Chem 162 - Ch 9 Thermochemistry (standard enthalpy of formation)

Chapter 1: Introduction to Standard Enthalpies of Formation

• Enthalpy Change (ΔH): Indicates the heat change associated with a chemical reaction.

• Methods to Determine Enthalpy Change:

  • Experimental Method: Conducted in a lab using a calorimeter where heat capacity or specific heat is measured. Related experiment will be done in class.

  • Hess's Law: This law allows for calculating enthalpy change by manipulating known thermochemical equations to derive the desired reaction. The manipulations on the equations are mirrored in the enthalpy values.

  • Standard Enthalpy of Formation (ΔH°_F):

    • Defined under standard conditions (25 °C or 298 K, and 1 atm pressure).

    • Notated with the symbol ΔH° (the degree symbol represents standard conditions).

Chapter 2: Definitions

• Allotropes: Different forms of the same element in the same physical state.

  • Examples:

    • Carbon: Diamond and graphite are allotropes.

    • Oxygen: O2 (oxygen gas) and O3 (ozone) are allotropes.

• Reference Form: The most stable form of an element at standard conditions, found in reference tables.

Chapter 3: Enthalpy Change

• Change in Enthalpy of Formation (ΔH°_F):

  • The enthalpy change for the formation of one mole of a substance from its elements in their reference state.

  • The ΔH° of formation of any element in its reference form equals zero.

• Example: Formation of Water (H2O):

  • Chemical equation: 2 H2 (g) + O2 (g) → 2 H2O (l)

  • Balanced for one mole: H2 (g) + 1/2 O2 (g) → H2O (l)

  • Standard enthalpy of formation of water is -285.8 kJ.

• Calculating ΔH of a Reaction:

  • Use the formula: [ ΔH°_{reaction} = \sum (n \times ΔH°_F(products)) - \sum (n \times ΔH°_F(reactants)) ]

  • Example with methanol (CH3OH) liquid enthalpy = -238.7 kJ/mol, gaseous methanol = -200.7 kJ/mol.

  • Compute heat of vaporization as standard enthalpy change when transitioning from liquid to gas.

  • Calculated ΔH for vaporization = +38.0 kJ.

Chapter 4: Another Example

• Reaction of Methanol and Oxygen:

  • Given reaction: 2 CH3OH (l) + O2 (g) → 2 HCHO (g) + 2 H2O (l)

• Standard Enthalpies of Formation:

  • Methanol: -245.9 kJ/mol

  • Formaldehyde (HCHO): -150.2 kJ/mol

  • Water: -285.8 kJ/mol

  • Oxygen: Zero (as it is in its elemental form).

• Calculation Steps:

  • Products:

    • Formaldehyde = 2 × -150.2 kJ = -300.4 kJ.

    • Water = 2 × -285.8 kJ = -571.6 kJ.

    • Total = -300.4 + -571.6 = -872.0 kJ.

  • Reactants:

    • Methanol: 2 × -245.9 kJ = -491.8 kJ.

    • Oxygen: 0 kJ.

    • Total = -491.8 kJ.

• Final Calculation:

  • ΔH°_{reaction} = Total products - Total reactants = -872.0 kJ - (-491.8 kJ) = -380.2 kJ.

• Conclusion: Third method of determining reaction enthalpy, alongside experimental setup and Hess’s Law.