Question

Which one of the following has \(\Delta S^{\circ}\) greater than zero? (a) \(\mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CaCO}_{3}(\mathrm{~g})\) (b) \(\mathrm{NaCl}(\mathrm{aq}) \rightleftharpoons \mathrm{NaCl}(\mathrm{s})\) (c) \(\mathrm{NaNO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq})\) (d) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\)

Short answer

Option (c) has ΔS° greater than zero.

Step-by-step solution

Understanding Entropy and ΔS°

Entropy ( abla S) is a measure of the disorder or randomness in a system. A positive abla S^{ ext{o}} indicates that the disorder has increased in the reaction. We can generally predict the sign of abla S^{ ext{o}} by considering changes in the state of matter and the number of molecules or ions in the reaction.

Analyzing Each Option

Let's analyze the reactions to determine the sign of abla S^{ ext{o}}. (a) abla S^{ ext{o}} < 0 because gaseous molecules are combining to form a solid, decreasing disorder. (b) abla S^{ ext{o}} < 0 as it involves the crystallization, going from dissolved ions (aq) to a solid (s). (c) abla S^{ ext{o}} > 0 because a solid is dissolving into ions in solution, increasing disorder. (d) abla S^{ ext{o}} < 0 because multiple gas molecules are combining to form fewer gas molecules, decreasing disorder.

Identifying the Reaction with ΔS° > 0

From the analysis, only reaction (c) ( ext{NaNO}_3(s) ightleftharpoons ext{Na}^+(aq) + ext{NO}_3^-(aq)), where a solid dissolves into ions, has an increase in disorder, resulting in abla S^{ ext{o}} > 0.

Key concepts

Entropy Change

Entropy, denoted as \( \Delta S \), is a fundamental concept in thermodynamics representing the degree of disorder or randomness in a system. In chemical reactions, the change in entropy \( \Delta S^{\circ} \) helps us understand how the disorder of the system changes as the reaction proceeds.

To determine \( \Delta S^{\circ} \), consider the states of the reactants and products:

• Gases generally have higher entropy than liquids, which in turn have higher entropy than solids.
• An increase in the number of gas molecules typically indicates an increase in entropy.
• Dissolution of solids into ions tends to increase entropy due to the greater distribution and movement of particles.

By evaluating the initial and final states of a reaction, entropy changes can help predict whether the abla S^{\circ} will be positive or negative, thus indicating an increase or decrease in disorder.

Reaction Entropy

Reaction entropy specifically refers to the change in entropy (\( \Delta S \)) that occurs during a chemical reaction. During a reaction, the arrangement and movement of atoms and molecules change, which then influences the system's entropy.

Consider the following when analyzing reaction entropy:

• If reactants combine to form a product with fewer molecules or a more ordered arrangement, \( \Delta S^{\circ} \) is likely negative.
• If products are formed from a greater number of reactant molecules, or if components dissolve forming a larger number of ions, \( \Delta S^{\circ} \) tends to be positive.
• The phase of each component in the reaction (solid, liquid, or gas) significantly affects the system's entropy.

This understanding of reaction entropy allows us to anticipate how the entropy of a system will change when a chemical reaction takes place, which can influence the feasibility and spontaneity of reactions.

Sign of Delta S

The sign of \( \Delta S^{\circ} \) is crucial in predicting the behavior of a chemical reaction. It informs us whether a process results in increased or decreased randomness in the system. Recognizing the signs of \( \Delta S \) helps students understand broader thermodynamic principles.

Here are key points to decide the sign of \( \Delta S^{\circ} \):

• Reactions where solid substances dissolve into ions or gases usually have \( \Delta S^{\circ} > 0 \), as they lead to more dispersed and free-moving particles.
• Conversely, reactions that produce fewer gas molecules from more gaseous reactants typically result in \( \Delta S^{\circ} < 0 \), indicating a decrease in disorder.
• The process of crystallization or condensing gases into solids often decreases entropy.

Ultimately, the sign of \( \Delta S \) guides us in determining the spontaneity of the reaction under certain conditions, aiding us in predicting whether the reaction will occur naturally.