Bond Enthalpies
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Core Concept
Bond enthalpy (or bond dissociation energy) is the amount of energy required to break one mole of a particular type of bond in the gas phase. Average bond enthalpies are used to estimate the enthalpy change of a reaction by considering the energy required to break bonds in the reactants and the energy released when new bonds form in the products.
$\Delta H_{\text{reaction}} \approx \sum \Delta H_{\text{bonds broken}} - \sum \Delta H_{\text{bonds formed}}$
Practice Tips
Use Average Values Carefully: Remember, these values are averages and may differ slightly depending on molecular environment.
Focus on Bonds Changed: Only consider bonds that are broken in the reactants and formed in the products.
Account for All Bonds: Double-check that all bonds in each molecule are accounted for, especially in large molecules.
Sign Conventions: Energy required to break bonds is positive, while energy released from forming bonds is negative.
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Steps for Calculating Reaction Enthalpy Using Bond Enthalpies
STEP 1: Draw out the reaction with Lewis Structures
Write the balanced chemical equation for the reaction to identify all bonds that will be broken and formed.
STEP 2: List Bonds Broken and Formed
Identify and list all the bonds in the reactants that will be broken.
Identify and list all the bonds in the products that will be formed.
STEP 3: Find Average Bond Enthalpies
Use a bond enthalpy table to find the average bond enthalpies for each type of bond involved.
STEP 4: Calculate the Total Energy for Bonds Broken and Formed
Multiply each bond enthalpy by the number of that bond type in the molecule, and sum for all bonds broken and all bonds formed.
STEP 5: Calculate $\Delta H_{\text{reaction}}$
Use the formula: $\Delta H_{\text{reaction}} \approx \sum \Delta H_{\text{bonds broken}} - \sum \Delta H_{\text{bonds formed}}$
This gives an estimate of the enthalpy change for the reaction.
Key Concepts
Bond Enthalpy (Bond Dissociation Energy):
The amount of energy needed to break 1 mole of bonds in a gaseous substance.
Represented in kJ/mol.
Always positive because energy is required to break bonds.
Average Bond Enthalpy:
Bond enthalpies can vary depending on the molecular environment, so the average bond enthalpy is used as an approximation.
For example, the C–H bond enthalpy in methane ($\text{CH}_4$) is slightly different from that in ethane ($\text{C}_2\text{H}_6$), so an average value is used for general calculations.
Bond Breaking and Bond Formation:
Bond Breaking: Endothermic process (ΔH > 0), energy is absorbed.
Bond Formation: Exothermic process (ΔH < 0), energy is released.
Using Bond Enthalpies to Estimate $\Delta H_{\text{reaction}}$:
The enthalpy change of a reaction can be estimated by subtracting the total bond energies of the bonds formed from the total bond energies of the bonds broken.
$\Delta H_{\text{reaction}} \approx \sum \Delta H_{\text{bonds broken}} - \sum \Delta H_{\text{bonds formed}}$
Example Problem
Calculate the enthalpy change ($\Delta H_{\text{rxn}}$) for the following reaction using the average bond energies:
$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$
Average Bond Energies (kJ/mol):
C–H: 412
O=O: 498
C=O (in $\text{CO}_2$): 799
O–H: 463
STEP 1: Write the Balanced Equation
The equation provided is already balanced: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$
The structural formulas clarify the bonds:
$$\text{H}-\underset{\text{H}}{\overset{\text{H}}{\mid}}\text{C}-\text{H} \quad + \quad 2(\text{O}=\text{O}) \quad \rightarrow \quad \text{O}=\text{C}=\text{O} \quad + \quad 2(\text{H}-\text{O}-\text{H})$$
STEP 2: List Bonds Broken and Formed
Bonds Broken (Reactants)
Four (4) $\text{C–H}$ single bonds (from the $\text{CH}_4$ molecule).
Two (2) $\text{O}=\text{O}$ double bonds (from the $2\text{O}_2$ molecules).
Bonds Formed (Products)
Two (2) $\text{C}=\text{O}$ double bonds (from the $\text{CO}_2$ molecule).
Four (4) $\text{O–H}$ single bonds (from the $2\text{H}_2\text{O}$ molecules).
STEP 3: Find Average Bond Enthalpies
We use the provided average bond energies (in $\text{kJ/mol}$):
$\text{C–H}: 412$
$\text{O}=\text{O}: 498$
$\text{C}=\text{O}$ (in $\text{CO}_2$): $799$
$\text{O–H}: 463$
STEP 4: Calculate the Total Energy for Bonds Broken and Formed
Energy for Bonds Broken ($\sum \Delta H_{\text{bonds broken}}$):
This is the energy input (positive value) required to break the reactant bonds.
$$\sum \Delta H_{\text{broken}} = [4 \times \Delta H(\text{C–H})] + [2 \times \Delta H(\text{O}=\text{O})]$$
$$\sum \Delta H_{\text{broken}} = [4 \times 412\ \text{kJ/mol}] + [2 \times 498\ \text{kJ/mol}]$$
$$\sum \Delta H_{\text{broken}} = 1648\ \text{kJ/mol} + 996\ \text{kJ/mol}$$
$$\mathbf{\sum \Delta H_{\text{broken}} = 2644\ \text{kJ/mol}}$$
Energy for Bonds Formed ($\sum \Delta H_{\text{bonds formed}}$):
This is the energy released (treated as positive for calculation, then subtracted) when the product bonds are formed.
$$\sum \Delta H_{\text{formed}} = [2 \times \Delta H(\text{C}=\text{O})] + [4 \times \Delta H(\text{O–H})]$$
$$\sum \Delta H_{\text{formed}} = [2 \times 799\ \text{kJ/mol}] + [4 \times 463\ \text{kJ/mol}]$$
$$\sum \Delta H_{\text{formed}} = 1598\ \text{kJ/mol} + 1852\ \text{kJ/mol}$$
$$\mathbf{\sum \Delta H_{\text{formed}} = 3450\ \text{kJ/mol}}$$
STEP 5: Calculate $\Delta H_{\text{reaction}}$
$$\Delta H_{\text{reaction}} = \sum \Delta H_{\text{bonds broken}} - \sum \Delta H_{\text{bonds formed}}$$
$$\Delta H_{\text{reaction}} = 2644\ \text{kJ/mol} - 3450\ \text{kJ/mol}$$
$$\mathbf{\Delta H_{\text{reaction}} = -806\ \text{kJ/mol}}$$