2nd Law of Thermodynamics

This lecture discusses the second law of thermodynamics, focusing on how universal processes work.
The driving force behind physical and chemical processes involves a combination of enthalpy and entropy.

In exothermic processes, energy (enthalpy) decreases, which is generally favorable.
However, many spontaneous processes are endothermic, indicating that enthalpy change alone doesn't fully explain spontaneity.
Entropy plays a crucial role in determining spontaneity.

Although increasing entropy is generally favorable, some spontaneous processes involve a decrease in entropy (e.g., crystallization, freezing).
These examples appear to contradict the idea that entropy must always increase for a spontaneous process.

The second law states that for a spontaneous change, the entropy of the universe must increase, not just the system.
The universe comprises both the system and the surroundings.
If the entropy of the system decreases, it must be accompanied by a larger increase in the entropy of the surroundings, ensuring that: \Delta S_{universe} > 0

\Delta S_{universe} > 0 : Spontaneous change in the forward direction.
\Delta S_{universe} < 0 : Non-spontaneous change in the forward direction; spontaneous in reverse.
$\Delta S_{universe} = 0$ : System is at equilibrium.

Direct measurement of the entropy change of the surroundings is impossible, as the surroundings encompass the entire universe.
A more useful form of the second law is needed, relating $\Delta S_{surroundings}$ to a measurable property of the system.

Exothermic processes release heat to the surroundings, increasing its entropy.
Endothermic processes absorb heat from the surroundings, decreasing its entropy.
At constant pressure, the heat exchanged (Q) is equal to the change in enthalpy ($\Delta H$).
$\Delta S{surroundings}$ is proportional to $-\Delta H{system}$ .

The impact on $\Delta S_{surroundings}$ is greater when the surroundings are initially at a lower temperature.
$\Delta S_{surroundings}$ is inversely proportional to temperature (T).
Analogy: Giving \$1,000 to someone with little money has a greater impact than giving it to a billionaire.
$\Delta S_{surroundings} \propto \frac{1}{T}$

Combining the relationships:
$\Delta S{surroundings} = -\frac{\Delta H{system}}{T}$

Substituting $\Delta S{surroundings}$ into the equation for $\Delta S{universe}$
\Delta S{universe} = \Delta S{system} - \frac{\Delta H_{system}}{T} > 0
This equation involves measurements only on the system.

Multiplying the previous equation by -T (and reversing the inequality):
-T\Delta S{universe} = -T\Delta S{system} + \Delta H_{system} < 0
Defining Gibbs Free Energy (G):
$G = H - TS$
At constant temperature:
$\Delta G = \Delta H - T\Delta S$

\Delta G < 0 : Spontaneous change.
The lowering of free energy ($\Delta G$) is the true driving force for processes in the universe.
Free energy accounts for both enthalpic and entropic contributions.
$\Delta G$ represents the maximum work obtainable from a process.
Delta G = -T * Delta S universe

Standard free energy change for a reaction: $\Delta G^\circ = \sum n\Delta Gf^\circ (products) - \sum m\Delta Gf^\circ (reactants)$
- n and m are the stoichiometric coefficients.
- Standard conditions: gases at 1 atm, solutions at 1 M, pure solids and liquids, and a standard temperature of 25°C.
The free energy of formation ($\Delta G_f^\circ$) of an element in its standard state is zero.
For compounds, $\Delta G_f^\circ$ is the free energy change when one mole of the compound is synthesized from its elements in their standard states.

If |\Delta G| < 10 \text{ kJ} , significant amounts of both reactants and products are present at equilibrium.
Large negative $\Delta G$ : Essentially a one-way, spontaneous forward reaction.
Large positive $\Delta G$ : Essentially no forward reaction (non-spontaneous).
This will be related to the equilibrium constant K in the next class.

Enthalpy: Exothermic ($\Delta H < 0$) is favorable; endothermic ($\Delta H > 0$) is unfavorable.
Entropy: Increase in entropy ($\Delta S > 0$) is favorable; decrease ($\Delta S < 0$) is unfavorable.

if the signs of delta H and delta S are opposite we don't need to calculate anything to figure out delta G
- If \Delta H > 0 and \Delta S < 0 , $\Delta G$ is always positive (non-spontaneous at all temperatures).
- If $\Delta H < 0$ and $\Delta S > 0$ , $\Delta G$ is always negative (spontaneous at all temperatures).
If $\Delta H$ and $\Delta S$ have the same sign:
- \Delta H > 0 (unfavorable) and \Delta S > 0 (favorable): High temperatures favor spontaneity.
- \Delta H < 0 (favorable) and \Delta S < 0 (unfavorable): Low temperatures favor spontaneity.

In freshman chemistry there are assumption that delta H and delta S are constant over a temp range
Partial differential equations are used in P-Chem
The calculation of the temperature required for spontaneity it's a good estimate, but it not an exact calculation

Entropy increases with temperature for solids, liquids, and gases.
Phase transitions (melting, boiling) involve significant increases in entropy.
At the melting point, the temperature remains constant while the solid melts to become a liquid
Gases > liquids > solids in terms of entropy due to increased molecular motion.

For the fusion of water at 0°C:
- $\Delta H_{fusion} = 6.02 \text{ kJ/mol}$
\Delta S_{fusion} = \frac{6.02 \times 10^3 \text{ J/mol}}{273 \text{ K}} = 22.1 \text{ J/(K \cdot mol)}
As expected, the entropy increases as ice melts to form liquid water.