Bond Enthalpy and Energy Changes in Reactions
Quick Notes
- Breaking bonds = endothermic (requires energy)
- Forming bonds = exothermic (releases energy)
- Use average bond enthalpy values to calculate ΔH of reactions:
- ΔH = Σ(bonds broken) − Σ(bonds formed)
- Bond enthalpies are average values from many compounds
– they aren't exact for a specific molecule. - Bond enthalpies are given in kJ mol⁻¹ and are found in the data booklet.
Full Notes
Understanding Bond Enthalpy
Bond enthalpy (sometimes also called bond dissociation energy) is:
“The energy required to break one mole of a particular bond in the gaseous state.”
Breaking bonds requires energy and is an Endothermic process (+ΔH).
Making bonds releases energy and is an Exothermic process (-ΔH).
Example Bond breaking and bond making
Breaking H–H bond: H2(g) → 2H(g), ΔH = +436 kJ mol⁻¹
Forming H–H bond: 2H(g) → H2(g), ΔH = −436 kJ mol⁻¹
Mean Bond Enthalpy
Mean bond enthalpy is the average energy required to break a bond, calculated using different molecules that have that bond type in.
For Example C–H bond
The C–H bond has a mean bond enthalpy of +412 kJ mol⁻¹. However, the exact bond enthalpy of a C–H bond will be slightly different depending on the exact environment (molecule) it’s in.
This means calculations using mean bond enthalpies to find enthalpy changes in reactions won’t be as accurate as those calculated using experimental data (such as from calorimetry).
Calculating Enthalpy Change Using Bond Enthalpies
The enthalpy change of a reaction can be estimated using:
Where:
- Bonds broken (reactants) means energy absorbed (endothermic, +ΔH).
- Bonds formed (products) means energy released (exothermic, -ΔH).
Remember bond enthalpies are for substances in gaseous states. It is really important to make sure the states of all substances are in gaseous phase when dealing with bond enthalpy calculations - sometimes you need to use enthalpy of vaporisation first in calculations.
Worked Example Calculation: Combustion of Methane (CH4)
Reaction and data
Given bond enthalpies:
C–H = +412 kJ mol⁻¹; O=O = +498 kJ mol⁻¹; C=O = +805 kJ mol⁻¹; O–H = +463 kJ mol⁻¹
- Step 1: Bonds Broken (Reactants — Energy Absorbed)
Bonds in CH4: 4 × C–H = 4 × 412 = 1648 kJ
Bonds in O2: 2 × O=O = 2 × 498 = 996 kJ
Total energy to break bonds = 1648 + 996 = 2644 kJ - Step 2: Bonds Formed (Products — Energy Released)
Bonds in CO2: 2 × C=O = 2 × 805 = 1610 kJ
Bonds in H2O: 4 × O–H = 4 × 463 = 1852 kJ
Total energy released = 1610 + 1852 = 3462 kJ - Step 3: Calculate Enthalpy Change
ΔH = Σ(bonds broken) − Σ(bonds formed)
ΔH = 2644 − 3462 = −818 kJ mol⁻¹ (exothermic reaction)
Accuracy of Bond Enthalpy Calculations
Bond enthalpy calculations are estimates because:
- They use mean bond enthalpies instead of exact bond enthalpies in a specific molecule.
- They assume all substances are in the gaseous state, which may not be true in real reactions.
Hence, experimental enthalpy values (from calorimetry) are often more accurate for a given reaction.
Summary
- Bond breaking is endothermic and bond making is exothermic.
- We can estimate reaction enthalpy change using average bond enthalpies with ΔH = bonds broken minus bonds formed.
- Mean bond enthalpies are approximate and assume gaseous states.
- Experimental methods such as calorimetry often give more accurate values.
Linked Course Questions
How would you expect bond enthalpy data to relate to bond length and polarity?
Bond length: Shorter bonds have greater orbital overlap, making them stronger and giving them higher bond enthalpies. For example, a C≡C triple bond has a higher bond enthalpy than a C=C double bond, which is stronger than a C–C single bond.
Bond polarity: Polar bonds often have stronger electrostatic attraction between atoms of differing electronegativity, which can increase bond enthalpy. For example, the H–F bond is both short and polar, leading to a very high bond enthalpy.
However, polar bonds can also be more reactive, especially in mechanisms like nucleophilic substitution, because the partial charges make them more susceptible to attack. So while polarity may increase bond enthalpy, it can also increase reactivity depending on the chemical context.
How does the strength of a carbon–halogen bond affect the rate of a nucleophilic substitution reaction?
The rate of nucleophilic substitution depends on how easily the carbon–halogen bond breaks. A weaker bond (i.e. lower bond enthalpy) is easier to break, leading to a faster reaction.
C–I bonds are the weakest, followed by C–Br, then C–Cl, because iodine atoms are larger and form longer, weaker bonds with carbon, while chlorine atoms are smaller, forming shorter, stronger bonds.
As a result, iodoalkanes react fastest, and chloroalkanes react slowest in nucleophilic substitution reactions.