Unit 9 AP chemistry ALL Rules

Unit 9: Applications of Thermodynamics

1. Laws of Thermodynamics

First Law of Thermodynamics (Law of Energy Conservation)

Energy cannot be created or destroyed; it can only be transformed from one form to another.

ΔU = q - w

Where:

• ΔU = change in internal energy
• q = heat added to the system
• w = work done by the system

Question 1: A system absorbs 200 J of heat and does 50 J of work. What is the change in internal energy of the system?

Using the first law: ΔU = q - w = 200 J - 50 J = 150 J.

Second Law of Thermodynamics

In any spontaneous process, the total entropy of the universe increases.

ΔS_universe = ΔS_system + ΔS_surroundings > 0

Question 2: A reaction has ΔS_system = -120 J/K and occurs at 300 K. What must be true about ΔS_surroundings for the reaction to be spontaneous?

For spontaneity, ΔS_universe = ΔS_system + ΔS_surroundings > 0. Therefore, ΔS_surroundings > 120 J/K.

Third Law of Thermodynamics

The entropy of a perfect crystalline substance approaches zero as the temperature approaches absolute zero (0 K).

Question 3: Why does the entropy of a perfect crystal approach zero at 0 K?

At 0 K, a perfect crystal has only one microstate (W=1), so S = k ln(W) = k ln(1) = 0.

2. Gibbs Free Energy and Spontaneity

The Gibbs free energy change determines the spontaneity of a process.

ΔG = ΔH - TΔS

Where:

• ΔG = change in Gibbs free energy
• ΔH = change in enthalpy
• T = temperature in Kelvin
• ΔS = change in entropy

Question 4: A reaction has ΔH = -50 kJ and ΔS = -0.1 kJ/K. At what temperatures is the reaction spontaneous?

ΔG = ΔH - TΔS. For spontaneity, ΔG < 0. Solving -50 kJ - T(-0.1 kJ/K) < 0 gives T < 500 K.

3. Relationship Between Gibbs Free Energy and Equilibrium Constant

The standard Gibbs free energy change is related to the equilibrium constant (K) by:

ΔG° = -RT ln K

Where:

• ΔG° = standard Gibbs free energy change
• R = universal gas constant (8.314 J/mol·K)
• T = temperature in Kelvin
• K = equilibrium constant

Question 5: At 298 K, a reaction has ΔG° = -5.7 kJ/mol. Calculate the equilibrium constant K.

ΔG° = -RT ln K ⇒ -5700 J/mol = -(8.314 J/mol·K)(298 K) ln K ⇒ ln K = 2.3 ⇒ K ≈ 10.

4. Van 't Hoff Equation

The Van 't Hoff equation relates the change in the equilibrium constant (K) with temperature (T):

ln(K₂/K₁) = (-ΔH°/R) * (1/T₂ - 1/T₁)

Where:

• K₁, K₂ = equilibrium constants at temperatures T₁ and T₂, respectively
• ΔH° = standard enthalpy change