Absolute Entropy and Entropy Change
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AP Chemistry › Absolute Entropy and Entropy Change
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 decomposition reactions always decrease entropy.
$\Delta S$ is positive because the process produces gaseous particles from a solid, increasing dispersal of matter.
$\Delta S$ is positive only if the reaction is endothermic.
$\Delta S$ is approximately zero because the container is closed.
$\Delta S$ is negative because a solid disappears.
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.
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 the temperature is constant.
$\Delta S$ is negative only if the compression releases heat.
$\Delta S$ is approximately zero because argon is monatomic.
$\Delta S$ is negative because the gas has fewer accessible positions (microstates) in a smaller volume.
$\Delta S$ is positive because work is done on the gas.
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.
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 reaction requires an electric spark.
$\Delta S$ is positive because forming a new substance increases disorder.
$\Delta S$ is negative because the number of moles of gas decreases from 3 to 2, reducing the number of accessible microstates.
$\Delta S$ is approximately zero because both reactant and product are gases.
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.
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 negative.
$\Delta S_1$ and $\Delta S_2$ are both positive.
$\Delta S_1$ is negative and $\Delta S_2$ is positive.
$\Delta S_1$ and $\Delta S_2$ are both approximately zero.
$\Delta S_1$ is positive and $\Delta S_2$ is 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.
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 positive because particles have greater freedom of motion in the gas phase than in the liquid phase.
$\Delta S$ is negative because gases are more ordered than liquids.
$\Delta S$ is approximately zero because the chemical identity of Br$_2$ does not change.
$\Delta S$ is positive only if the temperature increases; otherwise it must be zero.
$\Delta S$ is negative because energy is absorbed during vaporization.
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.
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 negative because the system becomes more ordered when a gas becomes a liquid.
$\Delta S$ is positive because the molecules move closer together.
$\Delta S$ is approximately zero because temperature is constant.
$\Delta S$ is approximately zero because no new substances are formed.
$\Delta S$ is positive because heat is released during condensation.
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.
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 container is rigid.
$\Delta S$ is positive because mixing increases the number of possible arrangements of particles.
$\Delta S$ is approximately zero because the total number of moles of gas is unchanged.
$\Delta S$ is approximately zero because the temperature is constant.
$\Delta S$ is negative because the gases collide with each other.
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.
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 approximately zero because the vessel is rigid.
$\Delta S$ is negative because the number of moles of gas decreases.
$\Delta S$ is positive because a new compound is formed.
$\Delta S$ is negative only if the reaction is exothermic.
$\Delta S$ is positive because gases always have high entropy.
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.
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 positive only if the temperature increases during the process.
$\Delta S$ is negative because the container is sealed and no matter enters or leaves.
$\Delta S$ is positive because particles have more accessible microstates in the gas phase than in the liquid phase.
$\Delta S$ is negative because forming a gas requires energy input.
$\Delta S$ 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.
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 positive and $\Delta S_2$ is negative.
$\Delta S_1$ is negative and $\Delta S_2$ is negative.
$\Delta S_1$ is positive and $\Delta S_2$ is positive.
$\Delta S_1$ is approximately zero and $\Delta S_2$ is approximately zero.
$\Delta S_1$ is negative and $\Delta S_2$ is positive.
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.