10.4

Concept Overview: Enthalpy of reaction can be calculated not only from enthalpies of formation but also from bond enthalpies.
- Bond Entropy Framework: The process involves deconstructing starting materials into their elemental bonds and constructing the products based on bonds formed.

Breaking and Forming Bonds
- Identify bonds broken in the reactants and bonds formed in the products.
- Differentiate between breaking bonds (requires energy) and forming bonds (releases energy).

Chemical Reaction:
- Methane ( ext{CH}4) reacts with water ( ext{H}2 ext{O}) to produce carbon dioxide ( ext{CO}2) and hydrogen ( ext{H}2).
 $ext{CH}4 + 2 ext{H}2 ext{O} ightarrow ext{CO}2 + 4 ext{H}2$
Bonds Involved:
- Bonds Broken:
- 4 carbon-hydrogen (C-H) bonds in methane.
- 4 oxygen-hydrogen (O-H) bonds from 2 water molecules.
- Bonds Formed:
- 2 carbon-oxygen double (C=O) bonds in carbon dioxide.
- 4 hydrogen-hydrogen (H-H) single bonds.
Calculating Enthalpy of Reaction ($ ext{Delta H}$):
- Use the equation:
  $ext{Delta H}_{ ext{reaction}} = ext{(Bonds Broken)} - ext{(Bonds Formed)}$
- Plugging in bond enthalpy values:
- 4 C-H: 4 * 414 kJ/mol = 1656 kJ/mol
- 4 O-H: 4 * 464 kJ/mol = 1856 kJ/mol
- Total Bonds Broken:
  $ext{Total Broken} = 1656 ext{ kJ/mol} + 1856 ext{ kJ/mol} = 3512 ext{ kJ/mol}$
Forming Bonds:
- 2 C=O: 2 * 799 kJ/mol = 1598 kJ/mol
- 4 H-H: 4 * 436 kJ/mol = 1744 kJ/mol
- Total Bonds Formed:
  $ext{Total Formed} = 1598 + 1744 = 3342 ext{ kJ/mol}$
Final Calculation:
- $ext{Delta H}_{ ext{reaction}} = 3512 - 3342 = 170 ext{ kJ/mol}$

Introduction to the Born-Haber Cycle:
- The Born-Haber cycle describes how ionic solids (salts like NaCl) form from elements, considering various energetic steps.
- Example Reaction: Sodium solid ( ext{Na}) + Chlorine gas ( ext{Cl}_2) → Sodium chloride solid ( ext{NaCl})
- The overall enthalpy change for the reaction is negative 411 kJ/mol.

Start with Sodium (Na):
- Is it gaseous? No (sublimation required to convert solid to gas).
  - Enthalpy of Sublimation: 108 kJ/mol
- Is it monoatomic? Yes.
- Proper charge? Convert Na gas → Na⁺ + e⁻ (ionization).
  - Ionization Energy: 496 kJ/mol.
Next for Chlorine (Cl):
- Gaseous? Yes.
- Monoatomic? No, convert Cl₂ → Cl (1/2 of bond energy required).
  - Bond Energy: Municipal energy for Cl-Cl bond is 244 kJ/mol, thus:
    $ext{Energy for half a bond} = 122 ext{ kJ/mol}$
- Proper charge? Convert Cl → Cl⁻ + e⁻ (electron affinity).
  - Electron Affinity: -349 kJ/mol.
Combining all steps:
- From gasses to ions ready for ionic bond formation:
  - Sodium in gaseous state: Na⁺, Chlorine in gaseous state: Cl⁻.
- Form Sodium Chloride (NaCl): Lattice formation.
- This is the enhalpy of lattice energy.

Overall Reaction:
$ext{Delta H reaction} = ext{Sublimation} + ext{Ionization} + ext{Half Bond Energy} + ext{Electron Affinity} + ext{Lattice Energy}$
Values to substitute:
- Reaction: -411 kJ/mol
- Sublimation: +108 kJ/mol
- Ionization: +496 kJ/mol
- Half Bond Energy: +122 kJ/mol
- Electron Affinity: -349 kJ/mol
Solving for Lattice Energy:
- Forward calculation using rearrangement yields:
  $ext{Lattice Energy} = -411 + 108 + 496 + 122 - 349$
  = 788 kJ/mol (as a negative value).
Conclusion:
- Lattice Energy can be seen as the energy released upon the formation of ionic bonds in salts.

Chemical Demonstration:
- Sodium stored in oil due to reactivity with air is placed into hexane. This step dissolves the oil for a cleaner chemical reaction.
Ionization:
- Upon ionization, calorimetric phenomena (color change) denote energizing sodium to Na⁺.
Chlorine Reaction:
- Reactive conditions shown, a vigorous reaction occurs releasing gas and light, showing the release of lattice energy during ionic compound formation.
Implications:
- This demonstration illustrates the principles of thermodynamics in ionic reactions, useful in various applications, highlighting the energetic stability associated with salt formation.