Every chemical reaction involves breaking bonds in the reactants and forming new bonds in the products. By using the known average strengths of these bonds, we can estimate the overall enthalpy change of a reaction before carrying it out in the laboratory.
🔑 Key Principle
Bond breaking is always an endothermic process because energy is required to overcome the electrostatic attraction between the nuclei and the shared pair of electrons (\( \Delta H > 0 \)). Conversely, bond making is always an exothermic process because energy is released when stable attractions are formed (\( \Delta H < 0 \)).
The enthalpy change required to break one mole of a specific covalent bond in the gaseous state, producing separate gaseous atoms or radicals under standard conditions.
The average value of the bond dissociation enthalpy for a specific type of covalent bond, averaged over a wide range of different compounds in the gaseous state.
Why We Use "Mean" Bond Enthalpies
The actual energy required to break a specific bond (like a C-H bond) depends on the surrounding atoms in the molecule. For example:
- Breaking the first C-H bond in methane (\( \text{CH}_4\text{(g)} \rightarrow \text{CH}_3\text{(g)} + \text{H(g)} \)) requires \( +435\text{ kJ mol}^{-1} \).
- Breaking the C-H bond in ethane (\( \text{C}_2\text{H}_6 \)) requires a slightly different amount of energy.
To make calculations practical, chemists use the mean bond enthalpy, which is the average value for that bond type across many different molecules. These values are always tabulated in data books.
The Bond Enthalpy Formula
Since breaking bonds is endothermic (positive) and making bonds is exothermic (negative), we calculate the overall enthalpy change using the following formula:
Or more simply:
\[ \Delta H = \sum \text{Reactants} - \sum \text{Products} \]
Energy Level Profile for Bond Cycles
We can model bond enthalpy calculations as a Hess's Law cycle where the intermediates are separate gaseous atoms. The reactants are broken down into gaseous atoms (endothermic step), which then combine to form the products (exothermic step).
- \( \text{C-H} = +413\text{ kJ mol}^{-1} \)
- \( \text{O=O} = +496\text{ kJ mol}^{-1} \)
- \( \text{C=O} = +805\text{ kJ mol}^{-1} \)
- \( \text{O-H} = +463\text{ kJ mol}^{-1} \)
Solution:
1. List the bonds broken in the reactants (left-hand side):
- \( 4 \times \text{C-H} \) bonds (in one mole of \( \text{CH}_4 \))
- \( 2 \times \text{O=O} \) bonds (in two moles of \( \text{O}_2 \))
\[ \sum \text{(bonds broken)} = (4 \times 413) + (2 \times 496) = 1652 + 992 = +2644\text{ kJ mol}^{-1} \]
2. List the bonds made in the products (right-hand side):
- \( 2 \times \text{C=O} \) bonds (in one mole of \( \text{CO}_2 \))
- \( 4 \times \text{O-H} \) bonds (in two moles of \( \text{H}_2\text{O} \))
\[ \sum \text{(bonds made)} = (2 \times 805) + (4 \times 463) = 1610 + 1852 = +3462\text{ kJ mol}^{-1} \]
3. Calculate the overall enthalpy change (\( \Delta H \)):
\[ \Delta H = \sum \text{(bonds broken)} - \sum \text{(bonds made)} \]
\[ \Delta H = 2644 - 3462 = -818\text{ kJ mol}^{-1} \]
- \( \text{C-H} = +413\text{ kJ mol}^{-1} \)
- \( \text{C-C} = +348\text{ kJ mol}^{-1} \)
- \( \text{H-H} = +436\text{ kJ mol}^{-1} \)
Solution:
1. Count the bonds on both sides of the equation. We can simplify by ignoring bonds that do not change (four C-H bonds exist in both ethene and ethane, so we can ignore them, or include them on both sides):
Reactants (Bonds Broken):
- \( 1 \times \text{C=C} \) bond (let this be \( x \))
- \( 4 \times \text{C-H} \) bonds
- \( 1 \times \text{H-H} \) bond
\[ \sum \text{broken} = x + (4 \times 413) + 436 = x + 1652 + 436 = x + 2088\text{ kJ mol}^{-1} \]
Products (Bonds Made):
- \( 1 \times \text{C-C} \) bond
- \( 6 \times \text{C-H} \) bonds
\[ \sum \text{made} = 348 + (6 \times 413) = 348 + 2478 = +2826\text{ kJ mol}^{-1} \]
2. Set up the equation using the given reaction enthalpy change (\( \Delta H = -124\text{ kJ mol}^{-1} \)):
\[ \Delta H = \sum \text{broken} - \sum \text{made} \]
\[ -124 = (x + 2088) - 2826 \]
\[ -124 = x - 738 \]
\[ x = -124 + 738 \]
\[ x = +614\text{ kJ mol}^{-1} \]
The mean bond enthalpy of the \( \text{C=C} \) double bond is \( +614\text{ kJ mol}^{-1} \).
Limitations of Mean Bond Enthalpies
Calculated bond enthalpy values are often slightly different from experimental enthalpies of reaction (measured using calorimetry and Hess's Law). In exams, you must provide two reasons for this divergence:
1. Values are Averages
Mean bond enthalpies are averaged across a wide range of different compounds. The actual strength of a bond in a specific molecule will vary slightly due to the electronic environment created by adjacent groups.
2. Gaseous State Limitation
Bond enthalpies are defined and measured exclusively for substances in the gaseous state. If any reactant or product is a liquid or solid in standard states (e.g. liquid water or liquid ethanol), standard enthalpy definitions must account for the energy associated with phase changes (e.g. condensation), whereas bond enthalpies do not.
A classic exam question asks: "Explain why the enthalpy of combustion of water vapour calculated using bond enthalpies differs from the standard enthalpy of combustion of hydrogen." The answer is that standard conditions require water to be in its standard state, which is a liquid, whereas bond enthalpies calculate water as a gas. Standard value includes the exothermic condensation of water vapour to liquid water, making it more exothermic.
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