AP Chemistry Quiz: Absolute Entropy And Entropy Change

20 questions · exam conditions

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Absolute Entropy And Entropy ChangeQuestion 1 of 20

Solid ammonium chloride is heated in a closed container and decomposes according to $\text{NH}_4\text{Cl}(s) \rightarrow \text{NH}_3(g) + \text{HCl}(g)$ . After heating, only gases are present. Which statement best describes the sign of $\Delta S$ for the system?

\Delta S

is negative because a solid disappears.

\Delta S

is approximately zero because the container is closed.

\Delta S

is positive because the process produces gaseous particles from a solid, increasing dispersal of matter.

\Delta S

is negative because decomposition reactions always decrease entropy.

\Delta S

is positive only if the reaction is endothermic.

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AP Chemistry Quiz

AP Chemistry Quiz: Absolute Entropy And Entropy Change

Practice Absolute Entropy And Entropy Change in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Absolute Entropy And Entropy Change, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$\Delta S$ is negative because a solid disappears.
$\Delta S$ is approximately zero because the container is closed.
$\Delta S$ is positive because the process produces gaseous particles from a solid, increasing dispersal of matter. (correct answer)
$\Delta S$ is negative because decomposition reactions always decrease entropy.
$\Delta S$ is positive only if the reaction is endothermic.

Explanation: This question assesses understanding of absolute entropy and entropy change. The decomposition of solid NH4Cl to gaseous NH3 and HCl involves a phase change from solid to gas, greatly increasing entropy. The number of particles increases as one solid unit produces two gas molecules, enhancing dispersal. This transition to gases allows for more freedom of motion and microstates, resulting in a positive entropy change. A tempting distractor is choice D, which says ΔS is negative because decomposition reactions always decrease entropy, stemming from the misconception that reaction type overrides phase and particle effects. To evaluate entropy changes in decomposition to gases, remember that entropy increases when matter or energy becomes more dispersed.

Question 2

In a sealed syringe, a fixed amount of Ar(g) is compressed by pushing the plunger inward at constant temperature, changing from a larger volume to a smaller volume. Which statement best describes the sign of $\Delta S$ for the gas in the syringe?

$\Delta S$ is positive because work is done on the gas.
$\Delta S$ is approximately zero because argon is monatomic.
$\Delta S$ is negative because the gas has fewer accessible positions (microstates) in a smaller volume. (correct answer)
$\Delta S$ is positive because the temperature is constant.
$\Delta S$ is negative only if the compression releases heat.

Explanation: This question assesses understanding of absolute entropy and entropy change. Compressing Ar gas at constant temperature reduces the volume without phase change, decreasing entropy. The number of particles stays the same, but dispersal is limited in the smaller volume, reducing accessible positions. This confinement leads to fewer microstates and a negative entropy change. A tempting distractor is choice A, which states ΔS is positive because work is done on the gas, confusing energy input with entropy increase regardless of volume effects. To evaluate entropy changes in compression, remember that entropy increases when matter or energy becomes more dispersed.

Question 3

A rigid container initially holds 1.0 mol of O $_2$ (g) at a uniform temperature. An electric spark converts it completely to ozone according to $3\text{O}_2(g) \rightarrow 2\text{O}_3(g)$ , with temperature returning to the original value. Which statement best describes the sign of $\Delta S$ for the system?

$\Delta S$ is positive because ozone has a higher molar mass than oxygen.
$\Delta S$ is negative because the number of moles of gas decreases from 3 to 2, reducing the number of accessible microstates. (correct answer)
$\Delta S$ is approximately zero because both reactant and product are gases.
$\Delta S$ is positive because forming a new substance increases disorder.
$\Delta S$ is negative because the reaction requires an electric spark.

Explanation: This question assesses understanding of absolute entropy and entropy change. The reaction converting 3 O2 to 2 O3 decreases the number of gas particles without phase change, reducing entropy. Fewer particles mean less dispersal and fewer microstates in the same volume. This reduction results in a negative entropy change for the system. A tempting distractor is choice D, which says ΔS is positive because forming a new substance increases disorder, based on the misconception that chemical change inherently increases entropy over particle count. To evaluate entropy changes in gas reactions, remember that entropy increases when matter or energy becomes more dispersed.

Question 4

Consider two processes occurring separately at the same temperature: Process 1: CO $_2$ (s) $\rightarrow$ CO $_2$ (g) (dry ice sublimation) Process 2: H $_2$ O(l) $\rightarrow$ H $_2$ O(s) (freezing) Which statement correctly compares the entropy changes $\Delta S_1$ and $\Delta S_2$ for the systems?

$\Delta S_1$ and $\Delta S_2$ are both approximately zero.
$\Delta S_1$ is positive and $\Delta S_2$ is negative. (correct answer)
$\Delta S_1$ is negative and $\Delta S_2$ is positive.
$\Delta S_1$ and $\Delta S_2$ are both positive.
$\Delta S_1$ and $\Delta S_2$ are both negative.

Explanation: This question assesses understanding of absolute entropy and entropy change. For Process 1, sublimation from solid to gas increases entropy due to greater particle dispersal in the gas phase. For Process 2, freezing from liquid to solid decreases entropy as particles become more ordered with less freedom. The changes in phase directly affect the number of microstates, with gas having more than solid. A tempting distractor is choice B, which says both are negative, based on the misconception that all phase changes to denser states decrease entropy without distinguishing directions. To compare entropy changes in phase transitions, remember that entropy increases when matter or energy becomes more dispersed.

Question 5

A sealed container initially holds a sample of liquid bromine, Br $_2$ (l), at room temperature. The container is warmed gently until all of the bromine becomes Br $_2$ (g), with no change in the amount of substance. Which statement best describes the sign of $\Delta S$ for this process?

$\Delta S$ is negative because energy is absorbed during vaporization.
$\Delta S$ is approximately zero because the chemical identity of Br $_2$ does not change.
$\Delta S$ is positive because particles have greater freedom of motion in the gas phase than in the liquid phase. (correct answer)
$\Delta S$ is negative because gases are more ordered than liquids.
$\Delta S$ is positive only if the temperature increases; otherwise it must be zero.

Explanation: This question assesses understanding of absolute entropy and entropy change. In the process of vaporizing liquid bromine to gas, the phase change from liquid to gas significantly increases entropy because particles in the gas phase have greater freedom of motion and can occupy more positions. The number of particles remains the same, but their dispersal increases as they transition from being closely packed in the liquid to spreading out in the gas phase. This greater dispersal leads to a higher number of accessible microstates, resulting in a positive entropy change. A tempting distractor is choice D, which incorrectly states that ΔS is negative because gases are more ordered than liquids, stemming from the misconception that order is confused with particle density rather than freedom of movement. To evaluate entropy changes in phase transitions, remember that entropy increases when matter or energy becomes more dispersed.

Question 6

At constant temperature, a sample of water vapor is cooled until it condenses completely: $\text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{O}(\ell)$ . What is the sign of $\Delta S$ for this process?

$\Delta S$ is approximately zero because no new substances are formed.
$\Delta S$ is approximately zero because temperature is constant.
$\Delta S$ is positive because heat is released during condensation.
$\Delta S$ is negative because the system becomes more ordered when a gas becomes a liquid. (correct answer)
$\Delta S$ is positive because the molecules move closer together.

Explanation: This question tests understanding of absolute entropy and entropy change. When water vapor condenses to liquid water, gas molecules with high freedom of motion become confined to a more ordered liquid state. The number of accessible microstates decreases significantly as molecules lose translational freedom and become more closely packed. Therefore, ΔS is negative because the system becomes more ordered. Choice D incorrectly associates heat release with positive entropy change—condensation is exothermic, but the heat flow direction doesn't determine entropy's sign. The key strategy is that entropy decreases when matter becomes less dispersed: gas → liquid → solid.

Question 7

At constant temperature, equal volumes of $\text{He}(g)$ and $\text{Ne}(g)$ are released into the same evacuated rigid container and allowed to mix, forming a uniform mixture of the two gases. What is the sign of $\Delta S$ for the mixing process (considering the gases as the system)?

$\Delta S$ is negative because the gases collide with each other.
$\Delta S$ is approximately zero because the temperature is constant.
$\Delta S$ is negative because the container is rigid.
$\Delta S$ is positive because mixing increases the number of possible arrangements of particles. (correct answer)
$\Delta S$ is approximately zero because the total number of moles of gas is unchanged.

Explanation: This question tests understanding of absolute entropy and entropy change. When helium and neon gases mix, the particles of each gas spread throughout the entire container volume, creating many more possible arrangements than when each gas was separate. The number of accessible microstates increases dramatically because each type of atom can now occupy any position in the container. Therefore, ΔS is positive for the mixing process. Choice C incorrectly assumes entropy depends only on the total moles—entropy actually increases due to the increased ways to arrange two different types of particles. Remember that mixing always increases entropy because it increases the number of possible particle arrangements.

Question 8

Consider the reaction occurring in a rigid, sealed vessel at constant temperature: $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$ . Based on the change in the number of gas particles, what is the sign of $\Delta S$ for the system?

$\Delta S$ is negative because the number of moles of gas decreases. (correct answer)
$\Delta S$ is positive because a new compound is formed.
$\Delta S$ is approximately zero because the vessel is rigid.
$\Delta S$ is positive because gases always have high entropy.
$\Delta S$ is negative only if the reaction is exothermic.

Explanation: This question tests understanding of absolute entropy and entropy change. In the reaction N₂(g) + 3H₂(g) → 2NH₃(g), we start with 4 moles of gas particles (1 N₂ + 3 H₂) and end with only 2 moles of gas particles (2 NH₃). When the number of gas particles decreases, the system has fewer ways to distribute energy and matter, resulting in fewer accessible microstates. Since entropy is a measure of the number of accessible microstates, reducing the particle count from 4 to 2 causes a decrease in entropy, making ΔS negative. Choice B incorrectly assumes that forming a new compound always increases entropy, but what matters for entropy is the dispersal of matter and energy, not the identity of the compounds. To predict entropy changes in gas-phase reactions, count the moles of gas on each side—entropy decreases when gas particles combine to form fewer particles.

Question 9

A sealed container initially holds a sample of pure $\mathrm{H_2O(l)}$ at constant temperature. The water is then completely converted to $\mathrm{H_2O(g)}$ in the same container. What is the sign of $\Delta S$ for the system?

$\Delta S$ is negative because the container is sealed and no matter enters or leaves.
$\Delta S$ is negative because forming a gas requires energy input.
$\Delta S$ is positive because particles have more accessible microstates in the gas phase than in the liquid phase. (correct answer)
$\Delta S$ is positive only if the temperature increases during the process.
$\Delta S$ is approximately zero because the identity of the substance does not change.

Explanation: This question tests understanding of absolute entropy and entropy change. When liquid water converts to water vapor at constant temperature, the molecules gain much more freedom of motion and can occupy many more positions in space. In the gas phase, water molecules move independently throughout the container with high kinetic energy, whereas in the liquid phase they are constrained by intermolecular forces and can only vibrate and rotate in place. This dramatic increase in molecular freedom means the gas phase has vastly more accessible microstates than the liquid phase, making ΔS positive. Choice A incorrectly confuses energy input (enthalpy) with entropy change—while vaporization does require energy, this doesn't determine the sign of ΔS. Remember that entropy increases whenever matter becomes more dispersed or disordered, regardless of whether the process requires or releases energy.

Question 10

Two different processes occur at the same temperature:

Process 1: $\mathrm{NaCl(s) \rightarrow Na^+(aq) + Cl^-(aq)}$ (solid sodium chloride dissolves in water)

Process 2: $\mathrm{H_2O(l) \rightarrow H_2O(s)}$ (liquid water freezes)

Which statement correctly compares the entropy changes of the systems, $\Delta S_1$ and $\Delta S_2$ ?

$\Delta S_1$ is approximately zero and $\Delta S_2$ is approximately zero.
$\Delta S_1$ is negative and $\Delta S_2$ is positive.
$\Delta S_1$ is positive and $\Delta S_2$ is negative. (correct answer)
$\Delta S_1$ is positive and $\Delta S_2$ is positive.
$\Delta S_1$ is negative and $\Delta S_2$ is negative.

Explanation: This question tests understanding of absolute entropy and entropy change. In Process 1, solid NaCl dissolves to form separated Na⁺ and Cl⁻ ions that move freely in solution, increasing from an ordered crystal lattice to dispersed aqueous ions—this increases disorder, making ΔS₁ positive. In Process 2, liquid water freezes to form ice, where molecules become locked in a rigid hexagonal crystal structure with much less freedom of motion—this decreases disorder, making ΔS₂ negative. The key is recognizing that dissolving increases particle dispersal while freezing decreases it. Choice A incorrectly reverses both signs, possibly confusing the energy changes (dissolving NaCl is endothermic, freezing is exothermic) with entropy changes. To determine entropy changes, focus on whether particles become more or less dispersed, not on whether heat is absorbed or released.

Question 11

Consider the reaction at constant temperature in a closed container: $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$ . Based only on the change in the number of gas particles, what is the sign of $\Delta S$ for the system?

$\Delta S$ is positive because a product forms.
$\Delta S$ is negative because the number of moles of gas decreases. (correct answer)
$\Delta S$ is approximately zero because all species are gases.
$\Delta S$ cannot be determined without $\Delta H$ .
$\Delta S$ is positive because bonds are formed.

Explanation: This question tests understanding of absolute entropy and entropy change. In the reaction N₂(g) + 3H₂(g) → 2NH₃(g), we start with 4 moles of gas particles (1 N₂ + 3 H₂) and end with only 2 moles of gas particles (2 NH₃). When the number of gas particles decreases, there are fewer ways to arrange the particles in the container, resulting in decreased disorder and lower entropy. The fact that all species are gases doesn't mean entropy doesn't change—what matters is the change in the number of particles and their freedom of motion. Some students might incorrectly think ΔS is positive because bonds are formed (choice E), confusing bond formation with entropy increase—actually, combining particles into fewer molecules decreases disorder. Remember that entropy decreases when the number of gas particles decreases because there are fewer independent units to disperse.

Question 12

At constant temperature, a sealed container initially contains 2 mol of $\mathrm{NO_2(g)}$ . The gas dimerizes to form $\mathrm{N_2O_4(g)}$ until the final state is 1 mol of $\mathrm{N_2O_4(g)}$ . What is the sign of $\Delta S$ for the system for this change from initial to final state?

$\Delta S$ is approximately zero because both states are gases.
$\Delta S$ cannot be determined without the container volume.
$\Delta S$ is negative because the number of gas particles decreases. (correct answer)
$\Delta S$ is positive because the reaction releases heat.
$\Delta S$ is positive because a larger molecule forms.

Explanation: This question tests understanding of absolute entropy and entropy change. When 2 moles of NO₂(g) dimerize to form 1 mole of N₂O₄(g), the number of independent gas particles decreases from 2 to 1. With fewer separate particles moving independently in the container, there are fewer possible arrangements and less disorder in the system. The entropy decreases because we go from having two molecules that can move independently to one larger molecule, reducing the overall freedom of motion and number of microstates. Some students might incorrectly think ΔS is positive because a larger molecule forms (choice A), confusing molecular size with entropy—what matters is the number of independent particles, not their size. Remember that entropy decreases when gas particles combine because fewer independent units means less dispersal and fewer possible arrangements.

Question 13

A mixture is prepared by combining 1 mol of $\mathrm{He(g)}$ and 1 mol of $\mathrm{Ne(g)}$ in the same container at the same temperature and pressure, producing a uniform gas mixture. Compared with the two gases kept in separate containers under the same conditions, what is the sign of $\Delta S$ for the system upon mixing?

$\Delta S$ is negative because the total pressure increases.
$\Delta S$ is approximately zero because both are noble gases.
$\Delta S$ is positive. (correct answer)
$\Delta S$ cannot be determined without volumes.
$\Delta S$ is negative because mixing forms a solution.

Explanation: This question tests understanding of absolute entropy and entropy change. When helium and neon gases are mixed, the two types of atoms become randomly distributed throughout the container rather than being separated. This mixing increases the number of possible arrangements of the particles—there are many more ways to arrange He and Ne atoms together than to keep them separated. The entropy of mixing is always positive because the mixed state has greater disorder than the separated state, even though both substances were already gases. Some students might incorrectly think ΔS is zero because both are noble gases (choice B), confusing chemical similarity with entropy effects—the identity of the gases doesn't matter for mixing entropy. Remember that entropy increases when different substances mix because there are more ways to arrange different types of particles together than separately.

Question 14

Two separate processes occur at the same temperature: Process 1: $\mathrm{NaCl(s) \rightarrow NaCl(aq)}$ (dissolving in water) Process 2: $\mathrm{H_2O(l) \rightarrow H_2O(s)}$ (freezing) Which process results in the greater increase in entropy of the system?

Process 1 results in a greater increase in entropy. (correct answer)
Process 2 results in a greater increase in entropy.
Both processes result in approximately the same increase in entropy.
Neither process changes entropy because temperature is constant.
Both processes decrease entropy by the same amount.

Explanation: This question tests understanding of absolute entropy and entropy change. Process 1 (NaCl dissolving) involves breaking apart a highly ordered crystal lattice and dispersing ions throughout the solution, resulting in a large entropy increase. Process 2 (water freezing) involves water molecules arranging into a more ordered ice crystal structure, resulting in an entropy decrease. Since Process 1 increases entropy while Process 2 decreases entropy, Process 1 clearly results in the greater increase in entropy (in fact, Process 2 doesn't increase entropy at all—it decreases it). Some students might think both processes have the same entropy change (choice C), failing to recognize that dissolving increases disorder while freezing decreases it. Remember that phase transitions to more dispersed states (solid→liquid→gas or solid→aqueous) increase entropy, while transitions to more ordered states decrease entropy.

Question 15

A beaker contains separate layers of hexane(l) and water(l) at room temperature. The liquids are vigorously stirred, forming a temporary dispersion, and then allowed to settle back into two distinct layers (no reaction occurs). Considering only the settling step (dispersion $\rightarrow$ separated layers), which statement best describes the sign of $\Delta S$ for the system?

$\Delta S$ is approximately zero because both substances remain liquids.
$\Delta S$ is negative because going from a mixed/dispersion state to separated layers reduces the number of possible arrangements. (correct answer)
$\Delta S$ is positive because the system returns to equilibrium.
$\Delta S$ is negative only if heat is released during settling.
$\Delta S$ is positive because separation increases purity.

Explanation: This question assesses understanding of absolute entropy and entropy change. The settling of mixed hexane and water into separate layers decreases entropy without phase change. The number of particles remains the same, but dispersal reduces as they separate into distinct phases. This separation leads to fewer possible arrangements and a negative entropy change. A tempting distractor is choice D, which claims ΔS is positive because separation increases purity, confusing purity with increased disorder rather than reduced mixing. To evaluate entropy changes in separation processes, remember that entropy increases when matter or energy becomes more dispersed.

Question 16

In a rigid, sealed container at constant temperature, the reaction $\text{N}_2\text{O}_4(g) \rightarrow 2\text{NO}_2(g)$ goes essentially to completion. Which statement best describes the sign of $\Delta S$ for the system?

$\Delta S$ is negative because a chemical bond is broken.
$\Delta S$ is approximately zero because the container is rigid.
$\Delta S$ is negative because one reactant forms two products.
$\Delta S$ is positive because the number of gas particles increases. (correct answer)
$\Delta S$ is positive only if the reaction absorbs heat.

Explanation: This question assesses understanding of absolute entropy and entropy change. In the decomposition of N2O4 gas to two NO2 gas molecules, there is no phase change, but the number of gas particles increases from one to two. This increase in particle number enhances the dispersal of matter in the fixed volume, leading to more accessible microstates. Consequently, the entropy change for the system is positive due to the greater disorder from more gas particles. A tempting distractor is choice C, which claims ΔS is negative because one reactant forms two products, confusing the increase in particle count with a decrease in entropy. To evaluate entropy changes in reactions involving gases, remember that entropy increases when matter or energy becomes more dispersed.

Question 17

Two gases are initially separated by a closed valve in a rigid container: the left side contains He(g) and the right side contains Ne(g), both at the same temperature. The valve is opened and the gases mix to form a uniform mixture, with no reaction. Which statement best describes the sign of $\Delta S$ for the mixing process?

$\Delta S$ is negative because the gases become more uniform.
$\Delta S$ is approximately zero because the temperature is unchanged.
$\Delta S$ is negative because the total pressure increases when gases mix.
$\Delta S$ is positive only if the process is exothermic.
$\Delta S$ is positive because mixing increases the number of possible particle arrangements. (correct answer)

Explanation: This question assesses understanding of absolute entropy and entropy change. When two gases like He and Ne mix, there is no phase change, but the dispersal of particles increases as they spread throughout the entire container. The number of particles remains constant, but mixing allows for more possible arrangements of the distinct particles, increasing the entropy. This increase in the number of microstates due to greater dispersal results in a positive entropy change for the process. A tempting distractor is choice A, which says ΔS is negative because total pressure increases, based on the misconception that pressure changes directly affect entropy sign rather than particle arrangements. To evaluate entropy changes in mixing processes, remember that entropy increases when matter or energy becomes more dispersed.

Question 18

A student dissolves a small amount of KBr(s) in water to form an aqueous solution, KBr(aq), and the solid disappears. Assume the solution is dilute and no precipitate forms. Which statement best describes the sign of $\Delta S$ for the system?

$\Delta S$ is positive because dissolving disperses ions throughout the solvent, increasing the number of accessible microstates. (correct answer)
$\Delta S$ is negative because ions in solution are more ordered than a crystal lattice.
$\Delta S$ is approximately zero because no gas is produced.
$\Delta S$ is negative because dissolving is always exothermic.
$\Delta S$ is positive only if the temperature increases during dissolving.

Explanation: This question assesses understanding of absolute entropy and entropy change. Dissolving solid KBr into aqueous solution involves a phase change from solid to dispersed ions in liquid, increasing entropy. The number of particles effectively increases as the crystal lattice breaks into individual ions, enhancing dispersal throughout the solvent. This greater dispersal leads to more accessible microstates and a positive entropy change. A tempting distractor is choice B, which claims ΔS is negative because ions in solution are more ordered than a crystal, confusing the rigidity of lattices with the freedom in solution. To evaluate entropy changes in dissolving processes, remember that entropy increases when matter or energy becomes more dispersed.

Question 19

A sample of water vapor, H $_2$ O(g), in a closed container is cooled until it completely condenses to liquid water, H $_2$ O(l), with no loss of material. Which statement best describes the sign of $\Delta S$ for the system?

$\Delta S$ is positive because condensation releases heat.
$\Delta S$ is negative because particles become less free to move in the liquid than in the gas. (correct answer)
$\Delta S$ is approximately zero because the number of moles of H $_2$ O is unchanged.
$\Delta S$ is positive because liquids are more disordered than gases.
$\Delta S$ is negative only if the temperature decreases; otherwise it must be zero.

Explanation: This question assesses understanding of absolute entropy and entropy change. The phase change from water vapor to liquid decreases entropy because particles in the liquid have less freedom of motion compared to the gas phase. Although the number of particles remains the same, their dispersal reduces as they condense into a more ordered liquid state. This reduction in accessible microstates leads to a negative entropy change for the system. A tempting distractor is choice A, which states ΔS is positive because condensation releases heat, based on the misconception that heat release is confused with increased disorder rather than phase-dependent freedom. To evaluate entropy changes in condensation, remember that entropy increases when matter or energy becomes more dispersed.

Question 20

At constant temperature, a gas-phase decomposition occurs in a sealed container: $\text{PCl}_5(g) \rightarrow \text{PCl}_3(g) + \text{Cl}_2(g)$ . What is the sign of $\Delta S$ for the system?

$\Delta S$ is approximately zero because temperature is constant.
$\Delta S$ is approximately zero because all species are gases.
$\Delta S$ is positive because the number of gas molecules increases, increasing the number of accessible microstates. (correct answer)
$\Delta S$ is negative because the container is sealed.
$\Delta S$ is negative because one reactant forms two products.

Explanation: This question tests understanding of absolute entropy and entropy change. In this decomposition, one PCl₅ molecule breaks apart to form one PCl₃ molecule and one Cl₂ molecule, increasing the total number of gas particles from 1 to 2. More gas molecules means more ways to arrange particles in the container, resulting in more accessible microstates. Therefore, ΔS is positive because the system becomes more disordered. Choice B incorrectly suggests negative entropy despite the increase in particle number—the formation of two products from one reactant always increases entropy in gas-phase reactions. Remember that entropy increases when the number of gas molecules increases.