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Energetics / Thermochemistry - Bond enthalpies

Grade 12IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Average bond enthalpy is defined as the energy required to break one mole of a specific type of bond in a gaseous molecule, averaged over several similar compounds. It is measured in kJ⋅mol−1kJ \cdot mol^{-1}.

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Bond breaking is an endothermic process because energy must be absorbed to overcome the electrostatic attraction between nuclei and the shared pair of electrons; thus, ΔH\Delta H is positive (++).

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Bond making is an exothermic process because energy is released when new chemical bonds form; thus, ΔH\Delta H is negative (−-).

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Calculations using bond enthalpies are approximations because they use 'average' values rather than specific environment values, and they assume all species are in the gaseous state (gg).

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The enthalpy change of a reaction is calculated by subtracting the total energy released by forming bonds (products) from the total energy required to break bonds (reactants): ΔH=∑BEbroken−∑BEformed\Delta H = \sum BE_{broken} - \sum BE_{formed}.

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Bond strength is influenced by bond length and bond order. Generally, triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds.

📐Formulae

ΔHreaction=∑(Bonds Broken)−∑(Bonds Formed)\Delta H_{reaction} = \sum (\text{Bonds Broken}) - \sum (\text{Bonds Formed})

X−Y(g)→X(g)+Y(g)ΔH=+BE(X−Y)X-Y(g) \rightarrow X(g) + Y(g) \quad \Delta H = +BE(X-Y)

ΔH≈∑BE(reactants)−∑BE(products)\Delta H \approx \sum BE(\text{reactants}) - \sum BE(\text{products})

💡Examples

Problem 1:

Calculate the enthalpy change (ΔH\Delta H) for the combustion of methane: CH4(g)+2O2(g)→CO2(g)+2H2O(g)CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g). Given bond enthalpies: C−H=414 kJ⋅mol−1C-H = 414 \, kJ \cdot mol^{-1}, O=O=498 kJ⋅mol−1O=O = 498 \, kJ \cdot mol^{-1}, C=O=804 kJ⋅mol−1C=O = 804 \, kJ \cdot mol^{-1}, and O−H=463 kJ⋅mol−1O-H = 463 \, kJ \cdot mol^{-1}.

Solution:

Bonds broken (reactants):\text{Bonds broken (reactants):} 4×(C−H)+2×(O=O)=4(414)+2(498)=1656+996=2652 kJ⋅mol−14 \times (C-H) + 2 \times (O=O) = 4(414) + 2(498) = 1656 + 996 = 2652 \, kJ \cdot mol^{-1} Bonds formed (products):\text{Bonds formed (products):} 2×(C=O)+4×(O−H)=2(804)+4(463)=1608+1852=3460 kJ⋅mol−12 \times (C=O) + 4 \times (O-H) = 2(804) + 4(463) = 1608 + 1852 = 3460 \, kJ \cdot mol^{-1} ΔH=∑BEbroken−∑BEformed=2652−3460=−808 kJ⋅mol−1\Delta H = \sum BE_{broken} - \sum BE_{formed} = 2652 - 3460 = -808 \, kJ \cdot mol^{-1}

Explanation:

To solve this, we first identify all bonds being broken in the reactants (4 C−H4 \, C-H bonds in methane and 2 O=O2 \, O=O double bonds in oxygen). Then identify all bonds being formed in the products (2 C=O2 \, C=O double bonds in carbon dioxide and 4 O−H4 \, O-H single bonds in two moles of water). The final enthalpy is the difference between energy absorbed and energy released.

Problem 2:

Explain why the calculated ΔH\Delta H for the reaction H2(g)+12O2(g)→H2O(l)H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) using bond enthalpies is usually different from the experimental value found in data booklets.

Solution:

The experimental value for H2O(l)H_2O(l) includes the enthalpy of condensation, whereas bond enthalpy calculations assume all reactants and products are in the gaseous state (H2O(g)H_2O(g)).

Explanation:

Bond enthalpy values are strictly defined for gaseous states. In the reaction provided, the product is liquid water (ll). To get the experimental value, one would have to account for the intermolecular forces (Hydrogen bonding) formed when water transitions from gas to liquid, which releases additional energy (exothermic).

Bond enthalpies - Revision Notes & Key Formulas | IB Grade 12 Chemistry