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Chemistry 2 Unit 6: Thermodynamics: entropy and free energy
Entropy and Gibbs free energy determine whether a reaction is thermodynamically favored and how it relates to equilibrium.
Covers: Entropy and the second law, standard entropy changes, Gibbs free energy, spontaneity, free energy and equilibrium, coupled reactions
Study guide
Entropy (S) measures the dispersal of energy and matter, often described as disorder. Entropy increases from solid to liquid to gas, when a substance dissolves, when the number of gas molecules increases, and when temperature rises. The second law of thermodynamics says the entropy of the universe increases in any spontaneous process. The third law says a perfect crystal at absolute zero has zero entropy. The standard entropy change of a reaction is ΔS° = ΣS°(products) − ΣS°(reactants).
Gibbs free energy combines enthalpy and entropy: ΔG = ΔH − TΔS. A reaction with ΔG 0 is not favored, and ΔG = 0 means equilibrium. The signs of ΔH and ΔS determine how temperature matters: if ΔH 0, the reaction is favored at all temperatures; if ΔH > 0 and ΔS < 0, it's never favored; if both are negative, it's favored at low temperatures; and if both are positive, it's favored at high temperatures. The crossover temperature is T = ΔH ÷ ΔS. Standard free energy can also be found from free energies of formation: ΔG° = ΣΔG°f(products) − ΣΔG°f(reactants). Be careful with units: ΔH is usually in kJ and ΔS in J/K.
Free energy connects to equilibrium: ΔG° = −RT ln K. If ΔG° 1 and products are favored; if ΔG° > 0, K < 1. Under nonstandard conditions, ΔG = ΔG° + RT ln Q. Thermodynamics says whether a reaction is favored but not how fast it goes; a favored reaction may be very slow because of a high activation energy (kinetic control), like diamond turning into graphite. Unfavorable reactions can be driven by coupling them with favorable ones, as cells do with ATP hydrolysis, or by an outside energy source like electricity.
At a phase transition temperature, the process is at equilibrium (ΔG = 0), so ΔS = ΔH/T. For example, ΔS of vaporization equals ΔH of vaporization divided by the boiling point in kelvins. The entropy change of the surroundings is ΔS(surr) = −ΔH(sys)/T, so exothermic reactions increase the entropy of the surroundings. A process is thermodynamically favored when ΔS(universe) = ΔS(sys) + ΔS(surr) is positive.
Standard molar entropies are larger for gases than for liquids and solids. Within the same phase, they are larger for heavier, more complex molecules with more atoms and more ways to move.
Key ideas
- Entropy increases from solid to liquid to gas and with more gas molecules.
- The entropy of the universe increases in spontaneous processes.
- ΔG = ΔH − TΔS; ΔG < 0 means thermodynamically favored.
- ΔG° = −RT ln K links free energy and equilibrium.
- Thermodynamics doesn't predict reaction speed.
Formulas and rules
Common mistakes
- Mixing kJ and J when using ΔG = ΔH − TΔS.
- Thinking spontaneous means fast; it only means thermodynamically favored.
Words to know
- Entropy (S)
- A measure of the dispersal of energy and matter.
- Second law of thermodynamics
- The entropy of the universe increases in spontaneous processes.
- Third law of thermodynamics
- A perfect crystal at absolute zero has zero entropy.
- Standard entropy change
- ΔS° for a reaction under standard conditions.
- Gibbs free energy (G)
- Energy available to do work: G = H − TS.
- Thermodynamically favored
- A process with ΔG < 0.
- Standard free energy of formation
- ΔG°f for forming one mole from elements.
- Crossover temperature
- Temperature where ΔG changes sign.
- Kinetic control
- A favored reaction that's slow because of high activation energy.
- Coupled reactions
- Pairing an unfavorable reaction with a favorable one.
- ATP hydrolysis
- A favorable reaction cells use to drive unfavorable ones.
Sample practice questions
15 more questions like these, from level 1 to 5, are in the practice set.
Which change increases entropy?
- H₂O(l) → H₂O(s)
- H₂O(g) → H₂O(l)
- 2 gas molecules combining into 1
- H₂O(l) → H₂O(g)
Show the answer and steps
Answer: D) H₂O(l) → H₂O(g)
- Gases are more dispersed than liquids.
- Vaporization increases entropy.
A reaction has ΔH 0. It is thermodynamically favored…
- Only at high temperatures
- Never
- Only at low temperatures
- At all temperatures
Show the answer and steps
Answer: D) At all temperatures
- ΔG = ΔH − TΔS.
- A negative ΔH and a positive ΔS both make ΔG negative.
If ΔG° for a reaction is negative, then K is…
- Greater than 1
- Equal to 1
- Less than 1
- Zero
Show the answer and steps
Answer: A) Greater than 1
- ΔG° = −RT ln K.
- A negative ΔG° means ln K > 0, so K > 1.