# Entropy and Gibbs Free Energy

> AP Chemistry · Applications of Thermodynamics
> Source: https://www.owlsprep.com/study/ap-chemistry-u9-entropy-and-gibbs-free-energy/

This subtopic covers entropy change prediction, standard entropy calculations, Gibbs free energy and spontaneity, temperature dependence of reaction favorability, and the relationship between ΔG° and the equilibrium constant K for AP Chemistry.

**Prerequisites:** Enthalpy change calculation from standard enthalpies of formation; Absolute temperature measurement in Kelvin; Equilibrium constant definition and interpretation

## Learning objectives

- Predict the sign of entropy change for chemical and physical processes
- Calculate standard entropy change of reaction from standard molar entropies
- Calculate Gibbs free energy change from enthalpy and entropy changes
- Determine the spontaneity of a reaction based on ΔG values
- Relate standard Gibbs free energy change to the equilibrium constant K

## Entropy, Second Law, and Predicting Entropy Changes

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.

**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 $\Delta S_\text{sys}$, 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 $\Delta S_\text{sys}$ for each process and justify your answer: (a) Freezing of liquid water to form ice (b) Reaction: $2 \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \text{SO}_3(g)$ (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 $\text{KNO}_3$ has an ordered, fixed structure; when dissolved, ions disperse throughout the solvent, increasing the number of available microstates. Disorder increases.
6. $$\Delta S_\text{sys} > 0 \text{ (positive)}$$

> **tip**
>
> Always check the change in moles of gas first when predicting $\Delta S$ sign. Changes in gas moles almost always dominate over changes in solids or liquids, so this will give you the correct sign 90% of the time on exam questions.

## Calculating Standard Entropy of Reaction

The standard entropy change for a reaction $\Delta S^\circ_{\text{rxn}}$ is calculated from tabulated standard molar entropies of reactants and products. A key difference from standard enthalpy of formation: unlike $\Delta H_f^\circ$, 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 $\text{J/(mol·K)}$, which is important for later Gibbs free energy calculations where enthalpy uses kJ units.

**Worked example:** Calculate $\Delta S^\circ_{\text{rxn}}$ for the reaction $2 \text{NO}(g) + \text{O}_2(g) \rightarrow 2 \text{NO}_2(g)$, given: $S^\circ(\text{NO}(g)) = 210.8 \text{ J/(mol·K)}$, $S^\circ(\text{O}_2(g)) = 205.2 \text{ J/(mol·K)}$, $S^\circ(\text{NO}_2(g)) = 240.1 \text{ J/(mol·K)}$

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 $\Delta S$ should be negative, matching our result. Round to 3 significant figures:
6. $$\Delta S^\circ_{\text{rxn}} = -147 \text{ J/(mol·K)}$$

> **tip**
>
> Always write down units for every value when doing entropy calculations. This will help you catch unit mismatch errors when you later calculate $\Delta G$.

## Gibbs Free Energy and Spontaneity

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:

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

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

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

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

> **tip**
>
> Memorize the four sign combinations: (1) ΔH -, ΔS + = always spontaneous; (2) ΔH +, ΔS - = never spontaneous; (3) ΔH -, ΔS - = spontaneous at low T; (4) ΔH +, ΔS + = spontaneous at high T. This is a very common MCQ question, so memorizing it saves time.

**Worked example:** For the decomposition of magnesium carbonate: $\text{MgCO}_3(s) \rightarrow \text{MgO}(s) + \text{CO}_2(g)$, $\Delta H^\circ = +117.3 \text{ kJ/mol}$ and $\Delta S^\circ = +174.8 \text{ J/(mol·K)}$. What temperature range is the reaction spontaneous?

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

## Relationship Between ΔG° and Equilibrium Constant K

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

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

> **tip**
>
> Always confirm that $R$ units match $\Delta G^\circ$ units: if $\Delta G^\circ$ is in kJ, use $R = 0.008314 \text{ kJ/(mol·K)}$ to avoid 1000x magnitude errors in $K$.

**Worked example:** At 298 K, $\Delta G^\circ$ 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 $\Delta G^\circ$ 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. $$K_a = e^{-10.94} \approx 1.8 \times 10^{-5}$$

**Check your understanding**

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

1. For which of the following reactions is $\Delta S^\circ$ negative?

   - A) $\text{BaCl}_2(s) \rightarrow \text{Ba}^{2+}(aq) + 2\text{Cl}^-(aq)$
   - B) $2\text{C}_2\text{H}_6(g) + 7\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 6\text{H}_2\text{O}(g)$
   - C) $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$
   - D) $2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g)$

   *Why:* To find the sign of $\Delta S^\circ$, check the change in moles of gas: 3 moles of gas reactants become 2 moles of gas products, so entropy decreases, $\Delta S^\circ$ is negative.

## Common pitfalls

- **Wrong:** Forgetting to convert $\Delta S$ from $\text{J/(mol·K)}$ to $\text{kJ/(mol·K)}$ before plugging into $\Delta G = \Delta H - T\Delta S$
  - Why it fails: $\Delta S$ is always reported in J per mol-K, while $\Delta H$ is almost always reported in kJ per mol, leading to a 1000x error in $\Delta G$
  - Correct: Write all units down for every value, and convert units so $\Delta H$ and $\Delta S$ match before doing any calculation
- **Wrong:** Treating $S^\circ$ of elements in standard state as zero, following the $\Delta H_f^\circ$ convention
  - Why it fails: Students confuse the convention for enthalpy of formation with entropy, which is non-zero for all substances above 0 K
  - Correct: Always include the full $S^\circ$ value of elements in your $\Delta S^\circ_{\text{rxn}}$ calculation, never assume it is zero
- **Wrong:** Predicting $\Delta S$ sign based on total moles of all species instead of moles of gas
  - Why it fails: Solids and liquids have negligible entropy compared to gases, so total moles can give the wrong sign
  - Correct: Always calculate the change in moles of gas first to get the sign of $\Delta S$
- **Wrong:** Concluding that a reaction with $\Delta G < 0$ will happen quickly
  - Why it fails: Students confuse thermodynamic spontaneity with reaction kinetics
  - Correct: Remember that $\Delta G$ only tells you if a reaction is thermodynamically favorable, not how fast it will proceed
- **Wrong:** Assuming a reaction with $\Delta G^\circ > 0$ can never produce products
  - Why it fails: Students confuse standard state $\Delta G^\circ$ with non-standard $\Delta G$, which depends on reactant/product concentrations
  - Correct: $\Delta G^\circ > 0$ only means $K < 1$, so reactants are favored at equilibrium, but products can still form if you start with pure reactants

## Cheatsheet

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

## 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.

- [Absolute entropy and the second law of thermodynamics](https://www.owlsprep.com/study/ap-chemistry-u9-absolute-entropy-and-the-second/)
- [Gibbs Free Energy and Thermodynamic Favorability](https://www.owlsprep.com/study/ap-chemistry-u9-gibbs-free-energy-and-thermodynamic/)
- [Thermodynamic favorability versus rate](https://www.owlsprep.com/study/ap-chemistry-u9-thermodynamic-favorability-versus-rate/)

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