Chapter 4: Chemical Reactions and Chemical Quantities – Study Notes

Chapter 4: Chemical Reactions and Chemical Quantities

Chemical Reactions

Chemical reactions are fundamental processes in chemistry where substances are transformed into new substances with different properties. Understanding these changes is essential for predicting product formation and quantifying reactants and products.

• Chemical reaction: A process in which one or more substances (reactants) are converted into one or more different substances (products).

• Chemical change: Involves the rearrangement of atoms, resulting in new chemical substances.

• Combustion reaction: A specific type of chemical reaction where a substance combines with oxygen to form one or more oxygen-containing compounds, often releasing heat.

Chemical equations represent these reactions, showing:

• Formulas of reactants and products

• States of reactants and products (e.g., solid, liquid, gas, aqueous)

• Relative numbers of molecules or moles involved

States of Reactants and Products

The physical state of each substance in a chemical equation is indicated by an abbreviation in parentheses. The state can influence the type of reaction and its observable characteristics.

Abbreviation	State
(g)	Gas
(l)	Liquid
(s)	Solid
(aq)	Aqueous (water solution)

• Solid product: Indicates a precipitation reaction.

• Gas product: Indicates a gas evolution reaction (often observed as bubbles).

• Aqueous reactions: Occur in water, very common in laboratory and biological contexts.

• Combustion reactions: Typically involve burning a compound in oxygen gas.

Types of Chemical Equations

Chemical reactions can be classified into several types based on the nature of the reactants and products:

• Combustion reactions: Organic compound + O2 → CO2 + H2O

  • Example: $\mathrm{C_3H_8} + 5\mathrm{O_2} \rightarrow 3\mathrm{CO_2} + 4\mathrm{H_2O}$

  • Organic compounds contain carbon and hydrogen, and possibly oxygen.

• Combination (synthesis) reactions: Two or more reactants form one product.

  • Example: $\mathrm{N_2} + 3\mathrm{H_2} \rightarrow 2\mathrm{NH_3}$

• Decomposition reactions: One reactant breaks down into two or more products.

  • Example: $2\mathrm{KClO_3} \rightarrow 2\mathrm{KCl} + 3\mathrm{O_2}$

• Other types: Net ionic reactions, acid-base reactions (covered in later chapters).

Combustion Reactions

Combustion reactions involve burning a hydrocarbon or other organic compound in air or pure oxygen. These reactions are important for energy production and environmental chemistry.

• Unbalanced example: $\mathrm{CH_4} (g) + \mathrm{O_2} (g) \rightarrow \mathrm{CO_2} (g) + \mathrm{H_2O} (g)$

• Balanced example: $\mathrm{CH_4} (g) + 2\mathrm{O_2} (g) \rightarrow \mathrm{CO_2} (g) + 2\mathrm{H_2O} (g)$

• Balancing the equation ensures the law of conservation of mass is satisfied.

Balancing Chemical Equations

Balancing chemical equations is essential to reflect the conservation of mass, ensuring the same number and type of atoms on both sides of the equation.

• Conservation of mass: Matter cannot be created or destroyed in a chemical reaction.

• When balanced, the equation shows the correct stoichiometric coefficients (whole numbers in front of formulas).

• Balanced equations predict the theoretical outcome, though real reactions may have side reactions.

Steps to Balance Chemical Equations

1. Write out the total number of atoms of each element on both sides of the equation.

2. Balance one element at a time, starting with elements that appear in only one substance on each side.

3. Balance the remaining elements one at a time.

4. Multiply through by appropriate integers to remove any fractions.

Example: $\mathrm{C_3H_8} + \mathrm{O_2} \rightarrow \mathrm{CO_2} + \mathrm{H_2O}$

Reaction Stoichiometry

Stoichiometry involves using the coefficients in a balanced chemical equation to determine the relative amounts of reactants and products in moles.

• Example equation: $\mathrm{C_8H_{18}} (l) + 25\mathrm{O_2} (g) \rightarrow 16\mathrm{CO_2} (g) + 18\mathrm{H_2O} (g)$

• 1 molecule (or mole) of C8H18 reacts with 25 molecules (or moles) of O2 to form 16 molecules (or moles) of CO2 and 18 molecules (or moles) of H2O.

Using Stoichiometry

Stoichiometry allows calculation of the amount of product formed from a given amount of reactant, or vice versa, using the balanced equation.

• From the equation: $\mathrm{C_8H_{18}} (l) + 25\mathrm{O_2} (g) \rightarrow 16\mathrm{CO_2} (g) + 18\mathrm{H_2O} (g)$

• Stoichiometric ratio (for C8H18 to CO2):

  • $\frac{16\ \mathrm{mol}\ \mathrm{CO_2}}{2\ \mathrm{mol}\ \mathrm{C_8H_{18}}} = \frac{8\ \mathrm{mol}\ \mathrm{CO_2}}{1\ \mathrm{mol}\ \mathrm{C_8H_{18}}}$

Converting Between Moles

To convert between moles of different substances in a reaction, use the stoichiometric coefficients from the balanced equation.

• General form: $n_A A \rightarrow n_B B$

• $n_A$ and $n_B$ are the stoichiometric coefficients for substances A and B.

• A and B can be either reactants or products.

Formula:

$\mathrm{Moles\ B} = \frac{n_B}{n_A} \times \mathrm{Moles\ A}$

The ratio $n_B/n_A$ is called the stoichiometric ratio.

Mole-to-Mole Calculations

These calculations determine how many moles of a product or reactant are involved, given a certain amount of another substance.

• Example: If 22.0 moles of C8H18 are burned, how many moles of CO2 are formed?

• Equation: $2\mathrm{C_8H_{18}} (l) + 25\mathrm{O_2} (g) \rightarrow 16\mathrm{CO_2} (g) + 18\mathrm{H_2O} (g)$

• Calculation: $22.0\ \mathrm{mol}\ \mathrm{C_8H_{18}} \times \frac{16\ \mathrm{mol}\ \mathrm{CO_2}}{2\ \mathrm{mol}\ \mathrm{C_8H_{18}}} = 176\ \mathrm{mol}\ \mathrm{CO_2}$

Similar calculations can be done for other products or reactants using their respective stoichiometric ratios.

Mass-to-Mass Calculations (Stoichiometry)

To determine the mass of a product formed from a given mass of reactant, convert mass to moles (using molar mass), use the stoichiometric ratio, then convert moles back to mass.

• Steps:

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

  2. Use the stoichiometric ratio to find moles of product.

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

• Example: What mass of H2O is produced from 1.00 g C8H18?

  • Molar mass (MM) of H2O = 18.016 g/mol

  • MM of C8H18 = 114.224 g/mol

  • Calculation:

    • $1.00\ \mathrm{g}\ \mathrm{C_8H_{18}} \times \frac{1\ \mathrm{mol}\ \mathrm{C_8H_{18}}}{114.224\ \mathrm{g}} \times \frac{18\ \mathrm{mol}\ \mathrm{H_2O}}{2\ \mathrm{mol}\ \mathrm{C_8H_{18}}} \times \frac{18.016\ \mathrm{g}\ \mathrm{H_2O}}{1\ \mathrm{mol}\ \mathrm{H_2O}} = 1.42\ \mathrm{g}\ \mathrm{H_2O}$

Limiting Reactants

In many reactions, one reactant is used up before the others, limiting the amount of product formed. This reactant is called the limiting reactant; the others are in excess.

• The reaction stops when the limiting reactant is consumed.

• To identify the limiting reactant:

  1. Balance the chemical equation.

  2. Calculate the moles (or mass) of product possible from each reactant, assuming the other is in excess.

  3. The reactant that produces the least amount of product is the limiting reactant.

  4. The other reactant(s) are in excess.

• Analogy: Making pizzas with limited cheese or sauce; the ingredient that runs out first limits the number of pizzas.

Yield of a Reaction

The theoretical yield is the maximum amount of product that can be formed from the given amounts of reactants, as predicted by stoichiometry. The actual yield is the amount actually obtained from the reaction, which is often less due to side reactions or incomplete reactions.

• Percent yield is calculated as:

$\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

• Example: If the theoretical yield of CO2 is 12.74 g, but only 10.01 g is produced, then:

$\text{Percent yield} = \frac{10.01\ \mathrm{g}}{12.74\ \mathrm{g}} \times 100\% = 78.57\%$