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Chapter 16: Problem 25

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 \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{aq})\)

Short Answer

Expert verified
(a) NaCl(s), (b) P鈧�(g), (c) NH鈧凬O鈧�(aq) have higher entropy.

Step by step solution

01

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.
02

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 \(\mathrm{CaO}(\mathrm{s})\).
03

Evaluating Part (b)

Compare the gases Cl鈧� and P鈧�. Both are in gaseous form, but Cl鈧� is a diatomic molecule, while P鈧� is a tetrahedral molecule. The more complex shape and higher number of atoms in P鈧� allow for greater vibrational and rotational modes. Therefore, P鈧�(g) has higher entropy than Cl鈧�(g).
04

Evaluating Part (c)

Compare NH鈧凬O鈧� 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 \(\mathrm{NH}_{4}\mathrm{NO}_{3}(\mathrm{s})\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.
In comparing NaCl and CaO, we observe that CaO's lattice adopts a complex structure because of the presence of calcium's divalent ions. This complexity restricts ion movement, thus resulting in lower entropy than the simpler lattice in NaCl.
Diatomic and Polyatomic Molecules
Molecules can be categorized based on the number of atoms they possess. Diatomic molecules have two atoms, like Cl鈧�, while polyatomic molecules contain more than two, such as P鈧�, which consists of four phosphorus atoms arranged in a tetrahedral shape. These structures influence their entropy levels:
  • Diatomic molecules like Cl鈧� are simpler and have fewer degrees of freedom, which means fewer ways to distribute energy among vibrational, rotational, and translational modes.
  • Polyatomic molecules like P鈧� are more intricate and can experience numerous vibrational and rotational movements, engendering more microstates.
As a result, P鈧�, with its more complex structure, demonstrates higher entropy than Cl鈧� because of its ability to engage in more diverse modes of motion and, consequently, attain greater disorder.
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 NH鈧凬O鈧� 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.
For example, NH鈧凬O鈧� in solid form has a fixed, orderly structure. When it dissolves in water, it dissociates into NH鈧勨伜 and NO鈧冣伝 ions, significantly boosting entropy through increased particle motion and distribution.

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Most popular questions from this chapter

For each process, tell whether the entropy change of the system is positive or negative. (a) Water vapor (the system) deposits as ice crystals on a cold windowpane. (b) A can of carbonated beverage loses its fizz. (Consider the beverage but not the can as the system. What happens to the entropy of the dissolved gas?) (c) A glassblower heats glass (the system) to its softening temperature.

Predict whether the reaction given is product-favored or reactant-favored by calculating \(\Delta_{\mathrm{r}} G^{\circ}\) from the entropy and enthalpy changes for the reaction at \(25^{\circ} \mathrm{C}\). \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g})\) $$ \Delta_{r} H^{\circ}=41.17 \mathrm{~kJ} / \mathrm{mol} \quad \Delta_{r} S^{\circ}=42.08 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} $$

Without doing a calculation, predict whether the entropy change is positive or negative when each reaction occurs in the direction it is written. (a) \(\mathrm{CH}_{3} \mathrm{OH}(\ell) \longrightarrow \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g})\) (b) \(\mathrm{Br}_{2}(\ell)+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HBr}(\mathrm{g})\) (c) \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{CH}_{4}(\mathrm{~g})\) (d) \(\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \longrightarrow \mathrm{AgI}(\mathrm{s})\)

Explain how the entropy of the universe increases when an aluminum metal can is made from aluminum ore. The first step is to extract the ore, which is primarily a form of \(\mathrm{Al}_{2} \mathrm{O}_{3},\) from the ground. After it is purified by freeing it from oxides of silicon and iron, aluminum oxide is changed to the metal by an input of electrical energy. $$ 2 \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s}) \stackrel{\text { electrical energy }}{\longrightarrow} 4 \mathrm{Al}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) $$

Another step in the metabolism of glucose, which occurs after the formation of glucose 6 -phosphate, is the conversion of fructose 6 -phosphate to fructose 1,6 -bisphosphate ("bis" means two): $$ \begin{aligned} \text { Fructose } 6 \text { -phosphate }(\mathrm{aq})+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(\mathrm{aq}) & \longrightarrow \\ & \text { fructose } 1,6 \text { -bisphosphate(aq) }+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{H}^{+}(\text {aq }) \end{aligned} $$ (a) This reaction has a Gibbs free energy change of \(+16.7 \mathrm{~kJ} / \mathrm{mol}\) of fructose 6 -phosphate. Is it endergonic or exergonic? (b) Write the equation for the formation of \(1 \mathrm{~mol}\) ADP from ATP, for which \(\Delta_{\mathrm{r}} G^{\circ}=-30.5 \mathrm{~kJ} / \mathrm{mol}\) (c) Couple these two reactions to get an exergonic process; write its overall chemical equation, and calculate the Gibbs free energy change.
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