Study Guide

Gibbs Free Energy and Thermodynamic Favorability

AP Chemistry· AP Chemistry CED — Applications of Thermodynamics· 14 min read

1. Core Concepts: Gibbs Free Energy and Favorability★★☆☆☆⏱ 3 min

Gibbs free energy () is a combined thermodynamic state function that links enthalpy () and entropy () to predict whether a process will be thermodynamically favorable for reactions at constant temperature and pressure — the condition that describes almost all reactions studied in AP Chemistry.

📘 Definition

Thermodynamically Favorable Process

A process that will proceed to form products at given conditions without continuous input of external energy, once initiated. This is the AP Chemistry term for what many sources call a "spontaneous" process.

By universal convention: a negative corresponds to a thermodynamically favorable process, while a positive corresponds to an unfavorable process. When , the process is at equilibrium with no net change.

✓ Quick check

Test your basic understanding:

  1. If a process has kJ/mol, what does this mean?

    • The process will occur very quickly

    • The process is thermodynamically favorable

    • The process is at equilibrium

    • The process will never occur

    Reveal answer
    1

    Correct! sign only indicates thermodynamic favorability, not reaction rate.

2. The Fundamental Equation: $\Delta G = \Delta H - T\Delta S$★★★☆☆⏱ 4 min

At constant temperature and pressure, the change in Gibbs free energy for any process is given by the equation:

ΔG=ΔHTΔS\Delta G = \Delta H - T\Delta S

Where = change in Gibbs free energy (kJ/mol), = enthalpy change (kJ/mol), = absolute temperature in Kelvin, and = entropy change of the system.

  • If (exothermic) and : at all temperatures (always favorable)

  • If (endothermic) and : at all temperatures (always unfavorable)

  • If and have matching signs: temperature determines the sign of and thus favorability

📐 Worked Example

For the evaporation of liquid ethanol to ethanol vapor at 1 atm, kJ/mol and J/mol·K. Is evaporation thermodynamically favorable at (a) 25°C, (b) 100°C?

  1. 1

    Convert temperatures from Celsius to Kelvin:

    Ta=25+273.15=298.15 K;Tb=100+273.15=373.15 KT_a = 25 + 273.15 = 298.15 \text{ K}; \quad T_b = 100 + 273.15 = 373.15 \text{ K}
  2. 2

    Convert to kJ/mol·K to match units of :

    110 J/mol\cdotpK=0.110 kJ/mol\cdotpK110 \text{ J/mol·K} = 0.110 \text{ kJ/mol·K}
  3. 3

    Calculate at 25°C:

    ΔG=38.6(298.15)(0.110)=+5.8 kJ/mol\Delta G = 38.6 - (298.15)(0.110) = +5.8 \text{ kJ/mol}
  4. 4

    Calculate at 100°C:

    ΔG=38.6(373.15)(0.110)=2.4 kJ/mol\Delta G = 38.6 - (373.15)(0.110) = -2.4 \text{ kJ/mol}
  5. 5

    Interpret: At 25°C, so evaporation is unfavorable (liquid ethanol is stable); at 100°C, so evaporation is favorable.

Exam tip:

Always convert from J/mol·K to kJ/mol·K before plugging into the equation. AP exam questions almost always give in joules and in kilojoules, so missing this conversion will give you a wrong sign and incorrect answer.

3. Standard Gibbs Free Energy Calculations★★★☆☆⏱ 3 min

Standard Gibbs free energy change () is the change in Gibbs free energy when reactants in their standard states (1 atm pressure, 1 M concentration, pure solid/liquid, 298 K by default) are converted to products in their standard states. There are two common methods to calculate :

📘 Definition

Standard Gibbs Free Energy of Formation

ΔGf\Delta G^\circ_f

The for formation of 1 mole of a compound from its constituent elements in their standard states. By definition, for any element in its standard state.

When using standard free energies of formation, the reaction is calculated as:

ΔGrxn=nΔGf(products)mΔGf(reactants)\Delta G^\circ_{\text{rxn}} = \sum n\Delta G^\circ_f(\text{products}) - \sum m\Delta G^\circ_f(\text{reactants})
📐 Worked Example

Calculate for the oxidation of iron (rust formation) at 298 K: . Use the values: kJ/mol, kJ/mol, kJ/mol.

  1. 1

    Write the formula for the balanced reaction:

    ΔGrxn=2ΔGf(Fe2O3)[4ΔGf(Fe)+3ΔGf(O2)]\Delta G^\circ_{\text{rxn}} = 2\Delta G^\circ_f(\text{Fe}_2\text{O}_3) - \left[4\Delta G^\circ_f(\text{Fe}) + 3\Delta G^\circ_f(\text{O}_2)\right]
  2. 2

    Substitute the given values:

    ΔGrxn=2(742.2)[4(0)+3(0)]\Delta G^\circ_{\text{rxn}} = 2(-742.2) - [4(0) + 3(0)]
  3. 3

    Calculate the result:

    ΔGrxn=1484.4 kJ/mol\Delta G^\circ_{\text{rxn}} = -1484.4 \text{ kJ/mol}
  4. 4

    Interpret: The negative confirms that rust formation is thermodynamically favorable under standard conditions.

Exam tip:

Remember the minus sign in the formula applies to the entire sum of reactants. If any of reactants is negative, you will subtract a negative which equals adding that value — always write out signs explicitly to avoid arithmetic errors.

4. $\Delta G$, $\Delta G^\circ$, and the Relationship to Equilibrium★★★★☆⏱ 4 min

only describes the Gibbs free energy change when the reaction is at standard state (all reactants and products at 1 M/1 atm, so ). For any non-standard conditions, we calculate using:

ΔG=ΔG+RTlnQ\Delta G = \Delta G^\circ + RT \ln Q

When a reaction reaches equilibrium, (no net driving force) and (the equilibrium constant). Substituting these values gives the key relationship connecting thermodynamics and equilibrium:

ΔG=RTlnK\Delta G^\circ = -RT \ln K
  • : Products are favored at equilibrium

  • : Reactants are favored at equilibrium

  • : Equal amounts of reactants and products at equilibrium

📐 Worked Example

For the dissolution of calcium hydroxide, , kJ/mol at 25°C. Calculate the solubility product constant for calcium hydroxide at 25°C.

  1. 1

    Convert temperature to Kelvin:

    T=25+273.15=298.15 KT = 25 + 273.15 = 298.15 \text{ K}
  2. 2

    Convert to J/mol to match units of :

    39.0 kJ/mol=39000 J/mol39.0 \text{ kJ/mol} = 39000 \text{ J/mol}
  3. 3

    Rearrange to solve for :

    lnKsp=ΔGRT\ln K_{\text{sp}} = -\frac{\Delta G^\circ}{RT}
  4. 4

    Substitute values ( J/mol·K):

    lnKsp=39000(8.314)(298.15)15.75\ln K_{\text{sp}} = -\frac{39000}{(8.314)(298.15)} \approx -15.75
  5. 5

    Exponentiate to get :

    Ksp=e15.755.6×107K_{\text{sp}} = e^{-15.75} \approx 5.6 \times 10^{-7}

Exam tip:

Always use J/mol·K for Gibbs free energy calculations, not L·atm/mol·K (the gas constant used for ideal gas law problems). Using the wrong R will give you a K that is orders of magnitude off.

5. Common Pitfalls

Wrong move:

Forgetting to convert ΔS from J/mol·K to kJ/mol·K in ΔG = ΔH - TΔS, leading to a ΔG with the wrong sign

Why:

AP questions almost always give ΔH in kJ and ΔS in J, so students plug in numbers without checking units

Correct move:

Always check units before plugging in; if they differ, convert ΔS to kJ to match ΔH

Wrong move:

Claiming a reaction with ΔG > 0 will never occur at any observable rate

Why:

Students confuse thermodynamic favorability with kinetic feasibility. ΔG only describes favorability, not how fast the reaction proceeds

Correct move:

State the forward reaction is thermodynamically unfavorable, and note that this tells you nothing about the rate of the reaction

Wrong move:

Claiming ΔG° = 0 at equilibrium

Why:

Students mix up the meaning of ΔG (any conditions) and ΔG° (only standard state)

Correct move:

Remember ΔG is always 0 at equilibrium; ΔG° is only 0 at equilibrium when K = 1

Wrong move:

Treating any process with a positive ΔS of the system as always favorable

Why:

Students forget the second law refers to the entropy of the universe, not just the system

Correct move:

Always use the sign of ΔG, not just ΔS of the system, to determine thermodynamic favorability

Wrong move:

Using R = 0.0821 L·atm/mol·K when calculating K from ΔG°

Why:

Students remember R from gas law problems and use it by mistake

Correct move:

Always reach for R = 8.314 J/mol·K for all Gibbs free energy calculations

Wrong move:

Subtracting a negative ΔG°f value incorrectly, getting a positive ΔG° when it should be negative

Why:

Students forget the formula is products minus reactants, so a negative reactant ΔG°f becomes a positive term

Correct move:

Write all negative signs explicitly before plugging in numbers, e.g., ΔG° = products - (-50) = products + 50

6. Quick Reference Cheatsheet

Category

Formula

Key Notes

Fundamental Gibbs free energy

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

Convert ΔS to kJ/mol·K to match ΔH units

Favorability rule

ΔG < 0 = Favorable; ΔG = 0 = Equilibrium; ΔG > 0 = Unfavorable

Applies to all constant T,P processes

ΔG° from standard formation

\Delta G^\circ_{\text{rxn}} = \sum n\Delta G^\circ_f(\text{products}) - \sum m\Delta G^\circ_f(\text{reactants})

ΔG°f = 0 for elements in standard state

ΔG for non-standard conditions

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

R = 8.314 J/mol·K, convert ΔG° to J

ΔG° and equilibrium constant

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

ΔG° < 0 → K > 1; ΔG° > 0 → K < 1

Boundary temperature for favorability

T = \frac{\Delta H}{\Delta S} (at ΔG = 0)

ΔH-,ΔS- → favorable below T; ΔH+,ΔS+ → favorable above T

Coupled reactions

\Delta G_{\text{total}} = \sum \Delta G_{\text{individual}}

Gibbs free energy adds for sequential reactions

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

  • 2023 · FRQ

    ΔG calculation from ΔH and ΔS

  • 2022 · MCQ

    Relate ΔG° to equilibrium K

  • 2021 · FRQ

    ΔG° from formation values

Going deeper

What's Next

This topic forms the foundational link connecting thermodynamics to two core AP Chemistry topics: chemical equilibrium and electrochemistry, which are heavily tested on both MCQ and FRQ sections. Next, you will apply the relationship between ΔG° and K to predict how equilibrium constants change with temperature, a common multi-part FRQ skill. You will also connect ΔG to cell potential in electrochemistry, using the relation ΔG = -nFE to convert between cell voltage and Gibbs free energy change. Without mastering sign rules and unit conversions for Gibbs free energy, both of these topics will be far more difficult to solve correctly on the exam. This topic also completes the framework of thermodynamics started in Unit 6, giving a complete picture of energy and favorability for chemical processes.