Entropy, Entropy Changes, and Gibbs Free Energy Study Notes

Entropy and Energy Dispersal

  • Entropic factors: energy dispersal reduces the likelihood that energy flows drive changes in the system; more dispersed energy → greater thermodynamic stability.

  • Energy is dispersed over sub-microscopic particle motions: translations, rotations, vibrations.

  • As the number of particles increases and interactions diversify, energy can be dispersed across more pathways.

Entropy (S) and Standard Molar Entropy (S°)

  • Entropy describes the extent to which energy can be dispersed across the sub-microscopic particles of a system.

  • Standard molar entropy S° for a substance represents the energy-dispersal extent in one mole of substance at 25°C (298 K) and 1 bar.

  • Units: S°\text{ (units)} = \mathrm{J\,mol^{-1}\,K^{-1}}

  • Standard state definitions (examples):

    • 1 M solution

    • 1 bar gas

    • liquid

    • pure liquid

    • solid

    • pure solid element

    • most stable allotrope at 1 bar and 25°C (e.g., \mathrm{O2(g)} is more stable than \mathrm{O(g)} or \mathrm{O3(g)})

Change in Entropy for Reactions

  • The change in entropy for a reaction (∆rS°) can be calculated from tabulated standard molar entropies:
    \Deltar S^{\circ} = \left(\sumi ni^{\text{products}} Si^{\circ}\right) - \left(\sumj nj^{\text{reactants}} S_j^{\circ}\right)

  • Like enthalpy, ∆rS° is obtained by summing product S° values minus reactant S° values using stoichiometric coefficients.

Third Law of Thermodynamics

  • A perfect crystal of a pure substance at 0 K has zero entropy: S = 0\quad \text{at } T = 0\ \text{K}

  • Entropy values are measured relative to a perfect crystal at 0 K.

Predicting Entropy: Largest Entropy and Trends

  • Which substance has the largest entropy? (given S° values)

    • A. \mathrm{H_2O(s)}

    • B. \mathrm{H_2O(l)}

    • C. \mathrm{H_2O(g)}

    • Given: S^{\circ}(\mathrm{H2O(s)}) = 41.0\ \mathrm{J\,mol^{-1}\,K^{-1}}, S^{\circ}(\mathrm{H2O(l)}) = 70.0\ \mathrm{J\,mol^{-1}\,K^{-1}}, S^{\circ}(\mathrm{H_2O(g)}) = 188.8\ \mathrm{J\,mol^{-1}\,K^{-1}}

  • Entropy increases with increasing temperature and with phase changes solid → liquid → gas.

  • Example data:

    • Gas phase generally has higher S° than liquid; liquid higher than solid.

  • Trends with mass and complexity: atoms/molecules with greater mass and more complex structures tend to have higher entropies.

  • Example trends (molar mass vs S°):

    • He(g): M\approx 4\ \mathrm{g\,mol^{-1}},\ S^{\circ}=126\ \mathrm{J\,mol^{-1}\,K^{-1}}

    • Ne(g): M\approx 20\ \mathrm{g\,mol^{-1}},\ S^{\circ}=146.2\ \mathrm{J\,mol^{-1}\,K^{-1}}

    • Ar(g): M\approx 40\ \mathrm{g\,mol^{-1}},\ S^{\circ}=154.8\ \mathrm{J\,mol^{-1}\,K^{-1}}

    • Kr(g): M\approx 84\ \mathrm{g\,mol^{-1}},\ S^{\circ}=164.0\ \mathrm{J\,mol^{-1}\,K^{-1}}

    • C3H8(g): M\approx 44\ \mathrm{g\,mol^{-1}},\ S^{\circ}=269.9\ \mathrm{J\,mol^{-1}\,K^{-1}}

    • C6H14(g): M\approx 86\ \mathrm{g\,mol^{-1}},\ S^{\circ}=388.82\ \mathrm{J\,mol^{-1}\,K^{-1}}

  • Takeaway: Greater mass, greater molecular complexity → higher entropy.

Entropy Changes in Chemical Reactions

  • Entropy changes can be predicted for reactions using changes in moles of gas and phase changes:

    • Example: 6 moles of gas on the reactant side vs 7 moles on the product side increases disorder.

    • Reaction example (gas): \mathrm{C3H8(g) + 5\,O2(g) \rightarrow 3\,CO2(g) + 4\,H_2O(g)}

  • Example (aqueous+solid): \mathrm{AgNO3(aq) + NaCl(aq) \rightarrow NaNO3(aq) + AgCl(s)}

  • Predicting positive ΔS°: certain reactions increase gas-phase moles or produce more dispersed species.

Positive ΔS° Reactions (Practice Question)

  • Which reactions would have a positive change in entropy, ΔrS°?
    1) \mathrm{CaCO3(s) + 2\,HCl(aq) \rightarrow CaCl2(aq) + CO2(g) + H2O(l)}
    2) \mathrm{2\ NO(g) + O2(g) \rightarrow 2\ NO2(g)}
    3) \mathrm{2\ KClO3(s) \rightarrow 2\ KCl(s) + 3\ O2(g)}

  • Answer: 1 and 3 (increase in gas moles or production of gas from solids). (Reason: 2 decreases gas moles; 1 and 3 increase disorder by producing gas or increasing gas-phase species.)

Spontaneity, Equilibrium, and the Reaction Quotient

  • Spontaneous vs. nonspontaneous: a process is spontaneous if its composition evolves toward equilibrium (lower free energy) without external driving forces.

  • A process can be spontaneous or nonspontaneous depending on conditions (e.g., temperature, pressure).

  • Example concept: K < 1 implies reactant-favored equilibrium.

  • A numeric example: K = 0.0313; for a given reaction, Q can be compared to K to assess spontaneity.

  • At Q = 1 (standard state), the standard Gibbs free energy change ΔrG° is defined and relates to K (see below).

  • Standard state reference: 0 atm vs 1 bar; 1 M solutions; these standards set the baseline for ΔrG°, ΔrG°, and K.

Gibbs Free Energy and Spontaneity

  • Gibbs free energy is at a minimum at equilibrium.

  • On a plot of Gibbs free energy G versus composition (Q), ΔrG is the slope of the curve at that Q:
    \Delta_r G = \text{(slope of G vs Q at that composition)}

  • For a given reaction, if ΔrG is positive, the system evolves to increase reactants; if negative, it evolves to increase products. In either case, G decreases as equilibrium is approached.

  • Example interpretation: If the forward process is condensation of water (liquid to solid, etc.), ΔrG is positive; if evaporation (liquid to gas) is favored, ΔrG is negative.

  • At equilibrium, ΔrG = 0.

Relating ΔrG to Enthalpy and Entropy

  • The relationship between ΔrG, ΔrH, and ΔrS:
    \Deltar G = \Deltar H - T\Delta_r S

  • ΔrG° and the equilibrium constant K are related by:
    \Delta_r G^{\circ} = -RT\ln K

  • ΔrG° is typically tabulated for standard composition (Q = 1).

  • Units: ΔrH° in kJ/mol; ΔrS° in J/mol·K; ΔrG° and ΔrG are in kJ/mol (consistent with R and T units in J, K).

  • For any composition, the Gibbs energy is given by:
    \Deltar G = \Deltar G^{\circ} + RT\ln Q

  • Where:

    • R = 8.3145\ \mathrm{J\,mol^{-1}\,K^{-1}}

    • T in kelvin, Q is the reaction quotient.

Standard State and Vaporization Example (Water)

  • When Q = 1 for the process \mathrm{H2O(l) \rightleftharpoons H2O(g)} (at 298 K):

    • The slide lists: \Delta_r G^{\circ} = 8.59\ \mathrm{kJ\,mol^{-1}}

    • This implies the forward process (liquid → gas) has a positive ΔrG° at standard state, so the reverse process (gas → liquid, condensation) would be favored at standard state.

  • Note: Real values for water vaporization give a ΔG° that is positive at 298 K, reflecting non-spontaneity of vaporization at room temperature; the exact value can depend on the data source and reference state.

  • Another slide repeats: at 298 K, for the same process, ΔrG° ≈ +8.59 kJ/mol and Q = P_H2O, with the interpretation that the forward process has positive ΔrG°.

ΔrG° and Temperature/Composition (Q) Examples

  • When Q = 5.00 × 10^−3 (298 K) for the liquid↔gas water process, ΔrG° is still given as 8.59 kJ/mol in the slide, and ΔrG is determined by including RT ln Q:
    \Deltar G = \Deltar G^{\circ} + RT\ln Q

  • When Q = 0.0313 (298 K) for the same process, ΔrG° remains 8.59 kJ/mol; ΔrG depends on Q, demonstrating how nonstandard composition shifts spontaneity.

  • Key point: ΔrG° relates to K via \Delta_r G^{\circ} = -RT\ln K\,.

Spontaneity and Temperature Dependence

  • Spontaneity can depend on temperature. A process may be spontaneous at one temperature and not at another.

  • Visual note (conceptual): some processes are spontaneous for T > 0 °C and others for T < 0 °C depending on the balance between enthalpy and entropy contributions.

  • Practical implication: cooling a hot substance, gas expansion into a vacuum, or phase changes can shift spontaneity with temperature.

Real-World and Conceptual Examples of Spontaneity

  • Spontaneous for T > 0 °C vs spontaneous for T < 0 °C: temperature can flip whether a process is spontaneous.

  • Coffee cooling down: generally spontaneous (exothermic flow of heat to surroundings with increasing entropy of the surroundings).

  • Expansion of a gas into a vacuum: spontaneous (increase in entropy due to more available microstates).

  • Melting of ice: spontaneous above 0 °C, non-spontaneous below 0 °C; depends on temperature relative to the phase boundary.

Gibbs Free Energy Minima and Equilibrium Slopes

  • At equilibrium, Gibbs free energy is at its minimum with respect to composition.

  • The slope of the G vs composition curve at a given Q equals ΔrG for the reaction: a direct link between thermodynamics and reaction progress.

  • For a process like H2O(l) ⇄ H2O(g) at a given T, the sign of ΔrG indicates whether the system moves toward more liquid (positive ΔrG would favor reactants) or more gas (negative ΔrG would favor products).

  • A representative value on the slide: ΔrG = 2.85 kJ mol^−1 at a specified Q, illustrating a nonzero driving force away from equilibrium.

Standard vs General Formulations and Units

  • ΔrG° is commonly tabulated for the standard composition (Q = 1) and relates to K via \Delta_r G^{\circ} = -RT\ln K.

  • For any composition (Q ≠ 1), the general relation is \Deltar G = \Deltar G^{\circ} + RT\ln Q.

  • Remember: to compare to standard states, ensure units are consistent: ΔH° in kJ/mol and ΔS° in J/mol·K, so ΔG° will be in kJ/mol when using R in J·mol^−1·K^−1 with T in K.

Quick Concept Checks and Reminders

  • Spontaneous does not imply fast: a reaction can be thermodynamically spontaneous but kinetically slow.

  • The standard composition (Q = 1) provides a baseline reference point for comparing spontaneity under standard conditions.

  • The concept of energy dispersal (entropy) works together with enthalpy to determine spontaneity via ΔG.

Homework and Study Reminders (From the Transcript)

  • By Saturday, 9/20 at 8:30 AM: Complete and Upload Week 3 Discussion packet (topics you worked on this week).

  • By Monday, 9/22 at 8:30 AM: Complete and Upload Homework 5-5; will review in class Tuesday.

Notes on Data and References from the Transcript

  • Standard molar entropy (S°) and its units: \text{S° (J mol^{-1} K^{-1})}

  • Example S° values (for illustration):

    • \mathrm{H_2O(s)\;:\;S^{\circ}=41.0}

    • \mathrm{H_2O(l)\;:\;S^{\circ}=70.0}

    • \mathrm{H_2O(g)\;:\;S^{\circ}=188.8}

  • Example data for noble gases and small molecules (S° values in J/mol·K):

    • He(g): 126

    • Ne(g): 146.2

    • Ar(g): 154.8

    • Kr(g): 164.0

    • C3H8(g): 269.9

    • C6H14(g): 388.82

  • Example reactions cited:

    • Gas-producing combustion/oxidation example: \mathrm{C3H8(g) + 5 O2(g) \rightarrow 3 CO2(g) + 4 H_2O(g)}

    • Aqueous + solid formation: \mathrm{AgNO3(aq) + NaCl(aq) \rightarrow NaNO3(aq) + AgCl(s)}$$

Final Remarks

  • Entropy captures how energy becomes distributed among many microstates; Gibbs free energy combines entropy with enthalpy to determine spontaneity and equilibrium.

  • Always relate ΔrG to K (thermodynamic equilibrium) and Q (current composition) to decide whether a process will proceed toward products or reactants under given conditions.