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Energetics of Reactions - Bond Enthalpy Calculations

Grade 9IB

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

🔑Concepts

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Bond Enthalpy (or Bond Energy) is the amount of energy required to break one mole of a specific covalent bond in the gaseous state.

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Bond breaking is an endothermic process because energy must be absorbed from the surroundings to overcome the electrostatic attraction between atoms (DeltaH\\Delta H is positive).

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Bond making is an exothermic process because energy is released to the surroundings when new stable bonds are formed (DeltaH\\Delta H is negative).

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The overall Enthalpy Change (DeltaH\\Delta H) of a reaction is determined by the balance between energy used to break reactant bonds and energy released when product bonds form.

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If the energy released during bond making is greater than the energy required for bond breaking, the reaction is exothermic.

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If the energy required for bond breaking is greater than the energy released during bond making, the reaction is endothermic.

📐Formulae

DeltaH=sum(textBondEnthalpiesofReactants)−sum(textBondEnthalpiesofProducts)\\Delta H = \\sum (\\text{Bond Enthalpies of Reactants}) - \\sum (\\text{Bond Enthalpies of Products})

DeltaH=textEnergyabsorbedtobreakbonds−textEnergyreleasedwhenbondsform\\Delta H = \\text{Energy absorbed to break bonds} - \\text{Energy released when bonds form}

💡Examples

Problem 1:

Calculate the enthalpy change (DeltaH\\Delta H) for the reaction between hydrogen and chlorine to form hydrogen chloride: H2(g)+Cl2(g)rightarrow2HCl(g)H_2(g) + Cl_2(g) \\rightarrow 2HCl(g). Given the following bond enthalpies: H−H=436textkJ/molH-H = 436 \\text{ kJ/mol}, Cl−Cl=242textkJ/molCl-Cl = 242 \\text{ kJ/mol}, and H−Cl=431textkJ/molH-Cl = 431 \\text{ kJ/mol}.

Solution:

  1. Energy required to break bonds (Reactants):
  • 1 mole of H−HH-H bonds: 1times436=436textkJ1 \\times 436 = 436 \\text{ kJ}
  • 1 mole of Cl−ClCl-Cl bonds: 1times242=242textkJ1 \\times 242 = 242 \\text{ kJ} Total Energy In = 436+242=678textkJ436 + 242 = 678 \\text{ kJ}
  1. Energy released forming bonds (Products):
  • 2 moles of H−ClH-Cl bonds: 2times431=862textkJ2 \\times 431 = 862 \\text{ kJ} Total Energy Out = 862textkJ862 \\text{ kJ}
  1. Calculate DeltaH\\Delta H: DeltaH=678−862\\Delta H = 678 - 862 beginarrayr678−862hline−184endarray\\begin{array}{r} 678 \\ -862 \\ \\hline -184 \\end{array} DeltaH=−184textkJ/mol\\Delta H = -184 \\text{ kJ/mol}

Explanation:

To solve this, we first sum the energy needed to break all bonds in the reactants (Hydrogen and Chlorine molecules). Then, we sum the energy released when the bonds in the products (two Hydrogen Chloride molecules) are formed. Finally, we subtract the energy of the products from the reactants. Since the result is negative (−184textkJ/mol-184 \\text{ kJ/mol}), the reaction is exothermic.