Chapter 5: Chemical Reactions – Principles, Calculations, and Applications

Introduction to Chemical Reactions

General Features of Physical and Chemical Changes

Chemical reactions are central to chemistry, involving the transformation of substances through the breaking and forming of chemical bonds. Understanding the distinction between physical and chemical changes is foundational:

• Physical Change: Alters the physical state (e.g., melting, boiling) without changing the chemical composition.

• Chemical Change (Chemical Reaction): Converts one substance into another by breaking bonds in reactants and forming new bonds in products.

Example: The reaction of methane (CH4) with oxygen (O2) produces carbon dioxide (CO2) and water (H2O).

Writing and Balancing Chemical Equations

Structure of Chemical Equations

A chemical equation uses formulas and symbols to represent the reactants and products of a reaction. The reactants are written on the left, products on the right, and coefficients indicate the number of molecules or moles involved.

• Law of Conservation of Mass: Atoms are neither created nor destroyed in a chemical reaction; equations must be balanced.

• Coefficients: Used to balance equations, ensuring the same number of each atom on both sides.

Symbols Used in Chemical Equations

Symbol	Meaning
→	Reaction arrow
Δ	Heat
(s)	Solid
(l)	Liquid
(g)	Gas
(aq)	Aqueous solution

Steps to Balance a Chemical Equation

1. Write the correct formulas for all reactants and products.

2. Balance one element at a time using coefficients (never change subscripts).

3. Check that the smallest set of whole numbers is used for coefficients.

Example: Balancing the combustion of propane (C3H8):

The Mole and Avogadro’s Number

Definition and Use

The mole is a counting unit in chemistry, representing 6.02 × 1023 items (Avogadro’s number). It allows chemists to relate the mass of substances to the number of particles (atoms, molecules).

• 1 mole of C atoms = 6.02 × 1023 C atoms

• 1 mole of CO2 molecules = 6.02 × 1023 CO2 molecules

Conversion factors:

• $1\ \text{mol} = 6.02 \times 10^{23}\ \text{atoms (or molecules)}$

Sample Calculation

How many molecules are in 5.0 moles of CO2?

• Given: 5.0 mol CO2

• Conversion: $5.0\ \text{mol} \times \frac{6.02 \times 10^{23}\ \text{molecules}}{1\ \text{mol}} = 3.0 \times 10^{24}\ \text{molecules}$

Mass to Mole Conversions

Formula Weight and Molar Mass

The formula weight is the sum of atomic weights of all atoms in a compound (in amu). The molar mass is the mass of one mole of a substance (in grams), numerically equal to the formula weight.

• Example: FeSO4 has a formula weight of 151.877 amu, so its molar mass is 151.877 g/mol.

Relating Grams to Moles

The molar mass is used as a conversion factor between grams and moles:

• $\text{moles} = \frac{\text{grams}}{\text{molar mass (g/mol)}}$

Example: How many moles are in 100 g of aspirin (C9H8O4, molar mass 180.2 g/mol)?

• $100\ \text{g} \times \frac{1\ \text{mol}}{180.2\ \text{g}} = 0.555\ \text{mol}$

Relating Grams to Number of Molecules

To find the number of molecules in a given mass:

• Convert mass to moles using molar mass.

• Convert moles to molecules using Avogadro’s number.

Example: How many molecules are in a 325-mg tablet of aspirin?

• Convert mg to g: $325\ \text{mg} \times \frac{1\ \text{g}}{1000\ \text{mg}} = 0.325\ \text{g}$

• Convert to moles: $0.325\ \text{g} \times \frac{1\ \text{mol}}{180.2\ \text{g}} = 0.00180\ \text{mol}$

• Convert to molecules: $0.00180\ \text{mol} \times 6.02 \times 10^{23}\ \text{molecules/mol} = 1.09 \times 10^{21}\ \text{molecules}$

Mole Calculations in Chemical Equations

Using Balanced Equations for Mole Ratios

Balanced chemical equations provide the mole ratios of reactants and products, which are used as conversion factors in stoichiometric calculations.

• Example: $\text{N}_2(g) + \text{O}_2(g) \rightarrow 2\ \text{NO}(g)$

• Mole ratios: $1\ \text{mol N}_2 : 1\ \text{mol O}_2 : 2\ \text{mol NO}$

Sample Calculation

How many moles of CO are produced from 3.5 moles of C2H6?

• Balanced equation: $2\ \text{C}_2\text{H}_6 + 5\ \text{O}_2 \rightarrow 4\ \text{CO} + 6\ \text{H}_2\text{O}$

• Conversion: $3.5\ \text{mol C}_2\text{H}_6 \times \frac{4\ \text{mol CO}}{2\ \text{mol C}_2\text{H}_6} = 7.0\ \text{mol CO}$

Mass Calculations in Chemical Equations

Converting Moles of Reactant to Grams of Product

To determine the mass of product formed from a given amount of reactant, use the following steps:

1. Convert moles of reactant to moles of product using the mole ratio from the balanced equation.

2. Convert moles of product to grams using the molar mass.

Example: How many grams of O3 are formed from 9.0 mol of O2?

• Balanced equation: $3\ \text{O}_2 \rightarrow 2\ \text{O}_3$

• Step 1: $9.0\ \text{mol O}_2 \times \frac{2\ \text{mol O}_3}{3\ \text{mol O}_2} = 6.0\ \text{mol O}_3$

• Step 2: $6.0\ \text{mol O}_3 \times 48.0\ \text{g/mol} = 290\ \text{g O}_3$

Converting Grams of Reactant to Grams of Product

To convert grams of reactant to grams of product:

1. Convert grams of reactant to moles using molar mass.

2. Convert moles of reactant to moles of product using the mole ratio.

3. Convert moles of product to grams using molar mass.

Example: How many grams of ethanol (C2H6O) are formed from 14 g of ethylene (C2H4)?

• $14\ \text{g C}_2\text{H}_4 \times \frac{1\ \text{mol}}{28.1\ \text{g}} \times \frac{1\ \text{mol C}_2\text{H}_6\text{O}}{1\ \text{mol C}_2\text{H}_4} \times 46.1\ \text{g} = 23\ \text{g C}_2\text{H}_6\text{O}$

Percent Yield

Theoretical vs. Actual Yield

The theoretical yield is the maximum amount of product expected based on stoichiometry. The actual yield is the amount actually obtained from the reaction. The percent yield is calculated as:

• $\text{Percent yield} = \frac{\text{actual yield (g)}}{\text{theoretical yield (g)}} \times 100\%$

Example: If the theoretical yield is 23 g and the actual yield is 15 g, percent yield is $\frac{15}{23} \times 100\% = 65\%$.

Limiting Reactants

Identifying the Limiting Reactant

The limiting reactant is the reactant that is completely consumed first, thus limiting the amount of product formed. To determine the limiting reactant:

1. Convert the mass of each reactant to moles.

2. Use the balanced equation to determine which reactant produces less product; this is the limiting reactant.

Example: For the reaction $\text{N}_2 + \text{O}_2 \rightarrow 2\ \text{NO}$, if 10.0 g of N2 and 10.0 g of O2 are used:

• Convert grams to moles: $10.0\ \text{g N}_2 \div 28.02\ \text{g/mol} = 0.357\ \text{mol N}_2$; $10.0\ \text{g O}_2 \div 32.00\ \text{g/mol} = 0.313\ \text{mol O}_2$

• From stoichiometry, 0.357 mol N2 would require 0.357 mol O2, but only 0.313 mol O2 is available, so O2 is limiting.

Oxidation and Reduction (Redox Reactions)

General Features

Oxidation is the loss of electrons, while reduction is the gain of electrons. These processes always occur together in redox reactions, involving electron transfer between substances.

• Oxidizing agent: Substance that is reduced (gains electrons) and causes oxidation.

• Reducing agent: Substance that is oxidized (loses electrons) and causes reduction.

Example: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$

• Zn is oxidized (loses electrons), acts as reducing agent.

• Cu2+ is reduced (gains electrons), acts as oxidizing agent.

Half-reactions:

• Oxidation: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$

• Reduction: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$

Examples of Redox Reactions

• Iron rusting: $4\ \text{Fe}(s) + 3\ \text{O}_2(g) \rightarrow 2\ \text{Fe}_2\text{O}_3(s)$

• Alkaline battery: $\text{Zn} + 2\ \text{MnO}_2 \rightarrow \text{ZnO} + \text{Mn}_2\text{O}_3$

Oxidation can also be described as gain of oxygen or loss of hydrogen, while reduction is loss of oxygen or gain of hydrogen.