Problem 25
Question
Comparing the formulas or states for each pair of substances, predict which has the higher entropy per mole at the same temperature, and explain why. (a) \(\mathrm{NaCl}(\mathrm{s})\) or \(\mathrm{CaO}(\mathrm{s})\) (b) \(\mathrm{Cl}_{2}(\mathrm{~g})\) or \(\mathrm{P}_{4}(\mathrm{~g})\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{~s})\) or \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{aq})\)
Step-by-Step Solution
Verified Answer
(a) NaCl(s), (b) P₄(g), (c) NH₄NO₃(aq) have higher entropy.
1Step 1: Understanding Entropy
Entropy measures the degree of disorder or randomness in a system. Solid substances tend to have lower entropy than liquids or gases, and complex molecules or mixtures typically have higher entropy than simpler ones.
2Step 2: Evaluating Part (a)
Compare the solids NaCl and CaO. Both are ionic compounds; however, NaCl has a simpler lattice structure with fewer interacting forces compared to CaO, which has a more complex lattice due to stronger ionic interactions (owing to divalent ions). More complex interactions generally mean lower freedom of motion for the ions, leading to lower entropy in CaO. So, \(\mathrm{NaCl}(\mathrm{s})\) has higher entropy than \(\mathrm{CaO}(\mathrm{s})\).
3Step 3: Evaluating Part (b)
Compare the gases Cl₂ and P₄. Both are in gaseous form, but Cl₂ is a diatomic molecule, while P₄ is a tetrahedral molecule. The more complex shape and higher number of atoms in P₄ allow for greater vibrational and rotational modes. Therefore, P₄(g) has higher entropy than Cl₂(g).
4Step 4: Evaluating Part (c)
Compare NH₄NO₃ in solid form and aqueous form. In aqueous solution, the compound dissociates into ions, increasing the number of particles and thus the disorder in the system. The greater number of microstates available leads to higher entropy. Hence, \(\mathrm{NH}_{4}\mathrm{NO}_{3}(\mathrm{aq})\) has higher entropy than \(\mathrm{NH}_{4}\mathrm{NO}_{3}(\mathrm{s})\).
Key Concepts
Ionic CompoundsDiatomic and Polyatomic MoleculesDissolution and Entropy
Ionic Compounds
Ionic compounds are formed from positively and negatively charged ions. They typically have high melting points because of the strong electrostatic forces between these oppositely charged ions. The structure of an ionic compound is a lattice, which is a systematic and repeating arrangement of ions. For instance, NaCl is a well-known ionic compound.
There are key factors affecting the entropy of ionic compounds.
- The complexity of the lattice structure: More complex lattices, with more intricate ion arrangements, generally exhibit lower entropy because ions are constrained by stronger interactions.
- Divalent versus monovalent ions: A lattice composed of divalent ions typically has stronger interactions and, hence, lowers entropy compared to one with monovalent ions.
Diatomic and Polyatomic Molecules
Molecules can be categorized based on the number of atoms they possess. Diatomic molecules have two atoms, like Cl₂, while polyatomic molecules contain more than two, such as P₄, which consists of four phosphorus atoms arranged in a tetrahedral shape.
These structures influence their entropy levels:
- Diatomic molecules like Cl₂ are simpler and have fewer degrees of freedom, which means fewer ways to distribute energy among vibrational, rotational, and translational modes.
- Polyatomic molecules like P₄ are more intricate and can experience numerous vibrational and rotational movements, engendering more microstates.
Dissolution and Entropy
The process of dissolution involves a solute being dispersed in a solvent, forming a solution. This change often leads to increased entropy, especially when it results in dissociation of ions, as seen with NH₄NO₃ transitioning from a solid to an aqueous state.
Here's why dissolution affects entropy:
- In a solid state, molecules are typically fixed in a rigid lattice with limited movement, leading to lower entropy.
- Upon dissolving, the lattice breaks apart, allowing ions to move freely in the solution. This increases the number of particles and possible microstates.
Other exercises in this chapter
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