9.2 Absolute Entropy and Entropy Change Student

9.2 Absolute Entropy and Entropy Change

Learning Objective

  • Calculate the entropy change for a chemical or physical process based on the absolute entropies of the species involved, which is essential in understanding thermodynamic processes.

Essential Knowledge

  • Entropy change ( ΔS) can be calculated using the absolute entropies of the species before and after a given process. It is an important concept in chemistry, reflecting the degree of disorder or randomness in a system, which impacts spontaneity and equilibrium.

Key Equation

  • The fundamental equation for determining the entropy change of a reaction is:

    [ ΔS°{reaction} = Σ∆S°{products} - Σ∆S°_{reactants} ]

Concepts

  • Entropy (∆S): A quantitative measure of the disorder or randomness in a system. An increase in entropy indicates a move toward greater disorder.

  • In section 9.1, entropy was qualitatively assessed, identifying when entropy increases or decreases.

Second Law of Thermodynamics

  • It states that:

    • ( ΔS°_{universe} > 0 ) for spontaneous processes, indicating that natural processes tend to progress in the direction of increasing total entropy.

    • ( ΔS°_{universe} = 0 ) at equilibrium, representing a state where no net change occurs in the system.

System Entropy Calculation

  • Focus on system entropy to derive reactions: ( ΔS°{system} = ΔS°{reaction} )

  • It is calculated by summing the standard entropies of products and reactants using the formula:

    [ ΔS°{reaction} = [(cΔS°{C}) + (dΔS°{D})] - [(aΔS°{A}) + (bΔS°_{B})] ]

    where a, b, c, and d are the coefficients in the balanced equation of the reaction, and A, B, C, and D are the reactants and products, respectively.

Example Calculation: Reaction of Ammonia with Oxygen

  • Given reaction: [ 4 ext{ NH}_3(g) + 7 ext{ O}_2(g) → 4 ext{ NO}_2(g) + 6 ext{ H}_2 ext{O}(g) ]

  • To calculate the total entropy change for this reaction, we use: [ ΔS°_{rxn} = [4 * 240.1 ext{ J/mol∙K} + 6 * 188.8 ext{ J/mol∙K}] - [4 * 192.8 ext{ J/mol∙K} + 7 * 205.3 ext{ J/mol∙K}] ]

  • Result:[ ΔS°_{rxn} = -114.4 ext{ J/mol∙K} ]

  • This negative value indicates a decrease in the disorder of the system during the reaction.

Standard Entropy Values

Compound

ΔS° (J/mol∙K)

NH3(g)

192.8

O2(g)

205.2

NO2(g)

240.1

H2O(g)

188.8

Practice Problems

WE DO: Example Calculation

  • Reaction:[ 2 ext{ A}_2 ext{B}_2(l) + ext{ C}_2 ext{A}_4(l) → ext{ C}_2(g) + 4 ext{ A}_2 ext{B}(g) ]

  • Standard Entropy Values:

    • ΔS° A2B2 = 109.6 J/mol∙K

    • ΔS° C2A4 = 121.1 J/mol∙K

    • ΔS° C2 = 191.6 J/mol∙K

    • ΔS° A2B = 188.1 J/mol∙K

YOU DO: Calculations

  • Reaction:[ 2 ext{ PCl}_3(g) + ext{ O}_2(g) → 2 ext{ POCl}_3(g) ]

  • Student A Calculation:[ ΔS°_{reaction} = [(325.45 ext{ J/mol∙K})] - [(311.78 ext{ J/mol∙K}) + (205.14 ext{ J/mol∙K})] ]

  • Student B Calculation:[ ΔS°_{reaction} = [(2)(311.78 ext{ J/mol∙K}) + (1)(205.14 ext{ J/mol∙K})] - [(2)(325.45 ext{ J/mol∙K})] ]

  • Correctness: Review calculations to verify accuracy and identify any discrepancies.

  • Additional Reaction Example:[ ext{C}_2 ext{H}_4(g) + ext{H}_2(g) → ext{C}_2 ext{H}_6(g) ]

  • Standard Entropy Values:

    • C2H4(g) | 219.4

    • H2(g) | 130.58

    • C2H6(g) | 229.5

  • Further Reaction Example:[ 2 ext{ CH}_3 ext{OH}(g) + 3 ext{ O}_2(g) → 2 ext{ CO}_2(g) + 4 ext{ H}_2 ext{O}(g) ]

  • Standard Entropy Values:

    • CH3OH(g) | 237.6

    • O2(g) | 205.138

    • CO2(g) | 213.6

    • H2O(g) | 188.83

Understanding these principles and practicing calculations will enhance proficiency in thermodynamics and chemical reactions.