Problem 40 - Chemistry: An Atoms-Focused Approach

Question

Use the standard molar entropies in Appendix 4 to calculate the \(\Delta S^{\circ}\) value for each of the following reactions of sulfur compounds. a. \(\mathrm{H}_{2} \mathrm{S}(g)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{SO}_{2}(g)\) b. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\) c. \(\mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) d. \(S(g)+O_{2}(g) \rightarrow S O_{2}(g)\).

Step-by-Step Solution

Answer

Question: Calculate the change in standard molar entropy, \(\Delta S^{\circ}\), for the following reactions: a. \(\mathrm{H}_{2} \mathrm{S}(g)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{SO}_{2}(g)\) b. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\) c. \(\mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) d. \(S(g)+O_{2}(g) \rightarrow S O_{2}(g)\) Answer: a. \(\Delta S^{\circ} = -78.20 \textrm{ J/mol K}\) b. \(\Delta S^{\circ} = -189.2 \textrm{ J/mol K}\) c. \(\Delta S^{\circ} = -169.25 \textrm{ J/mol K}\) d. \(\Delta S^{\circ} = -184.5 \textrm{ J/mol K}\)

Step 1: a. Calculation of \(\Delta S^{\circ}\) for the first reaction

For reaction: \(\mathrm{H}_{2} \mathrm{S}(g)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{SO}_{2}(g)\) From Appendix 4: \(S^{\circ}(\mathrm{H}_{2} \mathrm{S}(g))\) = 206.9 J/mol K \(S^{\circ}(\mathrm{O}_{2}(g))\) = 205.2 J/mol K \(S^{\circ}(\mathrm{H}_{2} \mathrm{O}(g))\) = 188.8 J/mol K \(S^{\circ}(\mathrm{SO}_{2}(g))\) = 248.2 J/mol K Applying the formula, we get: \(\Delta S^{\circ} = [(1\times 188.8) + (1\times 248.2)] - [(1\times 206.9) + (\frac{3}{2}\times 205.2)]\) \(\Delta S^{\circ} = 437.00 - 515.20 = -78.20 \textrm{ J/mol K}\)

Step 2: b. Calculation of \(\Delta S^{\circ}\) for the second reaction

For reaction: \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\) From Appendix 4: \(S^{\circ}(\mathrm{SO}_{2}(g))\) = 248.2 J/mol K \(S^{\circ}(\mathrm{O}_{2}(g))\) = 205.2 J/mol K \(S^{\circ}(\mathrm{SO}_{3}(g))\) = 256.2 J/mol K Applying the formula, we get: \(\Delta S^{\circ} = [(2\times 256.2)] - [(2\times 248.2) + (1\times 205.2)]\) \(\Delta S^{\circ} = 512.40 - 701.60 = -189.2 \textrm{ J/mol K}\)

Step 3: c. Calculation of \(\Delta S^{\circ}\) for the third reaction

For reaction: \(\mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) From Appendix 4: \(S^{\circ}(\mathrm{SO}_{3}(g))\) = 256.2 J/mol K \(S^{\circ}(\mathrm{H}_{2} \mathrm{O}(\ell))\) = 69.95 J/mol K \(S^{\circ}(\mathrm{H}_{2} \mathrm{SO}_{4}(a q))\) = 156.9 J/mol K Applying the formula, we get: \(\Delta S^{\circ} = [(1\times 156.9)] - [(1\times 256.2) + (1\times 69.95)]\) \(\Delta S^{\circ} = 156.9 - 326.15 = -169.25 \textrm{ J/mol K}\)

Step 4: d. Calculation of \(\Delta S^{\circ}\) for the fourth reaction

For reaction: \(S(g)+O_{2}(g) \rightarrow S O_{2}(g)\) From Appendix 4: \(S^{\circ}(S(g))\) = 227.5 J/mol K \(S^{\circ}(\mathrm{O}_{2}(g))\) = 205.2 J/mol K \(S^{\circ}(\mathrm{SO}_{2}(g))\) = 248.2 J/mol K Applying the formula, we get: \(\Delta S^{\circ} = [(1\times 248.2)] - [(1\times 227.5) + (1\times 205.2)]\) \(\Delta S^{\circ} = 248.2 - 432.7 = -184.5 \textrm{ J/mol K}\)

Key Concepts

Thermodynamics

Thermodynamics is the study of the principles governing energy and matter. It is an essential field in chemistry and physics. Imagine thermodynamics as the rulebook that nature follows when energy changes occur. For instance, any chemical reaction, like dissolving sugar in water or lighting a match, involves energy transformations. These transformations often involve changes in temperature, phase, and energy dispersal.

Thermodynamics has a few key laws that scientists use to understand these changes:

• The First Law (or the Law of Energy Conservation) - Energy cannot be created or destroyed, only transformed from one form to another.
• The Second Law of Thermodynamics - In any energy transfer, some energy becomes less organized over time, often increasing in randomness, known as entropy.
• The Third Law - As a system reaches absolute zero temperature, its entropy approaches zero, remaining constant beyond that point.

Entropy is a critical concept here, as it measures the amount of disorder or randomness in a system. Understanding thermodynamics can help interpret why reactions occur, predict the energy changes, and calculate properties like standard molar entropy.

Chemical Reactions

Chemical reactions are processes where substances called reactants transform into new substances with different properties. During a reaction, bonds between atoms break and form, leading to the rearrangement of atoms. Take the burning of coal as an example, where carbon combines with oxygen to form carbon dioxide, releasing energy.

In a balanced chemical reaction, the number of each type of atom on the reactant side equals the number on the product side, conserving mass. Reactions could be endothermic, absorbing energy, or exothermic, releasing energy. Thermodynamics aids in understanding these energy shifts.

• Endothermic reactions require heat, making surroundings cooler. Photosynthesis is a classical example.
• Exothermic reactions release heat, warming surroundings. Combustion, where fuel burns, is an exothermic process.

Standard molar entropy comes into the picture when understanding the energy balance in reactions, especially by calculating the change in entropy to predict reaction spontaneity.

Entropy Change Calculations

Entropy change (\(\Delta S^{\circ}\)) calculations are fundamental in understanding thermodynamic processes. Entropy quantifies the amount of disorder within a system, and during chemical reactions, it helps predict spontaneity. By calculating \(\Delta S^{\circ}\), one determines if a reaction increases or decreases the disorder.

The formula used is:\[\Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}}\]This formula subtracts the sum of entropies of reactants from the sum of entropies of products. When the result is positive, the reaction increases entropy, possibly spontaneous under the right conditions. A negative value indicates a decrease, often non-spontaneous without continuous input of energy.

• For example, in reaction step a: The values from the Appendix are plugged into the formula, carefully accounting for stoichiometric coefficients, leading to calculated results.
• Changes in states, such as solid to liquid, often involve significant entropy changes due to increased molecular freedom.

Learning to calculate entropy changes aids in understanding chemical equilibrium, reaction mechanisms, and the feasibility of reactions in different conditions.