AP Chemistry 9.1: Entropy and Unit 9 Review

Unit 9 connects several topics from earlier in AP Chemistry. You will combine ideas from:

• thermodynamics
• equilibrium
• redox reactions
• kinetics
• electrochemistry
• The major question throughout the unit is:
• Will a process be thermodynamically favored, and how can energy be connected to equilibrium and electrical work?

Entropy, represented by S, describes how widely energy and matter can be distributed among possible microscopic arrangements.

• A simple way to think about it is:
• more possible particle arrangements → greater entropy

People sometimes describe entropy as "disorder," but this can be misleading. For AP Chemistry, it is better to think about energy dispersal and the number of possible microstates.

A microstate is one possible microscopic arrangement of the positions and energies of the particles in a system.

More possible microstates generally means greater entropy.

The entropy change is:

• ΔS = Sproducts − Sreactants
• If:
• ΔS > 0
• the products have greater entropy.
• If:
• ΔS < 0
• the products have lower entropy.

For the same substance:

• solid < liquid < gas
• in entropy.
• Why?
• A solid's particles are held in relatively fixed positions.
• Liquid particles have more freedom of movement.
• Gas particles have the greatest freedom and can occupy a much larger number of possible arrangements.
• Therefore:

solid → liquid

• ΔS > 0

liquid → gas

• ΔS > 0

solid → gas

• ΔS > 0

liquid → solid

• ΔS < 0

gas → liquid

• ΔS < 0

The number of gas particles is often useful when predicting ΔS for a reaction.

• Example:
• N₂(g) + 3H₂(g) → 2NH₃(g)
• Reactants:
• 4 mol gas
• Products:
• 2 mol gas
• There are fewer gaseous particles after the reaction.
• Therefore:
• ΔS is likely negative.

CaCO₃(s) → CaO(s) + CO₂(g)

• A gas is produced from substances that were originally solids.
• The number of possible particle arrangements increases greatly.
• Therefore:
• ΔS is positive.

Entropy generally increases when:

• a solid becomes a liquid
• a liquid becomes a gas
• the number of gas particles increases
• particles become more spread out
• temperature increases
• substances mix or dissolve in many cases
• molecular complexity increases
• These are useful trends, but AP questions may require you to explain them at the particle level.

The standard molar entropy of a substance is represented by:

• S°
• Typical units:
• J/(mol·K)

Unlike standard enthalpy of formation, the standard entropy of an element in its standard state is not automatically zero at ordinary temperatures.

• That distinction is important.

A perfect pure crystal approaches zero entropy at:

• 0 K
• because there is essentially one perfectly ordered arrangement.

At temperatures above 0 K, substances generally have entropy greater than zero.

For a reaction:

• ΔS°rxn = ΣnS°products − ΣnS°reactants
• This looks similar to the standard enthalpy equation from Unit 6.

Remember to multiply each entropy value by its coefficient.

Suppose:

• A + 2B → C
• Then:
• ΔS°rxn = S°C − [S°A + 2S°B]

ΔS°rxn = ΣnS°products − ΣnS°reactants

ΔG° = ΔH° − TΔS°

ΔG°rxn = ΣnΔG°f(products) − ΣnΔG°f(reactants)

ΔG = ΔG° + RT ln Q

ΔG° = −RT ln K

E°cell = E°cathode − E°anode

ΔG° = −nFE°

E = E° − (0.0592/n)logQ

q = It

mol e⁻ = q/F

• with:
• F = 96,485 C/mol e⁻

ΔG < 0 → forward favored

• ΔG > 0 → reverse favored
• ΔG = 0 → equilibrium

E > 0 → forward favored

• E < 0 → forward unfavored
• E = 0 → equilibrium

K > 1 → products favored

• K < 1 → reactants favored

Memorize this connection:

• ΔG° < 0 ↔ E° > 0 ↔ K > 1
• and:
• ΔG° > 0 ↔ E° < 0 ↔ K < 1

At equilibrium:

• ΔG = 0
• E = 0
• Q = K

Thinking entropy only means "messiness."

• Better idea:
• entropy relates to energy dispersal and number of possible microstates.

Assuming exothermic means thermodynamically favored.

• Not always.
• You must consider:
• ΔG = ΔH − TΔS

Assuming favored means fast.

• Thermodynamics and kinetics are different.
• A favored process may still have a large activation barrier.

Using Celsius in Gibbs calculations.

• Use:
• Kelvin

Mixing joules and kilojoules.

• If ΔH is in kJ, convert ΔS into:
• kJ/K
• before multiplying by temperature.

Assuming ΔG° and ΔG are the same.

• ΔG° applies to standard-state comparison.
• ΔG depends on the actual conditions through Q.

Forgetting that at equilibrium:

• ΔG = 0
• not necessarily:
• ΔG° = 0

Thinking anode always means negative.

• False.
• The electrode signs change between galvanic and electrolytic cells.

The definition does not change:

• anode = oxidation

Thinking cathode always means positive.

• Again, only true for a galvanic cell.

Always use:

• cathode = reduction

Reversing electron flow.

• Electrons always flow through the external circuit:
• anode → cathode

Multiplying E° when multiplying a half-reaction.

Do not multiply standard electrode potentials.

Forgetting the salt bridge.

• The salt bridge maintains charge balance and allows the circuit to continue operating.

Using the wrong n in:

• ΔG° = −nFE°
• n is the number of moles of electrons transferred in the balanced overall redox reaction.

Using minutes directly with amperes.

• Convert time to seconds before:
• q = It

Do not simply say:

• "Entropy increases because disorder increases."
• A stronger explanation is:
• The particles can occupy a larger number of possible positions and energy arrangements, so the number of accessible microstates increases.

For:

• ΔH > 0 and ΔS > 0
• a good explanation is:
• At higher temperature, the favorable −TΔS term becomes large enough in magnitude to overcome the positive ΔH term, causing ΔG to become negative.

If asked why electrons flow:

• The thermodynamically favored redox reaction separates oxidation and reduction, causing electrons released at the anode to travel through the external circuit to the cathode, where they are consumed.

As a galvanic cell operates:

• reactants are consumed
• products form
• Q moves toward K
• driving force decreases
• E decreases

At equilibrium:

• Q = K
• and:
• E = 0

identify ΔH and ΔS → convert units → convert T to K → use ΔG = ΔH − TΔS → interpret sign

Set:

• ΔG = 0
• Then:
• T = ΔH/ΔS

products − reactants

• using:
• ΔG°rxn = ΣnΔG°f(products) − ΣnΔG°f(reactants)

Use:

• ΔG° = −RT ln K
• Then interpret:
• negative ΔG° → large K

identify oxidation

• identify reduction
• identify anode
• identify cathode
• balance electrons
• determine electron direction
• calculate E° if necessary

E°cell = E°cathode − E°anode

balance redox reaction → find n → use ΔG° = −nFE°

write overall reaction → calculate Q → determine n → use E = E° − (0.0592/n)logQ

• at 25°C.

I × t → coulombs → moles electrons → mole ratio → moles product → grams

Before a Unit 9 test, you should be able to:

explain entropy using microstates

predict the sign of ΔS

calculate ΔS°rxn

use ΔG = ΔH − TΔS

determine whether a process is thermodynamically favored

determine how temperature affects favorability

calculate a crossover temperature

calculate ΔG° from formation values

explain thermodynamics vs. kinetics

relate ΔG, Q, and K

use ΔG° = −RT ln K

explain coupled reactions

identify oxidation and reduction

identify anode and cathode

explain electron flow

explain the salt bridge

distinguish galvanic from electrolytic cells

calculate E°cell

use ΔG° = −nFE°

connect E°, ΔG°, and K

use the Nernst equation

calculate charge from current and time

use Faraday's constant

solve quantitative electrolysis problems

If you remember only a few things, remember these:

ΔG < 0 → favored

ΔG = ΔH − TΔS

ΔG° = −RT ln K

ΔG° = −nFE°

ΔG° < 0 ↔ E° > 0 ↔ K > 1

Anode = oxidation

• Cathode = reduction
• Electrons: anode → cathode

q = It

• then:
• charge → mol e⁻ → mol substance
• Unit 9 is basically the point where AP Chemistry ties together energy, equilibrium, redox chemistry, and electricity into one system.