Entropy, Free Energy, and Spontaneity (AP Chemistry Unit 9)

Introduction to Entropy

What entropy is

Entropy is a measure of how energy is spread out (dispersed) among the possible microscopic arrangements of particles. In chemistry, those microscopic arrangements are called microstates—different ways to distribute energy and position among particles while still matching the same overall, observable state (macrostate) like temperature, pressure, and volume.

A useful way to think about entropy is: systems tend to move toward states where energy is more widely distributed among more possible microstates. This is why gases expand to fill containers, why a drop of dye spreads through water, and why it’s hard to “unmix” something without doing work.

Why entropy matters

Entropy is one of the two major “drivers” of whether a process is thermodynamically favorable (spontaneous). The other driver is enthalpy (roughly, heat content). Many reactions are not “pulled” forward just because they release heat; they can also be driven forward because they increase entropy.

The most fundamental spontaneity criterion is based on the Second Law of Thermodynamics:

• A process is spontaneous if the entropy of the universe increases.

That idea is often expressed as:

$\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr}$

• $\Delta S_{sys}$ is the entropy change of the system (the chemicals you’re focusing on).
• $\Delta S_{surr}$ is the entropy change of the surroundings (everything else).

Spontaneous means:

$\Delta S_{univ} > 0$

A common misconception is that “spontaneous” means “fast.” It doesn’t. Spontaneous only means thermodynamically allowed; kinetics controls rate.

How entropy “behaves” in typical chemical situations

In AP Chemistry, you’ll often predict the sign of entropy changes using patterns:

• Phase changes: entropy increases going from solid to liquid to gas.
• Gas moles: increasing the number of moles of gas usually increases entropy.
• Mixing/dissolving: mixing tends to increase entropy because particles have more possible arrangements.
• Temperature: higher temperature generally corresponds to higher entropy because energy is spread across more accessible microstates.

These are not “magic rules”—they’re shortcuts based on microstates. Gas particles can occupy many more positions than particles locked in a crystal, so gases have far more microstates.

Example: qualitative sign of entropy change

Predict the sign of $\Delta S_{sys}$ for each.

1) $H_2O(l) \rightarrow H_2O(g)$

Liquid to gas increases disorder/energy dispersal, so $\Delta S_{sys}$ is positive.

2) $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

Gas moles go from 4 to 2. Fewer gas particles generally means fewer positional microstates, so $\Delta S_{sys}$ is negative.

Exam Focus

• Typical question patterns:
  • Predict the sign of $\Delta S_{sys}$ from a reaction or phase change.
  • Compare entropy of substances (solid vs liquid vs gas; larger molar mass; more atoms).
  • Explain, in words, why entropy increases or decreases in a process.
• Common mistakes:
  • Treating “spontaneous” as “rapid.” Remember: spontaneity is thermodynamics, not speed.
  • Assuming entropy always increases in the system; it’s the universe that must increase.
  • Forgetting to focus on gaseous moles when using the “moles of gas” heuristic.

Absolute Entropy and Entropy Change

Absolute entropy (Third Law perspective)

Absolute entropy refers to the entropy a substance has at a specific temperature, usually reported as a standard molar entropy value under standard conditions. The conceptual anchor is the Third Law of Thermodynamics: a perfectly ordered crystal at absolute zero has zero entropy. As temperature increases, substances gain accessible microstates, so entropy increases.

In practice, AP Chemistry most often uses tabulated standard molar entropies (often written as $S^\circ$) to compute reaction entropy changes. These values are typically in $J\,mol^{-1}\,K^{-1}$.

Key idea: absolute entropy is not “disorder points.” It’s a measurable thermodynamic quantity tied to energy distribution and microstates.

Calculating entropy change of a reaction

For a reaction under standard conditions, the standard entropy change is calculated from tabulated standard molar entropies:

$\Delta S^\circ_{rxn} = \sum nS^\circ(products) - \sum nS^\circ(reactants)$

• $n$ is the stoichiometric coefficient.
• $S^\circ$ values come from tables.

This parallels how you compute $\Delta H^\circ_{rxn}$ from standard enthalpies of formation, but here you use standard molar entropies directly.

Entropy change of surroundings (connecting to enthalpy)

You sometimes need $\Delta S_{surr}$ to judge spontaneity via $\Delta S_{univ}$. At constant temperature (a common AP assumption), the surroundings’ entropy change is related to heat flow:

$\Delta S_{surr} = -\frac{\Delta H_{sys}}{T}$

• $T$ is temperature in kelvins.
• Using $\Delta H_{sys}$ is appropriate at constant pressure (common for reactions open to atmosphere).

This equation encodes an important idea: if the system releases heat (negative enthalpy change), the surroundings gain heat, increasing the surroundings’ entropy.

Example (worked): computing $\Delta S^\circ_{rxn}$

Consider:

$2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$

Using the formula:

$\Delta S^\circ_{rxn} = \sum nS^\circ(products) - \sum nS^\circ(reactants)$

You would compute:

• Products: $2S^\circ(SO_3(g))$
• Reactants: $2S^\circ(SO_2(g)) + 1S^\circ(O_2(g))$

So:

$\Delta S^\circ_{rxn} = 2S^\circ(SO_3(g)) - \left(2S^\circ(SO_2(g)) + S^\circ(O_2(g))\right)$

Without even plugging numbers, you can often predict the sign: gas moles decrease from 3 to 2, so $\Delta S^\circ_{rxn}$ is likely negative.

What goes wrong: common reasoning traps

• Confusing “entropy of reaction” with “entropy of universe”: a reaction can have negative $\Delta S_{sys}$ and still be spontaneous if $\Delta S_{surr}$ is sufficiently positive.
• Forgetting coefficients: entropy is extensive—double the moles, double the entropy contribution.
• Unit mismatch: $\Delta S$ is usually in $J\,mol^{-1}\,K^{-1}$, while enthalpy is often in $kJ\,mol^{-1}$. Conversions matter later when combining with Gibbs free energy.

Exam Focus

• Typical question patterns:
  • Calculate $\Delta S^\circ_{rxn}$ from a table of $S^\circ$ values.
  • Predict sign/magnitude of $\Delta S$ using phase and gas-mole reasoning.
  • Use $\Delta S_{surr} = -\Delta H/T$ to connect heat flow to entropy.
• Common mistakes:
  • Mixing $kJ$ and $J$ without converting when needed.
  • Treating entropy as depending only on “messiness” rather than microstates (e.g., assuming all solids have the same entropy).
  • Using the “moles of gas” shortcut for reactions with no gases (where it’s much less predictive).

Gibbs Free Energy and Thermodynamic Favorability

Why we introduce Gibbs free energy

Using $\Delta S_{univ}$ is conceptually powerful, but it’s inconvenient experimentally because you’d need both system and surroundings. Gibbs free energy packages the Second Law into a criterion that depends only on the system (under common lab conditions).

Most reactions you study occur at roughly constant temperature and pressure. Under those conditions, Gibbs free energy tells you whether a process is thermodynamically favorable.

Definition and key equation

Gibbs free energy $G$ is a state function, and the change in Gibbs free energy is:

$\Delta G = \Delta H - T\Delta S$

• $\Delta G$ is the Gibbs free energy change of the system.
• $\Delta H$ is the enthalpy change.
• $T$ is temperature in kelvins.
• $\Delta S$ is the entropy change of the system.

This equation shows the competition:

• Enthalpy term favors exothermic processes (negative $\Delta H$).
• Entropy term favors processes that increase system entropy (positive $\Delta S$), especially at high temperature because of the factor $T$.

Spontaneity criteria using Gibbs free energy

At constant temperature and pressure:

• $\Delta G < 0$ means spontaneous (thermodynamically favorable).
• $\Delta G = 0$ means equilibrium.
• $\Delta G > 0$ means nonspontaneous as written (spontaneous in reverse).

A crucial conceptual connection (often tested) is that Gibbs free energy is directly tied to the entropy of the universe under these conditions:

$\Delta G = -T\Delta S_{univ}$

So $\Delta S_{univ} > 0$ corresponds exactly to $\Delta G < 0$.

Temperature dependence (the four sign cases)

Because $\Delta G = \Delta H - T\Delta S$, temperature can flip spontaneity depending on the signs of $\Delta H$ and $\Delta S$:

• $\Delta H < 0$ and $\Delta S > 0$: always spontaneous (both terms help).
• $\Delta H > 0$ and $\Delta S < 0$: never spontaneous (both terms oppose).
• $\Delta H < 0$ and $\Delta S < 0$: spontaneous at low $T$ (enthalpy dominates).
• $\Delta H > 0$ and $\Delta S > 0$: spontaneous at high $T$ (entropy term dominates).

Students often memorize these, but it’s better to reason from the equation: increasing $T$ increases the importance of the $-T\Delta S$ term.

Example (worked): finding the temperature where spontaneity changes

Suppose a process has $\Delta H = 25\,kJ\,mol^{-1}$ and $\Delta S = 75\,J\,mol^{-1}\,K^{-1}$.

1) Convert units so they match (use joules):

$\Delta H = 25000\,J\,mol^{-1}$

2) Set $\Delta G = 0$ to find the boundary temperature:

$0 = \Delta H - T\Delta S$

3) Solve for $T$:

$T = \frac{\Delta H}{\Delta S}$

$T = \frac{25000}{75} = 333\,K$

Interpretation: below about $333\,K$, $\Delta G$ is positive (not favorable); above $333\,K$, $\Delta G$ becomes negative (favorable). This is the classic “endothermic but entropy-increasing” situation.

Standard Gibbs free energy change

When reactants and products are in their standard states, you may see standard Gibbs free energy change $\Delta G^\circ$. It’s the free energy change under standard conditions.

Be careful with the phrase “standard”: it does not mean “at equilibrium.” In fact, many standard-state mixtures are not at equilibrium.

Exam Focus

• Typical question patterns:
  • Given $\Delta H$ and $\Delta S$, determine the sign of $\Delta G$ at a certain temperature.
  • Determine the temperature at which a process becomes spontaneous.
  • Conceptually justify why high temperature favors entropy-driven processes.
• Common mistakes:
  • Forgetting to convert $kJ$ to $J$ when combining with entropy values.
  • Using Celsius instead of kelvins for $T$.
  • Thinking $\Delta G^\circ < 0$ means the reaction “goes to completion.” It only implies products are favored at equilibrium, not 100% yield.

Thermodynamic and Kinetic Control

Thermodynamic vs kinetic: what’s the difference?

AP Chemistry often tests whether you can separate two ideas:

• Thermodynamic control is about which outcome is most stable (lowest Gibbs free energy). It answers: “Which products are favored at equilibrium?”
• Kinetic control is about which outcome forms fastest (lowest activation energy). It answers: “Which product forms first?”

A reaction can be thermodynamically favorable and still occur very slowly if it has a large activation energy barrier. Diamond turning into graphite is the classic idea: graphite is thermodynamically more stable at typical conditions, but diamond persists because the conversion is kinetically hindered.

Energy diagrams: connecting concepts visually

Reaction coordinate diagrams help you compare:

• Activation energy $E_a$: barrier height controlling rate.
• Overall free energy change (often shown as a vertical difference between reactants and products): indicates relative thermodynamic stability.

If there are multiple possible products, the kinetically controlled product typically has the smaller activation energy (lower barrier) even if it is not the lowest-energy product.

Catalysts: what they change and what they don’t

A catalyst lowers activation energy and increases rate, but it does not change thermodynamic quantities like $\Delta G$, $\Delta H$, or the equilibrium constant $K$.

This is a frequent exam target: catalysts help a system reach equilibrium faster, but they do not shift where equilibrium lies.

Example: “spontaneous but slow” vs “nonspontaneous but forced”

• Rusting of iron is thermodynamically favorable under many conditions (negative $\Delta G$) but may be slow without water/electrolytes.
• Electrolysis of water is not thermodynamically favorable as written (positive $\Delta G$) but can be driven by electrical work supplied from outside.

This highlights another misconception: “nonspontaneous” does not mean “impossible,” it means it requires continuous energy input.

Exam Focus

• Typical question patterns:
  • Given an energy diagram, identify which pathway is kinetically favored vs thermodynamically favored.
  • Explain why a negative $\Delta G$ reaction might not occur at a noticeable rate.
  • Predict what a catalyst changes (rate) versus what it doesn’t (equilibrium position).
• Common mistakes:
  • Claiming catalysts change $K$ or make $\Delta G$ more negative.
  • Confusing “most stable product” with “fastest formed product.”
  • Treating activation energy as a thermodynamic quantity; it’s kinetic.

Free Energy and Equilibrium

How free energy connects to equilibrium

Even if a reaction is thermodynamically favorable initially, it doesn’t necessarily go to completion. As the reaction proceeds, concentrations change, and the driving force changes.

To capture this, chemistry uses the relationship between free energy and the reaction quotient $Q$.

Nonstandard conditions: $\Delta G$ depends on $Q$

The free energy change at any moment is:

$\Delta G = \Delta G^\circ + RT\ln Q$

• $R$ is the gas constant.
• $T$ is temperature in kelvins.
• $Q$ is the reaction quotient (same form as $K$ but using current concentrations/pressures).

Interpretation:

• If $Q < K$, the reaction proceeds forward to make more products.
• If $Q > K$, the reaction proceeds backward to make more reactants.
• At equilibrium, $Q = K$ and $\Delta G = 0$.

The deep link: $\Delta G^\circ$ and $K$

At equilibrium, plug $\Delta G = 0$ and $Q = K$ into $\Delta G = \Delta G^\circ + RT\ln Q$:

$0 = \Delta G^\circ + RT\ln K$

So:

$\Delta G^\circ = -RT\ln K$

This is one of the most important equations in this unit because it connects a thermodynamic quantity (free energy) to an equilibrium quantity (the equilibrium constant).

Key consequences:

• If $\Delta G^\circ < 0$, then $K > 1$ (products favored at equilibrium).
• If $\Delta G^\circ > 0$, then $K < 1$ (reactants favored).
• If $\Delta G^\circ = 0$, then $K = 1$.

A common mistake is to interpret $K > 1$ as “all products.” It only means the equilibrium mixture has more products than reactants in the ratio dictated by $K$.

Example (worked): find $K$ from $\Delta G^\circ$

Suppose at $298\,K$ you are given $\Delta G^\circ = -5.70\,kJ\,mol^{-1}$.

1) Convert to joules:

$\Delta G^\circ = -5700\,J\,mol^{-1}$

2) Use:

$\Delta G^\circ = -RT\ln K$

3) Solve for $\ln K$:

$\ln K = -\frac{\Delta G^\circ}{RT}$

4) Substitute values (with $R = 8.314\,J\,mol^{-1}\,K^{-1}$ and $T = 298\,K$):

$\ln K = -\frac{-5700}{8.314 \cdot 298}$

$\ln K \approx 2.30$

5) Exponentiate:

$K = e^{2.30} \approx 10$

Interpretation: products are favored, but not exclusively; the equilibrium constant is moderate.

Example (worked): decide direction using $Q$

For a reaction where $K = 50$ at a given temperature, you calculate $Q = 200$ for the current mixture.

Because $Q > K$, the mixture has “too much product” compared with equilibrium. The reaction will proceed in the reverse direction until $Q$ decreases to 50.

Exam Focus

• Typical question patterns:
  • Use $\Delta G = \Delta G^\circ + RT\ln Q$ to predict reaction direction.
  • Convert between $\Delta G^\circ$ and $K$ using $\Delta G^\circ = -RT\ln K$.
  • Interpret what the sign/magnitude of $\Delta G^\circ$ implies about product favorability.
• Common mistakes:
  • Plugging Celsius temperature into equations.
  • Forgetting that $Q$ and $K$ must be built from the balanced equation (exponents matter).
  • Using $\Delta G^\circ$ to predict direction when conditions are nonstandard (you need $\Delta G$ or compare $Q$ to $K$).

Coupled Reactions

What coupling means and why it works

A coupled reaction is when an energetically unfavorable process (positive $\Delta G$) is driven by pairing it with a favorable process (negative $\Delta G$) so that the overall combined process has negative $\Delta G$.

This is not a “trick”—it follows from a key property of Gibbs free energy:

• When you add reactions to get an overall reaction, you add their Gibbs free energy changes.

Mathematically:

$\Delta G_{overall} = \Delta G_1 + \Delta G_2$

So if a nonspontaneous step has $\Delta G$ positive, it can still proceed if it is directly linked to another step with a more negative $\Delta G$ such that the sum is negative.

Why coupling is central in chemistry (and life)

Many important processes are uphill in free energy—building complex molecules, transporting ions against gradients, or synthesizing polymers. They happen because cells (and engineered systems) couple them to strongly downhill processes.

A famous biological example is coupling to ATP hydrolysis. Even if you don’t memorize ATP details for AP Chemistry, the thermodynamic point is what matters:

• An unfavorable reaction can become favorable when coupled to a sufficiently favorable one.

How coupling is implemented (conceptually)

For coupling to actually drive a process, the reactions must share intermediates or be mechanistically linked so that the favorable reaction “pays” for the unfavorable one. Simply writing two separate reactions on paper doesn’t guarantee coupling in reality.

In electrochemistry (closely related to Unit 9 themes), coupling is also the idea behind galvanic cells: a spontaneous redox reaction can be used to push electrons through a circuit to do electrical work.

Example (worked): determining whether coupling makes a process favorable

Imagine you want to drive Reaction A:

• Reaction A: $\Delta G_A = +12\,kJ\,mol^{-1}$ (unfavorable)

You can couple it to Reaction B:

• Reaction B: $\Delta G_B = -20\,kJ\,mol^{-1}$ (favorable)

Overall:

$\Delta G_{overall} = \Delta G_A + \Delta G_B$

$\Delta G_{overall} = 12 + (-20) = -8\,kJ\,mol^{-1}$

Because the overall $\Delta G$ is negative, the coupled process is thermodynamically favorable.

What goes wrong: misconceptions about coupling

• “If one step is spontaneous, everything becomes spontaneous.” Not necessarily—the favorable step must be favorable enough to overcome the unfavorable one.
• “Coupling always means equilibrium shifts.” Coupling is about overall free energy and feasibility; equilibrium considerations still apply, and the system may reach a new equilibrium state.
• “You can couple reactions just by adding equations.” Thermodynamically you can add $\Delta G$ values, but in reality the reactions must be mechanistically linked to transfer energy.

Exam Focus

• Typical question patterns:
  • Compute $\Delta G_{overall}$ by summing $\Delta G$ values for coupled steps.
  • Decide whether an unfavorable process can be driven by a given favorable reaction.
  • Explain, in words, how coupling allows nonspontaneous processes to proceed.
• Common mistakes:
  • Forgetting that reversing a reaction changes the sign of $\Delta G$.
  • Confusing coupling (thermodynamics) with catalysis (kinetics). Catalysts don’t change $\Delta G$.
  • Neglecting units and signs when adding free energy changes.