Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

AP Chemistry

AP Chemistry Help: Absolute Entropy And Entropy Change

Review real example questions for Absolute Entropy And Entropy Change in AP Chemistry.

Question 1

Solid ammonium chloride is heated in a closed container and decomposes according to NH4Cl(s)→NH3(g)+HCl(g)\text{NH}_4\text{Cl}(s) \rightarrow \text{NH}_3(g) + \text{HCl}(g)NH4​Cl(s)→NH3​(g)+HCl(g). After heating, only gases are present. Which statement best describes the sign of ΔS\Delta SΔS for the system?

  1. ΔS\Delta SΔS is negative because a solid disappears.
  2. ΔS\Delta SΔS is approximately zero because the container is closed.
  3. ΔS\Delta SΔS is positive because the process produces gaseous particles from a solid, increasing dispersal of matter.
  4. ΔS\Delta SΔS is negative because decomposition reactions always decrease entropy.
  5. ΔS\Delta SΔS 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 ΔS\Delta SΔS for the gas in the syringe?

  1. ΔS\Delta SΔS is positive because work is done on the gas.
  2. ΔS\Delta SΔS is approximately zero because argon is monatomic.
  3. ΔS\Delta SΔS is negative because the gas has fewer accessible positions (microstates) in a smaller volume.
  4. ΔS\Delta SΔS is positive because the temperature is constant.
  5. ΔS\Delta SΔS 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 O2_22​(g) at a uniform temperature. An electric spark converts it completely to ozone according to 3O2(g)→2O3(g)3\text{O}_2(g) \rightarrow 2\text{O}_3(g)3O2​(g)→2O3​(g), with temperature returning to the original value. Which statement best describes the sign of ΔS\Delta SΔS for the system?

  1. ΔS\Delta SΔS is positive because ozone has a higher molar mass than oxygen.
  2. ΔS\Delta SΔS is negative because the number of moles of gas decreases from 3 to 2, reducing the number of accessible microstates.
  3. ΔS\Delta SΔS is approximately zero because both reactant and product are gases.
  4. ΔS\Delta SΔS is positive because forming a new substance increases disorder.
  5. ΔS\Delta SΔS 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: CO2_22​(s) →\rightarrow→ CO2_22​(g) (dry ice sublimation) Process 2: H2_22​O(l) →\rightarrow→ H2_22​O(s) (freezing) Which statement correctly compares the entropy changes ΔS1\Delta S_1ΔS1​ and ΔS2\Delta S_2ΔS2​ for the systems?

  1. ΔS1\Delta S_1ΔS1​ and ΔS2\Delta S_2ΔS2​ are both approximately zero.
  2. ΔS1\Delta S_1ΔS1​ is positive and ΔS2\Delta S_2ΔS2​ is negative.
  3. ΔS1\Delta S_1ΔS1​ is negative and ΔS2\Delta S_2ΔS2​ is positive.
  4. ΔS1\Delta S_1ΔS1​ and ΔS2\Delta S_2ΔS2​ are both positive.
  5. ΔS1\Delta S_1ΔS1​ and ΔS2\Delta S_2ΔS2​ 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, Br2_22​(l), at room temperature. The container is warmed gently until all of the bromine becomes Br2_22​(g), with no change in the amount of substance. Which statement best describes the sign of ΔS\Delta SΔS for this process?

  1. ΔS\Delta SΔS is negative because energy is absorbed during vaporization.
  2. ΔS\Delta SΔS is approximately zero because the chemical identity of Br2_22​ does not change.
  3. ΔS\Delta SΔS is positive because particles have greater freedom of motion in the gas phase than in the liquid phase.
  4. ΔS\Delta SΔS is negative because gases are more ordered than liquids.
  5. ΔS\Delta SΔS 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: H2O(g)→H2O(ℓ)\text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{O}(\ell)H2​O(g)→H2​O(ℓ). What is the sign of ΔS\Delta SΔS for this process?

  1. ΔS\Delta SΔS is approximately zero because no new substances are formed.
  2. ΔS\Delta SΔS is approximately zero because temperature is constant.
  3. ΔS\Delta SΔS is positive because heat is released during condensation.
  4. ΔS\Delta SΔS is negative because the system becomes more ordered when a gas becomes a liquid.
  5. ΔS\Delta SΔS 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 He(g)\text{He}(g)He(g) and Ne(g)\text{Ne}(g)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 ΔS\Delta SΔS for the mixing process (considering the gases as the system)?

  1. ΔS\Delta SΔS is negative because the gases collide with each other.
  2. ΔS\Delta SΔS is approximately zero because the temperature is constant.
  3. ΔS\Delta SΔS is negative because the container is rigid.
  4. ΔS\Delta SΔS is positive because mixing increases the number of possible arrangements of particles.
  5. ΔS\Delta SΔS 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: N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}N2​(g)+3H2​(g)→2NH3​(g). Based on the change in the number of gas particles, what is the sign of ΔS\Delta SΔS for the system?

  1. ΔS\Delta SΔS is negative because the number of moles of gas decreases.
  2. ΔS\Delta SΔS is positive because a new compound is formed.
  3. ΔS\Delta SΔS is approximately zero because the vessel is rigid.
  4. ΔS\Delta SΔS is positive because gases always have high entropy.
  5. ΔS\Delta SΔS 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 H2O(l)\mathrm{H_2O(l)}H2​O(l) at constant temperature. The water is then completely converted to H2O(g)\mathrm{H_2O(g)}H2​O(g) in the same container. What is the sign of ΔS\Delta SΔS for the system?

  1. ΔS\Delta SΔS is negative because the container is sealed and no matter enters or leaves.
  2. ΔS\Delta SΔS is negative because forming a gas requires energy input.
  3. ΔS\Delta SΔS is positive because particles have more accessible microstates in the gas phase than in the liquid phase.
  4. ΔS\Delta SΔS is positive only if the temperature increases during the process.
  5. ΔS\Delta SΔS 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: NaCl(s)→Na+(aq)+Cl−(aq)\mathrm{NaCl(s) \rightarrow Na^+(aq) + Cl^-(aq)}NaCl(s)→Na+(aq)+Cl−(aq) (solid sodium chloride dissolves in water)

Process 2: H2O(l)→H2O(s)\mathrm{H_2O(l) \rightarrow H_2O(s)}H2​O(l)→H2​O(s) (liquid water freezes)

Which statement correctly compares the entropy changes of the systems, ΔS1\Delta S_1ΔS1​ and ΔS2\Delta S_2ΔS2​?

  1. ΔS1\Delta S_1ΔS1​ is approximately zero and ΔS2\Delta S_2ΔS2​ is approximately zero.
  2. ΔS1\Delta S_1ΔS1​ is negative and ΔS2\Delta S_2ΔS2​ is positive.
  3. ΔS1\Delta S_1ΔS1​ is positive and ΔS2\Delta S_2ΔS2​ is negative.
  4. ΔS1\Delta S_1ΔS1​ is positive and ΔS2\Delta S_2ΔS2​ is positive.
  5. ΔS1\Delta S_1ΔS1​ is negative and ΔS2\Delta S_2ΔS2​ 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.