Thermodynamics Notes: Free Energy, Enthalpy, Entropy, and Temperature Dependence

Key Concepts

• Free energy relation: $\Delta G = \Delta H - T\Delta S$

  • \Delta G determines spontaneity of a reaction at a given temperature.

  • Units: \Delta H typically in kJ (per mole), \Delta S in J/K (per mole). To combine with T in K, you may convert \Delta S to kJ/K by dividing by 1000 so that all terms are in kJ.

  • Temperature dependence: as T changes, the balance between enthalpy and entropy changes shifts the sign of \Delta G.

• Temperature units and conversions:

  • Kelvin vs Celsius: $T(K) = T(^{\circ}C) + 273.15$

  • When a problem asks for \Delta G at a specific temperature, ensure T is in Kelvin and convert any Celsius data appropriately.

• Different visual representations of the same thermodynamic information:

  • \Delta G vs T graphs: show how spontaneity changes with temperature.

  • H vs S (or PE-C diagrams) diagrams: show the same information from a different perspective (how enthalpy and entropy trade off).

  • Four qualitative scenarios arise from signs of \Delta H and \Delta S (see below).

• Practical lab connection:

  • Calorimetry in acid–base titrations: heat evolved or absorbed during neutralization (e.g., NaOH with HA) links to enthalpy changes and helps infer acid strength.

  • Real-world relevance: environmental context (e.g., smog) can be discussed in terms of energy changes and reaction directionality.

• Data sources for \Delta H and \Delta S:

  • Hess’s law: use standard heats of formation (\Delta Hf^\circ) to compute the reaction enthalpy: $\Delta H</em>{rxn}^\circ = \sum<em>i \nu</em>i\Delta H<em>f^{\circ}(products) - \sum</em>j \nu<em>j\Delta H</em>f^{\circ}(reactants)$

  • Standard molar entropies:
    $\Delta S<em>{rxn}^\circ = \sum</em>i \nu<em>i S^{\circ}(products) - \sum</em>j \nu_j S^{\circ}(reactants)$

  • Always pay attention to units and stoichiometry; misplacing units (e.g., writing 52.7 when you meant 5.27×10^1) costs points.

Core Equations and Conventions

• Fundamental relation for spontaneity:
  $\Delta G = \Delta H - T\Delta S$

• Temperature at which the reaction becomes equally favorable in both directions (\Delta G = 0):
  $0 = \Delta H - T^<em>\Delta S \quad\Rightarrow\quad T^</em> = \dfrac{\Delta H}{\Delta S}$

• Sign interpretation across cases (based on signs of \Delta H and \Delta S):

  • Case A: \Delta H < 0 and \Delta S > 0 → Product-favored at all temperatures (\Delta G < 0 for all T > 0).

  • Case B: \Delta H > 0 and \Delta S < 0 → Reactant-favored at all temperatures (\Delta G > 0 for all T > 0).

  • Case C: \Delta H and \Delta S have the same sign (both negative or both positive) → \Delta G crosses zero at a finite temperature T* (there is a transition between product- and reactant-favored as T crosses T*):

  • If \Delta H < 0 and \Delta S < 0: product favored at low T; at high T, reactant favored.

  • If \Delta H > 0 and \Delta S > 0: reactant favored at low T; at high T, product favored.

• Units and typical magnitudes:

  • Enthalpy: \Delta H in kJ per mole (kJ/mol).

  • Entropy: \Delta S in J/K per mole (J/K·mol).

  • To combine, either convert \Delta H to J or \Delta S to kJ/K; ensure consistent units throughout the calculation.

• Conceptual tip for interpretation:

  • If two terms have the same sign, there is a temperature at which \Delta G = 0; if the signs are opposite, one direction is favored at all temperatures.

  • The temperature dependence often reveals how a reaction may be product- or reactant-favored under different conditions (e.g., low vs high temperature).

Four Scenarios Based on Signs of \Delta H and \Delta S

• Scenario 1: \Delta H < 0, \Delta S < 0

  • \Delta G = \Delta H - T\Delta S; since \Delta S < 0, -T\Delta S > 0

  • At low T, product-favored (since \Delta H < 0 dominates);

  • At high T, reactant-favored (crosses the line at T* = \Delta H / \Delta S > 0 since both are negative).

  • Temperature crossover occurs at T* = \Delta H / \Delta S.

• Scenario 2: \Delta H > 0, \Delta S > 0

  • At low T, reactant-favored (since \Delta H > 0 dominates);

  • At high T, product-favored (crosses the line at T* = \Delta H / \Delta S).

• Scenario 3: \Delta H < 0, \Delta S > 0

  • Product favored at all temperatures (\Delta G < 0 for all T > 0).

• Scenario 4: \Delta H > 0, \Delta S < 0

  • Reactant favored at all temperatures (\Delta G > 0 for all T > 0).

Test-Taking and Problem-Solving Strategy (Partial-Credit Approach)

• Step 1: State the governing equation you intend to use

  • Identify and write down $\Delta G = \Delta H - T \Delta S$

• Step 2: Identify the variables and their values from the problem

  • Note which terms are given: \Delta H, \Delta S, temperature, and any required conversions (J vs kJ, Kelvin vs Celsius).

• Step 3: Convert temperatures to Kelvin when needed

  • If the problem provides temperature in Celsius, convert: $T(K) = T(^{\circ}C) + 273.15$

• Step 4: Plug values carefully and keep track of units

  • Example pitfall: mixing J and kJ; ensure consistent units (e.g., convert \Delta S to kJ/K if \Delta H is in kJ).

• Step 5: Interpret the sign of the result

  • If \Delta G < 0, reaction is product-favored at that temperature; if \Delta G > 0, reactant-favored.

• Step 6: Optional qualitative checks

  • Draw the corresponding \Delta G vs T or H vs S diagram to check which region the given temperature falls into.

• Step 7: If asked for a temperature where direction changes

  • Solve for $T^* = \dfrac{\Delta H}{\Delta S}$ and evaluate whether T^* is physically meaningful (positive) given the signs.

Worked Example from the Transcript: Dissociation of Dinitrogen Hydroxide

• Reaction described: dinitrogen hydroxide breaks into two nitrogen dioxide molecules.

  • Written as: $\text{N}<em>2\text{O}</em>\text{2?} \rightarrow 2\ \text{NO}_2$

  • Note: The transcript refers to a species called "dinitrogen hydroxide" breaking into two "nitrogen dioxides"; in practice this corresponds to a bond dissociation where one reactant molecule splits to yield 2 NO2 molecules (endothermic process).

• Key qualitative points from the description:

  • Bond breaking requires input of energy (positive \Delta H).

  • In this case, the reaction produces more gas molecules (1 → 2), so entropy increases (\Delta S > 0).

  • Therefore, \Delta H > 0 and \Delta S > 0, which places this in Scenario 2: reactant-favored at low T, product-favored at high T with a crossing at $T^* = \dfrac{\Delta H}{\Delta S}$.

• Illustrative calculation steps described (paraphrased):

  • Determine that at very low temperatures, since \Delta H > 0, the reaction is not favored; as temperature rises, the T\Delta S term grows and can overcome \Delta H, making \Delta G negative beyond T*.

  • The instructor mentions a specific example where the temperature at which \Delta G = 0 was around 273 K (≈ 0°C) for that set of numbers, implying a crossover in that vicinity.

  • Practical takeaway: use $T^* = \Delta H / \Delta S$ to find the temperature where the reaction switches from non-spontaneous to spontaneous (or vice versa).

• Important reminder embedded in the example:

  • For zero-temperature behavior, you rely on \Delta H; for high-temperature behavior, the sign of \Delta S dominates the direction beyond T*.

• Takeaway practice: given a reaction with endothermicity and increasing entropy, expect product-favoring tendency to emerge at higher temperatures; compute the exact crossover with the formula above and verify units.

Lab Context and Real-World Relevance

• Acid–base titrations and calorimetry in the lab:

  • The class is working on an acid–base titration lab in which sodium hydroxide (NaOH) reacts with a weak or strong acid HA to form the conjugate base A⁻ and water, releasing or absorbing heat.

  • The heat measured in the lab relates to the reaction enthalpy; comparing measured heat to expected enthalpies provides insight into acid strength and the reaction's thermodynamics.

• Practical measurement concept:

  • The heat released (or absorbed) can be used to infer the thermodynamics, consistent with the equation \Delta G = \Delta H - T\Delta S and calorimetric data.

• Smog and environmental context (qualitative example used in lecture):

  • Visual differences in smog (brownish vs. colorless) tied to temperature and air composition; while not a direct thermodynamics calculation, the discussion emphasizes how energy and entropy considerations play into real-world atmospheric chemistry (e.g., formation and breakdown of oxides of nitrogen).

Data-Driven Calculations: Using Heats of Formation and Standard Entropies

• When given tabulated data, the following are used:

  • Heats of formation: \Delta H_f^{\circ}

  • Standard molar entropies: S^{\circ}

• Step-by-step calculation:
  1) Write the balanced reaction with proper stoichiometric coefficients.
  2) Compute enthalpy change using Hess’s law:
  $\Delta H<em>{rxn}^{\circ} = \sum</em>i \nu<em>i\Delta H</em>f^{\circ}(products) - \sum<em>j \nu</em>j\Delta H<em>f^{\circ}(reactants)$ 3) Compute entropy change: $\Delta S</em>{rxn}^{\circ} = \sum<em>i \nu</em>i S^{\circ}(products) - \sum<em>j \nu</em>j S^{\circ}(reactants)$
  4) Pay attention to units; ensure that \Delta H is in the same energy units as used elsewhere (often kJ/mol) and that \Delta S is in J/(K·mol).
  5) If a problem provides data tables with multiple species, sum entropies and heats of formation with their stoichiometric coefficients.
  6) Check the sign of \Delta H and \Delta S; use the results to predict product vs reactant favored and, if needed, compute the crossover temperature via $T^* = \dfrac{\Delta H}{\Delta S}$.

• Common pitfalls discussed in the lecture:

  • Misreporting a value (e.g., writing 52.7 when the correct coefficient is 5.27×10^1) and confusing ice-cream-truck-type humor about misprints.

  • Mixing up units (J vs kJ) when plugging into formulas.

  • Forgetting to convert Celsius to Kelvin when a temperature is given in °C.

Quick Practical Takeaways

• Always start with the core equation: $\Delta G = \Delta H - T\Delta S$

• Identify the signs of \Delta H and \Delta S to determine temperature dependence and the possibility of a crossing at $T^* = \Delta H/\Delta S$.

• If \Delta H and \Delta S have opposite signs, there is no finite crossing; the reaction is spontaneously favored in one direction at all temperatures.

• If you’re only asked for the sign of \Delta G (positive or negative) at a given temperature, you can often decide by comparing the relative magnitudes of \Delta H and the product T\Delta S, with careful unit handling.

• When data is provided (heats of formation and standard entropies), use Hess’s law and entropy data to compute \Delta H and \Delta S before evaluating \Delta G at the target temperature.

• In lab contexts, relate calorimetry measurements to enthalpy changes and connect to qualitative conclusions about acid strength and reaction energetics.

Mini Summary for Exam Preparation

• Core formula: $\Delta G = \Delta H - T\Delta S$ with proper units and temperature in Kelvin.

• Temperature crossover: $T^* = \dfrac{\Delta H}{\Delta S}$ when \Delta H and \Delta S share the same sign.

• Four scenario guide:

  • (i) \Delta H < 0, \Delta S > 0: product-favored at all T.

  • (ii) \Delta H > 0, \Delta S < 0: reactant-favored at all T.

  • (iii) \Delta H and \Delta S < 0: product-favored at low T; crossing to reactant-favored at high T.

  • (iv) \Delta H and \Delta S > 0: reactant-favored at low T; crossing to product-favored at high T.

• Practical steps for problems: write the equation, convert units, plug numbers, decide sign of \Delta G, and optionally sketch the qualitative diagram or compute T^* for the crossover.

• Hess’s law and data interpretation: use standard enthalpies of formation and standard entropies to compute \Delta H{rxn}^\circ and \Delta S{rxn}^\circ, then apply the core equation to analyze the reaction.