Problem 52 - Chemistry The Central Science

Question

Calculate \(\Delta S^{\circ}\) values for the following reactions by using tabulated \(S^{\circ}\) values from Appendix \(\mathrm{C} .\) In each case, explain the sign of \(\Delta S^{\circ} .\) $$ \begin{array}{l}{\text { (a) } \mathrm{HNO}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{NO}_{3}(s)} \\ {\text { (b) } 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g)} \\ {\text { (c) } \mathrm{CaCO}_{3}(s, \text { calcite })+2 \mathrm{HCl}(g) \rightarrow} \\\ {\mathrm{CaCl}_{2}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)}\\\ {\text { (d) } 3 \mathrm{C}_{2} \mathrm{H}_{6}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)+6 \mathrm{H}_{2}(g)}\end{array} $$

Step-by-Step Solution

Answer

(a) \(\Delta S^{\circ} = -307.9 \: \text{J/mol K}\): The negative value indicates a decrease in disorder, as a gaseous molecule is condensed into a solid. (b) \(\Delta S^{\circ} = 414.8 \: \text{J/mol K}\): The positive value signifies an increase in disorder, due to the formation of gaseous O2 molecules from solid Fe2O3. (c) \(\Delta S^{\circ} = -104.8 \: \text{J/mol K}\): The negative value indicates a reduction in disorder, as a gaseous molecule (HCl) transforms into a solid species (CaCl2). (d) \(\Delta S^{\circ} = 50.4 \: \text{J/mol K}\): The positive value shows an increase in disorder, primarily due to the production of more gaseous H2 molecules, despite the formation of liquid C6H6.

Step 1: (a) HNO3(g) + NH3(g) → NH4NO3(s)

The standard molar entropies for the species in this reaction are as follows (in J/mol K): - HNO3(g): 266.4 - NH3(g): 192.8 - NH4NO3(s): 151.1 Using the formula, we calculate ∆S° for the reaction: \(\Delta S^{\circ} = (1 \times 151.1) - (1 \times 266.4 + 1 \times 192.8) = -307.9 \: \text{J/mol K}\) The negative value of ∆S° indicates that the disorder is decreasing, mainly because one of the gaseous molecules in the reactants is condensed into a solid.

Step 2: (b) 2Fe2O3(s) → 4Fe(s) + 3O2(g)

The standard molar entropies for the species in this reaction are as follows (in J/mol K): - Fe2O3(s): 87.4 - Fe(s): 27.3 - O2(g): 205.2 Calculating ∆S° for the reaction: \(\Delta S^{\circ} = (4 \times 27.3 + 3 \times 205.2) - (2 \times 87.4) = 414.8 \: \text{J/mol K}\) The positive ∆S° signifies that the disorder is increasing, largely due to the production of three moles of gaseous O2 molecules from two moles of solid Fe2O3.

Step 3: (c) CaCO3(s, calcite) + 2HCl(g) → CaCl2(s) + CO2(g) + H2O(l)

The standard molar entropies for the species in this reaction are as follows (in J/mol K): - CaCO3(s, calcite): 92.9 - HCl(g): 186.9 - CaCl2(s): 55.5 - CO2(g): 213.8 - H2O(l): 69.95 Calculating ∆S° for the reaction: \(\Delta S^{\circ} = (1 \times 55.5 + 1 \times 213.8 + 1 \times 69.95) - (1 \times 92.9 + 2 \times 186.9) = -104.8 \: \text{J/mol K}\) The decreasing entropy signified by the negative ∆S° indicates disorder reduction, primarily due to the transformation of a gaseous molecule (HCl) into a solid species (CaCl2).

Step 4: (d) 3C2H6(g) → C6H6(l) + 6H2(g)

The standard molar entropies for the species in this reaction are as follows (in J/mol K): - C2H6(g): 229.6 - C6H6(l): 173.4 - H2(g): 130.7 Calculating ∆S° for the reaction: \(\Delta S^{\circ} = (1 \times 173.4 + 6 \times 130.7) - (3 \times 229.6) = 50.4 \: \text{J/mol K}\) The positive ∆S° shows a disorder increase, primarily owing to the production of six moles of gaseous H2 molecules from three moles of gaseous C2H6 molecules. Although C6H6 forms as a liquid, the production of a higher number of gaseous molecules results in increased disorder.

Key Concepts

Standard Molar Entropy

Standard molar entropy, symbolized as \( S^\circ \), is a measure of the amount of thermal energy in a substance per mole per degree Kelvin at a standard state. It reflects the degree of disorder or randomness in a system. Different substances possess varying levels of standard molar entropy, influenced by molecular complexity and phase.

Standard molar entropy values are crucial for calculating the change in entropy during a chemical reaction. These values are typically found in tables, where they are listed for different substances, and they help in determining whether the reaction leads to an increase or decrease in system disorder.

Substances in a gaseous state generally have higher \( S^\circ \) values than those in a solid state due to greater disordered molecular motion. Understanding these values and comparing them across the reactants and products of a reaction is fundamental in assessing the change in entropy.

Entropy Change Calculation

Entropy change calculation, represented as \( \Delta S^\circ \), is carried out using standard molar entropy values from the reactants and products in a reaction. To find \( \Delta S^\circ \), the formula is used:
\[\Delta S^\circ = \sum S^\circ(\text{Products}) - \sum S^\circ(\text{Reactants})\]This formula helps in calculating the direction and magnitude of disorder change for a chemical reaction.

For each substance involved in the reaction, its coefficient in the balanced reaction equation is multiplied by its standard molar entropy. The total entropy of the products is then subtracted from the total entropy of the reactants. A positive \( \Delta S^\circ \) suggests an increase in disorder, while a negative \( \Delta S^\circ \) indicates a decrease.

Calculating \( \Delta S^\circ \) provides insight into the feasibility of a reaction and its alignment with entropy-driven processes.

Gaseous and Solid Species

Understanding the roles of gaseous and solid species is key to analyzing reactions regarding entropy.

Each physical state—solid, liquid, or gas—exhibits different degrees of molecular motion and arrangement, contributing to its entropy level. In general:

• Gases: High entropy due to random, high-speed molecular motion.
• Liquids: Moderate entropy, molecules are more constrained compared to gases.
• Solids: Low entropy as molecules are closely packed and vibrate minimally.

Reactions involving gaseous species often show significant entropy change because gases display more disorder than solids or liquids. For example, a reaction forming a gas from solid reactants usually increases \( \Delta S^\circ \), whereas converting a gas into a solid decreases it.

When analyzing chemical reactions, identifying the phases of reactants and products helps understand the overall change in entropy and the associated physical implications.

Significance of Entropy Change

Entropy change plays a significant role in predicting whether a reaction is spontaneous and its likely direction. A spontaneous reaction is one that occurs naturally without external influence, and in isolated systems, it is often regulated by an increase in entropy.

The sign of \( \Delta S^\circ \) (positive or negative) provides valuable information:

• Positive \( \Delta S^\circ \): Increased randomness, possibly favoring spontaneity.
• Negative \( \Delta S^\circ \): Decreased randomness, potentially non-spontaneous unless accompanied by other favorable conditions, like enthalpy changes.

Moreover, entropy change is also essential in computing the Gibbs free energy change (\( \Delta G^\circ \)), which further helps determine spontaneity, as it incorporates both entropy and enthalpy changes. Understanding the implications of \( \Delta S^\circ \) is key to grasping the thermodynamic motivations driving reactions.