Spontaneity, Entropy and Free Energy

Flashcards based on the key concepts from the lecture on spontaneity, entropy, and free energy.

Spontaneous Process

A process that occurs without outside intervention.

Entropy (S)

A measure of how dispersed a system's energy is; related to randomness.

Exothermicity

A reaction that releases heat; favors spontaneity but not the sole factor.

Microstates (Ω)

The number of ways a particular state can be achieved; related to the probability of that state.

Positional Probability

Probability based on the number of configurations in space that yield a particular state.

First Law of Thermodynamics

The energy of the universe is constant; used for energy bookkeeping.

Second Law of Thermodynamics

Entropy of the universe always increases in spontaneous processes.

Free Energy (G)

A thermodynamic function that predicts spontaneity at constant temperature and pressure.

Isothermal Process

A process where the temperature remains constant.

Adiabatic Process

A process where no heat is transferred in or out of the system.

ΔSuniv

The change in entropy of the universe; determines if a process is spontaneous.

Gibbs Free Energy Equation

ΔG = ΔH - TΔS; relates changes in enthalpy, entropy, and temperature.

Microstate vs. Macrostate

Microstate: a specific arrangement of particles; Macrostate: overall state defined by macroscopic properties.

Gibbs Free Energy (ΔG) Condition

For a process to be spontaneous, ΔG must be negative.

Standard Entropy (S°)

Entropy of a substance at standard conditions (298 K, 1 atm).

High Entropy State

A state where energy is highly dispersed; more probable.

Equilibrium Process

A process where the forward and reverse reaction rates are equal.

Relatively High Temperature Effect

At high temperatures, endothermic processes may become spontaneous.

T and P Relationship

Changes in pressure and temperature affect the spontaneity of reactions.

Calculation Problem 1: Gibbs Free Energy and Spontaneity

A reaction has a \Delta H = -120 kJ/mol and \Delta S = -30 J/(mol \cdot K) at 298 K. Calculate the Gibbs Free Energy change (\Delta G) and determine if the reaction is spontaneous under these conditions.
Solution:

1. Convert \Delta S to kJ/(mol \cdot K)
   -30 J/(mol \cdot K) = -0.030 kJ/(mol \cdot K)

2. Use the Gibbs Free Energy equation: \Delta G = \Delta H - T\Delta S
   \Delta G = -120 kJ/mol - (298 K)(-0.030 kJ/(mol \cdot K))
   \Delta G = -120 kJ/mol + 8.94 kJ/mol
   \Delta G = -111.06 kJ/mol
   Since \Delta G is negative, the reaction is spontaneous.

Calculation Problem 2: Temperature for Spontaneity

For a reaction in which \Delta H = +180 kJ/mol and \Delta S = +90 J/(mol \cdot K), at what temperature will the reaction become spontaneous? Assume standard conditions and constant \Delta H and \Delta S.
Solution:

1. Convert \Delta S to kJ/(mol \cdot K)
   +90 J/(mol \cdot K) = +0.090 kJ/(mol \cdot K)

2. For spontaneity, \Delta G < 0. At the edge of spontaneity (equilibrium), \Delta G = 0.

3. Set \Delta G = 0 in the Gibbs Free Energy equation: 0 = \Delta H - T\Delta S

4. Solve for T: T\Delta S = \Delta H \implies T = \frac{\Delta H}{\Delta S}

5. T = \frac{+180 kJ/mol}{+0.090 kJ/(mol \cdot K)}

6. T = 2000 K
   The reaction becomes spontaneous when T > 2000 K because for an endothermic reaction (\Delta H > 0) with increasing entropy (\Delta S > 0), high temperature favors spontaneity.

Calculation Problem 3: Predicting Spontaneity (Always Non-Spontaneous)

For a certain reaction, \Delta H = +50 kJ/mol and \Delta S = -100 J/(mol \cdot K). Determine if the reaction is spontaneous at 25 ^\circ C (298 K).

Solution:

1. Convert \Delta S to kJ/(mol \cdot K)
   

   -100 J/(mol \cdot K) = -0.100 kJ/(mol \cdot K)

2. Use the Gibbs Free Energy equation: \Delta G = \Delta H - T\Delta S
   

   \Delta G = +50 kJ/mol - (298 K)(-0.100 kJ/(mol \cdot K))
   

   \Delta G = +50 kJ/mol + 29.8 kJ/mol
   

   \Delta G = +79.8 kJ/mol
   

   Since \Delta G is positive (>0), the reaction is non-spontaneous at 25 ^\circ C. This type of reaction (\Delta H > 0 and \Delta S < 0) is never spontaneous at any temperature.
   

Calculation Problem 4: Determining \Delta H for Spontaneity

A reaction occurs at 350 K with an entropy change of \Delta S = +75 J/(mol \cdot K). What is the maximum value of \Delta H for this reaction to be spontaneous at this temperature?

Solution:

1. Convert \Delta S to kJ/(mol \cdot K)
   

   +75 J/(mol \cdot K) = +0.075 kJ/(mol \cdot K)

2. For a reaction to be spontaneous, \Delta G < 0. At the limit of spontaneity (equilibrium), \Delta G = 0.

3. Use the Gibbs Free Energy equation: 0 = \Delta H - T\Delta S

4. Solve for \Delta H: \Delta H = T\Delta S

5. Substitute values: \Delta H = (350 K)(+0.075 kJ/(mol \cdot K))

6. Calculate: \Delta H = +26.25 kJ/mol
   

   For the reaction to be spontaneous (\Delta G < 0), \Delta H must be less than +26.25 kJ/mol (i.e., \Delta H < +26.25 kJ/mol).