Chemistry Clarity · chemistryclarity.com

Hess's Law and Enthalpies of Formation: Flashcards

Revise Hess's law, combining equations, standard enthalpies of formation and calculating ΔH.

10 cardsIntermediate

10 cards

  1. Question
    State Hess's law.
    Answer

    The enthalpy change of a reaction is the same whatever route is taken from reactants to products.

  2. Question
    Why does Hess's law work?
    Answer

    Enthalpy is a state function: it depends only on the start and end, not the path (energy is conserved).

  3. Question
    An equation is reversed. What happens to ΔH?
    Answer

    It changes sign.

  4. Question
    An equation is multiplied by 3. What happens to ΔH?
    Answer

    ΔH is multiplied by 3.

  5. Question
    Define the standard enthalpy of formation, ΔHf°.
    Answer

    The enthalpy change when 1 mol of a compound forms from its elements in their standard states (1 bar, usually 25 °C).

  6. Question
    What is ΔHf° of O₂(g)?
    Answer

    0 kJ/mol: it is an element in its standard state.

  7. Question
    Formula for ΔH° from enthalpies of formation?
    Answer

    ΔH°=∑nΔHf°(products)−∑nΔHf°(reactants)\Delta H° = \sum n\Delta H_\text{f}°(\text{products}) - \sum n\Delta H_\text{f}°(\text{reactants})

  8. Question
    ΔH° for CH₄ + 2O₂ → CO₂ + 2H₂O(l)? (ΔHf°: CH₄ −74.6, CO₂ −393.5, H₂O(l) −285.8 kJ/mol)
    Answer

    [−393.5 kJ + 2(−285.8 kJ)] − (−74.6 kJ) = −890.5 kJ

  9. Question
    Why do ΔHf° of H₂O(l) and H₂O(g) differ?
    Answer

    By the enthalpy of vaporization: −241.8 kJ/mol − (−285.8 kJ/mol) = +44.0 kJ/mol.

  10. Question
    Use C + O₂ → CO₂ (−393.5 kJ) and CO + ½O₂ → CO₂ (−283.0 kJ) to find ΔH for C + ½O₂ → CO.
    Answer

    −393.5 kJ + 283.0 kJ = −110.5 kJ (the second equation is reversed)