Chapter 5 Principles of General, Organic, and Biological Chemistry: Chapter 5 - Chemical Reactions

Introduction to Chemical Reactions

• Definition of Chemical Change: A chemical reaction converts one substance into another.

• Process of a Reaction:

  • Breaking bonds in the reactants (starting materials).

  • Forming new bonds in the products.

  • Example: $CH_4$ and $O_2$ react to form $CO_2$ and $H_2O$.

• Chemical Equations:

  • An expression that uses chemical formulas and other symbols to illustrate what reactants constitute the starting materials and what products are formed.

  • Reactants are written on the left side of the equation.

  • Products are written on the right side of the equation.

• Law of Conservation of Mass:

  • Atoms cannot be created or destroyed in a chemical reaction.

  • A balanced equation must have the same number of atoms of each element on both sides.

• Coefficients:

  • Numbers placed in front of a formula to show the number of molecules of a given element or compound that react or are formed.

  • Coefficients are used to balance the equation.

• Symbols Used in Chemical Equations (Table 5.1):

  • $\rightarrow$: Reaction arrow.

  • $\Delta$: Heat.

  • $(s)$: Solid.

  • $(l)$: Liquid.

  • $(g)$: Gas.

  • $(aq)$: Aqueous solution.

Balancing Chemical Equations

• Standard Procedure for Balancing:

  1. Step [1]: Write the equation with the correct formulas: $C_3H_8 + O_2 \rightarrow CO_2 + H_2O$.

     • Red Warning: The subscripts in a formula can never be changed to balance an equation. Changing a subscript changes the identity of the compound.

  2. Step [2]: Balance the equation with coefficients one element at a time.

     • Balance C atoms first: Add a $3$ before $CO_2$ ($C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O$).

     • Balance H atoms next: Add a $4$ before $H_2O$ ($C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O$).

     • Balance O atoms finally: Count the oxygen atoms on the right ($3 \times 2 + 4 \times 1 = 10$ oxygen atoms). Add a $5$ before $O_2$ on the left ($C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$).

  3. Step [3]: Check to make sure that the smallest set of whole numbers is used as coefficients.

• Case Study: Chemistry of an Automobile Airbag:

  • A severe car crash triggers the ignition of sodium azide ($NaN_3$).

  • Conversion: $2NaN_3 \rightarrow 2Na + 3N_2$.

  • Function: The nitrogen gas ($N_2$) causes the bag to inflate fully in $40\,milliseconds$, protecting passengers from injury.

The Mole and Avogadro’s Number

• Definition of a Mole: A quantity that contains $6.02 \times 10^{23}$ items.

  • $1\,mole$ of $C$ atoms = $6.02 \times 10^{23}\,C\,atoms$.

  • $1\,mole$ of $H_2O$ molecules = $6.02 \times 10^{23}\,H_2O\,molecules$.

  • $1\,mole$ of vitamin C molecules = $6.02 \times 10^{23}\,vitamin\,C\,molecules$.

• Avogadro’s Number: The constant $6.02 \times 10^{23}$.

• Conversion Factors:

  • $\frac{1\,mol}{6.02 \times 10^{23}\,atoms}$ or $\frac{6.02 \times 10^{23}\,atoms}{1\,mol}$.

• Sample Problem 5.6: How many molecules are in $5.0\,moles$ of carbon dioxide ($CO_2$)?

  • Original Quantity: $5.0\,mol\,CO_2$.

  • Desired Quantity: Number of molecules of $CO_2$.

  • Conversion Factor: $\frac{6.02 \times 10^{23}\,molecules}{1\,mol}$.

  • Solution: $5.0\,mol \times \frac{6.02 \times 10^{23}\,molecules}{1\,mol} = 3.0 \times 10^{24}\,molecules\,CO_2$.

Mass to Mole Conversions

• Formula Weight: The sum of the atomic weights of all the atoms in a compound, reported in atomic mass units ($amu$).

• Sample Problem 5.8: Molar Mass of Nicotine ($C_{10}H_{14}N_2$):

  • Step [1]: Determine the number of atoms: $10\,C$, $14\,H$, and $2\,N$.

  • Step [2]: Multiply atoms by atomic weight:

    • $10\,C\,atoms \times 12.01\,amu = 120.1\,amu$.

    • $14\,H\,atoms \times 1.08\,amu = 14.11\,amu$.

    • $2\,N\,atoms \times 14.01\,amu = 28.02\,amu$.

  • Result: Formula weight of $C_{10}H_{14}N_2$ = $162.23\,amu$.

• Molar Mass:

  • The mass of one mole of any substance, reported in grams ($g$).

  • The value of molar mass in grams equals the value of formula weight in atomic mass units.

• Relating Grams to Moles:

  • Molar mass serves as a conversion factor between moles and grams.

• Sample Problem 5.10: Moles in 100. g of Aspirin ($C_9H_8O_4$, molar mass 180.2 g/mol):

  • Original Quantity: $100.\,g\,aspirin$.

  • Desired Quantity: $mol\,aspirin$.

  • Conversion Factor: $\frac{1\,mol}{180.2\,g\,aspirin}$.

  • Solution: $100.\,g\,aspirin \times \frac{1\,mol}{180.2\,g\,aspirin} = 0.555\,mol\,aspirin$.

Mole and Mass Calculations in Chemical Equations

• Mole Ratios: Balanced equations show the number of moles of each reactant that combine and products formed.

  • Example: $1\,N_{2(g)} + 1\,O_{2(g)} \rightarrow 2\,NO_{(g)}$ means $1\,mole$ of $N_2$ reacts with $1\,mole$ of $O_2$ to produce $2\,moles$ of $NO$.

  • Coefficients create mole ratios used as conversion factors (e.g., $\frac{1\,mol\,N_2}{2\,mol\,NO}$).

• Sample Problem 5.11: Moles Produced from a Given Quantity:

  • Equation: $2C_2H_{6(g)} + 7O_{2(g)} \rightarrow 4CO_{(g)} + 6H_2O_{(g)}$.

  • Problem: Find moles of $CO$ from $3.5\,moles$ of $C_2H_6$.

  • Conversion Factor: $\frac{4\,mol\,CO}{2\,mol\,C_2H_6}$.

  • Solution: $3.5\,mol\,C_2H_6 \times \frac{4\,mol\,CO}{2\,mol\,C_2H_6} = 7.0\,mol\,CO$.

• Converting Moles of Reactant to Grams of Product:

  • Example: Grams of $O_3$ formed from $9.0\,mol$ of $O_2$ via $3O_{2(g)} \rightarrow 2O_{3(g)}$.

  • Step [1]: Moles of reactant to moles of product (mole-mole factor: $\frac{2\,mol\,O_3}{3\,mol\,O_2}$).

  • Step [2]: Moles of product to grams of product (molar mass: $\frac{48.00\,g\,O_3}{1\,mol\,O_3}$).

  • Calculation: $9.0\,mol\,O_2 \times \frac{2\,mol\,O_3}{3\,mol\,O_2} \times \frac{48.00\,g\,O_3}{1\,mol\,O_3} = 290\,g\,O_3$.

• Converting Grams of Reactant to Grams of Product:

  • Example: Grams of ethanol ($C_2H_6O$, molar mass $46.07\,g/mol$) from $14\,g$ of ethylene ($C_2H_4$, molar mass $28.05\,g/mol$).

  • Sequence: Grams Reactant $\rightarrow$ Moles Reactant $\rightarrow$ Moles Product $\rightarrow$ Grams Product.

  • Calculation: $14\,g\,C_2H_4 \times \frac{1\,mol\,C_2H_4}{28.05\,g\,C_2H_4} \times \frac{1\,mol\,C_2H_6O}{1\,mol\,C_2H_4} \times \frac{46.07\,g\,C_2H_6O}{1\,mol\,C_2H_6O} = 23\,g\,C_2H_6O$.

Oxidation and Reduction (Redox)

• Core Concepts:

  • Oxidation: The loss of electrons ($e^-$) from an atom.

  • Reduction: The gain of electrons by an atom.

  • Redox Reaction: Oxidation and reduction always occur together in a single reaction involving electron transfer.

• Agents:

  • Reducing Agent: A compound that is oxidized while causing another compound to be reduced.

  • Oxidizing Agent: A compound that is reduced while causing another compound to be oxidized.

• Specific Examples:

  • Zinc and Copper Reaction: $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$.

    • $Zn$ loses $2\,e^-$ to form $Zn^{2+}$ (Oxidized / Reducing Agent).

    • $Cu^{2+}$ gains $2\,e^-$ to form $Cu$ (Reduced / Oxidizing Agent).

  • Half Reactions:

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

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

  • Iron Rusting: $4Fe_{(s)} + 3O_{2(g)} \rightarrow 2Fe_2O_{3(s)}$.

    • $Fe$ loses electrons and is oxidized.

    • $O$ gains electrons and is reduced.

  • Lithium-Iodine Battery (Pacemakers): $2Li + I_2 \rightarrow 2LiI$.

    • $Li$ is oxidized to $Li^+$.

    • $I$ is reduced to $I^-$ from neutral $I_2$.

Energy Changes in Reactions

• Bond Energy:

  • Bond breaking always requires an input of energy.

  • Bond formation always releases energy.

  • Example for $Cl-Cl$: To cleave this bond, $58\,kcal/mil$ must be added; to form it, $58\,kcal/mol$ is released.

• Enthalpy Change (Heat of Reaction), $\Delta H$:

  • Endothermic Reaction:

    • Energy is absorbed.

    • $\Delta H$ is positive ($+$).

    • Products are higher in energy than reactants.

    • Example: Photosynthesis ($6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2$), $\Delta H = +678\,kcal/mol$.

  • Exothermic Reaction:

    • Energy is released.

    • $\Delta H$ is negative ($-$).

    • Products are lower in energy than reactants.

    • Example: Methane combustion ($CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$), $\Delta H = -213\,kcal/mol$.

Energy Diagrams and Reaction Rates

• Collision Theory: For a reaction to occur, molecules must collide with enough kinetic energy to break bonds.

• Diagram Components:

  • Vertical Axis: Energy.

  • Horizontal Axis: Progress of reaction / Reaction coordinate.

  • Transition State: The peak of the energy curve.

  • Energy of Activation ($E_a$): The difference in energy between the reactants and the transition state. It is the "energy barrier" for the reaction.

• Reaction Rate Factors:

  • High $E_a$: Fewer molecules have enough energy; the reaction is slow.

  • Low $E_a$: Many molecules have enough energy; the reaction is fast.

  • Concentration: Increasing reactant concentration increases collisions and reaction rate.

  • Temperature: Increasing temperature increases kinetic energy and reaction rate.

• Catalysts:

  • Substances that speed up a reaction by lowering $E_a$.

  • They are recovered unchanged and do not appear in the product.

  • They do not affect $\Delta H$.

• Environmental Application: Catalytic Converters:

  • Uses metal catalysts (rhodium, platinum, or palladium) to clean up auto engine exhaust via three catalyzed reactions.

Equilibrium and Le Châtelier’s Principle

• Reversible Reactions: Can occur in either direction.

  • Forward: Reactants to products.

  • Reverse: Products to reactants.

• Chemical Equilibrium: A system is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction.

  • Net concentrations do not change at equilibrium.

• Le Châtelier’s Principle: If a chemical system at equilibrium is stressed, the system will react in a direction that counteracts the disturbance.

• Effect of Changes (Table 5.3):

  • Adding Reactant: Equilibrium favors the products (shifts right).

  • Removing Reactant: Equilibrium favors the reactants (shifts left).

  • Adding Product: Equilibrium favors the reactants (shifts left).

  • Removing Product: Equilibrium favors the products (shifts right).