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Energetics

Indirect enthalpy calculations using Hess's Law

  • 3 things to know
  • Unit AS 2

Overview

Recall questions on using standard enthalpies of formation and of combustion in a Hess cycle, worked calculations on methane, calcium carbonate, ethene, the thermite reaction, iron ore and sulfur dioxide, working backwards to an unknown enthalpy of formation, and checking the sign of the answer.

What you need to know

Answer each one before you reveal it.

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Key term

Describe how standard enthalpies of formation are used to find the enthalpy change of a reaction.

Key term

Describe how standard enthalpies of combustion are used to find the enthalpy change of a reaction.

Numbers

Using ΔfHӨ values of −824 kJ mol⁻¹ for Fe₂O₃(s) and −1676 kJ mol⁻¹ for Al₂O₃(s), calculate ΔH for Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(s).

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Describe how standard enthalpies of formation are used to find the enthalpy change of a reaction.

    Model answer

    Each value is multiplied by the balancing number of that substance, the values for the reactants are added, the values for the products are added, and the reactant sum is subtracted from the product sum.

  2. 2

    Describe how standard enthalpies of combustion are used to find the enthalpy change of a reaction.

    Model answer

    Each value is multiplied by the balancing number of that substance, and the sum for the products is subtracted from the sum for the reactants.

  3. 3

    Using ΔfHӨ values of −75 kJ mol⁻¹ for CH₄(g), −394 kJ mol⁻¹ for CO₂(g) and −286 kJ mol⁻¹ for H₂O(l), calculate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).

    Model answer

    The products total −394 + 2 × −286, which is −966 kJ mol⁻¹, and the reactants total −75 kJ mol⁻¹, so ΔH is −966 minus −75, which is −891 kJ mol⁻¹.

  4. 4

    Using ΔcHӨ values of −394 kJ mol⁻¹ for graphite, −286 kJ mol⁻¹ for hydrogen and −890 kJ mol⁻¹ for methane, calculate the standard enthalpy of formation of methane.

    Model answer

    The value is −394 + 2 × −286 − (−890), which is −394 − 572 + 890, giving −76 kJ mol⁻¹.

  5. 5

    Using ΔfHӨ values of −1207 kJ mol⁻¹ for CaCO₃(s), −635 kJ mol⁻¹ for CaO(s) and −394 kJ mol⁻¹ for CO₂(g), calculate ΔH for CaCO₃(s) → CaO(s) + CO₂(g).

    Model answer

    The products total −1029 kJ mol⁻¹ and the reactant is −1207 kJ mol⁻¹, so ΔH is −1029 minus −1207, which is +178 kJ mol⁻¹.

  6. 6

    Using ΔfHӨ values of +52 kJ mol⁻¹ for C₂H₄(g), −286 kJ mol⁻¹ for H₂O(l) and −278 kJ mol⁻¹ for C₂H₅OH(l), calculate ΔH for C₂H₄(g) + H₂O(l) → C₂H₅OH(l).

    Model answer

    The reactants total +52 − 286, which is −234 kJ mol⁻¹, so ΔH is −278 minus −234, which is −44 kJ mol⁻¹.

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