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Chemical Energetics - Bond breaking and bond forming calculation

Grade 11IGCSEChemistry

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

🔑Concepts

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Bond breaking is an endothermic process because energy must be absorbed from the surroundings to overcome the electrostatic forces of attraction between atoms. This corresponds to a positive enthalpy change (+ΔH+\Delta H).

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Bond forming is an exothermic process because energy is released to the surroundings when new chemical bonds are established and atoms reach a more stable state. This corresponds to a negative enthalpy change (−ΔH-\Delta H).

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The overall Enthalpy Change (ΔH\Delta H) of a reaction is the numerical 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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A reaction is exothermic if the energy released during bond forming is greater than the energy required for bond breaking (ΔH<0\Delta H < 0).

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A reaction is endothermic if the energy required for bond breaking is greater than the energy released during bond forming (ΔH>0\Delta H > 0).

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Bond energy is defined as the amount of energy required to break one mole of a specific covalent bond in the gaseous state, measured in kJ/molkJ/mol.

📐Formulae

ΔH=Total energy absorbed (breaking bonds)−Total energy released (forming bonds)\Delta H = \text{Total energy absorbed (breaking bonds)} - \text{Total energy released (forming bonds)}

ΔH=∑Bond Energies of Reactants−∑Bond Energies of Products\Delta H = \sum \text{Bond Energies of Reactants} - \sum \text{Bond Energies of Products}

💡Examples

Problem 1:

Calculate the enthalpy change (ΔH\Delta H) for the reaction between hydrogen and chlorine: H2(g)+Cl2(g)→2HCl(g)H_2(g) + Cl_2(g) \rightarrow 2HCl(g). Given bond energies: H−H=436 kJ/molH-H = 436 \text{ kJ/mol}, Cl−Cl=242 kJ/molCl-Cl = 242 \text{ kJ/mol}, and H−Cl=431 kJ/molH-Cl = 431 \text{ kJ/mol}.

Solution:

  1. Energy to break reactant bonds: (1×436)+(1×242)=678 kJ/mol(1 \times 436) + (1 \times 242) = 678 \text{ kJ/mol}. \n2. Energy released forming product bonds: 2×431=862 kJ/mol2 \times 431 = 862 \text{ kJ/mol}. \n3. ΔH=678−862=−184 kJ/mol\Delta H = 678 - 862 = -184 \text{ kJ/mol}.

Explanation:

Since the value of ΔH\Delta H is negative (−184 kJ/mol-184 \text{ kJ/mol}), the reaction is exothermic, meaning more energy is released when forming H−ClH-Cl bonds than is consumed breaking H−HH-H and Cl−ClCl-Cl bonds.

Problem 2:

Calculate the energy change for the combustion of methane: CH4+2O2→CO2+2H2OCH_4 + 2O_2 \rightarrow CO_2 + 2H_2O. Given: C−H=413C-H = 413, O=O=498O=O = 498, C=O=805C=O = 805, O−H=464O-H = 464 (all in kJ/molkJ/mol).

Solution:

  1. Reactants (Bond Breaking): (4×C−H)+(2×O=O)=(4×413)+(2×498)=1652+996=2648 kJ/mol(4 \times C-H) + (2 \times O=O) = (4 \times 413) + (2 \times 498) = 1652 + 996 = 2648 \text{ kJ/mol}. \n2. Products (Bond Forming): (2×C=O)+(4×O−H)=(2×805)+(4×464)=1610+1856=3466 kJ/mol(2 \times C=O) + (4 \times O-H) = (2 \times 805) + (4 \times 464) = 1610 + 1856 = 3466 \text{ kJ/mol}. \n3. ΔH=2648−3466=−818 kJ/mol\Delta H = 2648 - 3466 = -818 \text{ kJ/mol}.

Explanation:

The combustion of methane is highly exothermic because the bonds formed in CO2CO_2 and H2OH_2O are stronger/release more energy than the bonds broken in CH4CH_4 and O2O_2.

Bond breaking and bond forming calculation Revision - Grade 11 Chemistry IGCSE