Problem 74 - Chemistry: Principles and Reactions

Question

Theoretically, one can obtain zinc from an ore containing zinc sulfide, \(\mathrm{ZnS}\), by the reaction $$ \mathrm{ZnS}(s) \longrightarrow \mathrm{Zn}(s)+\mathrm{S}(s) $$ (a) Show by calculation that this reaction is not feasible at \(25^{\circ} \mathrm{C}\). (b) Show that by coupling the above reaction with the reaction $$ \mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) $$ the overall reaction, in which \(\mathrm{Zn}\) is obtained by roasting in oxygen, is feasible at \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

Based on the step-by-step solution above, we can determine that Reaction 1 (ZnS -> Zn + S) is not feasible at 25°C since ΔG° > 0. However, when coupled with Reaction 2 (S + O2 -> SO2), the overall reaction of obtaining Zn by roasting in oxygen becomes feasible at 25°C as the ΔG° for the overall reaction is negative (-92.1 kJ/mol).

Step 1: Gather the necessary data

Obtain the \(\Delta H^{\circ}_f\) and \(S^{\circ}_m\) values for the relevant substances involved in both reactions (from an appendix or chemistry database): Substance | \(\Delta H^{\circ}_f\) (kJ/mol) | \(S^{\circ}_m\) (J/mol·K) ------------------|-----------------------------|------------------------ \(\mathrm{ZnS}(s)\) | \(-206.0\) | \(80.5\) \(\mathrm{Zn}(s)\) | \(0\) | \(41.6\) \(\mathrm{S}(s)\) | \(0\) | \(31.8\) \(\mathrm{O}_{2}(g)\) | \(0\) | \(205.2\) \(\mathrm{SO}_{2}(g)\) | \(-296.9\) | \(248.2\)

Step 2: Compute the \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for Reaction 1

Use the \(\Delta H^{\circ}_f\) and \(S^{\circ}_m\) values to determine the standard enthalpy change \((\Delta H^{\circ})\) and standard entropy change \((\Delta S^{\circ})\) for the reaction \(\mathrm{ZnS}(s) \longrightarrow \mathrm{Zn}(s)+\mathrm{S}(s)\). The equation is: \(\Delta H^{\circ} = \Sigma \Delta H^{\circ}_{\text{products}} - \Sigma \Delta H^{\circ}_{\text{reactants}}\) \(\Delta S^{\circ} = \Sigma S^{\circ}_{\text{m, products}} - \Sigma S^{\circ}_{\text{m, reactants}}\) For Reaction 1: \(\Delta H^{\circ} = [(0) + (0)] - [(-206.0)] = 206.0\) kJ/mol \(\Delta S^{\circ} = [(41.6) + (31.8)] - [(80.5)] = -7.1\) J/mol·K

Step 3: Compute the \(\Delta G^{\circ}\) for Reaction 1

Now, we will compute the standard Gibbs free energy change \((\Delta G^{\circ})\) using the equation: \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\) For Reaction 1 at \(25^{\circ} \mathrm{C}\) (298 K): \(\Delta G^{\circ} = (206.0\,\text{kJ/mol}) - (298\,\text{K})(-7.1\,\text{J/mol·K})(\frac{1\,\text{kJ}}{1000\,\text{J}}) = 206.0\,\text{kJ/mol} - 2.1\text{ kJ/mol} = 208.1\,\text{kJ/mol}\) Since \(\Delta G^{\circ} > 0\), Reaction 1 is not feasible at \(25^{\circ} \mathrm{C}\).

Step 4: Compute the \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for Reaction 2

Next, compute the standard enthalpy change \((\Delta H^{\circ})\) and standard entropy change \((\Delta S^{\circ})\) for the reaction \(\mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)\): For Reaction 2: \(\Delta H^{\circ} = [(-296.9)] - [(0) + (0)] = -296.9\) kJ/mol \(\Delta S^{\circ} = [(248.2)] - [(31.8) + (205.2)] = 11.2\) J/mol·K

Step 5: Compute the \(\Delta G^{\circ}\) for Reaction 2

Compute the standard Gibbs free energy change \((\Delta G^{\circ})\) for Reaction 2 at \(25^{\circ} \mathrm{C}\) (298 K): \(\Delta G^{\circ} = (-296.9\,\text{kJ/mol}) - (298\,\text{K})(11.2\,\text{J/mol·K})(\frac{1\,\text{kJ}}{1000\,\text{J}}) = -296.9\,\text{kJ/mol} - 3.3\,\text{kJ/mol} = -300.2\,\text{kJ/mol}\)

Step 6: Compute the overall reaction by coupling Reaction 1 and Reaction 2

Couple Reaction 1 and Reaction 2 to obtain the following overall reaction: \(\mathrm{ZnS}(s) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{Zn}(s) + \mathrm{SO}_{2}(g)\) To determine the \(\Delta G^{\circ}\) for the overall reaction, sum the \(\Delta G^{\circ}\) values from Reaction 1 and Reaction 2: \(\Delta G^{\circ}_{\text{overall}} = \Delta G^{\circ}_{1} + \Delta G^{\circ}_{2} = 208.1\,\text{kJ/mol} - 300.2\,\text{kJ/mol} = -92.1\,\text{kJ/mol}\) Since \(\Delta G^{\circ}_{\text{overall}} < 0\), the overall reaction of obtaining \(\mathrm{Zn}\) by roasting in oxygen is feasible at \(25^{\circ} \mathrm{C}\).

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, often denoted as \( \Delta G \), is a fundamental concept in thermodynamics. It helps predict the direction and feasibility of a chemical reaction at constant temperature and pressure. The equation used for calculating Gibbs Free Energy is \( \Delta G = \Delta H - T \Delta S \), where \( \Delta H \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy.
A negative \( \Delta G \) indicates that a reaction is spontaneous, meaning it can occur without the input of additional energy. Conversely, if \( \Delta G \) is positive, the reaction is non-spontaneous or not feasible under the given conditions.
It’s important to note that \( \Delta G \) does not inform us about the rate of reaction but only the thermodynamic favorability.

Standard Enthalpy Change

Standard Enthalpy Change, symbolized as \( \Delta H^{\circ} \), refers to the heat absorbed or released during a chemical reaction at standard conditions (298 K and 1 atm). It is calculated as the difference between the enthalpies of the products and the reactants, represented by the equation \( \Delta H^{\circ} = \Sigma \Delta H^{\circ}_{\text{products}} - \Sigma \Delta H^{\circ}_{\text{reactants}} \).
Enthalpy changes are crucial for understanding whether a reaction is endothermic or exothermic. If \( \Delta H^{\circ} \) is negative, the reaction releases heat and is exothermic. On the contrary, a positive \( \Delta H^{\circ} \) signifies an endothermic process where heat is absorbed.
In the exercise, calculating \( \Delta H^{\circ} \) helped determine that the reaction of zinc sulfide decomposition is endothermic, making it non-spontaneous at room temperature.

Standard Entropy Change

Standard Entropy Change, denoted as \( \Delta S^{\circ} \), measures the change in disorder or randomness during a chemical reaction. The standard entropy change can be calculated with the formula \( \Delta S^{\circ} = \Sigma S^{\circ}_{\text{m, products}} - \Sigma S^{\circ}_{\text{m, reactants}} \).
Entropy is a crucial factor in determining reaction spontaneity, along with enthalpy. An increase in entropy (positive \( \Delta S^{\circ} \)) typically contributes to a reaction being favorable.
For the initial reaction in the exercise, the calculated \( \Delta S^{\circ} \) was negative, indicating a decrease in disorder, which further disfavors the reaction's feasibility when combined with the positive \( \Delta H^{\circ} \).

Chemical Reaction Feasibility

Chemical Reaction Feasibility involves determining whether a reaction can proceed under given conditions. Gibbs Free Energy is the key criterion; a reaction is feasible if \( \Delta G < 0 \).
Feasibility depends on both \( \Delta H \) and \( \Delta S \). A negative \( \Delta H \) (exothermic) and positive \( \Delta S \) (increase in disorder) create favorable conditions. Conversely, a positive \( \Delta H \) and negative \( \Delta S \) often result in \( \Delta G > 0 \), indicating non-feasibility.
In the exercise, the initial zinc extraction reaction was not feasible at 25°C due to a positive \( \Delta G \), aligning the discussion around its non-spontaneity and leading to coupling with another reaction to achieve feasibility.

Zinc Extraction Process

The Zinc Extraction Process typically involves the conversion of zinc sulfide (\( \text{ZnS} \)) into metallic zinc (\( \text{Zn} \)). However, the direct thermal decomposition of zinc sulfide is not feasible at standard conditions due to non-spontaneity characterized by a positive Gibbs Free Energy.
To overcome this, an effective strategy is to couple it with another spontaneous reaction, such as the oxidation of sulfur to sulfur dioxide, which is highly exothermic. By supplementing the first reaction with a feasible reaction, the overall \( \Delta G \) becomes negative. This combined process, known as "roasting," efficiently produces zinc by offsetting the energy deficits, thus achieving a viable extraction method.

• Roasting involves adding oxygen to the reaction, making an overall exothermic process.
• The method utilizes thermodynamic principles to enhance efficiency and output.

Consequently, this process highlights the practical application of thermodynamics principles in metallurgical operations.