Problem 63 - Chemistry The Molecular Nature of Matter

Question

Predict the sign of \(\Delta S\) for the following reactions. (a) The sublimation of \(\mathrm{I}_{2}(s)\) (b) \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)\) (c) \(2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NaCl}(s)\)

Step-by-Step Solution

Answer

\(\Delta S\) is positive for (a), negative for (b), and negative for (c).

Step 1 - Understanding \(\Delta S\)

The symbol \(\Delta S\) represents the change in entropy for a reaction. A positive \(\Delta S\) indicates an increase in randomness or disorder, while a negative \(\Delta S\) signifies a decrease. To predict the sign of \(\Delta S\), consider the states of the reactants and products and how they represent disorder.

Step 2 - Analyzing Sublimation of \(\mathrm{I}_{2}(s)\)

Sublimation is the phase change from solid to gas, which generally increases entropy as the gas state is more disordered than the solid state. Since \(\mathrm{I}_{2}\) is changing from solid to gas form, there will be an increase in disorder. Therefore, the sign of \(\Delta S\) for the sublimation of \(\mathrm{I}_{2}(s)\) is positive.

Step 3 - Evaluating the Synthesis of \(\mathrm{CH}_{3}\mathrm{OH}(g)\)

This reaction involves the combination of one molecule of \(\mathrm{CO}(g)\) and two molecules of \(\mathrm{H}_{2}(g)\) to form a single molecule of \(\mathrm{CH}_{3}\mathrm{OH}(g)\). There's a decrease in the number of gas molecules, which usually means a decrease in disorder. Hence, the sign of \(\Delta S\) for this synthesis is likely to be negative.

Step 4 - Considering the Formation of \(\mathrm{NaCl}(s)\) from Elements

In this reaction, solid sodium and chlorine gas react to form solid sodium chloride. The products are in a more ordered state than the reactants due to the solidification and the reduction in the number of gas particles. This ordering signifies a negative \(\Delta S\) because the disorder decreases as a result of the reaction.

Key Concepts

Predicting Entropy Changes

Entropy, denoted as \( \Delta S \), is a measure of randomness or disorder within a system. When predicting whether entropy increases or decreases during a chemical reaction, one must consider the physical states of reactants and products, as well as the complexity of the molecules involved.

• Phase Changes: Transitioning from solid to liquid to gas usually increases entropy, with gases displaying the most disorder.
• Molecular Complexity: Larger and more complex molecules generally have higher entropy compared to simpler ones.
• Number of Particles: A reaction that produces more particles (atoms or molecules) will likely lead to increased entropy.

If a reaction leads to more organized and fewer particles, expect a decrease in entropy (\( \Delta S < 0 \)). Conversely, if the reaction results in less organized, more complex, or a greater number of particles, the entropy is likely to increase (\( \Delta S > 0 \)).

Sublimation and Entropy

Sublimation—the process where a solid changes directly into a gas—significantly increases the entropy of a substance. This leap from a tightly packed solid structure to the freedom of gaseous form allows particles to move more independently and occupy a larger volume.

Example of Sublimation:

In the case of iodine (\( \mathrm{I}_{2}(s) \) subliming, the tightly packed solid structure becomes a spread out gas, creating far more microstates in which the iodine can exist. Since entropy is tied directly to the number of microstates, sublimation of iodine leads to a higher entropy, shown by a positive \( \Delta S \).

Synthesis Reaction Entropy

In synthesis reactions, multiple reactants combine to form a single product. The change in entropy depends on the states of reactants and products, and their complexity. Generally, if a synthesis reaction results in fewer gas molecules, it means the entropy likely decreases, as gases have higher entropy than liquids and solids.

Synthesis Example:

The synthesis of methanol (\( \mathrm{CH}_{3} \mathrm{OH}(g) \) from \( \mathrm{CO}(g) \) and \( 2 \mathrm{H}_{2}(g) \) results in fewer gas molecules. Originally, there are three gas molecules, which become a single molecule of methanol gas. Even though the product is still a gas, the decrease in the number of gas particles signalizes a negative \( \Delta S \) due to reduced disorder.

Chemical Reaction Phase Changes

During a chemical reaction, a phase change can result in a significant alteration in entropy. When substances convert from gas to liquid or solid, there is a decrease in entropy because the particles are more restricted in their movement.

Phase Change in Formation of NaCl:

When sodium metal (\( 2 \mathrm{Na}(s) \) reacts with chlorine gas (\( \mathrm{Cl}_{2}(g) \) to form solid sodium chloride (\( 2 \mathrm{NaCl}(s) \), the overall disorder is reduced. The gas particles are constrained into a solid lattice structure, which reduces the number of possible microstates. Therefore, the formation of a solid from a gas and a solid typically involves a negative change in entropy (\( \Delta S < 0 \)), signaling a move towards a more ordered system.