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Energetics / Thermochemistry - Bond enthalpies

Grade 11IBChemistry

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

🔑Concepts

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Average Bond Enthalpy is the energy required to break one mole of a specific type of bond in a gaseous molecule, averaged over several similar compounds.

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Bond breaking is an endothermic process, meaning it requires energy (+ΔH+\Delta H).

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Bond making is an exothermic process, meaning it releases energy (−ΔH-\Delta H).

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The enthalpy change of a reaction (ΔHrxn\Delta H_{rxn}) can be estimated using the formula: ΔH=∑BE(bonds broken)−∑BE(bonds formed)\Delta H = \sum BE(\text{bonds broken}) - \sum BE(\text{bonds formed}).

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Limitations: Bond enthalpy values are averages and may not be exact for specific molecules. Furthermore, these values are only strictly valid for substances in the gaseous state (gg).

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If the calculated ΔH\Delta H is negative, the reaction is exothermic; if it is positive, the reaction is endothermic.

📐Formulae

ΔHrxn=∑BE(reactants)−∑BE(products)\Delta H_{rxn} = \sum BE(\text{reactants}) - \sum BE(\text{products})

ΔH=∑BE(bonds broken)−∑BE(bonds formed)\Delta H = \sum BE(\text{bonds broken}) - \sum BE(\text{bonds formed})

💡Examples

Problem 1:

Calculate the enthalpy change 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 average bond enthalpies: C−H=414 kJ mol−1C-H = 414\text{ kJ mol}^{-1}, O=O=498 kJ mol−1O=O = 498\text{ kJ mol}^{-1}, C=O=804 kJ mol−1C=O = 804\text{ kJ mol}^{-1}, O−H=463 kJ mol−1O-H = 463\text{ kJ mol}^{-1}.

Solution:

ΔH=[4(C−H)+2(O=O)]−[2(C=O)+4(O−H)]\Delta H = [4(C-H) + 2(O=O)] - [2(C=O) + 4(O-H)] ΔH=[4(414)+2(498)]−[2(804)+4(463)]\Delta H = [4(414) + 2(498)] - [2(804) + 4(463)] ΔH=[1656+996]−[1608+1852]\Delta H = [1656 + 996] - [1608 + 1852] ΔH=2652−3460=−808 kJ mol−1\Delta H = 2652 - 3460 = -808\text{ kJ mol}^{-1}

Explanation:

First, identify all bonds in the reactants (4 C−HC-H bonds and 2 O=OO=O double bonds) and the products (2 C=OC=O double bonds and 4 O−HO-H bonds). Sum the energy required to break the reactant bonds and subtract the energy released when product bonds are formed. The negative result indicates an exothermic reaction.

Problem 2:

Explain why the calculated ΔH\Delta H using bond enthalpies for the reaction H2O(g)→H2O(l)H_2O(g) \rightarrow H_2O(l) would be inaccurate if the state change is not accounted for.

Solution:

Bond enthalpies are defined for substances in the gaseous state. The transition from H2O(g)H_2O(g) to H2O(l)H_2O(l) involves the formation of intermolecular forces (hydrogen bonds), which releases energy (ΔHcondens\Delta H_{condens}), and is not accounted for in covalent bond enthalpy values.

Explanation:

Standard bond enthalpy calculations assume all species are gases. If a product is a liquid, the calculated ΔH\Delta H will be less exothermic (less negative) than the experimental value because the energy released during condensation is ignored.