Problem 40 - Chemistry: The Central Science in SI Units, Global Edition

Question

For each of the following pairs, predict which substance possesses the larger entropy per mole: (a) \(1 \mathrm{~mol}\) of \(\mathrm{O}_{2}(g)\) at \(300^{\circ} \mathrm{C}, 1.013 \mathrm{kPa},\) or \(1 \mathrm{~mol}\) of \(\mathrm{O}_{3}(g)\) at \(300^{\circ} \mathrm{C}, 1.013 \mathrm{kPa} ;\) (b) \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(g)\) at \(100^{\circ} \mathrm{C}, 101.3 \mathrm{kPa}\), or \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(100^{\circ} \mathrm{C}, 101.3 \mathrm{kPa} ;(\mathbf{c}) 0.5 \mathrm{~mol}\) of \(\mathrm{N}_{2}(g)\) at \(298 \mathrm{~K}, 20-\mathrm{L}\). vol- ume, or \(0.5 \mathrm{~mol} \mathrm{CH}_{4}(g)\) at \(298 \mathrm{~K}, 20-\mathrm{L}\) volume; (d) \(100 \mathrm{~g}\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(s)\) at \(30^{\circ} \mathrm{C}\) or \(100 \mathrm{~g} \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) at \(30^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

(a) 1 mole of O3(g) at 300°C and 1.013kPa has the larger entropy per mole. (b) 1 mole of H2O(g) at 100°C and 101.3kPa has the larger entropy per mole. (c) 0.5 mole of CH4(g) at 298K and 20L volume has the larger entropy per mole. (d) 100g Na2SO4(aq) at 30°C has the larger entropy per mole.

Step 1: Compare the molecular complexity

Both O2 and O3 are gases. O3 molecule has a more complex structure than O2 molecule which results in a larger entropy per mole for O3. Therefore, 1 mole of O3(g) at 300°C and 1.013kPa has the larger entropy per mole. (b) Comparing 1 mole of H2O(g) and 1 mole of H2O(l)

Step 2: Compare the molecular phase

Here, we have one mole of the same substance but in different phases. Gaseous phase has more entropy per mole than the liquid phase as the molecules are more disordered. Therefore, 1 mole of H2O(g) at 100°C and 101.3kPa has the larger entropy per mole. (c) Comparing 0.5 mole of N2(g) and 0.5 mole of CH4(g)

Step 3: Compare the molecular complexity once again

Both N2 and CH4 are in gaseous phase and both have 0.5 mole under the same condition of temperature and volume. However, CH4 molecule is more complex than N2 molecule which results in a larger entropy per mole for CH4. Therefore, 0.5 mole of CH4(g) at 298K and 20L volume has the larger entropy per mole. (d) Comparing 100g Na2SO4(s) and 100g Na2SO4(aq)

Step 4: Compare the ionic solvation and solution formation

In this case, Na2SO4 is in a solid-state and an aqueous solution. When a solute dissolves in a solvent forming a solution, the entropy of the solution gets increased. In the aqueous solution, the ions get hydrated and increase their entropy. Therefore, 100g Na2SO4(aq) at 30°C has the larger entropy per mole.

Key Concepts

Molecular Complexity

When comparing molecules to determine which has more entropy per mole, molecular complexity plays a key role. Take oxygen, for example. The molecule \( \text{O}_{3} \) appears more complex than \( \text{O}_{2} \) because it consists of more atoms and shows a non-linear structure. More complex molecules have more ways of rotating and vibrating, leading to higher entropy.

Here are some reasons why complexity leads to higher entropy:

• More atoms create more vibrational modes.
• Non-linear structures offer more rotational freedom.
• Complex molecules can exhibit a greater number of microstates.

In thermodynamic terms, the more microstates available to a system, the greater the entropy. Hence, for gases like \( \text{O}_{3}(g) \), the larger arrangement of atoms results in greater entropy per mole than for less complex molecules like \( \text{O}_{2}(g) \).

Gaseous vs. Liquid Phases

Understanding the difference between gas and liquid phases is crucial when comparing their entropies. In general, gases have higher entropy than liquids. Why? Because the molecules in gases move more freely and randomly spread throughout a given volume.

Let's explore why gases exhibit higher entropy:

• Gas molecules are dispersed, giving them more space to move around.
• Liquid molecules are more ordered due to closer interactions.
• The energetic freedom in gases allows for more possible rearrangements.

Taking water as an example: \( \text{H}_{2}\text{O}(g) \) will have a greater entropy per mole than \( \text{H}_{2}\text{O}(l) \), even if they are at the same temperature, due to the inherently greater freedom and disorder of gaseous molecules.

Ionic Solvation

Ionic solvation refers to the process of dissolving ionic compounds in a solvent, which increases the system's entropy. When ions are placed in a solvent like water, they become surrounded by solvent molecules. This formation is called a solvation shell. The solvation shell leads to greater disorder, which enhances the entropy of the solution.

The entropy increases in ionic solvation due to:

• Ion-dipole interactions between ions and solvent molecules.
• The liberation of previously ordered water molecules.
• New configurations emerging in the solution.

Sodium sulfate, for instance, when dissolved in water, moves from a solid-state with organized ionic structures to an aqueous state with dispersed hydrated ions, significantly boosting the system's entropy.

Solution Formation

Forming solutions, particularly with ionic compounds, involves changes in entropy. Dissolving a solid in a liquid breaks the ordered lattice of the solid, increasing randomness as solute particles disperse in the solvent. This results in higher entropy for the solution compared to the solid form.

Why does solution formation increase entropy?

• Disruption of the solid lattice leads to more disordered arrangements.
• Extraction from the fixed positions in the solid phase permits more movement.
• Intermixing of solute and solvent creates many new potential configurations.

As demonstrated by sodium sulfate, transitioning from solid to solution enhances randomness and increases entropy per mole. This process illustrates how solution formation serves as a key driver of entropic change in chemical systems.