Different Types of Enthalpies in Reactions

🔥 Standard Enthalpy of Combustion (ΔcH°)

This is the heat released per mole when a substance burns completely in oxygen, with all reactants and products in their standard states. Combustion is always exothermic (releases heat)!

Examples:

  • Butane (cooking gas): \[ \ce{C4H10(g) + \frac{13}{2}O2(g) -> 4CO2(g) + 5H2O(l)} \quad \Delta_c H^\circ = -2658.0 \text{kJ mol}^{-1} \]
  • Glucose (energy in food): \[ \ce{C6H12O6(s) + 6O2(g) -> 6CO2(g) + 6H2O(l)} \quad \Delta_c H^\circ = -2802.0 \text{kJ mol}^{-1} \]

⚛️ Enthalpy of Atomization (ΔaH°)

This is the heat needed to break 1 mole of bonds completely to form gaseous atoms.

Examples:

  • Dihydrogen: \[ \ce{H2(g) -> 2H(g)} \quad \Delta_a H^\circ = 435.0 \text{kJ mol}^{-1} \]
  • Methane: \[ \ce{CH4(g) -> C(g) + 4H(g)} \quad \Delta_a H^\circ = 1665 \text{kJ mol}^{-1} \]
  • Sodium (same as sublimation): \[ \ce{Na(s) -> Na(g)} \quad \Delta_a H^\circ = 108.4 \text{kJ mol}^{-1} \]

🔗 Bond Enthalpy (ΔbondH°)

This measures the strength of chemical bonds:

  • Bond dissociation enthalpy: Heat to break 1 mole of a specific bond in a gaseous molecule (e.g., \(\ce{H-H}\) bond: 435.0 kJ/mol).
  • Mean bond enthalpy: Average heat to break 1 mole of a bond type across different molecules. For methane (\(\ce{CH4}\)), the mean \(\ce{C-H}\) bond enthalpy is: \[ \Delta_{\ce{C-H}} H^\circ = \frac{1665}{4} = 416 \text{kJ mol}^{-1} \]

Calculating reaction heat using bonds: \[ \Delta_r H^\circ = \sum (\text{Bond enthalpies of reactants}) – \sum (\text{Bond enthalpies of products}) \]

Note: Only valid for gas-phase reactions!

🧂 Lattice Enthalpy

This is the heat needed to split 1 mole of an ionic solid into gaseous ions. Example for NaCl: \[ \ce{NaCl(s) -> Na+(g) + Cl^{-}(g)} \quad \Delta_{\text{lattice}} H^\circ = +788 \text{kJ mol}^{-1} \]

Born-Haber Cycle (used to calculate lattice enthalpy indirectly):
Enthalpy diagram for NaCl lattice enthalpy Steps for NaCl:

  1. Sublimation of Na: +108.4 kJ/mol
  2. Ionization of Na: +496 kJ/mol
  3. Dissociation of Cl₂: +121 kJ/mol
  4. Electron gain by Cl: -348.6 kJ/mol
  5. Formation of NaCl(s): -411.2 kJ/mol
\[ \Delta_{\text{lattice}} H^\circ = 411.2 + 108.4 + 496 + 121 – 348.6 = +788 \text{kJ mol}^{-1} \]

💧 Enthalpy of Solution (ΔsolH°)

This is the heat change when 1 mole of a substance dissolves in a solvent. For ionic compounds: \[ \Delta_{\text{sol}} H^\circ = \Delta_{\text{lattice}} H^\circ + \Delta_{\text{hyd}} H^\circ \]

Example (NaCl): \[ \Delta_{\text{sol}} H^\circ = (+788) + (-784) = +4 \text{kJ mol}^{-1} \] (Slightly endothermic → little heat absorbed)

📉 Enthalpy of Dilution

This is the heat change when extra solvent is added to a solution. Example for HCl: \[ \ce{HCl.25aq + 15aq -> HCl.40aq} \quad \Delta H = -0.76 \text{kJ mol}^{-1} \] (Heat is released when diluting concentrated HCl!)

🚀 Important Concepts for NEET

  1. 🔥 Combustion Enthalpy Calculations
    Use combustion data to find formation enthalpies (Hess’s Law).
    Ex: Find ΔfH° of benzene from its combustion heat (-3267 kJ/mol).
  2. 🔗 Bond Enthalpy Applications
    Calculate reaction enthalpies using: \[ \Delta_r H^\circ = \sum (\text{Bond energies})_{\text{reactants}} – \sum (\text{Bond energies})_{\text{products}} \]
  3. 🧂 Born-Haber Cycle
    Deduce lattice enthalpy using sublimation, ionization, bond dissociation, and electron gain enthalpies.
  4. 💧 Enthalpy of Solution
    Relate ΔsolH° to lattice enthalpy and hydration enthalpy: \[ \Delta_{\text{sol}} H^\circ = \Delta_{\text{lattice}} H^\circ + \Delta_{\text{hyd}} H^\circ \]

Keep practicing these concepts—they’re the building blocks of thermodynamics! 💪✨