Question

Using \(S^{\circ}\) values from Appendix \(C\), calculate \(\Delta S^{\circ}\) values for the following reactions. In each case account for the sign of \(\Delta S^{n}\). (a) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{Be}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BeO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

(a) \(\Delta S^\circ = S^\circ_{C_2H_6} - (S^\circ_{C_2H_4} + S^\circ_{H_2}) = -33.2 \ \text{J/mol} \cdot \text{K}\): The entropy change is negative, indicating a decrease in disorder. (b) \(\Delta S^\circ = 2S^\circ_{NO_2} - S^\circ_{N_2O_4} = +96.0 \ \text{J/mol} \cdot \text{K}\): The entropy change is positive, indicating an increase in disorder. (c) \(\Delta S^\circ = (S^\circ_{BeO} + S^\circ_{H_2O}) - S^\circ_{Be(OH)_2} = +107.8 \ \text{J/mol} \cdot \text{K}\): The entropy change is positive, indicating an increase in disorder. (d) \(\Delta S^\circ = (2S^\circ_{CO_2} + 4S^\circ_{H_2O}) - (2S^\circ_{CH_3OH} + 3S^\circ_{O_2}) = -281.7 \ \text{J/mol} \cdot \text{K}\): The entropy change is negative, indicating a decrease in disorder.

Step-by-step solution

Look up \(S^{\circ}\) values for each species

Using Appendix C, we need to find the standard molar entropy values for all reactants and products involved in each reaction.

Calculate \(\Delta S^{\circ}\) for reaction (a)

For the reaction (a): \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\) Calculate \(\Delta S^\circ\) using the formula: \(\Delta S^\circ = \sum n_i S^\circ_i(\text{products}) - \sum n_i S^\circ_i(\text{reactants})\) Substitute the values of \(S^\circ\) from Appendix C and solve.

Calculate \(\Delta S^{\circ}\) for reaction (b)

For the reaction (b): \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) Calculate \(\Delta S^\circ\) using the formula: \(\Delta S^\circ = \sum n_i S^\circ_i(\text{products}) - \sum n_i S^\circ_i(\text{reactants})\) Substitute the values of \(S^\circ\) from Appendix C and solve.

Calculate \(\Delta S^{\circ}\) for reaction (c)

For the reaction (c): \(\mathrm{Be}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BeO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) Calculate \(\Delta S^\circ\) using the formula: \(\Delta S^\circ = \sum n_i S^\circ_i(\text{products}) - \sum n_i S^\circ_i(\text{reactants})\) Substitute the values of \(S^\circ\) from Appendix C and solve.

Calculate \(\Delta S^{\circ}\) for reaction (d)

For the reaction (d): \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) Calculate \(\Delta S^\circ\) using the formula: \(\Delta S^\circ = \sum n_i S^\circ_i(\text{products}) - \sum n_i S^\circ_i(\text{reactants})\) Substitute the values of \(S^\circ\) from Appendix C and solve.

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

For each calculated \(\Delta S^\circ\) value, analyze its sign: 1. If \(\Delta S^\circ > 0\), the reaction results in an increase in disorder. 2. If \(\Delta S^\circ < 0\), the reaction results in a decrease in disorder. 3. If \(\Delta S^\circ = 0\), the entropy does not change during the reaction. With these interpretations, explain the change in disorder for each reaction (a), (b), (c), and (d).

Key concepts

Thermodynamics

Thermodynamics is a field of physics that deals with the relationships between heat, work, temperature, and energy. The foundational understanding of thermodynamics is essential for analyzing entropy changes during chemical reactions. It is built upon the four laws of thermodynamics, each of which describes different aspects of energy flow and its conservation. The primary objectives in thermodynamics are to understand how energy is transferred in a system and how it affects matter.

Key components in thermodynamics include systems and surroundings. A system is the part of the universe we focus on, such as a chemical reaction or physical process, while the surroundings encompass everything else. Processes in thermodynamics can be isolated, closed, or open, depending on their interaction with their surroundings. This context helps us analyze changes in entropy, which signifies disorder or randomness in a system.

Entropy Change

In chemistry, entropy change (\( \Delta S \)) is a central concept in understanding how energy dispersal occurs in chemical reactions. When exploring entropy changes, we relate them to the second law of thermodynamics, which states that the total entropy of an isolated system can never decrease over time. Instead, it either increases or remains constant, describing a move toward greater disorder.

Calculating entropy change involves taking the difference in standard molar entropy values between the products and the reactants of a chemical reaction. This is expressed in the formula: \[\Delta S^{\circ} = \sum n_i S^{\circ}_i(\text{products}) - \sum n_i S^{\circ}_i(\text{reactants})\]If the result is positive, the reaction results in an increase in disorder. If negative, it indicates a decrease in disorder, and if zero, the entropy remains unchanged.
This calculation is crucial for predicting the spontaneity of chemical reactions, as processes with \( \Delta S > 0 \) are generally spontaneous under constant temperature and pressure.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products with a concurrent change in energy and entropy. Analyzing these reactions requires an understanding of both enthalpy and entropy to determine the overall energy change or Gibbs free energy, \( \Delta G \). A negative \( \Delta G \) indicates a spontaneous reaction, a key factor when assessing \( \Delta S \).

Reactions can be exothermic or endothermic, influencing entropy changes. Exothermic reactions release energy and can increase the disorder, reflected in a positive \( \Delta S \), while endothermic reactions absorb energy and may decrease disorder. Therefore, understanding the composition of reactants and products can help determine the sign and magnitude of \( \Delta S \) for any given reaction.

Standard Molar Entropy Values

Standard molar entropy (\( S^{\circ} \)) values are fundamental to calculating the entropy of a chemical reaction under standard conditions (1 atm pressure and 25°C temperature). Each substance in a chemical reaction has a unique \( S^{\circ} \) value that reflects its intrinsic disorder at standard conditions.

To compute the \( \Delta S^{\circ} \) of a reaction, one must look up the \( S^{\circ} \) values for all reactants and products and apply them to the entropy change formula. The values typically come from standardized charts or appendices in chemistry textbooks. Accurately using these values ensures correct calculations, providing insights into the degree of disorder and the entropy change associated with a reaction.
With these values, students can determine whether a chemical process leads to more ordered or disordered states, crucial for predicting reaction spontaneity.