Question

Use Appendix \(\mathrm{C}\) to compare the standard entropies at \(25^{\circ} \mathrm{C}\) for the following pairs of substances: (a) \(\mathrm{Sc}(s)\) and \(\mathrm{Sc}(g)\), \(\mathrm{NH}_{3}(g)\) and \(\mathrm{NH}_{3}(a q)\) (c) \(1 \mathrm{~mol} \mathrm{P}_{4}(g)\) and \(2 \mathrm{~mol} \mathrm{P}_{2}(g)\), (d) C(graphite) and C(diamond). In each case explain the difference in the entropy values.

Short answer

Comparing the standard entropies at 25°C for the given pairs of substances, we find that: (a) Sc(g) has a higher entropy (176.1 J/mol K) than Sc(s) (42.7 J/mol K) due to more freedom of motion in the gaseous state. (b) NH3(g) has a higher entropy (193.3 J/mol K) than NH3(aq) (80.8 J/mol K) due to the greater freedom of motion in the gaseous state compared to the dissolved state. (c) 2 mol P2(g) has a higher entropy (437.2 J/mol K) than 1 mol P4(g) (287.7 J/mol K) because there are more P2 molecules, resulting in more ways to arrange them. (d) C(graphite) has a higher entropy (5.7 J/mol K) than C(diamond) (2.4 J/mol K) due to the more flexible arrangement of carbon atoms in graphite compared to the rigid structure of diamond.

Step-by-step solution

(a) Compare standard entropies of Sc(s) and Sc(g)

First, we need to find the standard entropies for Sc(s) and Sc(g) from Appendix C. On consulting Appendix C, we find that the standard entropy for Sc(s) is 42.7 J/mol K and for Sc(g) is 176.1 J/mol K. Now, we can compare them: since the standard entropy of Sc(g) is greater than Sc(s), we can conclude that Sc(g) has a higher randomness at 25°C compared to Sc(s). When a substance changes from a solid to a gaseous state, its molecules are more spread out and have greater freedom of motion, which results in a higher entropy.

(b) Compare standard entropies of NH3(g) and NH3(aq)

First, we need to find the standard entropies for NH3(g) and NH3(aq) from Appendix C. On consulting Appendix C, we find that the standard entropy for NH3(g) is 193.3 J/mol K and for NH3(aq) is 80.8 J/mol K. Now, we can compare them: since the standard entropy of NH3(g) is greater than NH3(aq), we can conclude that NH3(g) has a higher randomness at 25°C compared to NH3(aq). When a substance changes from a gaseous state to a dissolved state in a solvent, its freedom of motion is somewhat more restricted due to interactions with the solvent molecules, which results in lower entropy.

(c) Compare standard entropies of 1 mol P4(g) and 2 mol P2(g)

First, we need to find the standard entropies for P4(g) and P2(g) from Appendix C. On consulting Appendix C, we find that the standard entropy for P4(g) is 287.7 J/mol K and for P2(g) is 218.6 J/mol K. Now, let's figure out the entropy of the system with 2 moles of P2(g): Entropy of 2 mol P2(g) = 2 * (Entropy of 1 mol P2(g)) = 2 * 218.6 J/mol K = 437.2 J/mol K We can then compare the entropies for 1 mol P4(g) and 2 mol P2(g): since the standard entropy of 2 mol P2(g) is greater than that of 1 mol P4(g), we can conclude that 2 mol of P2(g) has a higher randomness at 25°C compared to 1 mol of P4(g). This can be explained by the fact that there are twice as many P2 molecules in the gaseous state, which results in more possible ways to arrange them and hence, a higher entropy.

(d) Compare standard entropies of C(graphite) and C(diamond)

First, we need to find the standard entropies for C(graphite) and C(diamond) from Appendix C. On consulting Appendix C, we find that the standard entropy for C(graphite) is 5.7 J/mol K and for C(diamond) is 2.4 J/mol K. Now, we can compare them: since the standard entropy of C(graphite) is greater than C(diamond), we can conclude that C(graphite) has a higher randomness at 25°C compared to C(diamond). Carbon atoms in graphite are arranged in layers, which have more freedom to slide past each other. In contrast, diamond has a rigid tetrahedral lattice structure that allows less freedom of movement, leading to a lower entropy value.

Key concepts

Entropy and States of Matter

Understanding the concept of entropy in relation to the states of matteris crucial when delving into thermodynamics. Simply put, entropy can be described as a measure of the disorder or randomness in a system.

Consider a substance like scandium, with its standard entropies provided for both the solid and gaseous states. The greater entropy of scandium in the gaseous state (\(176.1 \text{J/mol K}\) compared to the solid state (\(42.7 \text{J/mol K}\) , illustrates a foundational principle: as matter transitions from solid to liquid, and then to gas, entropy increases. This increase in randomness is due to molecules in the gaseous state having more freedom to move and occupy more positions.

When students encounter difficulties grasping this concept, a helpful visual might be imagining a crowd in a confined room (solid) moving into a larger hall (liquid) and finally spreading out into an open park (gas). Just as people have more freedom of movement in the park, molecules in the gaseous state have more ways to disperse and hence exhibit higher entropy.

Entropy and Physical Changes

Physical changes in matter impact entropy significantly. Compare, for example, the entropies of ammonia in the gaseous state and dissolved in water (aqueous). Gaseous ammonia possesses a higher standard entropy (\(193.3 \text{J/mol K}\) than its aqueous counterpart (\(80.8 \text{J/mol K}\) . This is because gaseous molecules have more freedom, whereas being in a solution, the ammonia molecules are partially restricted by interactions with water molecules.

The impact of physical changes on entropy is often a point of confusion, but thinking about it in the context of freedom can clarify things. Dissolving a gas into a liquid is like moving dancers from a stage with free choreography into a coordinated ballet; the gas has more possibilities for movement (higher entropy) before it is limited by the solvent (lower entropy).

It is important to remember that entropy's relationship with physical changes is not just confined to states of matter but also includes dissolution, mixing, and other transformations that impact molecular randomness.

Entropy and Molecular Complexity

The complexity of a molecule can influence its entropy. When comparing phosphorus molecules, we see that having more molecules, even of simpler types, contributes to an increase in entropy. Two moles of diatomic phosphorus (\(2 \times 218.6 \text{J/mol K}\) have a collective higher entropy than one mole of tetraatomic phosphorus (\(287.7 \text{J/mol K}\) due to the increased number of ways to arrange those molecules.

To better understand this concept, consider a jigsaw puzzle. One large, complex puzzle piece (like the P4 molecule) has only a few orientations, while numerous smaller pieces (like the P2 molecules) can be arranged in many more ways. This diversity in arrangements is essentially what contributes to higher entropy in the system with more, simpler molecules.

When faced with the challenge of understanding molecular complexity and entropy, it's key to consider not just the size or complexity of a single molecule, but the number of molecules and the variety of ways they can be arranged within a system.