Entropy (S) is a measure of the number of possible arrangements (microstates) of a system. More microstates = greater disorder = higher entropy.
The Second Law of Thermodynamics:
For any spontaneous process, the total entropy of the universe increases:
ΔSuniverse = ΔSsystem + ΔSsurroundings > 0
A process is spontaneous if ΔSuniverse > 0.
Predicting the Sign of ΔS
ΔS is positive (entropy increases) when:
• Solid → liquid → gas (phase changes that increase disorder)
• Fewer moles of gas → more moles of gas
• A solid or liquid dissolves in a solvent
• Temperature increases (more kinetic energy, more microstates)
• A molecule breaks into smaller fragments
ΔS is negative (entropy decreases) when:
• Gas → liquid → solid
• More moles of gas → fewer moles of gas
• Precipitation (ions come together into ordered solid)
The Third Law of Thermodynamics:
The entropy of a perfect crystalline substance at absolute zero (0 K) is exactly zero. This provides a reference point for measuring absolute entropy values.
9.2 — Absolute Entropy and Entropy Change
Unlike enthalpy, we CAN measure absolute entropy values (S°) for substances because of the Third Law. These are tabulated as standard molar entropy values in J/(mol·K).
Standard entropy change of a reaction:
ΔS°rxn = Σ nS°(products) − Σ nS°(reactants)
where n = stoichiometric coefficients and S° = standard molar entropy values.
Trends in Absolute Entropy
Gases > liquids > solids (gases have far more microstates)
Larger/more complex molecules have higher S° (more ways to vibrate and rotate)
More moles of substance = higher total entropy (S is extensive)
Negative ΔS° makes sense: 3 moles of gas → 2 moles of gas (fewer gas molecules = less disorder).
9.3 — Gibbs Free Energy and Thermodynamic Favorability
Gibbs free energy (G) combines enthalpy and entropy into a single criterion for spontaneity at constant temperature and pressure.
ΔG = ΔH − TΔS
• ΔG < 0: Process is thermodynamically favorable (spontaneous) in the forward direction.
• ΔG = 0: System is at equilibrium.
• ΔG > 0: Process is not favorable in the forward direction (spontaneous in reverse).
Just like ΔH°f: ΔG°f of elements in their standard states = 0.
Finding the crossover temperature:
When ΔG = 0: ΔH = TΔS → T = ΔH / ΔS
This is the temperature where the reaction switches from spontaneous to non-spontaneous (or vice versa).
Example: If ΔH = −100 kJ and ΔS = −200 J/K:
T = (−100,000 J) / (−200 J/K) = 500 K
Below 500 K: ΔG < 0 (spontaneous). Above 500 K: ΔG > 0 (not spontaneous).
9.4 — Thermodynamic and Kinetic Control
A reaction can be thermodynamically favorable (ΔG < 0) but still occur extremely slowly if it has a high activation energy barrier. Thermodynamics tells us whether a reaction can happen; kinetics tells us how fast.
Key distinction:
• Thermodynamic favorability (ΔG < 0): The reaction CAN proceed spontaneously — products are more stable than reactants.
• Kinetic feasibility (Ea): Determines HOW FAST the reaction actually proceeds.
Examples:
• Diamond → Graphite: ΔG < 0 (thermodynamically favorable), but Ea is enormous → essentially never happens at room temperature.
• Combustion of glucose: ΔG is very negative, but glucose can sit in air indefinitely until a spark (activation energy) is provided.
• A catalyst lowers Ea but does NOT change ΔG or the position of equilibrium.
AP Exam tip: “Thermodynamically favorable” is the preferred AP term instead of “spontaneous” because “spontaneous” implies the reaction happens immediately, which is misleading. A favorable reaction might take millions of years without a catalyst.
9.5 — Free Energy and Equilibrium
There is a direct mathematical relationship between ΔG° and the equilibrium constant K.
ΔG° = −RT ln K
• R = 8.314 J/(mol·K)
• T = temperature in Kelvin
• K = equilibrium constant
• ln = natural logarithm
ΔG°
K value
Meaning
ΔG° < 0
K > 1
Products favored at equilibrium
ΔG° = 0
K = 1
Neither side favored
ΔG° > 0
K < 1
Reactants favored at equilibrium
ΔG under nonstandard conditions:
ΔG = ΔG° + RT ln Q
• Q = reaction quotient (current conditions)
• At equilibrium: Q = K, so ΔG = ΔG° + RT ln K = 0
• This equation tells you whether a reaction is favorable at any specific set of concentrations.
Example: Calculate K at 298 K for a reaction with ΔG° = −30.0 kJ/mol.
ΔG° = −RT ln K
−30,000 = −(8.314)(298) ln K
ln K = 30,000 / 2478 = 12.1
K = e12.1 = 1.8 × 10⁵
A large K confirms that products are strongly favored, consistent with the negative ΔG°.
9.6 — Coupled Reactions
A thermodynamically unfavorable reaction (ΔG > 0) can be driven forward by coupling it with a highly favorable reaction (ΔG << 0), so that the overall ΔG is negative.
How coupling works:
If reaction 1 has ΔG₁ > 0 (unfavorable) and reaction 2 has ΔG₂ < 0 (favorable), and |ΔG₂| > |ΔG₁|, then:
ΔGoverall = ΔG₁ + ΔG₂ < 0 (favorable!)
The coupled process is thermodynamically favorable even though one individual step is not.
Biological example — ATP hydrolysis:
Many biological reactions are endergonic (ΔG > 0). Cells couple them with ATP hydrolysis:
ATP → ADP + Pi ΔG° = −30.5 kJ/mol
If a reaction needs +20 kJ/mol, coupling with ATP hydrolysis gives:
ΔGoverall = +20 + (−30.5) = −10.5 kJ/mol (now favorable!)
Industrial example — Extraction of metals:
Extracting iron from Fe₂O₃ is unfavorable alone, but coupling with carbon oxidation makes the overall process favorable:
Fe₂O₃ + 3CO → 2Fe + 3CO₂ (ΔG < 0 overall at high temperatures)
9.7 — Galvanic (Voltaic) Cells
A galvanic cell (also called a voltaic cell) converts chemical energy into electrical energy through a spontaneous redox reaction. The two half-reactions occur in separate half-cells connected by a salt bridge.
A galvanic (voltaic) cell showing the anode, cathode, salt bridge, and electron flow. Oxidation occurs at the anode; reduction at the cathode. (Wikimedia Commons)
Key components:
• Anode: Where oxidation occurs (loses electrons). Negative terminal. Remember: AN OX (anode = oxidation).
• Cathode: Where reduction occurs (gains electrons). Positive terminal. Remember: RED CAT (reduction = cathode).
• Salt bridge: Allows ions to flow between half-cells to maintain electrical neutrality. Cations migrate toward the cathode; anions toward the anode.
• External wire: Electrons flow from anode to cathode through the wire.
• Cell potential (E°cell): The “voltage” driving the electrons. Must be positive for a galvanic cell.
Calculating E°cell:
E°cell = E°cathode − E°anode
Use standard reduction potentials from the table. Do NOT flip the sign when writing the equation — always subtract anode from cathode using the tabulated reduction potentials.
Example: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
E°(Cu²⁺/Cu) = +0.34 V (cathode, reduction)
E°(Zn²⁺/Zn) = −0.76 V (anode, oxidation)
E°cell = +0.34 − (−0.76) = +1.10 V
Large negative ΔG° confirms this is a strongly favorable reaction.
9.9 — Cell Potential Under Nonstandard Conditions
Standard cell potentials assume all concentrations are 1 M and all gas pressures are 1 atm. Under nonstandard conditions, the Nernst equation is used.
Nernst Equation:
E = E° − (RT / nF) ln Q
At 25°C (298 K), this simplifies to: E = E° − (0.0592 / n) log Q
• E = cell potential under actual conditions
• Q = reaction quotient (current concentrations/pressures)
• n = moles of electrons transferred
Key insights from the Nernst equation:
• If Q < K (Q < 1 for standard conditions): E > E° (cell potential is higher).
• If Q = K: E = 0 (the cell is “dead” — at equilibrium, no voltage).
• If Q > K: E < E° (cell potential is lower).
• Increasing [reactants] or decreasing [products] increases E (shifts Q lower).
Example: The Zn-Cu cell with [Zn²⁺] = 0.10 M and [Cu²⁺] = 2.0 M:
Q = [Zn²⁺] / [Cu²⁺] = 0.10 / 2.0 = 0.050
E = 1.10 − (0.0592/2) log(0.050) = 1.10 − (0.0296)(−1.30) = 1.10 + 0.039 = 1.14 V
E > E° because Q < 1 (more reactant, less product than standard).
Concentration Cells
A concentration cell has the same electrode material on both sides but different ion concentrations. E° = 0 (identical electrodes), but E ≠ 0 because Q ≠ 1. The cell works to equalize concentrations: the dilute side is oxidized (anode) and the concentrated side is reduced (cathode).
9.10 — Electrolysis and Faraday’s Law
Electrolysis uses electrical energy to drive a non-spontaneous reaction. It is the reverse of a galvanic cell — an external power source forces the reaction to proceed.
Galvanic vs. Electrolytic Cells:
Feature
Galvanic
Electrolytic
ΔG
< 0 (spontaneous)
> 0 (non-spontaneous)
Ecell
Positive
Negative (requires external voltage)
Energy conversion
Chemical → electrical
Electrical → chemical
Anode
Negative (−)
Positive (+)
Cathode
Positive (+)
Negative (−)
Oxidation at anode?
Yes
Yes (always!)
Faraday’s Law of Electrolysis
The amount of substance produced is proportional to the charge passed:
q = It
• q = charge (Coulombs)
• I = current (Amperes)
• t = time (seconds)
Conversion pathway:
Coulombs → moles e⁻ → moles substance → grams substance
q / F = mol e⁻ then use stoichiometry to find moles of substance deposited.
Example: How many grams of Cu are deposited when a 3.00 A current passes through a CuSO₄ solution for 2.00 hours?
Cu²⁺ + 2e⁻ → Cu(s)
q = It = (3.00 A)(2.00 × 3600 s) = 21,600 C
mol e⁻ = 21,600 / 96,485 = 0.2239 mol
mol Cu = 0.2239 / 2 = 0.1120 mol (2 electrons per Cu)
mass Cu = 0.1120 × 63.55 = 7.12 g
Applications of Electrolysis
Electroplating: Depositing a thin layer of metal (e.g., chrome plating, silver plating)
Aluminum refining: Hall-Héroult process uses electrolysis to extract Al from Al₂O₃
Chlor-alkali process: Electrolysis of brine to produce Cl₂, NaOH, and H₂
📝 Practice Multiple-Choice Questions
Test your knowledge of Unit 9. Click “Show Answer” to reveal the correct choice and explanation.
1. Which of the following changes results in an increase in entropy?
(A) H₂O(g) → H₂O(l)
(B) 2H₂(g) + O₂(g) → 2H₂O(g)
(C) NaCl(s) → Na⁺(aq) + Cl⁻(aq)
(D) N₂(g) + 3H₂(g) → 2NH₃(g)
Answer: (C)
Dissolving a solid into aqueous ions increases disorder (ΔS > 0). (A) is gas→liquid (decrease). (B) and (D) have fewer moles of gas as products (decrease).
2. A reaction has ΔH = +50 kJ and ΔS = +150 J/K. At what temperature does it become thermodynamically favorable?
(A) 0.33 K
(B) 333 K
(C) Above 333 K
(D) Below 333 K
Answer: (C)
T = ΔH / ΔS = 50,000 / 150 = 333 K. With ΔH > 0 and ΔS > 0, the reaction is favorable when TΔS > ΔH, which means above 333 K.
3. If ΔG° = −40 kJ/mol for a reaction, which is true about K?
(A) K < 1
(B) K = 1
(C) K > 1
(D) K cannot be determined
Answer: (C)
ΔG° = −RT ln K. If ΔG° is negative, then ln K must be positive, which means K > 1 (products favored at equilibrium).
4. Diamond spontaneously converting to graphite has ΔG < 0 but does not occur at room temperature. This is because:
(A) ΔH is positive
(B) The activation energy is extremely high
(C) K < 1 for the reaction
(D) ΔS is negative
Answer: (B)
The reaction is thermodynamically favorable (ΔG < 0, K > 1), but the activation energy is extremely high, so the reaction is kinetically hindered. Thermodynamics says it CAN happen; kinetics says it won’t happen fast enough to observe.
5. In a galvanic cell, electrons flow from:
(A) Cathode to anode through the wire
(B) Anode to cathode through the salt bridge
(C) Anode to cathode through the wire
(D) Cathode to anode through the solution
Answer: (C)
Electrons are produced at the anode (oxidation) and flow through the external wire to the cathode (reduction). Ions (not electrons) flow through the salt bridge/solution.
6. Given E°(Ag⁺/Ag) = +0.80 V and E°(Fe²⁺/Fe) = −0.44 V. What is E°cell for Fe | Fe²⁺ || Ag⁺ | Ag?
(A) +0.36 V
(B) +1.24 V
(C) −1.24 V
(D) +2.04 V
Answer: (B)
E°cell = E°cathode − E°anode = (+0.80) − (−0.44) = +1.24 V. Ag is reduced (cathode), Fe is oxidized (anode).
7. For a galvanic cell with E°cell = +0.50 V and n = 2, what is ΔG°?
8. According to the Nernst equation, increasing the concentration of reactants in a galvanic cell will:
(A) Decrease Ecell
(B) Increase Ecell
(C) Have no effect on Ecell
(D) Make Ecell negative
Answer: (B)
Increasing [reactants] decreases Q. Since E = E° − (0.0592/n) log Q, a smaller Q (smaller log Q, possibly negative) makes the subtracted term smaller or negative, so E increases.
9. In electrolysis, the anode is:
(A) Negative and the site of reduction
(B) Positive and the site of oxidation
(C) Positive and the site of reduction
(D) Negative and the site of oxidation
Answer: (B)
In an electrolytic cell, the anode is connected to the positive terminal of the power source and is the site of oxidation (always!). The sign of the anode flips from galvanic to electrolytic, but oxidation ALWAYS occurs at the anode.
10. A current of 5.00 A is passed through a solution of AgNO₃ for 1.00 hour. How many grams of Ag are deposited? (Ag⁺ + e⁻ → Ag; MAg = 107.87 g/mol)
(A) 5.58 g
(B) 10.1 g
(C) 20.1 g
(D) 40.3 g
Answer: (C)
q = It = (5.00)(3600) = 18,000 C
mol e⁻ = 18,000 / 96,485 = 0.1866 mol
Ag⁺ requires 1 e⁻, so mol Ag = 0.1866 mol
mass = 0.1866 × 107.87 = 20.1 g
💡 Study Tip for Unit 9
Unit 9 ties together thermodynamics, equilibrium, and electrochemistry. Memorize the ΔG = ΔH − TΔS and ΔG° = −RT ln K = −nFE° relationships — they connect everything. Know the four-case table for spontaneity (signs of ΔH and ΔS). For electrochemistry, master E°cell = E°cathode − E°anode and the Nernst equation. Remember that oxidation always occurs at the anode in BOTH galvanic and electrolytic cells (the sign of the electrode flips, but not the reaction type). Practice Faraday’s Law calculations — the conversion pathway is always: Coulombs → mol e⁻ → mol substance → grams.
