9.3 Gibbs Free Energy and Thermodynamic Favorability

AP Chemistry 9.3 Thermodynamic Favorability Summary

Gibbs free energy (ΔG°) tells you whether a process is thermodynamically favored. When ΔG° is negative, the process is favored; when it is positive, it is not. You can find ΔG° either from ΔG° = ΔH° − TΔS° or from standard free energies of formation, and the signs of ΔH° and ΔS° (plus temperature) control the outcome.

Why This Matters for the AP Chemistry Exam

This topic shows up when you need to predict and justify whether a reaction or phase change will happen on its own. AP questions often ask you to reason about which factor, enthalpy or entropy, is driving a process, not just plug into a formula. You will connect particle-level behavior (dispersal of matter and energy) to macroscopic outcomes (does the reaction proceed under standard conditions?), and you may need to explain your reasoning in writing instead of only calculating a number.

A common exam move is asking you to decide favorability at different temperatures or to estimate the sign of ΔG° without a full calculation. Being fluent here also sets up later topics like free energy and equilibrium, coupled reactions, and electrochemical cell potential.

Key Takeaways

• ΔG° < 0 means the process is thermodynamically favored; ΔG° > 0 means it is not.
• Use ΔG° = ΔH° − TΔS°, and keep ΔH° and ΔS° in matching energy units (watch the kJ vs J trap).
• You can also find ΔG° from ΔG°reaction = ΣΔG°f(products) − ΣΔG°f(reactants); ΔG°f of an element in its standard state is 0.
• If ΔH° < 0 and ΔS° > 0, the process is favored at all temperatures with no calculation needed; if ΔH° > 0 and ΔS° < 0, it is never favored.
• When ΔH° and ΔS° have the same sign, temperature decides favorability.
• Standard state means pure substances, 1.0 M solutions, and gases at 1.0 atm (or 1.0 bar).

What Gibbs Free Energy Measures

Gibbs free energy change (ΔG°) is the value you use to decide if a chemical or physical process is thermodynamically favored. When ΔG° < 0, the process is favored. When ΔG° > 0, it is not favored under standard conditions.

A quick note on wording: you may have heard favored processes called "spontaneous." The better phrase is "thermodynamically favored," because "spontaneous" can make it sound like the reaction happens suddenly or fast. Favorability says nothing about speed. A favored reaction can still be extremely slow, which is the idea behind kinetic control in the next topic.

Standard state, shown by the degree symbol in ΔG°, means each substance is in a defined reference condition: pure substances, solutions at 1.0 M, and gases at a pressure of 1.0 atm (or 1.0 bar).

Enthalpy and Entropy: The Two Inputs

Two quantities feed into favorability.

Enthalpy change (ΔH°) is the heat exchanged during a process. A negative ΔH° means the system releases heat (exothermic). A positive ΔH° means the system absorbs heat (endothermic).

Entropy change (ΔS°) is the change in how dispersed matter and energy are. A positive ΔS° means matter or energy spreads out more (for example, solid to liquid to gas, or more moles of gas produced). A negative ΔS° means the system becomes more ordered.

Neither factor alone decides favorability in every case. That is why you combine them into one value.

Calculating ΔG° from ΔH° and ΔS°

The core equation is:

ΔG° = ΔH° − TΔS°

This equation is provided on the AP Chemistry reference sheet. Temperature T must be in Kelvin.

The biggest unit trap: ΔH° is usually in kJ/mol while ΔS° is usually in J/(mol·K). Convert so they match before subtracting.

Worked Example

Consider: 2H₂ + N₂ ⇌ N₂H₄ at 25°C, with ΔH° = 50.6 kJ/mol and ΔS° = −0.332 kJ/(mol·K). Find ΔG° and decide if the reaction is favored.

First convert temperature: T = 25 + 273 = 298 K.

ΔG° = ΔH° − TΔS° ΔG° = (50.6) − (298)(−0.332) ΔG° = 50.6 + 98.9 = 149.5 kJ/mol

Since ΔG° > 0, this reaction is not thermodynamically favored at 298 K.

Calculating ΔG° from Free Energies of Formation

You can also find ΔG° using standard free energies of formation, the same way you handle ΔH° with formation values:

ΔG°reaction = ΣΔG°f(products) − ΣΔG°f(reactants)

Remember that ΔG°f for an element in its standard state is zero, just like with ΔH°f.

Worked Example

Calculate ΔG° for the combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Given ΔG°f values (kJ/mol):

• CH₄(g): −50.8
• O₂(g): 0 (element in standard state)
• CO₂(g): −394.4
• H₂O(g): −228.6

ΔG°reaction = [ΔG°f(CO₂) + 2ΔG°f(H₂O)] − [ΔG°f(CH₄) + 2ΔG°f(O₂)] ΔG°reaction = [(−394.4) + 2(−228.6)] − [(−50.8) + 2(0)] ΔG°reaction = [−394.4 − 457.2] − [−50.8] ΔG°reaction = −851.6 + 50.8 = −800.8 kJ/mol

Since ΔG° < 0, the combustion of methane is thermodynamically favored at 25°C.

Thermodynamic Favorability Chart: ΔH°, ΔS°, and Temperature

Look at ΔG° = ΔH° − TΔS° and think about the signs.

• If ΔH° < 0 (exothermic) and ΔS° > 0 (entropy increases), ΔG° is negative at every temperature. Favored at all T, no calculation needed.
• If ΔH° > 0 (endothermic) and ΔS° < 0 (entropy decreases), ΔG° is positive at every temperature. Never favored, no calculation needed.
• If both are negative (ΔH° < 0, ΔS° < 0), the reaction is favored only at low temperatures, where the small TΔS° term cannot overcome the negative ΔH°.
• If both are positive (ΔH° > 0, ΔS° > 0), the reaction is favored only at high temperatures, where the large TΔS° term outweighs the positive ΔH°.

When ΔH° and ΔS° share a sign, you can find the changeover temperature by setting ΔG° = 0, which gives T = ΔH°/ΔS°.

Enthalpy-Driven vs Entropy-Driven Processes

When AP questions ask which factor "drives" a process, they want you to compare the two terms, not just say both matter.

A process is entropy-driven when the increase in dispersal is what makes ΔG° negative, even if enthalpy works against it. A process is enthalpy-driven when the heat released makes ΔG° negative, even if entropy works against it.

Freezing of Water

H₂O(l) → H₂O(s)

• ΔH° < 0 (exothermic, heat is released)
• ΔS° < 0 (entropy decreases, liquid to solid)
• Favored only below 0°C (273 K)

Above 0°C, the TΔS° term grows large enough to make ΔG° positive, so ice does not form. This is an enthalpy-driven case that only wins at low temperature.

Dissolution of Sodium Nitrate

NaNO₃(s) → Na⁺(aq) + NO₃⁻(aq)

• ΔH° > 0 (endothermic, the solution gets cold)
• ΔS° > 0 (entropy increases, ordered solid to mobile ions)
• Favored at room temperature even though it is endothermic

Here the large entropy increase from the crystal breaking into free-moving ions outweighs the unfavorable enthalpy, so dissolution is favored. This is an entropy-driven process.

How to Use This on the AP Chemistry Exam

Problem Solving

• Identify which method fits: use ΔG° = ΔH° − TΔS° when given enthalpy, entropy, and temperature; use formation values when given a table of ΔG°f.
• Convert temperature to Kelvin and align energy units before subtracting.
• Carry units through and report ΔG° in kJ/mol or J/mol, then state whether the process is favored based on the sign.

Free Response

• When asked which factor drives a process, name enthalpy or entropy specifically and explain why that term controls the sign of ΔG°. Saying "both, because of the equation" usually does not earn the reasoning.
• Connect particle-level changes (matter or energy spreading out) to the macroscopic result (does the process happen under standard conditions).
• For temperature questions, justify your answer using the signs of ΔH° and ΔS°, not just a plugged-in number.

Common Trap

• If ΔH° < 0 and ΔS° > 0, or ΔH° > 0 and ΔS° < 0, you do not need to calculate ΔG° at all; the signs alone settle it.

Common Misconceptions

• "Thermodynamically favored means fast." Favorability only tells you the direction the process is favored, not the rate. A favored reaction can be extremely slow.
• "Spontaneous means it happens instantly or on its own with no cause." This is why the preferred term is thermodynamically favored. It describes the sign of ΔG°, nothing about speed or suddenness.
• "Both enthalpy and entropy always drive a reaction." When the two terms push in opposite directions, one of them controls the sign of ΔG°. Compare their effects instead of citing the equation.
• "Mixing kJ and J is fine." ΔH° in kJ and ΔS° in J will give a wildly wrong ΔG°. Always match units first.
• "ΔG°f of an element is some lookup value." For an element in its standard state, ΔG°f is zero.
• "A positive ΔG° means the reaction can never go." It means the forward process is not favored under standard conditions, but changing temperature (when the signs allow) or coupling it to a favored reaction can change the outcome.

Related AP Chemistry Guides

• 9.1 Introduction to Entropy
• 9.2 Absolute Entropy and Entropy Change
• 9.8 Galvanic (Voltaic) and Electrolytic Cells
• 9.4 Thermodynamic and Kinetic Control
• Unit 9 Review: Thermodynamics and Electrochemistry
• 9.5 Free Energy and Equilibrium