Absolute Entropy and Entropy Change
Help Questions
AP Chemistry › Absolute Entropy and Entropy Change
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 bonds are formed.
$\Delta S$ is negative because the number of moles of gas decreases.
$\Delta S$ cannot be determined without $\Delta H$.
$\Delta S$ is positive because a product forms.
$\Delta S$ is approximately zero because all species are gases.
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.
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 mixing forms a solution.
$\Delta S$ is approximately zero because both are noble gases.
$\Delta S$ cannot be determined without volumes.
$\Delta S$ is negative because the total pressure increases.
$\Delta S$ is positive.
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.
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?
Neither process changes entropy because temperature is constant.
Process 2 results in a greater increase in entropy.
Process 1 results in a greater increase in entropy.
Both processes result in approximately the same increase in entropy.
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.
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?
$\Delta S$ cannot be determined without the container volume.
$\Delta S$ is negative because the number of gas particles decreases.
$\Delta S$ is positive because the reaction releases heat.
$\Delta S$ is positive because a larger molecule forms.
$\Delta S$ is approximately zero because both states are gases.
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.
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 positive because the system returns to equilibrium.
$\Delta S$ is negative only if heat is released during settling.
$\Delta S$ is negative because going from a mixed/dispersion state to separated layers reduces the number of possible arrangements.
$\Delta S$ is approximately zero because both substances remain liquids.
$\Delta S$ 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.
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 positive only if the reaction absorbs heat.
$\Delta S$ 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.
$\Delta S$ is negative because a chemical bond is broken.
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.
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 positive because the process produces gaseous particles from a solid, increasing dispersal of matter.
$\Delta S$ is approximately zero because the container is closed.
$\Delta S$ is positive only if the reaction is endothermic.
$\Delta S$ is negative because decomposition reactions always decrease entropy.
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 negative because the gas has fewer accessible positions (microstates) in a smaller volume.
$\Delta S$ is positive because the temperature is constant.
$\Delta S$ is positive because work is done on the gas.
$\Delta S$ is negative only if the compression releases heat.
$\Delta S$ is approximately zero because argon is monatomic.
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 positive because forming a new substance increases disorder.
$\Delta S$ is negative because the reaction requires an electric spark.
$\Delta S$ 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?
$\Delta S_1$ is positive and $\Delta S_2$ is negative.
$\Delta S_1$ and $\Delta S_2$ are both positive.
$\Delta S_1$ and $\Delta S_2$ are both negative.
$\Delta S_1$ is negative and $\Delta S_2$ is positive.
$\Delta S_1$ and $\Delta S_2$ are both approximately zero.
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.