Unit 9 · Topic 9.3 Beta

Gibbs Free Energy and Thermodynamic Favorability

Gibbs free energy, ΔG° = ΔH° − TΔS°, decides thermodynamic favorability: a process is favored when ΔG° < 0.

Practice 4: Model AnalysisPractice 5: Mathematical RoutinesPractice 6: Argumentation

Question set for this topic

Part 1 · Hook

Why this matters

A cold pack in a first-aid kit gets icy the moment you squeeze it, yet nothing pushes the salt inside to dissolve. Ice melts on a warm day even though melting absorbs heat. Neither enthalpy nor entropy alone predicts these. Gibbs free energy combines the two into one number that does.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the sign of ΔH for an exothermic reaction?

  1. Negative
  2. Positive
  3. Zero
  4. It depends on the temperature
Show the answer

Exothermic means the system releases heat to the surroundings, so ΔH < 0.

  • Correct: Negative:
  • Positive:
  • Zero:
  • It depends on the temperature:

2. What is the unit of a standard entropy change, ΔS°?

  1. J/(mol·K)
  2. kJ/mol
  3. K/mol
  4. J/mol
Show the answer

Entropies are tabulated in joules per mole per kelvin.

  • Correct: J/(mol·K):
  • kJ/mol:
  • K/mol:
  • J/mol:

3. Convert 25 °C to kelvin.

  1. 298 K
  2. 248 K
  3. 25 K
  4. −248 K
Show the answer

K = °C + 273.15, so 25 °C is 298 K.

  • Correct: 298 K:
  • 248 K:
  • 25 K:
  • −248 K:

Part 4 · See it

See it first

A two by two grid of the signs of ΔH° and ΔS°. ΔH° negative with ΔS° positive: favored at all temperatures. ΔH° negative with ΔS° negative: favored at low temperature, below ΔH°/ΔS°. ΔH° positive with ΔS° positive: favored at high temperature, above ΔH°/ΔS°. ΔH° positive with ΔS° negative: not favored at any temperature.
The signs of ΔH° and ΔS° decide whether a process is favored at all temperatures, at low T, at high T or at none. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Releasing heat (ΔH < 0) spreads energy into the surroundings and raises their entropya negative ΔH° makes a process more likely to be favored
  2. Dispersing the system’s own matter or energy (ΔS > 0) raises the system’s entropya positive ΔS° also makes a process more likely to be favored
  3. Gibbs free energy weighs both, ΔG° = ΔH° − TΔS°a process is thermodynamically favored when ΔG° < 0
  4. The entropy term is multiplied by Ttemperature decides which term wins when the two disagree, and T = ΔH°/ΔS° is where they balance

Part 6 · Key ideas

Key ideas

  • ΔG° = ΔH° − TΔS°. A process is thermodynamically favored when ΔG° < 0.
  • Make units match: ΔS° from J/(mol·K) to kJ/(mol·K) (÷ 1000), and T in kelvin.
  • ΔH° < 0, ΔS° > 0: favored at all T. ΔH° > 0, ΔS° < 0: not favored at any T. Same signs: temperature decides.
  • When the signs agree, the crossover is T = ΔH°/ΔS°, where ΔG° = 0.
  • ΔG°rxn = Σ nΔG°f(products) − Σ nΔG°f(reactants); ΔG°f of an element in its standard state is 0.

Part 7 · Misconception

A common mistake

The wrong idea: An endothermic process is never favored, or an exothermic one always is.

What actually happens: Favorability depends on ΔG° = ΔH° − TΔS°. An endothermic process is favored when TΔS° > ΔH° (ice melting above 0 °C); an exothermic one with ΔS° < 0 stops being favored at high temperature.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Four processes

A student collects standard enthalpy and entropy changes for four processes. Assume ΔH° and ΔS° do not change with temperature.

ΔH° and ΔS° for four processes
ProcessΔH° (kJ/mol)ΔS° (J/(mol·K))
P−85.0+42.0
Q−118.6−154.0
R+58.4+152
S+40.0−25.0

1. Which process is thermodynamically favored at every temperature?

  1. Q
  2. P
  3. R
  4. S
Show the answer

With ΔH° < 0 and ΔS° > 0, ΔG° = ΔH° − TΔS° is negative whatever the temperature.

  • Q: ΔS° < 0, so −TΔS° is positive and grows with T. Q is favored only at low temperature.
  • Correct: P: Right: ΔH° < 0 and ΔS° > 0, so both ΔH° and −TΔS° are negative and ΔG° < 0 at every temperature.
  • R: ΔH° > 0, so R is favored only when T is high enough for −TΔS° to outweigh ΔH°.
  • S: ΔH° > 0 and ΔS° < 0 make ΔG° positive at every temperature: S is not favored at any temperature.

2. Which processes are thermodynamically favored at 298 K? Select ALL that apply.

  1. P
  2. Q
  3. R
  4. S
Show the answer

Convert ΔS° to kJ/(mol·K) and compute ΔG° = ΔH° − TΔS° at 298 K for each process; only P and Q give negative values.

  • Correct: P: ΔG° = −85.0 − 298(0.0420) = −97.5 kJ/mol < 0.
  • Correct: Q: ΔG° = −118.6 − 298(−0.1540) = −72.7 kJ/mol < 0.
  • R: ΔG° = 58.4 − 298(0.152) = 13.1 kJ/mol > 0: 298 K is below the temperature where R becomes favored.
  • S: ΔG° = 40.0 − 298(−0.0250) = 47.5 kJ/mol > 0.

3. Above what temperature does process R become thermodynamically favored?

Type a number and its unit.

Show the answer

R becomes favored where ΔG° = 0: T = ΔH°/ΔS° = (58,400 J/mol)/(152 J/(mol·K)) = 384.2 K, which is 384 K to three significant figures. Above this, −TΔS° outweighs ΔH°.

  • Answer: 384 K

4. Calculate ΔG° for process R at 500. K.

Type a number and its unit.

Show the answer

ΔG° = ΔH° − TΔS° = 58.4 kJ/mol − (500. K)(0.152 kJ/(mol·K)) = 58.4 − 76.0 = −17.6 kJ/mol. Negative, so R is favored at 500. K, as expected above 384 K.

  • Answer: -17.6 kJ/mol

5. For CaCO₃(s) → CaO(s) + CO₂(g), ΔH° = +178.3 kJ/mol and ΔS° = +160.2 J/(mol·K). Calculate the lowest temperature at which the reaction is thermodynamically favored.

Type a number and its unit.

Show the answer

Set ΔG° = 0: T = ΔH°/ΔS° = (178,300 J/mol)/(160.2 J/(mol·K)) = 1113 K. Above this temperature, ΔG° < 0. (At 298 K, ΔG° = 130.6 kJ/mol, strongly unfavored, which is why limestone does not fall apart on its own.)

  • Answer: 1113 K

6. For ice melting, ΔH° = +6.01 kJ/mol and ΔS° = +22.0 J/(mol·K). Which statement about a block of ice at −10 °C (263 K) is supported by these data?

  1. It melts: ΔS° is positive, and any process with a positive ΔS° is favored.
  2. It melts: ΔG° = 6.01 − (−10)(0.0220) = +6.23 kJ/mol, which is close to zero.
  3. It does not melt: ΔG° = 6.01 − 263(0.0220) = +0.22 kJ/mol, so melting is not favored.
  4. It does not melt: melting absorbs heat, and heat-absorbing processes are not favored.
Show the answer

At 263 K the entropy term is too small to overcome ΔH°, so melting is not favored. Above 273 K it is.

  • It melts: ΔS° is positive, and any process with a positive ΔS° is favored.: A positive ΔS° helps but must outweigh ΔH°. At 263 K it does not.
  • It melts: ΔG° = 6.01 − (−10)(0.0220) = +6.23 kJ/mol, which is close to zero.: Temperature must be in kelvin, and a positive ΔG° is not favored anyway.
  • Correct: It does not melt: ΔG° = 6.01 − 263(0.0220) = +0.22 kJ/mol, so melting is not favored.: Right: at 263 K, TΔS° (5.79 kJ/mol) is smaller than ΔH°, so ΔG° > 0.
  • It does not melt: melting absorbs heat, and heat-absorbing processes are not favored.: The conclusion is right for the wrong reason. Ice does melt above 273 K even though melting absorbs heat, because TΔS° then outweighs ΔH°.

7. A reaction has ΔH° = −50.0 kJ/mol and ΔS° = −100. J/(mol·K). A student calculates ΔG° at 300. K as −50.0 − (300.)(−100.) = +29,950 kJ/mol. What went wrong?

  1. The temperature should be in °C, so T = 27 and ΔG° = −47.3 kJ/mol instead.
  2. ΔS° was not converted to kJ/(mol·K); with −0.100 kJ/(mol·K), ΔG° = −20.0 kJ/mol.
  3. The sign of ΔS° should have been ignored, giving ΔG° = −80.0 kJ/mol.
  4. ΔH° should have been multiplied by T, giving ΔG° = −15,000 kJ/mol.
Show the answer

ΔH° is in kJ/mol and ΔS° in J/(mol·K). Convert one so the units match: ΔG° = −50.0 − (300.)(−0.100) = −20.0 kJ/mol.

  • The temperature should be in °C, so T = 27 and ΔG° = −47.3 kJ/mol instead.: Thermodynamic equations need kelvin, so 300. K was right.
  • Correct: ΔS° was not converted to kJ/(mol·K); with −0.100 kJ/(mol·K), ΔG° = −20.0 kJ/mol.: Right: −50.0 − (300.)(−0.100) = −50.0 + 30.0 = −20.0 kJ/mol.
  • The sign of ΔS° should have been ignored, giving ΔG° = −80.0 kJ/mol.: The sign of ΔS° matters; with ΔS° < 0, −TΔS° is positive.
  • ΔH° should have been multiplied by T, giving ΔG° = −15,000 kJ/mol.: In ΔG° = ΔH° − TΔS°, only ΔS° is multiplied by T.

Part 9 · Summary

Summary

Gibbs free energy, ΔG° = ΔH° − TΔS°, decides thermodynamic favorability: a process is favored when ΔG° < 0. Match the units (kJ and J) and use kelvin. The signs of ΔH° and ΔS° tell you whether a process is favored at all temperatures, none, or only above or below T = ΔH°/ΔS°.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections