Thermodynamics: Temperature Dependence of Gibbs Free Energy, Entropy, and Enthalpy (Lecture Notes)
Exam logistics and resources
Announcement about APM 1 and exam location: check the posted announcement; location is assigned by your TA. Discussion/lab numbers correspond to your room: Discussion 4, Lab 7.
Do not come to the usual classroom for the exam.
Allowed materials:
Calculator: graphing calculator is allowed; it must not transmit or receive data (no phone functionality); programmable calculators are allowed.
Note card: a 3 by 5 inch handwritten note card, or an equivalent piece of paper of the same size.
The equation sheet (labeled on the discussion packet page as “equation sheet” but containing the equation sheet) is available and should be reviewed now; there is a link to the exam resources page that leads to the same information.
If you have testing accommodations (through Rosemary in the undergrad office and the testing center): check your email for instructions to register. If you did not receive the accommodations email, email the instructor and/or upload your accommodation letter in McBurney Connect so the instructor can assist.
Questions about the exam? This today’s material is the last topic on the exam. Thursday’s material will not be on the exam. Unit 6 begins Friday.
Today’s plan and scope
Cover the last two entropy-related questions from the previous homework.
Discuss the temperature dependence of Gibbs free energy in equilibrium (Gibbs free energy vs temperature and its implications).
Introduce/solidify the idea that Gibbs free energy is temperature-dependent and that equilibrium constants also depend on temperature, but in a different way.
Use these ideas to connect entropy, enthalpy, and entropy changes with spontaneity across temperatures.
Key concepts: entropy changes during phase transitions
General rule: entropy tends to increase when going to less condensed phases; it tends to decrease when going to more condensed phases.
Gas → liquid or liquid → solid: entropy decreases (fewer microstates available in condensed phase).
Solid → liquid or liquid → gas: entropy increases (more freedom of motion and more accessible microstates).
In the homework example, the transition involved phase changes (no net energy storage in intramolecular changes here beyond phase change effects).
Intuition for entropy change (ΔS):
Melting (solid to liquid) and boiling (liquid to gas) involve overcoming intermolecular forces, increasing disorder.
Freezing (liquid to solid) and condensation (gas to liquid) involve forming stronger interactions and decreasing disorder.
Practical takeaway: when asked which change has the greatest positive ΔS, focus on the step that increases disorder the most (usually gas → more disordered state), and compare solid↔gas, liquid↔gas, etc.
How to assess entropy changes qualitatively in reactions
Look at reactants vs products in terms of phase and number of gas moles.
A reaction that increases the number of gas moles or converts condensed phases to less condensed phases tends to have a positive ΔS.
A reaction that reduces gas moles or converts to more condensed phases tends to have a negative ΔS.
In the example from the lecture, the change involved going from condensed to less condensed phases, and the instructor highlighted that the greatest positive ΔS would occur for transitions involving significant disruption of intermolecular attractions (e.g., solid to gas), whereas going to a more condensed phase (gas → liquid, liquid → solid) reduces entropy.
Gibbs free energy vs entropy: why we use ΔG
Entropy of the universe would require accounting for surroundings; this is very difficult to measure.
Chemists focus on the system of interest. Gibbs free energy combines enthalpy and entropy changes of the system to assess spontaneity, without needing to quantify the surroundings’ entropy change.
Central relation: oxed{ \Delta G = \Delta H - T\Delta S }
Sign of ΔG determines spontaneity at a given T for the system.
If ΔG < 0, the process is spontaneous at that temperature.
If ΔG > 0, the process is non-spontaneous at that temperature (the reverse process may be spontaneous).
Practical implication: Gibbs free energy provides a system-centric measure that integrates heat (enthalpy) and disorder (entropy) effects.
Temperature dependence of Gibbs free energy and related ideas
At a given temperature, ΔG = ΔH − TΔS determines spontaneity.
For phase changes like ice ⇄ water, signs of ΔH and ΔS can be reasoned qualitatively across temperatures:
At low temperatures (well below freezing), condensation/solidification is favored (negative ΔS and often negative ΔG for the written direction).
At the freezing point, ΔG for the phase-change process is zero (equilibrium).
At temperatures above the freezing point, melting/boiling can become spontaneous in the opposite direction, depending on ΔH and ΔS.
Key takeaway: Exothermic and endothermic processes can both be spontaneous; spontaneity also depends on the sign and magnitude of ΔS and the temperature T.
Temperature dependence trends (qualitative table)
Endothermic + Decreasing ΔS: ΔG is always positive (never spontaneous in the written direction).
Exothermic + Increasing ΔS: ΔG is always negative (always spontaneous in the written direction).
Endothermic + Increasing ΔS: Temperature-dependent spontaneity; at sufficiently high T, ΔG can become negative and the reaction becomes spontaneous.
Exothermic + Decreasing ΔS: Temperature-dependent spontaneity; at sufficiently low T, ΔG can become negative and the reaction becomes spontaneous in the written direction; at high T it may become non-spontaneous.
These trends are system-dependent via ΔH and ΔS values; the temperature thresholds can be calculated from ΔG = 0:
For endothermic + increasing ΔS: threshold T = ΔH/ΔS (in Kelvin) when ΔG changes sign.
For exothermic + decreasing ΔS: threshold T = ΔH/ΔS (in Kelvin) in the opposite temperature regime.
Important caveat: while qualitative trends can guide intuition, the exact threshold temperatures depend on the specific ΔH and ΔS of the reaction and must be computed with consistent units.
The temperature dependence of K (equilibrium constant) and ΔG
ΔG° and K° are related at a given temperature by oxed{ \Delta G^\ = -RT \ln K }
However, you cannot generalize the ΔG vs T relationship to predict how K changes with T in a single simple way, because ΔG°(T) and K(T) have different temperature dependences.
The correct temperature dependence of K is given by the Van’t Hoff relation (integrated form): oxed{ \ln K = -\frac{\Delta H^\circ}{R}\frac{1}{T} + \frac{\Delta S^\circ}{R} }
Also: \frac{d\ln K}{dT} = \frac{\Delta H^\circ}{RT^2}
Consequences:
For endothermic reactions (ΔH° > 0), increasing T generally increases K (more products favored at higher T) regardless of sign of ΔS°.
For exothermic reactions (ΔH° < 0), increasing T generally decreases K (more reactants favored at higher T).
Entropy (ΔS°) determines whether a reaction is ever spontaneous at a given temperature, but not the direction of the K(T) trend with temperature on its own; enthalpy governs the slope in the Van’t Hoff plot.
Practical implication: You can plot K vs T (Van’t Hoff plot) and read off ΔH° (from the slope) and ΔS° (from the intercept, after converting units properly).
Important formulae and their usage
Gibbs free energy of a process: oxed{ \Delta G = \Delta H - T \\Delta S }
Temperature at which a process becomes spontaneous (ΔG = 0):oxed{ T_{\text{spont}} = \frac{\Delta H}{\Delta S} }
This assumes consistent units (e.g., J, kJ) for ΔH and ΔS.
Relation between ΔG° and equilibrium constant K at a fixed temperature:oxed{ \Delta G^\circ = -RT \ln K }
Van’t Hoff relationship (temperature dependence of K):oxed{ \ln K = -\frac{\Delta H^\circ}{R} \\frac{1}{T} + \\frac{\Delta S^\circ}{R} }
Reaction coordinate perspective (qualitative): changes in enthalpy and entropy determine the shape and direction of the energy landscape; the portion of the curve near the transition state governs rate, while ΔG vs T indicates spontaneity across temperatures.
Worked example: end-to-end flow for an exam-style problem (qualitative and quantitative)
Given a reaction, first determine qualitative signs:
Is the process endothermic or exothermic? What is ΔH° sign?
Does the reaction increase or decrease entropy (ΔS° sign)? This can be inferred from phase changes and gas moles.
Use ΔG° = ΔH° − TΔS° to reason about spontaneity at a given T, and compute ΔG° to determine if the process is spontaneous (ΔG° < 0) or not (ΔG° > 0).
To determine the temperature at which the process becomes spontaneous (ΔG goes from positive to negative): set ΔG = 0 and solve for T:
0 = ΔH^° − TΔS^° \Rightarrow T = \frac{ΔH^°}{ΔS^°}
To relate this to the equilibrium constant K(T): use the Van’t Hoff equation to understand how K changes with T, and note that the temperature at which ΔG = 0 corresponds to K = 1; this should be consistent with the Van’t Hoff plot when ΔH° and ΔS° are used correctly.
Example 1: Phase-change and ΔG sign intuition for water (ice ↔ liquid water)
At −10°C: entropy change for liquid to solid is negative; ΔH is negative (exothermic solidification). Predict ΔG for the written direction to be negative (ice formation) given that the temperature is below freezing.
At 0°C (freezing point): ΔG = 0 for the phase transition; system is at equilibrium.
At temperatures above 0°C: ice melting is spontaneously favored in the reverse direction (liquid to gas may become spontaneous at higher temperatures). The sign of ΔG for the written direction flips depending on T.
These signs illustrate that temperature can drive the direction of a phase-change reaction even when enthalpy or entropy signs suggest a direction.
Example 2: A specific reaction and its ΔG vs T behavior (NOCl example from the lecture)
Given a reaction NOCl(g) → NO(g) + Cl2(g) with a reported ΔG° (at standard conditions) of +40.94 kJ/mol.
A Gibbs vs T plot for this reaction shows the crossing point where ΔG = 0 occurring at a certain temperature (indicating the reaction becomes spontaneous above/below that temperature depending on the sign of ΔH° and ΔS°).
In the lecture, a computed crossing temperature ~362.8 K was mentioned, consistent with using the equation ΔG° = ΔH° − TΔS° and solving for T when ΔG° = 0, or equivalently using the Van’t Hoff approach with the appropriate ΔH° and ΔS° values.
The slide also highlighted that using a straight line for ΔG vs T can be rearranged to resemble y = mx + b form, where:
Slope m = −ΔS° (since ΔG = ΔH − TΔS, plotting ΔG vs T gives slope −ΔS°).
Intercept b = ΔH° (value at T = 0 K).
The key takeaway: the temperature at which ΔG crosses zero matches the temperature at which K crosses 1 (K = 1 at ΔG = 0), but the two dependences (ΔG vs T and K vs T) are governed by different thermodynamic parameters (ΔS° vs ΔH°).
In the same discussion, it was emphasized that you should not rely on the simple ΔG = ΔH − TΔS to predict K(T) universally; the correct dependency of K on T is given by Van’t Hoff, which depends on ΔH° primarily for the slope, with ΔS° dictating the intercept in the ln K vs 1/T plot.
Example 3: The Van’t Hoff plot in lab context
In a typical lab activity, you collect data for temperature and equilibrium constant K, then generate a Van’t Hoff plot (ln K vs 1/T).
The slope of the line equals −ΔH°/R.
The intercept equals ΔS°/R.
The lab exercises illustrate how the same reaction can show different qualitative behavior in K(T) depending on ΔH° and ΔS°, and how the enthalpy controls the slope while the entropy controls the intercept.
The lecture notes also warn that while you can obtain the same ΔH° and ΔS° by different routes (e.g., directly from ΔG = ΔH − TΔS at a known T, or from Van’t Hoff plot), you must use the correct theoretical framework for temperature dependence, not just plug-and-play algebra.
Example 4: Combustion of sugar and coupling unfavourable reactions
Combustion of sugar is exothermic and typically increases entropy (more gas products, more disorder).
As a result, the forward combustion is typically spontaneous (ΔG° < 0).
In biological systems (plants, photosynthesis, etc.), such energetically favorable reactions are often coupled to unfavorable reactions to drive them forward, using the energy released in one process to power another process.
Conceptual takeaway: free energy describes the energy available to do work; coupling reactions is a common strategy to drive energetically unfavorable processes in real systems.
Practical implications and connections
When predicting spontaneity across temperatures, always check both ΔH° and ΔS°; temperature can shift the balance between enthalpy and entropy contributions.
Use ΔG = ΔH − TΔS for a given temperature to decide spontaneity; use the Van’t Hoff relation to understand how K varies with temperature and to connect to ΔH° and ΔS° from experimental data.
For exams and problem-solving:
If asked for the boiling point from thermodynamic data, set ΔG = 0 and solve for T: T = \frac{ΔH^°}{ΔS^°}
When given ΔH° and ΔS° and asked for how K changes with temperature, use Van’t Hoff: \ln K = -\frac{ΔH^°}{R}\frac{1}{T} + \frac{ΔS^°}{R} and interpret the slope and intercept.
Remember units consistency: ΔH° in J/mol or kJ/mol and ΔS° in J/(mol·K) or kJ/(mol·K); convert as needed to keep units consistent in calculations.
Quick recap of the main ideas
Entropy changes depend on phase changes and gas moles; phase transitions can either increase or decrease ΔS.
Gibbs free energy bridges enthalpy and entropy to predict spontaneity: \Delta G = \Delta H - T\Delta S.
Temperature can flip spontaneity for reactions that have opposing energetic factors; some cases are temperature-independent (spontaneous in all conditions for the given direction), while others are strongly temperature-dependent.
The equilibrium constant K and ΔG are related at a fixed temperature via \Delta G^\circ = -RT\ln K\;, but their temperature dependences are governed by different thermodynamic quantities; the Van’t Hoff equation captures how K varies with temperature via ΔH° and ΔS°.
Practical lab applications include generating Van’t Hoff plots to extract ΔH° and ΔS° and understanding how to drive reactions by temperature changes or coupling.