Writing Balanced Chemical Equations, Reaction Types, and Collision Energy

Key Principles of Chemical Equations and Conservation of Mass

• Key Learning Idea: Balanced chemical equations display the precise numerical ratio of reactants and products participating in a chemical reaction, directly reflecting the Law of Conservation of Mass.

• The Law of Conservation of Mass:

  • Formulated historically by Antoine Lavoisier: "Nothing is lost, nothing is created, everything is transformed."

  • States that matter is neither created nor destroyed during any chemical reaction.

  • Implies that the total mass of the reactants before a chemical reaction occurs must equal the total mass of the products formed after the reaction.

  • Requires that the total numbers of each individual type of atom are identical before and after the reaction takes place.

• Nature of Chemical Transformations:

  • The fundamental difference between reactants and products lies solely in how the constituent atoms are arranged.

  • Chemical reactions involve the breaking of existing chemical bonds between atoms in the starting materials and the formation of new chemical bonds to generate new substances.

• Balanced vs. Unbalanced Equations:

  • Unbalanced Chemical Equation: An equation that does not display equal numbers of each type of atom on both the reactant and product sides.

  • Balanced Chemical Equation: An equation where full-number coefficients are placed in front of chemical formulas, resulting in equal numbers of each type of atom on both sides.

  • Balancing equations reflects the actual quantitative ratios (stoichiometric coefficients) of reactants consumed and products yielded.

Atomic Tallies and Visual Representations of Reactions

• Visualizing Molecular Transformations:

  • Molecules can be represented as linked atomic spheres to verify atom conservation visually.

  • Example: Synthesis of Ammonia from Nitrogen and Hydrogen Gas:

    • $\text{N}_2 + 3\,\text{H}_2 \rightarrow 2\,\text{NH}_3$

    • Reactants side: One diatomic nitrogen molecule ($\text{N}_2$ = 2 Nitrogen atoms) plus three diatomic hydrogen molecules ($3\,\text{H}_2$ = 6 Hydrogen atoms).

    • Products side: Two ammonia molecules ($2\,\text{NH}_3$ = 2 Nitrogen atoms + 6 Hydrogen atoms total).

• Step-by-Step Diagram Construction for Water Synthesis:

  • Initial Unbalanced State:

    • Diagram: One diatomic hydrogen molecule ($\text{H}_2$) plus one diatomic oxygen molecule ($\text{O}_2$) forming one water molecule ($\text{H}_2\text{O}$).

    • Formula equation: $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$

    • Tally Table:

      • Hydrogen ($\text{H}$): Reactants = 2, Products = 2

      • Oxygen ($\text{O}$): Reactants = 2, Products = 1

      • Status: Unbalanced.

  • Adding an Extra Water Molecule to Products:

    • Diagram: Adding a second $\text{H}_2\text{O}$ molecule to the product side.

    • Formula equation: $\text{H}_2 + \text{O}_2 \rightarrow 2\,\text{H}_2\text{O}$

    • Tally Table:

      • Hydrogen ($\text{H}$): Reactants = 2, Products = 4

      • Oxygen ($\text{O}$): Reactants = 2, Products = 2

      • Status: Unbalanced (Oxygen is balanced, but Hydrogen is now unequal).

  • Adding an Extra Hydrogen Molecule to Reactants:

    • Diagram: Adding a second $\text{H}_2$ molecule to the reactant side.

    • Formula equation: $2\,\text{H}_2 + \text{O}_2 \rightarrow 2\,\text{H}_2\text{O}$

    • Tally Table:

      • Hydrogen ($\text{H}$): Reactants = 4, Products = 4

      • Oxygen ($\text{O}$): Reactants = 2, Products = 2

      • Status: Balanced (Atom counts match completely on both sides).

• Molecular Diagram Equations:

  • Methane combustion: $\text{C}\text{H}_4 + 2\,\text{O}_2 \rightarrow \text{C}\text{O}_2 + 2\,\text{H}_2\text{O}$

  • Nitrogen monoxide oxidation: $2\,\text{N}\text{O} + \text{O}_2 \rightarrow 2\,\text{N}\text{O}_2$

  • Ammonia formation: $\text{N}_2 + 3\,\text{H}_2 \rightarrow 2\,\text{NH}_3$

• Completing Diagrams by Drawing Molecules:

  • Sulfur dioxide oxidation: $2\,\text{S}\text{O}_2 + \text{O}_2 \rightarrow 2\,\text{S}\text{O}_3$

  • Ammonia synthesis: $\text{N}_2 + 3\,\text{H}_2 \rightarrow 2\,\text{NH}_3$

  • Nitrogen monoxide reduction: $2\,\text{N}\text{O} + 2\,\text{H}_2 \rightarrow \text{N}_2 + 2\,\text{H}_2\text{O}$

  • Hydrochloric acid oxidation: $4\,\text{H}\text{Cl} + \text{O}_2 \rightarrow 2\,\text{Cl}_2 + 2\,\text{H}_2\text{O}$

Rules and Step-by-Step Procedure for Balancing Chemical Equations

• Key Terminology:

  • Reactants: Starting substances that undergo chemical transformation.

  • Products: New chemical substances formed as a result of the reaction.

  • Coefficient: The large number written directly in front of a chemical formula indicating the total quantity of entire molecules or formula units.

  • Subscript: The small subscript number inside a chemical formula indicating the exact number of atoms of an element bonded within that individual molecule.

• The Golden Rule of Equation Balancing:

  • You can ONLY change coefficients.

  • NEVER change subscripts.

  • Rationale: Changing a subscript alters the actual chemical identity and molecular structure of the substance (e.g., changing $\text{H}_2\text{O}$ to $\text{H}_2\text{O}_2$ transforms water into hydrogen peroxide).

• The 5 Systematic Steps to Balance Any Chemical Equation:

  1. Write the equation: Obtain or construct the correct chemical formula equation.

  2. Count atoms on each side: Perform an inventory tallies for every distinct element present in reactants and products.

  3. Add coefficients: Place stoichiometric coefficients in front of formulas to equalize element counts starting with non-hydrogen and non-oxygen elements.

  4. Recount atoms: Update tallies across both sides of the equation after applying coefficients.

  5. Repeat until balanced: Continue adjusting coefficients and recounting until the atomic counts for every single element match identically on both sides.

Comprehensive Balancing Practice and Worked Examples

• Detailed Worked Example 1: Balancing Water Formation

  • Step 1 (Write): $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$

  • Step 2 (Count): Left side: H = 2, O = 2; Right side: H = 2, O = 1. Balanced? No.

  • Step 3 (Add coefficients): Place a coefficient of 2 in front of $\text{H}_2\text{O}$ $\rightarrow \text{H}_2 + \text{O}_2 \rightarrow 2\,\text{H}_2\text{O}$

  • Step 4 (Recount): Left side: H = 2, O = 2; Right side: H = 4, O = 2. Oxygen is fixed, but Hydrogen is unbalanced.

  • Step 5 (Repeat): Place a coefficient of 2 in front of $\text{H}_2$ on the left $\rightarrow 2\,\text{H}_2 + \text{O}_2 \rightarrow 2\,\text{H}_2\text{O}$

  • Final Check: Left H = 4, Right H = 4; Left O = 2, Right O = 2. Fully balanced.

• Detailed Worked Example 2: Guided Practice - Making Ammonia

  • Unbalanced equation: $\text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3$

  • Nitrogen (N) count on left = 2. The coefficient required to make N equal on the right is 2 ($2\,\text{NH}_3$).

  • Rewrite: $\text{N}_2 + \text{H}_2 \rightarrow 2\,\text{NH}_3$

  • Recount Hydrogen (H) on the right: H = 6. The coefficient required to make H equal on the left is 3 ($3\,\text{H}_2$).

  • Final balanced equation: $\text{N}_2 + 3\,\text{H}_2 \rightarrow 2\,\text{NH}_3$

• Single Step Verification Examples:

  1. $\text{C} + \text{O}_2 \rightarrow \text{C}\text{O}_2$

     • Check: C (R: 1, P: 1), O (R: 2, P: 2). Status: Already balanced.

  2. $\text{Mg} + \text{O}_2 \rightarrow \text{Mg}\text{O}$

     • Check: Mg (R: 1, P: 1), O (R: 2, P: 1).

     • Balanced equation: $2\,\text{Mg} + \text{O}_2 \rightarrow 2\,\text{Mg}\text{O}$

  3. $\text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3$

     • Check: Al (R: 1, P: 2), O (R: 2, P: 3).

     • Balanced equation: $4\,\text{Al} + 3\,\text{O}_2 \rightarrow 2\,\text{Al}_2\text{O}_3$

• Categorization and Atom Tallies for Selected Reactions:

  • (a) $\text{Fe}_2\text{O}_3 \rightarrow \text{Fe} + \text{O}_2$

    • Atom Tallies: Fe (Reactants = 2, Products = 1); O (Reactants = 3, Products = 2).

    • Categorization: Unbalanced.

  • (b) $\text{H}\text{N}\text{O}_3 + \text{Na}\text{O}\text{H} \rightarrow \text{Na}\text{N}\text{O}_3 + \text{H}_2\text{O}$

    • Atom Tallies: H (Reactants = 2, Products = 2); N (Reactants = 1, Products = 1); O (Reactants = 4, Products = 4); Na (Reactants = 1, Products = 1).

    • Categorization: Balanced.

  • (c) $\text{Cu}\text{C}\text{O}_3 + \text{H}\text{Cl} \rightarrow \text{Cu}\text{Cl}_2 + \text{C}\text{O}_2 + \text{H}_2\text{O}$

    • Atom Tallies: Cu (R: 1, P: 1); C (R: 1, P: 1); O (R: 3, P: 3); H (R: 1, P: 2); Cl (R: 1, P: 2).

    • Categorization: Unbalanced.

  • (d) $\text{S}_8 + 16\,\text{O}_2 \rightarrow 8\,\text{S}\text{O}_2$

    • Atom Tallies: S (Reactants = 8, Products = 8); O (Reactants = 32, Products = 16).

    • Categorization: Unbalanced.

  • (e) $\text{H}_2\text{S}\text{O}_4 + \text{Mg}(\text{O}\text{H})_2 \rightarrow \text{Mg}\text{S}\text{O}_4 + 2\,\text{H}_2\text{O}$

    • Atom Tallies: H (R: 4, P: 4); S (R: 1, P: 1); O (R: 6, P: 6); Mg (R: 1, P: 1).

    • Categorization: Balanced.

  • (f) $\text{C}_7\text{H}_8 + 3\,\text{H}\text{N}\text{O}_3 \rightarrow \text{C}_7\text{H}_5\text{N}_3\text{O}_6 + 3\,\text{H}_2\text{O}$

    • Atom Tallies: C (R: 7, P: 7); H (R: 11, P: 11); N (R: 3, P: 3); O (R: 9, P: 9).

    • Categorization: Balanced.

• Before and After Atom Count Analysis:

  • (a) $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$

    • Before Balancing: H (R: 2, P: 2); O (R: 2, P: 1).

    • Balanced Equation: $2\,\text{H}_2 + \text{O}_2 \rightarrow 2\,\text{H}_2\text{O}$

    • After Balancing: H (R: 4, P: 4); O (R: 2, P: 2).

  • (b) $\text{P}_4 + \text{O}_2 \rightarrow \text{P}_2\text{O}_5$

    • Before Balancing: P (R: 4, P: 2); O (R: 2, P: 5).

    • Balanced Equation: $\text{P}_4 + 5\,\text{O}_2 \rightarrow 2\,\text{P}_2\text{O}_5$

    • After Balancing: P (R: 4, P: 4); O (R: 10, P: 10).

  • (c) $\text{Al}\text{F}_3 + \text{Mg}\text{Cl}_2 \rightarrow \text{Mg}\text{F}_2 + \text{Al}\text{Cl}_3$

    • Before Balancing: Al (R: 1, P: 1); F (R: 3, P: 2); Mg (R: 1, P: 1); Cl (R: 2, P: 3).

    • Balanced Equation: $2\,\text{Al}\text{F}_3 + 3\,\text{Mg}\text{Cl}_2 \rightarrow 3\,\text{Mg}\text{F}_2 + 2\,\text{Al}\text{Cl}_3$

    • After Balancing: Al (R: 2, P: 2); F (R: 6, P: 6); Mg (R: 3, P: 3); Cl (R: 6, P: 6).

  • (d) $\text{C}_2\text{H}_5\text{O}\text{H} + \text{O}_2 \rightarrow \text{C}\text{O}_2 + \text{H}_2\text{O}$

    • Before Balancing: C (R: 2, P: 1); H (R: 6, P: 2); O (R: 3, P: 3).

    • Balanced Equation: $\text{C}_2\text{H}_5\text{O}\text{H} + 3\,\text{O}_2 \rightarrow 2\,\text{C}\text{O}_2 + 3\,\text{H}_2\text{O}$

    • After Balancing: C (R: 2, P: 2); H (R: 6, P: 6); O (R: 7, P: 7).

• Complete Collection of Practice Chemical Equations:

  • (a) $\text{N}_2 + 3\,\text{H}_2 \rightarrow 2\,\text{NH}_3$

  • (b) $2\,\text{N}\text{O} + \text{O}_2 \rightarrow 2\,\text{N}\text{O}_2$

  • (c) $\text{Ba}\text{Cl}_2 + (\text{NH}_4)_2\text{S}\text{O}_4 \rightarrow \text{Ba}\text{S}\text{O}_4 + 2\,\text{NH}_4\text{Cl}$

  • (d) $\text{S}_8 + 12\,\text{O}_2 \rightarrow 8\,\text{S}\text{O}_3$

  • (e) $4\,\text{H}\text{Cl} + \text{O}_2 \rightarrow 2\,\text{Cl}_2 + 2\,\text{H}_2\text{O}$

  • (f) $\text{C}\text{H}_4 + 2\,\text{O}_2 \rightarrow \text{C}\text{O}_2 + 2\,\text{H}_2\text{O}$

  • (g) $\text{Ni} + \text{Cu}\text{Cl}_2 \rightarrow \text{Ni}\text{Cl}_2 + \text{Cu}$

  • (h) $\text{Mg} + \text{S}\text{O}_2 \rightarrow \text{Mg}\text{S}\text{O}_4 + \text{Mg}\text{S}$

  • (i) $2\,\text{H}_2\text{O}_2 \rightarrow 2\,\text{H}_2\text{O} + \text{O}_2$

  • (j) $2\,\text{Ag}\text{N}\text{O}_3 \rightarrow 2\,\text{Ag} + 2\,\text{N}\text{O}_2 + \text{O}_2$

  • (k) $\text{F}_2 + 2\,\text{Li}\text{Cl} \rightarrow 2\,\text{Li}\text{F} + \text{Cl}_2$

  • (l) $4\,\text{Al} + 3\,\text{O}_2 \rightarrow 2\,\text{Al}_2\text{O}_3$

  • (m) $2\,\text{H}\text{N}\text{O}_3 + \text{Na}_2\text{C}\text{O}_3 \rightarrow 2\,\text{Na}\text{N}\text{O}_3 + \text{C}\text{O}_2 + \text{H}_2\text{O}$

  • (n) $\text{Fe}\text{Cl}_3 + 3\,\text{Na}\text{O}\text{H} \rightarrow \text{Fe}(\text{O}\text{H})_3 + 3\,\text{Na}\text{Cl}$

  • (o) $2\,\text{Al}\text{Br}_3 + 3\,\text{K}_2\text{S}\text{O}_4 \rightarrow 6\,\text{K}\text{Br} + \text{Al}_2(\text{S}\text{O}_4)_3$

  • (p) $2\,\text{Al} + 3\,\text{H}_2\text{S}\text{O}_4 \rightarrow \text{Al}_2(\text{S}\text{O}_4)_3 + 3\,\text{H}_2$

  • (q) $2\,\text{Fe}_2\text{O}_3 + 3\,\text{C} \rightarrow 4\,\text{Fe} + 3\,\text{C}\text{O}_2$

  • (r) $\text{Ca}_3(\text{P}\text{O}_4)_2 + 3\,\text{S}\text{i}\text{O}_2 \rightarrow \text{P}_4\text{O}_{10} + 3\,\text{Ca}\text{S}\text{i}\text{O}_3$

  • (s) $4\,\text{NH}_3 + 5\,\text{O}_2 \rightarrow 4\,\text{N}\text{O} + 6\,\text{H}_2\text{O}$

  • (t) $2\,\text{N}_2\text{O}_4 + \text{O}_2 + 2\,\text{H}_2\text{O} \rightarrow 4\,\text{H}\text{N}\text{O}_3$

  • (u) $4\,\text{NH}_3 + 3\,\text{O}_2 \rightarrow 2\,\text{N}_2 + 6\,\text{H}_2\text{O}$

  • (v) $\text{C}_8\text{H}_{20} + 9\,\text{O}_2 \rightarrow 8\,\text{C}\text{O} + 10\,\text{H}_2\text{O}$

States of Matter in Chemical Equations

• Standard Symbols for Physical States:

  • Solid: $(s)$

  • Liquid: $(l)$

  • Gas: $(g)$

  • Aqueous solution: $(aq)$ (denotes a chemical substance dissolved completely in water)

• Annotating Chemical Equations with State Symbols:

  • (a) $\text{C}(s) + \text{O}_2(g) \rightarrow \text{C}\text{O}_2(g)$

  • (b) $\text{H}_2\text{S}\text{O}_4(aq) + 2\,\text{Na}\text{O}\text{H}(aq) \rightarrow \text{Na}_2\text{S}\text{O}_4(aq) + 2\,\text{H}_2\text{O}(l)$

  • (c) $\text{Zn}(s) + 2\,\text{H}\text{Cl}(aq) \rightarrow \text{Zn}\text{Cl}_2(aq) + \text{H}_2(g)$

  • (d) $3\,\text{Ca}(s) + 2\,\text{Al}(\text{N}\text{O}_3)_3(aq) \rightarrow 3\,\text{Ca}(\text{N}\text{O}_3)_2(aq) + 2\,\text{Al}(s)$

  • (e) $\text{S}\text{O}_2(g) + 2\,\text{Na}\text{O}\text{H}(aq) \rightarrow \text{Na}_2\text{S}\text{O}_3(aq) + \text{H}_2\text{O}(l)$

  • (f) $2\,\text{Pb}(\text{N}\text{O}_3)_2(s) \rightarrow 2\,\text{Pb}\text{O}(s) + 4\,\text{N}\text{O}_2(g) + \text{O}_2(g)$

  • (g) $2\,\text{N}_2\text{O}(g) \rightarrow 2\,\text{N}_2(g) + \text{O}_2(g)$

  • (h) $\text{Cl}_2(g) + 2\,\text{Li}\text{Br}(aq) \rightarrow 2\,\text{Li}\text{Cl}(aq) + \text{Br}_2(l)$

The Six Major Types of Chemical Reactions

• 1. Synthesis Reactions (Composition):

  • General Equation: $\text{A} + \text{B} \rightarrow \text{AB}$

  • Word Equation: hydrogen + oxygen $\rightarrow$ water

  • Chemical Equation: $2\,\text{H}_2 + \text{O}_2 \rightarrow 2\,\text{H}_2\text{O}$

  • How to Recognize: Two separate individual elements or simple compounds combine to form 1 single compound.

• 2. Decomposition Reactions:

  • General Equation: $\text{AB} \rightarrow \text{A} + \text{B}$

  • Word Equation: water $\rightarrow$ hydrogen + oxygen

  • Chemical Equation: $2\,\text{H}_2\text{O} \rightarrow 2\,\text{H}_2 + \text{O}_2$

  • How to Recognize: One complex compound separates into 2 or more simpler components or constituent elements.

• 3. Single Displacement Reactions (Single Replacement):

  • General Equation (metal replacement): $\text{A} + \text{BC} \rightarrow \text{AC} + \text{B}$

  • General Equation (nonmetal replacement): $\text{A} + \text{BC} \rightarrow \text{BA} + \text{C}$

  • Word Equation: aluminum + copper chloride $\rightarrow$ copper + aluminum chloride

  • Chemical Equation: $2\,\text{Al} + 3\,\text{Cu}\text{Cl}_2 \rightarrow 3\,\text{Cu} + 2\,\text{Al}\text{Cl}_3$

  • How to Recognize: A pure uncombined element reacts with a compound, replacing one component to form a single new element and a new compound.

• 4. Double Displacement Reactions (Double Replacement):

  • General Equation: $\text{AB} + \text{CD} \rightarrow \text{AD} + \text{CB}$

  • Word Equation: potassium iodide + lead nitrate $\rightarrow$ potassium nitrate + lead iodide

  • Chemical Equation: $2\,\text{KI} + \text{Pb}(\text{N}\text{O}_3)_2 \rightarrow 2\,\text{K}\text{N}\text{O}_3 + \text{Pb}\text{I}_2$

  • How to Recognize: Two reacting ionic compounds switch compound partners with each other.

• 5. Combustion Reactions (Complete Combustion of Hydrocarbons):

  • General Equation: Fuel + Oxygen $\rightarrow \text{C}\text{O}_2 + \text{H}_2\text{O}$

  • Reactants: ALWAYS include a hydrocarbon fuel plus oxygen ($\text{O}_2$).

  • Products: ALWAYS yield carbon dioxide ($\text{C}\text{O}_2$) and water ($\text{H}_2\text{O}$).

• 6. Neutralization (Acid-Base) Reactions:

  • Specialized Sub-type: A specific class of double displacement reaction occurring between an acid and a base.

  • General Equation: Acid + Base $\rightarrow$ Salt + Water

  • Word Equation: hydrochloric acid + sodium hydroxide $\rightarrow$ sodium chloride + water

  • Chemical Equation: $\text{H}\text{Cl} + \text{Na}\text{O}\text{H} \rightarrow \text{Na}\text{Cl} + \text{H}_2\text{O}$

  • How to Recognize: Reactants consist of an acid (containing $\text{H}^+$) and a base (containing $\text{OH}^-$), producing a salt ionic compound and neutral liquid water.

• Reaction Type Identification Worksheet Solutions:

  1. $\text{Na}\text{Br} + \text{H}_3\text{P}\text{O}_4 \rightarrow \text{Na}_3\text{P}\text{O}_4 + \text{H}\text{Br}$: Double displacement

  2. $\text{Ca}(\text{O}\text{H})_2 + \text{Al}_2(\text{S}\text{O}_4)_3 \rightarrow \text{Ca}\text{S}\text{O}_4 + \text{Al}(\text{O}\text{H})_3$: Double displacement

  3. $\text{Mg} + \text{Fe}_2\text{O}_3 \rightarrow \text{Fe} + \text{Mg}\text{O}$: Single displacement

  4. $\text{C}_2\text{H}_4 + \text{O}_2 \rightarrow \text{C}\text{O}_2 + \text{H}_2\text{O}$: Combustion

  5. $\text{Pb}\text{S}\text{O}_4 \rightarrow \text{Pb}\text{S}\text{O}_3 + \text{O}_2$: Decomposition

  6. $\text{NH}_3 + \text{I}_2 \rightarrow \text{N}_2\text{I}_6 + \text{H}_2$: Single displacement

  7. $\text{H}_2\text{O} + \text{S}\text{O}_3 \rightarrow \text{H}_2\text{S}\text{O}_4$: Synthesis

  8. $\text{H}_2\text{S}\text{O}_4 + \text{NH}_4\text{O}\text{H} \rightarrow \text{H}_2\text{O} + (\text{NH}_4)_2\text{S}\text{O}_4$: Double displacement (Neutralization)

Collision Theory, Energy Changes, and Reaction Kinetics

• Fundamental Definitions in Reaction Kinetics:

  • Chemical Reaction: The process of transforming one or more chemical substances into entirely new chemical substances with distinct properties.

  • Elements: Pure substances composed of only one single type of atom (e.g., elemental iron $\text{Fe}$, metallic gold $\text{Au}$).

  • Compounds: Substances composed of two or more different elements chemically bonded together in fixed proportions (e.g., $\text{C}\text{O}_2$, $\text{Na}\text{Cl}$).

• Bond Re-arrangement Mechanism:

  • During chemical transformations, pre-existing chemical bonds binding reactant atoms must be broken, and new chemical bonds are established to construct product molecules.

  • Worked Example: Molecules of diatomic nitrogen ($\text{N}_2$) and diatomic oxygen ($\text{O}_2$) have their internal bonds broken to yield isolated nitrogen and oxygen atoms, which reorganize into molecules of nitrogen oxide ($\text{N}\text{O}$).

• Collision Theory Requirements:

  • For any chemical reaction to proceed, reacting particles must physically collide with one another.

  • Not every collision leads to a successful chemical change.

  • The Two Criteria for a Successful (Effective) Collision:

    1. Sufficient Energy: Reacting particles must collide with energy equal to or exceeding the activation energy, which is defined as the minimum energy threshold required to break pre-existing chemical bonds.

    2. Correct Orientation: Colliding particles must line up with the correct spatial alignment at the exact moment of collision.

  • Collision Outcomes:

    • Ineffective Collision: Particles lack activation energy or proper alignment; they bounce apart unchanged.

    • Effective Collision: Particles satisfy both energy and orientation criteria; bonds break and new product molecules form.

• Energy and Chemical Bonds:

  • Energy must be absorbed by starting substances to break existing chemical bonds (input of activation energy).

  • Energy is released when new chemical bonds form.

  • Chemical potential energy is stored directly within chemical bonds. Whether a reaction absorbs or releases energy overall depends on the net difference between energy stored in reactants versus products.

• Exothermic vs. Endothermic Reactions Summary:

  • Exothermic Reactions:

    • Energy Behavior: Energy is released overall to the surrounding environment.

    • Energy Comparison: Energy stored in products is LOWER than in reactants.

    • Surroundings Temperature: Temperature of surroundings will RISE / INCREASE.

    • Everyday Example: Combustion (burning hydrocarbon fuel).

  • Endothermic Reactions:

    • Energy Behavior: Energy is absorbed overall from the surrounding environment.

    • Energy Comparison: Energy stored in products is HIGHER than in reactants.

    • Surroundings Temperature: Temperature of surroundings will FALL / DECREASE.

    • Everyday Example: Photosynthesis.

• Energy Profile Diagrams:

  • Exothermic Profile Diagram:

    • Reactants start at a higher potential energy level.

    • Energy rises to a peak representing the activation energy required for bond breaking.

    • Energy drops significantly as new bonds form, ending with products at a lower potential energy level.

    • Net energy difference is released into surroundings as heat.

  • Endothermic Profile Diagram:

    • Reactants start at a lower potential energy level.

    • Energy climbs to a high peak representing activation energy.

    • Energy drops slightly as bonds form, but products remain at a higher potential energy level than reactants.

    • Net energy difference is absorbed from surroundings.

Comprehensive Multiple-Choice and Conceptual Review Questions

• Question 1: Is the generalized equation $\text{H}_2\text{S}(g) \rightarrow \text{H}_2(g) + \text{S}(s)$ classified as a decomposition reaction?

  • Answer: True (a single compound breaks down into two constituent elements).

• Question 2: What type of reaction is represented by $\text{C} + \text{O}_2 \rightarrow \text{C}\text{O}_2$?

  • Answer: Synthesis Reaction (or combination).

• Question 3: What type of reaction is $\text{Cu} + 2\,\text{Ag}\text{N}\text{O}_3 \rightarrow 2\,\text{Ag} + \text{Cu}(\text{N}\text{O}_3)_2$?

  • Answer: Single Displacement Reaction.

• Question 4: Consider the chemical equation $\text{C}\text{H}_4 + 2\,\text{O}_2 \rightarrow \text{C}\text{O}_2 + 2\,\text{H}_2\text{O}$. In this equation, $\text{C}\text{H}_4$ is a:

  • Answer: reactant.

• Question 5: Consider the equation $4\,\text{Fe} + 3\,\text{O}_2 \rightarrow 2\,\text{Fe}_2\text{O}_3$. In this equation, $3\,\text{O}_2$ is a:

  • Answer: reactant.

• Question 6: Which generalized equation represents a synthesis reaction?

  • Answer: $\text{A} + \text{B} \rightarrow \text{AB}$.

• Question 7: Which generalized equation represents a decomposition reaction?

  • Answer: $\text{AB} \rightarrow \text{A} + \text{B}$.

• Question 8: Which generalized equation represents a single displacement reaction?

  • Answer: $\text{A} + \text{BC} \rightarrow \text{AC} + \text{B}$.

• Question 9: Identify the reaction type: $2\,\text{NH}_3 + 1\,\text{H}_2\text{S}\text{O}_4 \rightarrow 1\,(\text{NH}_4)_2\text{S}\text{O}_4$

  • Answer: Synthesis (or combination).

• Question 10: Identify the reaction type: $\text{Pb} + \text{Fe}\text{S}\text{O}_4 \rightarrow \text{Pb}\text{S}\text{O}_4 + \text{Fe}$

  • Answer: Single displacement.

• Question 11: Identify the reaction type: $\text{P}_4 + 3\,\text{O}_2 \rightarrow 2\,\text{P}_2\text{O}_3$

  • Answer: Synthesis (or combination).

• Question 12: The reaction $\text{Na}\text{Cl} + \text{Ag}\text{F} \rightarrow \text{Na}\text{F} + \text{Ag}\text{Cl}$ simulates what type of reaction?

  • Answer: Double Replacement.

• Question 13: What type of reaction involves 2 substances combining to form 1 new compound?

  • Answer: Synthesis Reaction.

• Question 14: What type of reaction involves one element replacing another element in a compound?

  • Answer: Single Displacement Reaction.

• Question 15: What are the reactants in the chemical equation $\text{C}\text{H}_4 + 2\,\text{O}_2 \rightarrow \text{C}\text{O}_2 + 2\,\text{H}_2\text{O}$?

  • Answer: $\text{C}\text{H}_4 + 2\,\text{O}_2$.

• Question 16: What are the products of the chemical reaction $\text{C}\text{H}_4 + 2\,\text{O}_2 \rightarrow \text{C}\text{O}_2 + 2\,\text{H}_2\text{O}$?

  • Answer: $\text{C}\text{O}_2$ and $\text{H}_2\text{O}$.

• Question 17: The right side of the equation $\text{Na} + \text{Cl}_2 \rightarrow \text{Na}\text{Cl}$ is called the:

  • Answer: products.

• Question 18: The reaction type $2\,\text{N}\text{O}_2 \rightarrow 2\,\text{O}_2 + \text{N}_2$ is called:

  • Answer: Decomposition.