Problem 54 - Chemistry The Central Science

Question

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

Step-by-Step Solution

Answer

For reaction (a), we have \(\Delta S^{\circ} = -307.8\: \mathrm{J \cdot K^{-1} \cdot mol^{-1}}\), indicating a decrease in entropy and a more ordered state. Follow the same 3-step process to find the \(\Delta S^{\circ}\) and interpret the entropy change for the other reactions.

Step 1: Find \(S^{\circ}\) values for molecules involved in the reaction

Using Appendix C, look up the \(S^{\circ}\) values for the involved molecules: - \(S^{\circ}_{\mathrm{HNO}_{3}(g)} = 266.4\: \mathrm{J \cdot K^{-1} \cdot mol^{-1}}\) - \(S^{\circ}_{\mathrm{NH}_{3}(g)} = 192.5\: \mathrm{J \cdot K^{-1} \cdot mol^{-1}}\) - \(S^{\circ}_{\mathrm{NH}_{4} \mathrm{NO}_{3}(s)} = 151.1\: \mathrm{J \cdot K^{-1} \cdot mol^{-1}}\)

Step 2: Calculate \(\Delta S^{\circ}\) using the formula

Plug the \(S^{\circ}\) values into the formula: \(\Delta S^{\circ} = S^{\circ}_{\mathrm{NH}_{4}\mathrm{NO}_{3}(s)} - (S^{\circ}_{\mathrm{HNO}_{3}(g)} + S^{\circ}_{\mathrm{NH}_{3}(g)})\) \(\Delta S^{\circ} = 151.1 - (266.4 + 192.5) = -307.8\: \mathrm{J \cdot K^{-1} \cdot mol^{-1}}\)

Step 3: Interpret the entropy change

Since \(\Delta S^{\circ}\) is negative, the entropy of the system decreases, meaning the reaction leads to a more ordered state. For the other reactions, follow the same 3-step process to find the \(\Delta S^{\circ}\) and interpret the entropy change.

Key Concepts

Calculating ΔS°

Understanding how to calculate the standard entropy change (ΔS°) in chemical reactions is crucial for students studying thermodynamics. ΔS° is computed by subtracting the sum of the standard molar entropies (S°) of the reactants from the sum of the S° of the products. The formula applied is: ΔS° = ΣS°(products) — ΣS°(reactants). The standard molar entropy values of substances can often be found in a textbook appendix or a reliable database.

Let's take the reaction where HNO3(g) and NH3(g) combine to form NH4NO3(s) as an example. By looking up the standard entropy values and inserting them into the formula, students arrive at the entropy change for this specific reaction. It’s essential to account for the coefficients of each compound in the reaction, which represent the number of moles, when summing the entropies.

The calculation of ΔS° allows students to predict whether a reaction will result in a more ordered (negative ΔS°) or a more disordered (positive ΔS°) system. Ensuring that you carefully perform these calculations, often necessary for identifying spontaneous reactions, is key to mastering thermodynamics in chemistry.

Standard Molar Entropy Values

Standard molar entropy values (S°) are a set of thermodynamic properties that quantify the amount of disorder in a mole of substance at a standard condition — typically 1 bar or 1 atmosphere of pressure. These values can be thought of as the entropic 'fingerprints' of substances, providing insight into their thermodynamic behavior.

For instance, gases tend to have higher S° values compared to solids, reflecting the higher degree of disorder in the gaseous state. In the educational exercise, students see this principle illustrated clearly when comparing the S° of NH3(g) to that of NH4NO3(s). A higher entropy value for a gas underscores that gas particles are more dispersed and have more translational motion.

Using standard molar entropy values allows us to compare different substances and predict the entropic outcome of reactions. Data from tables like Appendix C are integral in these calculations. When analyzing a reaction, students learn why some reactions intuitively lead to a positive or negative ΔS°, reinforcing the relevancy of these abstract numbers in concrete contexts.

Entropy Interpretation

Interpreting entropy changes in reactions offers rich learning experiences for students. The sign and magnitude of the entropy change (ΔS°) communicate much about the molecular events occurring during a reaction. A negative ΔS°, as seen in the reaction forming NH4NO3(s), indicates that the products are more ordered than the reactants. This could be due to a gas transforming to a solid or a decrease in the number of molecules, both instances of increased order.

Students should note that entropy is only one piece of the puzzle. While a negative ΔS° might suggest non-spontaneity, other factors such as enthalpy changes and temperature also play significant roles in determining a reaction's spontaneity via the Gibbs free energy equation.

The use of realistic examples, like those provided in the educational exercise, encourages students to relate abstract thermodynamic concepts to observable chemical phenomena, thus deepening their understanding of entropy and its implications in chemistry.

Thermodynamics in Chemistry

Thermodynamics in chemistry is the branch of the science that deals with the energy changes associated with chemical reactions and physical transformations. It gives us the tools to predict whether processes will occur spontaneously in nature.

The central concept in this area is the Gibbs free energy equation: G = H - TΔS, where G is the Gibbs free energy, H is the enthalpy, T is the temperature, and ΔS is the entropy change of the system. This equation synergizes the concepts of enthalpy and entropy and introduces the influence of temperature on spontaneous behavior.

In the classroom, emphasizing the importance of thermodynamic principles in everyday chemical processes can help students appreciate the subject's scope and utility. By combining calculated entropy changes with enthalpy and temperature data, students can assess the energetics of reactions. They learn that while negative ΔS° and ΔH° values usually indicate non-spontaneity and exothermic reactions respectively, additional temperature info is required to make definitive predictions about spontaneous behavior. Illustrating these interrelations provides key insights into the dynamism of chemistry.