Thermodynamics

THERMODYNAMICS

INTRODUCTION TO THERMODYNAMICS

  • Chapter 16: AP Class

    • Covers units 9.1 - 9.5.

ENTROPY (ΔS) – DISPERSAL OF MATTER & ENERGY

  • Definition: Entropy is a measure of the disorder or dispersal of matter and energy in a system.

  • Key Points:

    • If ΔS > 0: Indicates that the system has more disorder (entropy increases).

    • If ΔS < 0: Indicates that the system has less disorder (entropy decreases).

  • Second Law of Thermodynamics:

    • The universe is constantly increasing the dispersal of matter and energy.

  • Relationship Between Entropy and Arrangements:

    • The greater the number of possible arrangements in a system, the higher the entropy.

  • Units of Entropy:

    • The units used to express entropy are J/(mol·rxn·K).

FORMULAS AND CALCULATIONS

  • Entropy Change Calculation:

    • Formula:
      <br>ΔS°<em>RXN=ΣS°</em>PRODUCTS−ΣS°REACTANTS<br><br>ΔS°<em>{RXN} = ΣS°</em>{PRODUCTS} - ΣS°_{REACTANTS} <br>

  • Factors Affecting Entropy:

    • Entropy increases:

    • From solid to liquid to gas state.

    • When a pure solid or liquid dissolves in a solvent.

    • Gas escaping from a solvent.

    • With increasing molecular complexity due to increased moving electrons (Example: KCl vs CaCl₂).

    • Reactions that result in an increase in the number of moles of particles, particularly gas, typically increase entropy.

PREDICTING ENTROPY CHANGE (ΔS)

  • Examples to consider for sign of ΔS:

    • Solid sugar dissolving in water: ΔS > 0 (increase in disorder).

    • Iodine vapor condensing into crystals: ΔS < 0 (decrease in disorder).

    • Formation of water from hydrogen and oxygen gases: Generally ΔS < 0 (though it may depend on the specific conditions).

CALCULATING ENTROPY CHANGE

  • Example Reaction:

    • 2 SO₂(g) + O₂(g) → 2 SO₃(g)

    • Given Data:

    • SO₂ (g) = 248.1 J/(mol·rxn·K)

    • O₂ (g) = 205.3 J/(mol·rxn·K)

    • SO₃ (g) = 256.6 J/(mol·rxn·K)

    • Task: Calculate the entropy change at 25°C in J/(mol·rxn·K).

PHASE CHANGES

  • Key Characteristics:

    • Occur at constant temperature and represent a system at equilibrium where ΔG = 0.

    • Enables the determination of absolute temperature at which a process becomes favorable.

  • Gibbs Free Energy Equation:
    <br>ΔG=ΔH−TΔS<br><br>ΔG = ΔH - TΔS <br>

GIBBS FREE ENERGY (ΔG)

  • Formula:
    <br>ΔG°<em>RXN=ΣG°</em>PRODUCTS−ΣG°REACTANTS<br><br>ΔG°<em>{RXN} = ΣG°</em>{PRODUCTS} - ΣG°_{REACTANTS} <br>

  • Significance of ΔG:

    • The value of ΔG determines the thermodynamic favorability of a reaction.

    • If ΔG > 0: Reaction is thermodynamically unfavorable.

    • If ΔG = 0: The reaction is in equilibrium.

    • If ΔG < 0: Reaction is thermodynamically favorable.

ECOLOGICAL IMPLICATIONS OF ΔG

  • Exothermic reactions often lead to negative ΔG.

  • More disorder (positive ΔS) generally favors a reaction's progress.

PRACTICE PROBLEMS

  • Calculate the thermodynamic boiling point of water vaporization given:

    • ΔHvap=+44extkJ/mol⋅rxnΔH_{vap} = +44 ext{ kJ/mol·rxn}

    • ΔSvap=118.8extJ/(K⋅mol⋅rxn)ΔS_{vap} = 118.8 ext{ J/(K·mol·rxn)}

SIGN AND PREDICTION PROBLEMS

  • Key Observations:

    • If ΔH (enthalpy) is negative and ΔS is positive, ΔG will likely be negative and the reaction will be favorable.

    • Various scenarios can yield positive or negative changes in these thermodynamic variables and thus influence ΔG.

RELATIONSHIPS BETWEEN ΔG, K, AND E

  • Fundamental Relationships:

    • ΔGextandK:ΔG ext{ and } K:

    • If ΔG < 0: K > 1 (products favored).

    • If ΔG > 0: K < 1 (reactants favored).

    • Gibbs Free Energy Equations:

    • General Formula:
      ΔG°=−nFEΔG° = -nFE

    • Relating Gibbs Free Energy and equilibrium constant:
      ΔG°=−RTextlnKΔG° = -RT ext{ln}K (where R = 8.314 J/(mol·K)).

PRACTICE EXAMPLES

  • For the reaction 2 H₂O(l) + O₂(g) → 2 H₂O₂(l):

    • Calculate the free energy of formation from provided ΔG°f values:

    • H₂O(l): -56.7 kcal/mol·rxn

    • O₂(g): 0 kcal/mol·rxn

    • H₂O₂(l): -27.2 kcal/mol·rxn

  • Example Reaction:

    • 2 SO₂(g) + O₂(g) → 2 SO₃(g):

      • Calculate ΔH°, ΔS°, and ΔG° using provided data for substances with their corresponding enthalpy and standard entropy values.

ADDITIONAL PRACTICE PROBLEMS

  • Problem involving rusting of iron:

    • Reaction:
      4Fe(s)+3O2(g)→2Fe2O3(s)4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s)

    • Task: Calculate the equilibrium constant using given data (enthalpy and entropy) for each reactant and product.

  • Explore the equilibrium constant for the reaction of carbon allotropes:

  1. Cdiamond(s)+O2(g)→CO2(g):ΔG°=−397kJC_{diamond}(s) + O₂(g) → CO₂(g): ΔG°= −397 kJ

  2. Cgraphite(s)+O2(g)→CO2(g):ΔG°=−394kJC_{graphite}(s) + O₂(g) → CO₂(g): ΔG°=-394 kJ

  3. C<em>diamond(s)→C</em>graphite(s)C<em>{diamond}(s) → C</em>{graphite}(s)