In the last section, we introduced the idea of entropy as a measure of disorder in a system. Unlike enthalpy, which was discussed in unit 5, entropy can be measured both absolutely and as a change. This differs from enthalpy, which can only be represented as a change (we can never find H° for a reaction but we can find ΔH°).
Comparing S° and ΔS°
Entropy can be thought of both in its absolute form (S°) and as representing a change in entropy (ΔS°) (note that the ° symbol in each simply means that we are at standard conditions, that is 1atm of pressure and 273K). Absolute entropies, also known as standard entropies, describe the number of possible states a molecule can take, a measure of its disorder. Calculating these values is incredibly complex, but you will be given any absolute entropies you need on the exam. Similarly, most if not all chemistry textbooks have tables of thermodynamic data in the back of the book with a chart of standard entropies. These tables may also contain other data such as enthalpies of formation like the one below. For now, lets look at the far right column that says S°.
Change in entropy (ΔS°) can be thought of the same way we thought about changes in enthalpy (ΔH°) in unit 6 just thinking about entropy instead of heat changes. In Unit 6, we learned about Hess’s Law, which told us that enthalpy was a state function, meaning enthalpy changes are pathway independent. The same is true about entropy.
State Functions
Let’s take a quick side note to explain what a state function means and why it’s important. In essence, a state function is a function that has the property of pathway independence. This means that whatever “path” you take to get to the end result, the end result will be exactly the same. An example of a state function is the change in altitude when climbing a mountain, whereas an example of a non-state function is the distance traveled. Since no matter which way you go, you will end up at the top of the mountain, your change in altitude will always be the altitude. Whether you went straight up, zigzagged, went curvily, or any other pathway, your change in altitude will be the same at the end. This makes this function pathway independent. On the other hand, the distance you travel up the mountain does depend on the path you take. If you go straight up, the distance will differ from if you zigzag and for any other path you take. Therefore, the distance traveled is not a state function and is not pathway independent.
Applying the idea of a state function to entropy, we can logically find the following equation (note that it is strikingly similar to an equivalent statement about enthalpy change):
This equation tells us that for a reaction the overall entropy change is the sum of the changes in the entropies of the products minus the sum of the entropies in the reactants. This can be applied the same way we did in unit 5 with enthalpy, so you can simply use the table and plug in numbers for each product/reactant.
🔥 Pro Tip: Remember your stoichiometric coefficients! The real equation has you multiplying by the stoichiometric coefficient of each product/reactant but the CollegeBoard omits it from the formula for some unknown reason.
Interpreting The Sign of ΔS°
Another important way of looking at entropy is by observing the sign of ΔS° for a reaction and drawing conclusions as to whether or not the reaction became more or less ordered and vice versa—predicting the sign of ΔS° by looking at properties of a reaction. When ΔS° is positive, a reaction creates disorder and thus increases the entropy of the system. Conversely, when ΔS° is negative, the reaction reduces disorder or in other words creates order. Therefore the entropy of the system decreases.
By using these rules we can also predict the opposite: the sign of ΔS° of a reaction given the reaction. We’ll take a look at a few of these in the practice problems.
Practice Problems
In this problem, we’re given a reaction along with the entropies of various substances at 298K. For this problem we can simply use our equation for S° to find the ΔS° for the reaction:
= ((175) + 2(150)) - (250 + 2(125)) = -25 J/molrxnK ⇒ A.
Note that we multiplied 150 and 125 by 2 because the stoichiometric coefficients on KI and KNO3 are 2.
2. Consider the reaction: 2Na (s) + Cl2 (g) → 2NaCl (s). Predict the sign of ΔS° for this reaction and explain
Looking at this reaction, we see two reactants, sodium and chlorine, forming one solid product, sodium chloride. Let’s think about the disorder of the reactants versus the disorder of the products. On the reactants side we have 3 total moles (2 + 1 = 3) of reactants, one of which is a gas. These react to form fewer moles of a solid (less chaotic) product. Therefore, our reaction is more chaotic at the beginning and becomes less chaotic. This means in the process we “lose” entropy (we don’t actually destroy entropy because of the first law of thermodynamics, but we can think of it this way). Therefore, we can predict that ΔS° for this reaction is negative.
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| absolute entropies | The total entropy of a substance at a given temperature and pressure, measured relative to zero entropy at absolute zero. |
| entropy change | The difference in entropy between the final and initial states of a system during a chemical or physical process. |
| standard entropy change | The change in entropy for a chemical or physical process under standard conditions, calculated as the difference between the absolute entropies of products and reactants. |
| standard molar entropies | The absolute entropy of one mole of a substance under standard conditions, typically used to calculate entropy changes in reactions. |
Frequently Asked Questions
What is absolute entropy and how is it different from regular entropy?
Absolute entropy (standard molar entropy, S°) is the measured entropy of a pure substance in its standard state (1 bar) at a given temperature—a real number you can find in thermodynamic tables. It’s grounded in the Third Law of Thermodynamics: a perfect crystal at 0 K has S° = 0, and S° for other substances comes from counting microstates (statistical ideas like the Sackur–Tetrode equation for ideal gases) and experimental measurements. “Regular” entropy often refers to an entropy change, ΔS, for a process. For AP problems you’ll use absolute entropies to get the standard entropy change for reactions: ΔS°reaction = ΣS°(products) − ΣS°(reactants). That’s an AP CED skill (9.2.A). For a focused review see the Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG) and more unit resources (https://library.fiveable.me/ap-chemistry/unit-9). Need practice? Try the practice set (https://library.fiveable.me/practice/ap-chemistry).
How do I calculate entropy change for a reaction using the formula?
Use the formula from the CED: ΔS°reaction = ΣS°(products) − ΣS°(reactants). Steps: 1. Get each species’ standard molar entropy S° (usually in J·mol⁻¹·K⁻¹) from a thermodynamic table (standard state = 1 bar). 2. Multiply each S° by its stoichiometric coefficient. 3. Sum the S° values for products, sum for reactants, then subtract: ΔS° = (sum products) − (sum reactants). 4. Report units as J·mol⁻¹·K⁻¹ and keep proper significant figures. Example: for A + 2B → C, ΔS° = S°(C) − [S°(A) + 2·S°(B)]. On the AP exam you’ll often calculate standard ΔS° this way (CED 9.2.A.1). For help finding S° values and extra practice, see the Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG) and the practice problem set (https://library.fiveable.me/practice/ap-chemistry).
Why do we use absolute entropies instead of just regular entropy values?
We use absolute (standard molar) entropies because entropy has a true zero (Third Law) at 0 K for a perfect crystal, so S° values are meaningful as absolute quantities and can be tabulated. That lets you compute ΔS° directly with ΔS°reaction = ΣS°products − ΣS°reactants (CED 9.2.A.1). If we only used relative or arbitrary "entropy changes," you couldn't add/subtract values from different species or phases consistently. Absolute S° values incorporate all microscopic contributions (translational, rotational, vibrational, electronic and any residual disorder), so they reflect real differences between reactants and products—essential for AP problems that ask you to calculate ΔS° and then use ΔG° = ΔH° − TΔS° (see Topic 9.3). For practice, review the Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG) and try problems from the Unit 9 page (https://library.fiveable.me/ap-chemistry/unit-9) or the practice set (https://library.fiveable.me/practice/ap-chemistry).
What's the difference between ΔS°reaction and just ΔS?
ΔS°reaction means the entropy change for a reaction when every species is in its standard state (usually 1 bar for gases) and is calculated from standard molar (absolute) entropies: ΔS°reaction = ΣS°(products) − ΣS°(reactants) (this is exactly what the CED gives). ΔS (no °) is the actual entropy change for the process under the specific conditions you have—different pressures, concentrations, or nonstandard states—and so can differ from ΔS°. Remember entropy is a state function, so ΔS depends only on initial and final states, but those states might not be “standard.” For example for an ideal gas: S = S° − R ln(P/P°), so if pressures aren’t 1 bar you’ll get ΔS ≠ ΔS°. On the AP exam you’ll usually compute ΔS° from S° tables (use ΔS° = ΣS°products − ΣS°reactants)—practice that on the Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG). For extra problems, try the AP practice set (https://library.fiveable.me/practice/ap-chemistry).
I'm confused about the entropy change equation - do I subtract reactants from products or products from reactants?
You subtract reactants from products: ΔS°reaction = ΣS°(products) − ΣS°(reactants). That means add up the standard molar entropies (S°) of each product multiplied by its coefficient, add up the S° for all reactants (with their coefficients), then take product sum minus reactant sum. Watch units (J·mol−1·K−1) and include physical states—tables list S° for solids, liquids, gases separately. Quick tip: if ΔS° is positive, entropy increases (more disorder); if negative, entropy decreases. This exact formula is listed in the CED for Topic 9.2 and is what you should use on the AP exam when you’re asked to calculate standard entropy change. For a refresher and worked examples, see the Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG) and try practice problems at (https://library.fiveable.me/practice/ap-chemistry).
How do I find the standard molar entropy values for different compounds?
Look up S° (standard molar entropy) values in thermodynamic tables (chemistry textbooks, the appendix on the AP exam formula sheet, or reliable online tables). Standard S° units are J·mol⁻¹·K⁻¹ and refer to the standard state (1 bar). To calculate a reaction’s standard entropy change use the CED equation: ΔS°rxn = ΣS°(products) − ΣS°(reactants). Make sure to multiply each S° by its stoichiometric coefficient. Use the Third Law idea: S° values are absolute (zero for a perfect crystal at 0 K), so tabulated S° are experimentally determined or from statistical mechanics. On the AP exam you’ll be expected to perform these ΣS° calculations (Topic 9.2), so practice extracting values and computing ΔS° on free-response and multiple-choice problems. For targeted review, see the Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why does entropy change matter in chemical reactions?
Entropy change matters because it tells you how the disorder (number of microstates) of a system changes during a reaction—and that affects whether a process is thermodynamically favored. Use the AP formula ΔS°reaction = ΣS°products − ΣS°reactants to calculate standard entropy change from tabulated S° values (units: J·mol−1·K−1). Positive ΔS° often comes from creating more gas, melting, vaporization, or mixing; negative ΔS° happens when gas is consumed or a gas condenses. ΔS° combines with ΔH° in ΔG° = ΔH° − TΔS° to determine spontaneity on the exam (Topic 9.3 linkage). So even if a reaction is exothermic, a large negative ΔS° at high T can make it nonspontaneous. Practice calculating ΔS° and relating it to ΔG° and equilibrium for AP free-response problems. For a focused review and examples, see the Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).
What does the S° symbol mean and why is there a degree sign?
S° means the standard molar (absolute) entropy of a substance—the entropy of one mole measured in the standard state. The little degree symbol (°) indicates “standard conditions” (by convention for entropy, the standard state is 1 bar pressure for gases/solutions). So S° for H2O(l) is the entropy of 1 mol of liquid water at the chosen standard state and a specified temperature (usually 298.15 K on tables). Units are J·mol⁻¹·K⁻¹. You use S° values in the AP equation ΔS°reaction = ΣS°(products) − ΣS°(reactants) (CED 9.2.A.1) to calculate standard entropy changes. The concept ties to absolute entropy from the Third Law (entropy of a perfect crystal at 0 K = 0), which lets tabulated S° values be meaningful. For more review and examples see the Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG) and the Unit 9 overview (https://library.fiveable.me/ap-chemistry/unit-9). For extra practice, check Fiveable’s AP Chem practice questions (https://library.fiveable.me/practice/ap-chemistry).
Can someone explain step by step how to calculate ΔS°reaction for a combustion reaction?
Step-by-step (quick)—use ΔS°rxn = ΣS°(products) − ΣS°(reactants) (CED 9.2.A.1): 1. Write the balanced combustion equation (include physical states). Example form: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l). 2. Look up standard molar entropies S° (units J·mol−1·K−1) from a thermodynamic table for each species (gas vs. liquid matters). 3. Multiply each S° by its stoichiometric coefficient. - ΣS°products = 1·S°(CO2,g) + 2·S°(H2O,l) - ΣS°reactants = 1·S°(CH4,g) + 2·S°(O2,g) 4. Compute ΔS°rxn = ΣS°products − ΣS°reactants. Keep units J·mol−1·K−1. If table gives kJ·mol−1·K−1, convert to J. 5. Interpret sign: positive ΔS° → net increase in entropy; negative → decrease. Report to proper sig figs. On the exam you must show balanced equation, S° values with units, the algebraic sum, and the final ΔS° with units (AP free-response expects those steps). For a guided walkthrough, see the Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG) and unit resources (https://library.fiveable.me/ap-chemistry/unit-9). For extra practice problems, use Fiveable’s practice set (https://library.fiveable.me/practice/ap-chemistry).
I don't understand why we sum up all the products and reactants separately in the entropy equation.
Think of S° as an absolute, per-mole measure of how many microstates (disorder) each substance has. Entropy is an extensive property, so for a reaction you add up the total entropy contributed by every mole of product and subtract the total entropy of every mole of reactant—that gives the net change in entropy for the process. Mathematically: ΔS°reaction = Σ (ν·S°)products − Σ (ν·S°)reactants, where ν are stoichiometric coefficients. Why subtraction? Because you’re comparing “after” minus “before” (final minus initial), the same idea used for ΔH° or ΔG°. Use standard molar (absolute) entropies from tables, multiply by coefficients, then sum. That’s exactly what the CED expects you to be able to calculate (9.2.A). For a refresher and worked examples, see the Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
What happens to entropy when a solid melts into a liquid?
When a solid melts to a liquid, entropy increases. Melting (fusion) converts a more ordered phase (solid) to a less-ordered phase (liquid), so the number of accessible microstates rises and molecular freedom (translational/rotational motion) increases. On the AP Chem CED this shows up as a positive entropy of fusion: ΔS°fusion = S°(liquid) − S°(solid) > 0. You can calculate such changes using standard molar entropies and the formula ΔS°reaction = ΣS°products − ΣS°reactants (CED 9.2.A.1). Conceptually this follows the Third Law idea that entropy measures the multiplicity of microstates (Boltzmann). For practice calculating ΔS° values and seeing fusion examples, check the Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do I know if an entropy change is favorable or unfavorable for a reaction?
Look at ΔS°reaction = ΣS°products − ΣS°reactants. If ΔS° > 0, entropy increased (more microstates)—that change is “entropy-favorable.” If ΔS° < 0, entropy decreased and that change is “entropy-unfavorable.” On the AP you’ll often spot trends without numbers: forming more gas molecules, vaporization, or mixing → ΔS° positive; freezing, condensation, or forming fewer gas moles → ΔS° negative. Topic 9.2 expects you to calculate ΔS° from tabulated S° values (CED eqn and standard state 1 bar). Important: entropy alone doesn’t guarantee spontaneity—use Gibbs free energy ΔG° = ΔH° − TΔS°. A positive ΔS° helps make ΔG° more negative (more favorable) at higher T if ΔS° is positive. For full thermodynamic favorability, combine ΔH, ΔS, and T (see Topic 9.3 on Gibbs; Fiveable unit overview (https://library.fiveable.me/ap-chemistry/unit-9) and this Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG)). For practice, try problems at (https://library.fiveable.me/practice/ap-chemistry).
Why do gases usually have higher absolute entropy values than solids?
Gases usually have higher absolute entropies than solids because their particles can access far more microstates (ways to arrange energy and position). In a gas molecules have large translational freedom (they can move throughout the volume), plus accessible rotational and often vibrational states; that huge positional and energetic freedom raises S (Sackur–Tetrode shows S for an ideal monatomic gas depends on ln(V) and T). Solids lock atoms into fixed lattice sites, so positional options and disorder are much smaller (fewer microstates). By the Third Law we assign zero entropy to a perfect crystal at 0 K, and measured standard molar entropies S° (standard state 1 bar) reflect these differences—gases at 298 K typically have S° values much larger than corresponding solids. On the AP exam you’ll use these tabulated S° values to calculate ΔS°reaction = ΣS°products − ΣS°reactants (CED 9.2.A.1). For more review see the Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG) and lots of practice problems (https://library.fiveable.me/practice/ap-chemistry).
What's the connection between entropy change and spontaneous reactions?
Entropy change (ΔS) tells you how disorder or the number of microstates changes during a process. For calculations on the exam use the CED equation ΔS°reaction = ΣS°products − ΣS°reactants (use standard molar entropies from tables). A positive ΔS (more disorder) tends to make a process more likely to occur, but spontaneity isn’t decided by ΔS alone—the AP CED expects you to connect entropy with Gibbs free energy: ΔG = ΔH − TΔS. A reaction is spontaneous when ΔG < 0. So: - If ΔS > 0 it helps make ΔG more negative (favors spontaneity), especially at high T. - If ΔS < 0 it works against spontaneity (can be overcome by a sufficiently negative ΔH). Remember to consider sign and temperature together on free-response problems (they often ask you to justify spontaneity using ΔG). For more Topic 9.2 review see the Fiveable study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG) and practice questions (https://library.fiveable.me/practice/ap-chemistry).
I missed the lab on entropy - how do we actually measure absolute entropy in real life?
Absolute (standard molar) entropies S° aren't measured in one direct shot—they’re built up experimentally using the Third Law: S = 0 for a perfect crystal at 0 K. In practice you measure heat capacity Cp(T) from low T up to 298 K (calorimetry) and integrate S(T) = ∫0→T Cp/T dT, adding ΔS for phase changes (fusion, vaporization) at their T (ΔS = ΔH/T). For some solids there’s residual entropy (frozen disorder) you add separately. For gases, statistical formulas like the Sackur–Tetrode equation (Boltzmann’s approach) can give absolute S for ideal monoatomic gases. Experimental S° values are compiled in thermodynamic tables and those standard S° values are what you use on the AP to compute ΔS°reaction = ΣS°products − ΣS°reactants. If you missed the lab, review the Topic 9.2 study guide (https://library.fiveable.me/ap-chemistry/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG) and practice problems (https://library.fiveable.me/practice/ap-chemistry).