Problem 48 - Chemistry: The Central Science in SI Units, Global Edition

Question

Predict which member of each of the following pairs has the greater standard entropy at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) or \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\), (b) \(\mathrm{CO}(g)\) or \(\mathrm{CO}_{2}(g)\) (c) \(1 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{4}(g)\) or \(2 \mathrm{~mol} \mathrm{NO}_{2}(g)\) (d) \(\mathrm{HCl}(g)\) or \(\mathrm{HCl}(a q) .\) Use Appendix \(\mathrm{C}\) to find the standard entropy of each substance.

Step-by-Step Solution

Answer

In summary, the substances with greater standard entropy at $25^{\circ} \mathrm{C}$ in each pair are: (a) \( C_6H_6(g) \) (b) \( CO_2(g) \) (c) \( 2 \thinspace mol \thinspace NO_2(g) \) (d) \( HCl(g) \)

Step 1: Examine phase changes

In general, the gas phase of a substance has a higher standard entropy than the liquid or solid phase, as gas particles are more spread out and have more potential to move freely. (a) C6H6(l) or C6H6(g) Since C6H6(g) is in the gas phase and C6H6(l) is in liquid phase, the standard entropy of C6H6(g) will be greater.

Step 2: Compare molecular complexity

When comparing molecules, more complex or larger molecules tend to have a greater standard entropy due to increased molecular vibration. (b) CO(g) or CO2(g) CO2(g) is a larger and more complex molecule than CO(g). Therefore, the standard entropy of CO2(g) will be greater.

Step 3: Consider stoichiometry

If a reaction involves more moles of gaseous products than reactants, the products will generally have a greater standard entropy. (c) 1 mol N2O4(g) or 2 mol NO2(g) The reaction involves the decomposition of N2O4(g) to form 2 moles of NO2(g). Since there are more moles of gas in the products, the standard entropy of 2 mol NO2(g) will be greater.

Step 4: Examine a combination of phase change and dissociation in solution

When a gaseous substance dissolves in a liquid, the standard entropy will generally decrease due to increased intermolecular interactions and a decrease in randomness. (d) HCl(g) or HCl(aq) HCl(aq) involves the dissociation of HCl(g) in water. Since the gas particles become restrained in the solution, the standard entropy of HCl(g) will be greater. In summary, the substances with greater standard entropy in each pair are: (a) C6H6(g) (b) CO2(g) (c) 2 mol NO2(g) (d) HCl(g)

Key Concepts

Phase Changes and Entropy

Understanding the role of phase changes is crucial when predicting entropy. Entropy, a measure of disorder, is often higher in gases compared to liquids and solids. This is because gas particles can move freely, filling any available space, which maximizes randomness. In liquids, molecules are more closely packed, which restricts movement and lowers entropy.
To illustrate, let's consider benzene ( C_{6}H_{6} ):

• Benzene in gas form ( C_{6}H_{6}(g) ) exhibits higher entropy than its liquid form ( C_{6}H_{6}(l) ).
• This is because the gas form allows particles more freedom and spread compared to the liquid form, which is more structured.

Hence, when comparing different phases of the same substance, always expect the gaseous state to have greater entropy due to its higher molecular freedom.

Molecular Complexity and Entropy

The complexity of a molecule affects its standard entropy. Larger and more complex molecules have many more ways to vibrate, rotate, and be disordered, which tend to increase their entropy. A molecule's size and atomic composition can significantly influence these possible arrangements.
Consider the molecules CO and CO2:

• Carbon dioxide ( CO_{2}(g) ) is more complex and larger compared to carbon monoxide ( CO(g) ).
• Due to carbon dioxide's additional atom, it possesses a higher molecular complexity, leading to greater entropy than carbon monoxide.

So, for molecules with similar states, those with greater molecular complexity tend to have higher entropy. This is because they have more internal degrees of freedom.

Stoichiometry and Entropy

In stoichiometry, the comparison often involves reactions and their products versus reactants. When analyzing stoichiometric relationships, understand that more moles of gas usually increase the system's entropy. This is because more particles mean more disorder.
Consider the reaction of nitrogen dioxide and dinitrogen tetroxide:

• The decomposition of 1 mole of N_{2}O_{4}(g) into 2 moles of NO_{2}(g) results in an increase in the number of gas molecules.
• The increase from 1 to 2 moles indicates greater randomness and, therefore, more entropy in the system of 2 ext{ mol } NO_{2}(g) .

Thus, more moles of gaseous products generally correspond to an increase in standard entropy compared to the reactants in a chemical reaction.

Dissolution in Chemistry: Impact on Entropy

The dissolution, particularly of a gas in a liquid, can affect a substance's entropy significantly. When a gas dissolves in a liquid, the potential for movement (and hence entropy) typically decreases. This is due to the increased intermolecular interactions that restrict gaseous movement, thus reducing disorder.
Let's look at hydrochloric acid:

• When HCl(g) becomes dissolved in water to form HCl(aq) , the ions become surrounded by water molecules, reducing freedom of movement.
• Compared to its gaseous state, the dissolved state HCl(aq) has lower entropy because the water's structured environment limits ionic movement.

As a result, when comparing a dissolved gas with its pure gaseous state, expect the gas to have a higher standard entropy due to its capacity for more random molecular motion.