Question

Predict the sign of \(\Delta S^{\circ}\) for each of the following reactions. (a) \(2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NaCl}(s)\) (b) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g)\)

Short answer

Based on the analysis and prediction of the signs of \(\Delta S^{\circ}\) for each reaction: (a) For the reaction \(2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NaCl}(s)\), we predict that \(\Delta S^{\circ}<0\) (negative). (b) For the reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\), we predict that \(\Delta S^{\circ}<0\) (negative). (c) For the reaction \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(\mathrm{~g})\longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)\), we estimate that \(\Delta S^{\circ}<0\) (negative). (d) For the reaction \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)\longrightarrow 2 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g)\), we predict that \(\Delta S^{\circ}>0\) (positive).

Step-by-step solution

Determine the change in the number of particles and phases

In this reaction, there are 3 particles on the reactant side (2 Na atoms and 1 Cl2 molecule) and 2 particles on the product side (2 NaCl formula units). Also, we have 1 gas molecule on the reactant side and no gas molecules on the product side. So, there is a decrease in the number of particles and gases which indicates a decrease in entropy.

Predict the sign of \(\Delta S^{\circ}\)

Based on the decrease in particles and gases, we can predict that the change in entropy would be negative. So, \(\Delta S^{\circ}<0\) for this reaction. (b) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\)

Determine the change in the number of particles and phases

In this reaction, there are 3 particles on the reactant side (2 NO molecules and 1 O2 molecule) and 2 particles on the product side (2 NO2 molecules). All species are in the gaseous phase. So, there is a decrease in the number of particles but no change in the number of gases.

Predict the sign of \(\Delta S^{\circ}\)

Since we only have a decrease in particles and no change in gases, the change in entropy would likely be negative. Therefore, \(\Delta S^{\circ}<0\) for this reaction. (c) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(\mathrm{~g})\longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)\)

Determine the change in the number of particles and phases

In this reaction, there are 4 particles on the reactant side (1 C2H4 molecule and 3 O2 molecules) and 5 particles on the product side (2 CO2 molecules and 3 H2O molecules). We also have 4 gases on the reactant side and 2 gases on the product side. While we have an increase in the number of particles, we have a decrease in the number of gases.

Predict the sign of \(\Delta S^{\circ}\)

Due to the opposing factors of increase in particles and decrease in gases, predicting the sign of \(\Delta S^{\circ}\) is not as straightforward. However, since we have fewer gases in the products, we can estimate that the entropy would decrease, making \(\Delta S^{\circ}<0\) for this reaction. (d) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)\longrightarrow 2 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g)\)

Determine the change in the number of particles and phases

In this reaction, we have 2 particles on the reactant side (1 NH4NO3 formula unit and 1 H2O molecule) and 3 particles on the product side (2 NH3 molecules and 1 O2 molecule). We also have no gases on the reactant side and 3 gases on the product side. In this case, we have an increase in the number of particles and gases.

Predict the sign of \(\Delta S^{\circ}\)

Given the increase in both particles and gases indicating an increase in entropy, we predict that \(\Delta S^{\circ}>0\) for this reaction.

Key concepts

Entropy Change

When we delve into the concept of entropy change in chemical reactions, we're observing how disorder or randomness within a system increases or decreases. Entropy, symbolized by ‘S', is a thermodynamic quantity representing the unavailability of a system's energy to do work. It's also a measure of disorder or randomness.

An increase in entropy (positive \( \Delta S^\circ \)) implies a system becomes more disordered after the reaction, typical of endothermic reactions where heat is absorbed from the surroundings. This can include the formation of more molecules or a change from a more ordered phase (solid or liquid) to a less ordered phase (gas). In contrast, a decrease in entropy (negative \( \Delta S^\circ \)) is exemplified by exothermic reactions where heat is released, and the system becomes more ordered, such as when gas molecules combine to form fewer solid or liquid products.

Chemical Thermodynamics

Chemical thermodynamics is the study of energy transformations accompanying chemical reactions and physical changes. It combines the concepts of energy, enthalpy, and entropy to determine the favorability of reactions. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. This fundamental law influences how we predict the spontaneity of chemical processes; reactions where the total entropy of the system and its surroundings increases (\(+\Delta S_{total}\)) are generally spontaneous. However, understanding this broad concept requires applying it to specific situations, such as predicting the entropy change of various reactions, which is often done by considering changes in molecule count and states of matter.

Predicting Entropy Sign

Predicting the sign of an entropy change (\(*\Delta S^\circ*\)) in a chemical reaction is crucial for understanding the direction in which the disorder of the system changes. Several factors influence this prediction:

• The number of molecules: An increase in the total number of gas molecules usually leads to a positive entropy change.
• Phase changes: The transformation from solid to liquid to gas increases entropy, while the reverse decreases it.
• Temperature and pressure conditions: Higher temperatures can increase molecular motion and randomness, thus increasing entropy.

Combining these indicators can help estimate whether the entropy change for a reaction is positive or negative. However, it's essential to consider the overall context of the reaction, as sometimes opposing factors may complicate the prediction.

Phase Changes and Entropy

Phase changes considerably influence entropy in a chemical system. During a phase transition, such as melting, boiling, or sublimation, a substance undergoes a transition between different states of matter. Each of these changes is associated with an increase in entropy because the molecules in the system spread out and occupy a larger volume, which leads to a greater degree of randomness.

For instance, when ice melts into water, its molecules, previously arranged in a structured lattice, gain energy and transition into a less organized, liquid state. This is why melting and boiling are associated with an increase in entropy (\(+\Delta S^\circ\)). Conversely, when a gas condenses into a liquid or a liquid freezes into a solid, the entropy decreases (\(-\Delta S^\circ\)) as the molecules lose energy and become more ordered.