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Chemical Energetics - Bond energies

Grade 12IGCSEChemistry

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

🔑Concepts

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Bond Energy (or Bond Enthalpy) is defined as the amount of energy required to break one mole of a specific covalent bond in the gaseous state, measured in kJ⋅mol−1kJ \cdot mol^{-1}.

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Bond Breaking is always an endothermic process (DeltaH>0\\Delta H > 0) because energy must be absorbed from the surroundings to overcome the electrostatic forces of attraction between the atoms.

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Bond Making is always an exothermic process (DeltaH<0\\Delta H < 0) because energy is released when atoms form a stable lower-energy state in a bond.

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The overall enthalpy change (DeltaH\\Delta H) of a reaction is the difference between the total energy required to break the bonds in the reactants and the total energy released when new bonds are formed in the products.

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Average Bond Enthalpies are used in calculations. These are mean values taken across a range of different compounds (e.g., the C−HC-H bond energy is an average of C−HC-H strengths in various hydrocarbons) and may differ slightly from specific experimental results.

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If DeltaH\\Delta H is negative, the reaction is exothermic (more energy released than absorbed). If DeltaH\\Delta H is positive, the reaction is endothermic (more energy absorbed than released).

📐Formulae

ΔH=∑(bond energies of reactants)−∑(bond energies of products)\Delta H = \sum (\text{bond energies of reactants}) - \sum (\text{bond energies of products})

ΔH=∑Ebroken−∑Eformed\Delta H = \sum E_{\text{broken}} - \sum E_{\text{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). Bond energies: C−H=413 kJ⋅mol−1C-H = 413\, kJ \cdot mol^{-1}, O=O=498 kJ⋅mol−1O=O = 498\, kJ \cdot mol^{-1}, C=O=805 kJ⋅mol−1C=O = 805\, kJ \cdot mol^{-1}, O−H=463 kJ⋅mol−1O-H = 463\, kJ \cdot mol^{-1}.

Solution:

  1. Energy to break bonds (reactants):
  • 4×(C−H)=4×413=1652 kJ4 \times (C-H) = 4 \times 413 = 1652\, kJ
  • 2×(O=O)=2×498=996 kJ2 \times (O=O) = 2 \times 498 = 996\, kJ
  • Total ∑Ebroken=1652+996=2648 kJ⋅mol−1\sum E_{\text{broken}} = 1652 + 996 = 2648\, kJ \cdot mol^{-1}.
  1. Energy released by forming bonds (products):
  • 2×(C=O)=2×805=1610 kJ2 \times (C=O) = 2 \times 805 = 1610\, kJ
  • 4×(O−H)=4×463=1852 kJ4 \times (O-H) = 4 \times 463 = 1852\, kJ
  • Total ∑Eformed=1610+1852=3462 kJ⋅mol−1\sum E_{\text{formed}} = 1610 + 1852 = 3462\, kJ \cdot mol^{-1}.
  1. Calculate ΔH\Delta H:
  • ΔH=2648−3462=−814 kJ⋅mol−1\Delta H = 2648 - 3462 = -814\, kJ \cdot mol^{-1}.

Explanation:

The negative value of ΔH=−814 kJ⋅mol−1\Delta H = -814\, kJ \cdot mol^{-1} indicates that the reaction is exothermic. More energy is released when forming the bonds in CO2CO_2 and H2OH_2O than is required to break the bonds in CH4CH_4 and O2O_2.

Problem 2:

Calculate the enthalpy change for the reaction: H2(g)+Cl2(g)→2HCl(g)H_2(g) + Cl_2(g) \rightarrow 2HCl(g). Given bond energies: H−H=436 kJ⋅mol−1H-H = 436\, kJ \cdot mol^{-1}, Cl−Cl=242 kJ⋅mol−1Cl-Cl = 242\, kJ \cdot mol^{-1}, H−Cl=431 kJ⋅mol−1H-Cl = 431\, kJ \cdot mol^{-1}.

Solution:

  1. Bonds Broken: 1×(H−H)+1×(Cl−Cl)=436+242=678 kJ⋅mol−11 \times (H-H) + 1 \times (Cl-Cl) = 436 + 242 = 678\, kJ \cdot mol^{-1}.
  2. Bonds Formed: 2×(H−Cl)=2×431=862 kJ⋅mol−12 \times (H-Cl) = 2 \times 431 = 862\, kJ \cdot mol^{-1}.
  3. ΔH=678−862=−184 kJ⋅mol−1\Delta H = 678 - 862 = -184\, kJ \cdot mol^{-1}.

Explanation:

The reaction is exothermic. The energy required to break the diatomic molecules H2H_2 and Cl2Cl_2 is less than the energy released when forming two moles of HClHCl.

Bond energies - Revision Notes & Key Formulas | IGCSE Grade 12 Chemistry