Entropy and Gibbs Free Energy

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

1. Entropy, Second Law, and Predicting Entropy Changes★★☆☆☆⏱ 4 min

Entropy ( $S$ ) is a thermodynamic state function that quantifies the degree of disorder, or the number of possible microstates available to a system. It is defined by Boltzmann’s formula:

S = k \ln W

where $k$ is the Boltzmann constant and $W$ is the number of distinct microstates a system can occupy. More microstates = higher entropy = greater disorder.

📘 Definition

Second Law of Thermodynamics

For any spontaneous process, the total entropy change of the universe is always positive

Example:

All spontaneous processes increase the total disorder of the universe

The third law of thermodynamics tells us that a perfect crystalline substance at 0 K has zero entropy, so all pure substances above 0 K have a positive standard molar entropy $S^{\circ}$ (entropy of 1 mole at 1 atm and 298 K). To predict the sign of $Δ S_{sys}$ , follow these general rules:

Gases have far higher entropy than liquids, which have higher entropy than solids
An increase in total moles of gas increases entropy
Dissolving a crystalline solid increases entropy
Increasing temperature increases entropy

📐 Worked Example

Predict the sign of $Δ S_{sys}$ for each process and justify your answer: (a) Freezing of liquid water to form ice (b) Reaction: $2 SO_{2} (g) + O_{2} (g) \to 2 SO_{3} (g)$ (c) Dissolving solid potassium nitrate in pure water

1
For (a): Freezing converts liquid water to solid ice. Solids have a highly ordered molecular arrangement with fewer microstates than liquids, so disorder decreases.
2
$\Delta S_\text{sys} < 0 \text{ (negative)}$
3
For (b): Count moles of gas on each side: 3 moles of gaseous reactants form 2 moles of gaseous products. A decrease in total moles of gas reduces the number of available microstates, so disorder decreases.
4
$\Delta S_\text{sys} < 0 \text{ (negative)}$
5
For (c): Crystalline $KNO_{3}$ has an ordered, fixed structure; when dissolved, ions disperse throughout the solvent, increasing the number of available microstates. Disorder increases.
6
$\Delta S_\text{sys} > 0 \text{ (positive)}$

2. Calculating Standard Entropy of Reaction★★☆☆☆⏱ 3 min

The standard entropy change for a reaction $Δ S_{rxn}^{\circ}$ is calculated from tabulated standard molar entropies of reactants and products. A key difference from standard enthalpy of formation: unlike $Δ H_{f}^{\circ}$ , which is zero for elements in their standard state, $S^{\circ}$ is always positive for all substances (including elements) above 0 K. Never skip including $S^{\circ}$ for elements in your calculation. The formula is:

\Delta S^\circ_{\text{rxn}} = \sum nS^\circ_{\text{products}} - \sum mS^\circ_{\text{reactants}}

where $n$ and $m$ are the stoichiometric coefficients of products and reactants, respectively. $S^{\circ}$ is almost always reported in units of $J/(mol \cdotp K)$ , which is important for later Gibbs free energy calculations where enthalpy uses kJ units.

📐 Worked Example

Calculate $Δ S_{rxn}^{\circ}$ for the reaction $2 NO (g) + O_{2} (g) \to 2 NO_{2} (g)$ , given: $S^{\circ} (NO (g)) = 210.8 J/(mol \cdotp K)$ , $S^{\circ} (O_{2} (g)) = 205.2 J/(mol \cdotp K)$ , $S^{\circ} (NO_{2} (g)) = 240.1 J/(mol \cdotp K)$

1
Write the formula with stoichiometric coefficients substituted:
2
$\Delta S^\circ_{\text{rxn}} = \left[2 \times S^\circ(\text{NO}_2(g))\right] - \left[2 \times S^\circ(\text{NO}(g)) + 1 \times S^\circ(\text{O}_2(g))\right]$
3
Substitute the given values:
4
$\Delta S^\circ_{\text{rxn}} = (2 \times 240.1) - (2 \times 210.8 + 205.2) = 480.2 - 626.8 = -146.6 \text{ J/(mol·K)}$
5
Check the sign against prediction: 3 moles of gas react to form 2 moles of gas, so $Δ S$ should be negative, matching our result. Round to 3 significant figures:
6
$\Delta S^\circ_{\text{rxn}} = -147 \text{ J/(mol·K)}$

3. Gibbs Free Energy and Spontaneity★★★☆☆⏱ 4 min

At constant temperature and pressure (the conditions for most chemical reactions), Gibbs free energy change combines enthalpy and entropy into a single value that directly predicts spontaneity. The core formula is:

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

where $T$ is absolute temperature in Kelvin (always positive). The spontaneity rules are:

$Δ G < 0$ : Spontaneous in the forward direction
$Δ G = 0$ : Reaction is at equilibrium
$Δ G > 0$ : Non-spontaneous in the forward direction (spontaneous in reverse)

For standard state conditions, the formula becomes $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ . We can find the threshold temperature where a reaction switches spontaneity by setting $Δ G^{\circ} = 0$ , giving:

T = \frac{\Delta H^\circ}{\Delta S^\circ}

This is only useful when $Δ H$ and $Δ S$ have the same sign; if they have opposite signs, the reaction is always or never spontaneous regardless of temperature.

📐 Worked Example

For the decomposition of magnesium carbonate: $MgCO_{3} (s) \to MgO (s) + CO_{2} (g)$ , $Δ H^{\circ} = + 117.3 kJ/mol$ and $Δ S^{\circ} = + 174.8 J/(mol \cdotp K)$ . What temperature range is the reaction spontaneous?

1
Convert $Δ S^{\circ}$ to kJ to match the units of $Δ H^{\circ}$ :
2
$174.8 \text{ J/(mol·K)} = 0.1748 \text{ kJ/(mol·K)}$
3
The reaction has $Δ H^{\circ}$ positive and $Δ S^{\circ}$ positive, so it is spontaneous at high temperatures. Find the threshold temperature by setting $Δ G^{\circ} = 0$ :
4
$T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{117.3}{0.1748} \approx 671 \text{ K}$
5
Confirm: Above 671 K (≈ 398°C), $Δ G^{\circ} < 0$ , so the decomposition is spontaneous. Below 671 K, $Δ G^{\circ} > 0$ and the reaction is non-spontaneous.

4. Relationship Between ΔG° and Equilibrium Constant K★★★☆☆⏱ 3 min

Gibbs free energy connects thermodynamics to chemical equilibrium: the standard Gibbs free energy change is directly related to the equilibrium constant $K$ of a reaction by:

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

where $R = 8.314 J/(mol \cdotp K)$ (or $0.008314 kJ/(mol \cdotp K)$ ) and $T$ is absolute temperature. The key interpretations are:

If $Δ G^{\circ} < 0$ , $ln K > 0$ , so $K > 1$ : Products are favored at equilibrium
If $Δ G^{\circ} = 0$ , $ln K = 0$ , so $K = 1$ : Products and reactants are equally favored at equilibrium
If $Δ G^{\circ} > 0$ , $ln K < 0$ , so $K < 1$ : Reactants are favored at equilibrium

📐 Worked Example

At 298 K, $Δ G^{\circ}$ for the dissociation of acetic acid in water is +27.1 kJ/mol. Calculate $K_{a}$ for acetic acid at 298 K.

1
Rearrange the formula to solve for $ln K_{a}$ :
2
$\ln K_a = -\frac{\Delta G^\circ}{RT}$
3
Convert $Δ G^{\circ}$ to J to match the units of $R$ :
4
$+27.1 \text{ kJ/mol} = +27100 \text{ J/mol}$
5
Substitute the values:
6
$\ln K_a = -\frac{27100}{(8.314)(298)} \approx -10.94$
7
Exponentiate to get $K_{a}$ :
8
$K_a = e^{-10.94} \approx 1.8 \times 10^{-5}$

✓ Quick check

Test your understanding with this AP-style multiple choice question:

For which of the following reactions is $Δ S^{\circ}$ negative?
- A) $BaCl_{2} (s) \to Ba^{2 +} (a q) + 2 Cl^{-} (a q)$
- B) $2 C_{2} H_{6} (g) + 7 O_{2} (g) \to 4 CO_{2} (g) + 6 H_{2} O (g)$
- C) $CaCO_{3} (s) \to CaO (s) + CO_{2} (g)$
- D) $2 SO_{2} (g) + O_{2} (g) \to 2 SO_{3} (g)$
Reveal answer
D —
To find the sign of $Δ S^{\circ}$ , check the change in moles of gas: 3 moles of gas reactants become 2 moles of gas products, so entropy decreases, $Δ S^{\circ}$ is negative.

5. Common Pitfalls

Wrong move:

Forgetting to convert $Δ S$ from $J/(mol \cdotp K)$ to $kJ/(mol \cdotp K)$ before plugging into $Δ G = Δ H - T Δ S$

Why:

$Δ S$ is always reported in J per mol-K, while $Δ H$ is almost always reported in kJ per mol, leading to a 1000x error in $Δ G$

Correct move:

Write all units down for every value, and convert units so $Δ H$ and $Δ S$ match before doing any calculation

Wrong move:

Treating $S^{\circ}$ of elements in standard state as zero, following the $Δ H_{f}^{\circ}$ convention

Why:

Students confuse the convention for enthalpy of formation with entropy, which is non-zero for all substances above 0 K

Correct move:

Always include the full $S^{\circ}$ value of elements in your $Δ S_{rxn}^{\circ}$ calculation, never assume it is zero

Wrong move:

Predicting $Δ S$ sign based on total moles of all species instead of moles of gas

Why:

Solids and liquids have negligible entropy compared to gases, so total moles can give the wrong sign

Correct move:

Always calculate the change in moles of gas first to get the sign of $Δ S$

Wrong move:

Concluding that a reaction with $Δ G < 0$ will happen quickly

Why:

Students confuse thermodynamic spontaneity with reaction kinetics

Correct move:

Remember that $Δ G$ only tells you if a reaction is thermodynamically favorable, not how fast it will proceed

Wrong move:

Assuming a reaction with $Δ G^{\circ} > 0$ can never produce products

Why:

Students confuse standard state $Δ G^{\circ}$ with non-standard $Δ G$ , which depends on reactant/product concentrations

Correct move:

$Δ G^{\circ} > 0$ only means $K < 1$ , so reactants are favored at equilibrium, but products can still form if you start with pure reactants

6. Quick Reference Cheatsheet

Category	Formula	Notes
Standard Reaction Entropy Change	$Δ S_{rxn}^{\circ} = \sum n S_{products}^{\circ} - \sum m S_{reactants}^{\circ}$	$S^{\circ}$ is always positive above 0 K; units = J/(mol·K)
Gibbs Free Energy Change	$Δ G = Δ H - T Δ S$	$T$ = absolute temperature (Kelvin); $Δ H$ and $Δ S$ units must match
Spontaneity Criteria (constant T,P)	$Δ G < 0$ : spontaneous forward; $Δ G = 0$ : equilibrium; $Δ G > 0$ : non-spontaneous forward	No information about reaction rate, only thermodynamic favorability
Spontaneity Threshold Temperature	$T = \frac{Δ H ^{\circ}}{Δ S ^{\circ}}$	Only used when $Δ H$ and $Δ S$ have the same sign
ΔG° and Equilibrium Constant	$Δ G^{\circ} = - R T ln K$	$R = 8.314$ J/(mol·K) = 0.008314 kJ/(mol·K); match units to ΔG°
ΔG°-K Relationship	$Δ G^{\circ} < 0 \to K > 1$ ; $Δ G^{\circ} > 0 \to K < 1$	ΔG° applies only to standard state conditions
Second Law of Thermodynamics	$Δ S_{univ} = Δ S_{sys} + Δ S_{surr} > 0$ (spontaneous)	Total entropy of the universe always increases for spontaneous processes

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 · MCQ
Predict entropy change sign
2022 · FRQ
Calculate ΔG and equilibrium K

What's Next

Entropy and Gibbs free energy form the foundation for connecting thermodynamics to equilibrium and other core AP Chemistry topics. Mastery of sign conventions, unit conversions, and the core relationships between ΔG, ΔH, ΔS, and K is required for downstream topics including solubility equilibria, electrochemistry, and non-standard free energy changes. This subtopic also unites the two major pillars of AP Chemistry: thermodynamics and equilibrium, showing that equilibrium is a natural consequence of the second law of thermodynamics.