Entropy, Free Energy & Reaction Feasibility — Print · Grade 11 Science
Chapter: Chemical Energy & ThermodynamicsEntropy, Free Energy & Reaction Feasibility
Learning goals
- I can calculate ΔG with consistent energy units and kelvin temperature.
- I can distinguish thermodynamic favorability from reaction speed.
Readings for this unit
Before this unit
Energy signs, kelvin temperature, and algebra.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
Can a process absorb heat yet be thermodynamically favored?
Illustrative evidence case. A selected transformation has positive ΔH and a sufficiently positive ΔS at its stated temperature.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
Can a process absorb heat yet be thermodynamically favored?
Your first model
Compare the enthalpy term with TΔS using compatible units and a defined system.
Build the language from the idea
- entropy
- A measure related to accessible microscopic arrangements and energy dispersal.
- Gibbs free energy
- A thermodynamic potential useful for constant-temperature, constant-pressure processes.
- spontaneous
- Thermodynamically favored under stated conditions; not necessarily rapid.
Enthalpy alone does not determine whether a change is thermodynamically favored. Some endothermic processes occur readily because the total entropy of system and surroundings increases. Entropy counts accessible microscopic arrangements and energy dispersal; simply calling it disorder can hide what is being counted.
For a system at constant temperature and pressure, Gibbs free-energy change is ΔG=ΔH−TΔS. Negative ΔG indicates a thermodynamic driving tendency in the specified conditions; positive ΔG favors the reverse; zero corresponds to equilibrium. Use temperature in kelvin and matching energy units.
A larger number of gas particles often increases entropy under comparable conditions, and dispersing a substance or warming it can increase accessible arrangements. These are useful tendencies, not universal replacements for data. Entropy changes in surroundings must also be considered to judge total entropy.
If ΔH<0 and ΔS>0, ΔG is negative at all positive temperatures in the simple constant-value model. If both are positive, sufficiently high temperature can favor the process. The crossover estimate T=ΔH/ΔS assumes those values are approximately constant across the interval.
Thermodynamically favorable does not mean fast. A large activation barrier can make a favored reaction proceed imperceptibly slowly. Catalysts change kinetic pathways but do not change the overall equilibrium constant or reaction ΔG for the same reactants and products.
For independent study, label enthalpy and entropy values as supplied at stated conditions. Convert J to kJ before subtracting. Describe the result as favorability under a model, not a prediction of completion time or a guarantee of laboratory outcome.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Explain entropy beyond a disorder slogan
Entropy S is a thermodynamic state quantity connected with the number and distribution of accessible microscopic arrangements. For suitable equal-probability microstate counting, S=kB ln W provides a useful bridge, while macroscopic changes can be evaluated through reversible heat transfer and temperature. “Disorder” is an imperfect analogy, not a measurement procedure. The same particle count can have different accessible arrangements when volume or other constraints change; assess the full system and surroundings rather than judging entropy from a messy-looking drawing.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why does a more spread-out dot picture not by itself determine a numerical entropy change?
Visual reasoning — after your attempt
The relevant energy, constraints and accessible microstate model are not specified quantitatively. Appearance alone is insufficient.
Entropy relates to the number of accessible microscopic arrangements and energy dispersal in a thermodynamic description. Gas formation, mixing, and phase changes can influence it, but visual messiness is not a complete definition. Distinguish system entropy from total entropy including surroundings.
A system’s entropy can decrease while surrounding entropy increases sufficiently. Looking only at the system cannot establish the spontaneity of the full process.
Pause and explain
Can a spontaneous process have a negative system entropy change?
Check your reasoning after answering
Yes. A sufficiently positive surroundings contribution can make total entropy increase even if the system’s entropy decreases.
Part 2 · Use free energy under its conditions
At constant temperature and pressure under the appropriate constraints, Gibbs free energy helps assess the thermodynamic direction through ΔG=ΔH−TΔS. A process can be endothermic yet favored when the entropy term is sufficiently positive. Convert entropy units before multiplying by absolute T; joules and kilojoules cannot be subtracted without conversion. The result describes the specified state change under its conditions, not every possible mixture or a measured rate. State any approximation that ΔH and ΔS remain constant over a temperature range.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why can the selected positive-ΔH transformation have negative ΔG at 300 K?
Visual reasoning — after your attempt
The positive entropy contribution TΔS=30.0 kJ/mol exceeds the 20.0 kJ/mol enthalpy cost.
At constant temperature and pressure, ΔG=ΔH−TΔS provides a feasibility criterion under suitable thermodynamic conditions. Use Kelvin and compatible energy units. Negative ΔG favors forward change for the stated actual conditions; zero corresponds to equilibrium; positive favors the reverse.
If ΔH is in kJ/mol and ΔS in J/mol·K, convert entropy to kJ/mol·K before multiplying by T.
Pause and explain
What common error changes a free-energy calculation by a factor of one thousand?
Check your reasoning after answering
Combining kJ enthalpy with J entropy without conversion. The two terms must use compatible energy units.
Part 3 · Distinguish actual and standard-state favorability
The standard reaction free energy ΔG° refers to a defined reference-state account. The actual driving tendency also depends on mixture composition through Q, the reaction quotient, using consistent dimensionless activities. At equilibrium Q=K and the reaction’s ΔG is zero. A negative ΔG° does not mean every possible starting mixture moves forward, because the RT ln Q contribution can change the actual sign. Concentration-only classroom expressions approximate activities and require an explicitly consistent convention.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Why can a reaction with negative ΔG° fail to proceed net forward from a very product-rich starting mixture?
Visual reasoning — after your attempt
A sufficiently large Q can make RT ln Q positive enough to change the actual ΔG; the mixture composition matters.
ΔG° describes standard-state conditions, while actual ΔG depends on reaction composition through ΔG=ΔG°+RT ln Q. A positive standard value does not prove no forward conversion can occur at any composition. Equilibrium links ΔG° to K rather than forcing all reactions to stop.
A reaction mixture far from standard ratios can have a different driving direction from the standard-state criterion.
Pause and explain
Why can actual composition change direction without changing ΔG°?
Check your reasoning after answering
The RT ln Q term reflects the mixture’s state. ΔG° is a reference property, while actual ΔG includes current activities.
Part 4 · Separate feasibility from rate
Thermodynamic feasibility and kinetics answer different questions. A lower product free-energy state can be favored while an activation barrier makes conversion slow on the observed timescale. A catalyst changes available pathways and rates without changing equilibrium for the same reaction and conditions. A free-energy curve is a model, not a stopwatch. Evidence for practical performance must include reaction time, pathway, transport and possible competing changes rather than only the endpoint sign.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Does a negative reaction free-energy change imply a reaction will finish quickly at room temperature?
Visual reasoning — after your attempt
No. The activation pathway and other kinetic factors can make the change slow even when its direction is thermodynamically favored.
Thermodynamic favorability does not determine how quickly a process occurs. Activation barriers can make a favored reaction slow, and a catalyst can accelerate approach to equilibrium without changing ΔG° or K for the same temperature. State whether a conclusion concerns direction or timescale.
A calculated negative ΔG does not justify a claim of instant conversion or complete product recovery. Kinetics and equilibrium composition still matter.
Pause and explain
Does negative ΔG guarantee a fast process?
Check your reasoning after answering
No. It indicates thermodynamic direction under stated conditions, while rate depends on the kinetic pathway and barriers.
Apply the whole unit · Find a temperature-dependent sign change
Illustrative approximate values: ΔH=+40.0 kJ/mol and ΔS=+100 J/mol·K, treated as constant over the compared range. Compare 300 K and 500 K.
Your task
Calculate ΔG at both temperatures and the zero-crossing temperature, explain the competing terms, and state why the result is not a rate prediction or a universal standard-state/actual-state equivalence.
What your response should show
- Choose the appropriate chemical model or relationship and state its conditions.
- Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
- Use supplied evidence to justify the conclusion; distinguish observation from inference.
- Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt
Convert ΔS to 0.100 kJ/mol·K.
At 300 K: ΔG=40.0−300×0.100=+10.0 kJ/mol.
At 500 K: ΔG=40.0−500×0.100=−10.0 kJ/mol.
Zero crossing: T=ΔH/ΔS=400 K in the constant-value approximation.
The entropy term outweighs the positive enthalpy term above the modeled crossing. The calculation applies to the conditions associated with the supplied changes, with no guarantee they stay constant indefinitely. Standard-state values would describe ΔG° and actual composition adds another term. Neither sign establishes a reaction timescale.
Scientific references and source-model foundations
Parent / teacher teaching notes
Require Kelvin and an explicit J-to-kJ conversion. Ask direction, conditions, and rate as three separate questions. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.
The key idea
Free energy connects enthalpy and entropy; kinetics determines how quickly change can occur.
Worked examples
Example 1
- Use ΔG=ΔH−TΔS.
- 20−300(.100)=−10.
Example 2 · Worked application
- Convert entropy to 0.050 kJ/(mol·K).
- ΔG=ΔH−TΔS=10−300×0.050.
- Negative ΔG indicates thermodynamic favorability under the specified state, not rapid reaction.
Units and absolute temperature must match.
Example 3
- At ΔG=0, T=ΔH/ΔS.
- 40/.080.
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: Units and absolute temperature must match.
- Thermodynamic favorability says nothing by itself about kinetic speed; barriers can make favorable changes slow.
Units and absolute temperature must match.
Example 5 · Reason from the visual model
- The standard reaction free energy ΔG° refers to a defined reference-state account. The actual driving tendency also depends on mixture composition through Q, the reaction quotient, using consistent dimensionless activities. At equilibrium Q=K and the reaction’s ΔG is zero. A negative ΔG° does not mean every possible starting mixture moves forward, because the RT ln Q contribution can change the actual sign. Concentration-only classroom expressions approximate activities and require an explicitly consistent convention.
- A sufficiently large Q can make RT ln Q positive enough to change the actual ΔG; the mixture composition matters.
Original thermodynamic relationships; no logarithm of a dimensional raw concentration is intended.
Avoid these mistakes
- Spontaneous is a thermodynamic term, not a promise that a process is immediate.
Check your understanding
Distinguish favorable, fast, and complete.
Discuss the answer
Free energy addresses direction, activation barriers address rate, and equilibrium composition addresses extent.
Return to the phenomenon
Return to the initial case. Can a process absorb heat yet be thermodynamically favored? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
The free-energy balance can be negative despite positive enthalpy. This still needs the actual mixture conditions and a separate kinetic account.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. Can a process absorb heat yet be thermodynamically favored? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the chemical mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep species identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
The free-energy balance can be negative despite positive enthalpy. This still needs the actual mixture conditions and a separate kinetic account.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
For a selected process use ΔH=+30.0 kJ/mol and ΔS=+0.150 kJ/(mol·K), treated constant. Find ΔG at 250 K, the modeled crossover temperature and one limit of the prediction.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
At 250 K, ΔG=30.0−250(0.150)=−7.50 kJ/mol. The modeled crossover is 30.0/0.150=200 K. Actual ΔH/ΔS can vary with temperature and phases, and the calculation does not establish a rate.
Words to know
- entropy
- A measure related to accessible microscopic arrangements and energy dispersal.
- Gibbs free energy
- A thermodynamic potential useful for constant-temperature, constant-pressure processes.
- spontaneous
- Thermodynamically favored under stated conditions; not necessarily rapid.
In the real world
Use the chapter investigation to connect this unit to measured evidence and a practical decision.
Exit check
Distinguish favorable, fast, and complete.
Check your explanation after answering
Free energy addresses direction, activation barriers address rate, and equilibrium composition addresses extent.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
-20 kJ/mol
−50−300(−.100)=−20.
Problem 2
Answer and explanation
Yes. A high activation-energy barrier can limit its rate.
Response criteria
- State a scientifically accurate conclusion.
- Show the calculation or mechanism that supports it.
- Include units, conditions, or limits where relevant.
Problem 3
Hints
List the given quantities, select a relationship, and check units before substituting.
Answer and explanation
-3 kJ/mol
12−300(.050)=−3.
Problem 4
Hints
Write the relationship first; substitute with consistent units.
Answer and explanation
600 K
30/.050=600.
Problem 5
Hints
Units and absolute temperature must match.
Answer and explanation
−20−300×(−0.040)=−8 kJ/mol.
Problem 6
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
A sufficiently positive entropy change makes TΔS exceed ΔH, producing negative ΔG under appropriate conditions.
Problem 7
Answer and explanation
A sufficiently large Q can make RT ln Q positive enough to change the actual ΔG; the mixture composition matters.
The standard reaction free energy ΔG° refers to a defined reference-state account. The actual driving tendency also depends on mixture composition through Q, the reaction quotient, using consistent dimensionless activities. At equilibrium Q=K and the reaction’s ΔG is zero. A negative ΔG° does not mean every possible starting mixture moves forward, because the RT ln Q contribution can change the actual sign. Concentration-only classroom expressions approximate activities and require an explicitly consistent convention.
A sufficiently large Q can make RT ln Q positive enough to change the actual ΔG; the mixture composition matters.
Response criteria
- Interpret labels, structures, axes, units and directions correctly.
- Explain the chemistry relationship or mechanism.
- Respect stated assumptions and distinguish observation from inference.
Problem 8
Answer and explanation
At 250 K, ΔG=30.0−250(0.150)=−7.50 kJ/mol. The modeled crossover is 30.0/0.150=200 K. Actual ΔH/ΔS can vary with temperature and phases, and the calculation does not establish a rate.
At 250 K, ΔG=30.0−250(0.150)=−7.50 kJ/mol. The modeled crossover is 30.0/0.150=200 K. Actual ΔH/ΔS can vary with temperature and phases, and the calculation does not establish a rate.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the chemical mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.