Course Title: General Chemistry 1B 2025
Instructor: Dr. A. J. Shaka
Institution: University of California, Irvine
Quarter: Winter Quarter 2025
Course Code: 40120
Lecture 13:
Enthalpy of Formation
Third Law Entropy
Free Energy
Definition:
The enthalpy of formation is the heat absorbed or released when making one mole of a compound from its elements in their standard states.
By convention, enthalpy of formation of any element is set to zero.
Analogy:
Similar to how sea level is assigned a height of zero.
Without this convention, differences in height would have to be recorded for every location.
Example:
For the reaction: [ N_2(g) + O_2(g) \rightarrow N_2O(g) ]
The enthalpy of formation, ( \Delta H_f^\circ ) for N2O is +82.05 kJ/mol.
Assumed conditions: 298.15 K and 1 atm pressure.
Note: This reaction may not occur under standard lab conditions.
Warnings about Enthalpy Values:
Enthalpy values apply strictly to the written reaction.
If a reaction is multiplied by a number, multiply the ( \Delta H ) by that same number.
Enthalpy is described per mole of product formed; sometimes fractions are needed to balance reactions.
Physical State Considerations:
Enthalpy of formation values vary depending on the physical state of the substance, e.g.:
Liquid water: ( H_2O(l) \Delta H_f^\circ = -285.8 , \text{kJ/mol} )
Water vapor: ( H_2O(g) \Delta H_f^\circ = -241.8 , \text{kJ/mol} )
Scenario:
Decomposition of ammonia: [ 4NH_3(g) \rightarrow 2N_2(g) + 6H_2(g) ]
Ammonia’s formation reaction serves as a point for decomposition calculations.
Methodology:
Find the standard enthalpy of formation of ammonia (given: ( \Delta H_f^\circ = -46 , \text{kJ/mol} )).
Convert forward reaction to enthalpy values for product formation reactions.
Results:
Total calculated enthalpy change for the reaction: ( \Delta H = -812 , \text{kJ} )
Ludwig Boltzmann's Contribution:
Developed a statistical theory defined by the equation:[ S = k , ln W ]
Where ( k ) is Boltzmann’s constant, ( S ) is absolute entropy, and ( W ) is the number of possible microstates.
Absolute Zero and Perfect Crystals:
At absolute zero (0 Kelvin), there is no thermal energy available for a perfect crystal, yielding:[ S = k , ln 1 = 0 ]
Thus, the entropy of a perfect crystal at absolute zero is defined as zero, which gives rise to the Third Law.
Understanding ( \Delta G ):
At constant temperature, the surroundings can be treated as an infinite heat bath.
Heat exchange occurs while maintaining constant temperature.
Spontaneity Criterion:
The criterion for spontaneity is given by ( \Delta S_{universe} \geq 0 ).
Connects system changes to thermodynamic potentials.
Equating Enthalpy and Entropy Changes:
At constant pressure and temperature:[ \Delta G = \Delta H - T \Delta S \leq 0 ]
Third Law Application:
Entropy can be directly calculated from the behavior of a perfect crystal at absolute zero, advancing further to relate heat added.
Phase Change Considerations:
In phase transitions, entropy changes can be evaluated using: [ \Delta S = \frac{\Delta H_{Transition}}{T_{Transition}} ]
An example includes melting and boiling points as equilibria with ( \Delta G = 0 ).
Example Problem 30:
Calculate changes in entropy for water heating from 50.0 °C to 150.0 °C using specific heat capacities and the enthalpy of vaporization.
Final result reflects a positive entropy change (( \Delta S = 248.96 , J/K )).
General Rule:
Any real process has ( \Delta S_{universe} > 0 ).
Example:
Two copper blocks come to a common temperature demonstrating an increase in total entropy of the universe through heat transfer leading to equalized temperatures.
Implication:
Entropy changes reflect the direction of spontaneous processes, emphasizing that irreversible processes remain non-reversible in nature.