrow H_{6}P_{4}O_{13}\)

Step 3: Balance the chemical equation

To balance the equation, we need to adjust the coefficients for the reactants and products until the number of atoms for each element is the same on both sides of the equation. First, balance the phosphorus (P) atoms. There are 4 P atoms in the product molecule and only 1 in the reactant molecule. Add a coefficient of 4 in front of the reactant to balance the P atoms: \(4H_{3}PO_{4} \rightarrow H_{6}P_{4}O_{13}\) Now, balance the hydrogen (H) atoms. There are 12 H atoms on the reactant side and 6 H atoms on the product side. Since the number of hydrogen atoms on the product side is already balanced, we do not need to make any changes in this case. Next, balance the oxygen (O) atoms. There are 16 O atoms on the reactant side and 13 O atoms on the product side. Add a coefficient of .5 or 1/2, in front of the product to balance the O atoms: \(4H_{3}PO_{4} \rightarrow 1/2H_{6}P_{4}O_{13}\) However, having fractional coefficients is not ideal in a balanced chemical equation. Therefore, to eliminate the fractional coefficient, we can multiply the entire equation by 2: \(2(4H_{3}PO_{4} \rightarrow 1/2H_{6}P_{4}O_{13})\) This results in the final balanced chemical equation: \(8H_{3}PO_{4} \rightarrow H_{6}P_{4}O_{13}\)

Key Concepts

Chemical Equation Balancing

Chemical equation balancing is a crucial step in representing a chemical reaction accurately. It ensures that the number of atoms for each element is equal on the reactant and product sides. This is essential because according to the Law of Conservation of Mass, matter cannot be created or destroyed in a chemical reaction. Balancing equations allows chemists to predict the amounts of substances consumed and produced.

Here's a simple way to approach balancing:

• Identify each type of atom or element present.
• Count the number of atoms for each element on both sides of the equation.
• Adjust the coefficients (the numbers placed before compounds) to balance the atoms.

Usually, it helps to start with the element that appears in the least number of molecules. In our phosphoric acid example, balancing phosphorus first was key. Double-check your balanced equation by recounting the atoms to ensure accuracy.

Phosphoric Acid

Phosphoric acid, with the chemical formula \(H_{3}PO_{4}\), is a triprotic acid meaning it can release three hydrogen ions. Each molecule consists of hydrogen (H), phosphorus (P), and oxygen (O) atoms.

In inorganic chemistry, phosphoric acid is used in various applications like soft drinks, cleaning agents, and fertilizers. It plays a significant role in chemical synthesis and manufacturing.

When phosphoric acid undergoes a condensation reaction, two or more molecules join, releasing a small molecule like water. This reaction is an important way to form complex compounds, such as in the production of ATP (adenosine triphosphate), vital for energy in biological cells.

Inorganic Chemistry

Inorganic chemistry is the branch of chemistry that focuses on inorganic compounds, which are not based on carbon-hydrogen bonds, unlike organic chemistry. This field covers a wide array of substances, including metals, minerals, and acids like phosphoric acid.

The study involves:

• Chemical bonding of inorganic compounds.
• Study of metal complexes and catalysts.
• Understanding the behavior and application of acids and bases.

Inorganic chemistry is integral in developing materials used in industry, manufacturing, and technology. It helps us understand and manipulate substances that form the basis of modern-day life. For example, it assists in creating fertilizers that improve agricultural productivity and the synthesis of compounds used in electronics and healthcare.