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., $\mathrm{O<em>2(g)}$ is more stable than $\mathrm{O(g)}$ or $\mathrm{O</em>3(g)}$)

Change in Entropy for Reactions

• The change in entropy for a reaction (∆rS°) can be calculated from tabulated standard molar entropies:
  $\Delta<em>r S^{\circ} = \left(\sum</em>i n<em>i^{\text{products}} S</em>i^{\circ}\right) - \left(\sum<em>j n</em>j^{\text{reactants}} S_j^{\circ}\right)$

• 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: $S^{\circ}(\mathrm{H<em>2O(s)}) = 41.0\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $S^{\circ}(\mathrm{H</em>2O(l)}) = 70.0\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $S^{\circ}(\mathrm{H_2O(g)}) = 188.8\ \mathrm{J\,mol^{-1}\,K^{-1}}$

• 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): $\mathrm{C<em>3H</em>8(g) + 5\,O<em>2(g) \rightarrow 3\,CO</em>2(g) + 4\,H_2O(g)}$

• Example (aqueous+solid): $\mathrm{AgNO<em>3(aq) + NaCl(aq) \rightarrow NaNO</em>3(aq) + AgCl(s)}$

• 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) $\mathrm{CaCO<em>3(s) + 2\,HCl(aq) \rightarrow CaCl</em>2(aq) + CO<em>2(g) + H</em>2O(l)}$
  2) $\mathrm{2\ NO(g) + O<em>2(g) \rightarrow 2\ NO</em>2(g)}$
  3) $\mathrm{2\ KClO<em>3(s) \rightarrow 2\ KCl(s) + 3\ O</em>2(g)}$

• 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:
  $\Delta<em>r G = \Delta</em>r H - T\Delta_r S$

• 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:
  $\Delta<em>r G = \Delta</em>r G^{\circ} + RT\ln Q$

• 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 $\mathrm{H<em>2O(l) \rightleftharpoons H</em>2O(g)}$ (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:
  $\Delta<em>r G = \Delta</em>r G^{\circ} + RT\ln Q$

• 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 $\Delta<em>r G = \Delta</em>r G^{\circ} + RT\ln Q.</p></li><li><p>Remember: to compare to standard states, ensure units are consistent: ΔH° in kJ/mol and ΔS° in J/mol·K, so ΔG° will be in kJ/mol when using R in J·mol^−1·K^−1 with T in K.</p></li></ul><h3 id="5447a4a0-b815-48c3-b792-e6bb671a6d06" data-toc-id="5447a4a0-b815-48c3-b792-e6bb671a6d06" collapsed="false" seolevelmigrated="true">Quick Concept Checks and Reminders</h3><ul><li><p>Spontaneous does not imply fast: a reaction can be thermodynamically spontaneous but kinetically slow.</p></li><li><p>The standard composition (Q = 1) provides a baseline reference point for comparing spontaneity under standard conditions.</p></li><li><p>The concept of energy dispersal (entropy) works together with enthalpy to determine spontaneity via ΔG.</p></li></ul><h3 id="11bff629-f0b3-42b2-851c-9189323f75cd" data-toc-id="11bff629-f0b3-42b2-851c-9189323f75cd" collapsed="false" seolevelmigrated="true">Homework and Study Reminders (From the Transcript)</h3><ul><li><p>By Saturday, 9/20 at 8:30 AM: Complete and Upload Week 3 Discussion packet (topics you worked on this week).</p></li><li><p>By Monday, 9/22 at 8:30 AM: Complete and Upload Homework 5-5; will review in class Tuesday.</p></li></ul><h3 id="9d18334b-47bf-44df-9f06-2c36a157b0bb" data-toc-id="9d18334b-47bf-44df-9f06-2c36a157b0bb" collapsed="false" seolevelmigrated="true">Notes on Data and References from the Transcript</h3><ul><li><p>Standard molar entropy (S°) and its units:$\text{S° (J mol^{-1} K^{-1})}$</p></li><li><p>Example S° values (for illustration):</p><ul><li><p>$\mathrm{H_2O(s)\;:\;S^{\circ}=41.0}$</p></li><li><p>$\mathrm{H_2O(l)\;:\;S^{\circ}=70.0}$</p></li><li><p>$\mathrm{H_2O(g)\;:\;S^{\circ}=188.8}$</p></li></ul></li><li><p>Example data for noble gases and small molecules (S° values in J/mol·K):</p><ul><li><p>He(g): 126</p></li><li><p>Ne(g): 146.2</p></li><li><p>Ar(g): 154.8</p></li><li><p>Kr(g): 164.0</p></li><li><p>C3H8(g): 269.9</p></li><li><p>C6H14(g): 388.82</p></li></ul></li><li><p>Example reactions cited:</p><ul><li><p>Gas-producing combustion/oxidation example:$\mathrm{C3H8(g) + 5 O2(g) \rightarrow 3 CO2(g) + 4 H_2O(g)}$</p></li><li><p>Aqueous + solid formation:$\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.