Bond enthalpies Knowledge-Based Study

Study notes

1. What is Bond Enthalpy?

   Bond enthalpy (also referred to as bond dissociation energy) is the amount of energy required to break one mole of a specific covalent bond in a gaseous molecule into gaseous atoms under standard conditions.

   Because breaking chemical bonds requires an input of energy, bond breaking is always an endothermic process, meaning bond enthalpy values are always positive ($\Delta H > 0$). Conversely, forming bonds releases energy and is an exothermic process ($\Delta H < 0$).

   For diatomic molecules, bond enthalpy is unambiguous. For example, for hydrogen gas:

   $$\text{H}_2\text{(g)} ightarrow 2\text{H(g)} \quad \Delta H = +436\text{ kJ/mol}$$

   Mean (Average) Bond Enthalpy

   In polyatomic molecules, the energy required to break a specific type of bond varies slightly depending on the surrounding chemical environment. For example, breaking the four individual $\text{C–H}$ bonds in methane ($\text{CH}_4$) requires slightly different energies for each successive step:

   • First bond: $\text{CH}_4\text{(g)} ightarrow \text{CH}_3\text{(g)} + \text{H(g)}$
   • Second bond: $\text{CH}_3\text{(g)} ightarrow \text{CH}_2\text{(g)} + \text{H(g)}$
   • Third bond: $\text{CH}_2\text{(g)} ightarrow \text{CH(g)} + \text{H(g)}$
   • Fourth bond: $\text{CH(g)} ightarrow \text{C(g)} + \text{H(g)}$

   To provide practical reference data, chemists use mean bond enthalpy, which is the average value taken over a range of different gaseous molecules containing that specific bond.

   Note: Calculations using mean bond enthalpies are approximations because they average out specific molecular environments and strictly apply only to species in the gas phase.

   Enthalpy Change of Reaction from Bond Enthalpies

   During a chemical reaction, bonds in the reactants are broken, and new bonds in the products are formed. The overall standard enthalpy change of reaction ($\Delta H_{\text{rxn}}^\circ$) can be estimated using the following relationship:

   $$\Delta H_{\text{rxn}}^\circ = \sum (\text{Bond enthalpies of bonds broken}) - \sum (\text{Bond enthalpies of bonds formed})$$

   This can also be written as:

   $$\Delta H_{\text{rxn}}^\circ = \sum D(\text{reactants}) - \sum D(\text{products})$$

   where $D$ represents the bond enthalpy per mole of bonds.

   • If the energy released during bond formation is greater than the energy consumed during bond breaking, $\Delta H_{\text{rxn}}^\circ$ is negative, and the reaction is exothermic.
   • If more energy is required to break bonds than is released upon forming bonds, $\Delta H_{\text{rxn}}^\circ$ is positive, and the reaction is endothermic.

   Applying the Bond Enthalpy Formula

   Consider the complete combustion of methane in the gas phase:

   $$\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} ightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}$$

   Given Bond Enthalpies (in kJ/mol):

   • $\text{C–H} = 413$
   • $\text{O=O} = 498$
   • $\text{C=O (in CO}_2\text{)} = 804$
   • $\text{O–H} = 463$

   Step 1: Calculate Energy to Break Bonds

   Bonds broken (reactants):

   • $4 \times \text{C–H} = 4 \times 413 = 1652\text{ kJ}$
   • $2 \times \text{O=O} = 2 \times 498 = 996\text{ kJ}$
   • Total broken = $1652 + 996 = +2648\text{ kJ}$

   Step 2: Calculate Energy Released Forming Bonds

   Bonds formed (products):

   • $2 \times \text{C=O} = 2 \times 804 = 1608\text{ kJ}$
   • $4 \times \text{O–H} = 4 \times 463 = 1852\text{ kJ}$
   • Total formed = $1608 + 1852 = 3460\text{ kJ}$

   Step 3: Calculate $\Delta H_{\text{rxn}}^\circ$

   $$\Delta H_{\text{rxn}}^\circ = 2648\text{ kJ} - 3460\text{ kJ} = -812\text{ kJ/mol}$$

   The negative value confirms that this combustion reaction is strongly exothermic.