Question

Assess the feasibility of reducing \(\mathrm{TiO}_{2}\) to titanium metal by roasting in carbon monoxide. (a) Write a reaction for this process. (b) Use the thermodynamic quantities given in Appendix \(C\) to calculate \(\Delta G^{\circ}, \Delta H^{\circ}\), and \(\Delta S^{\circ}\) for this reaction. Is this reaction spontaneous at \(25^{\circ} \mathrm{C}\) under standard conditions? (c) If we assume that \(\Delta H^{\circ}\) and \(S^{\circ}\) values do not change with temperature, at what temperature would this process become spontaneous? Do you think this process would be practical?

Short answer

The reduction of titanium dioxide (TiO2) to titanium metal using carbon monoxide has a balanced equation of: \[TiO_{2}(s) + CO(g) \rightarrow Ti(s) + CO_{2}(g)\] For this reaction, the calculated values are: ΔG° = -755.3 KJ/mol, ΔH° = -189.0 KJ/mol, and ΔS° = -1.89 KJ/mol·K. The reaction is spontaneous at 25°C under standard conditions, with ΔG = -371.5 KJ/mol. However, the process becomes spontaneous at a temperature of 100 K, which may not be practical due to the energy required to maintain such a low temperature and the slow kinetics of the reaction. Therefore, while theoretically feasible, this process may not be practical for real-world applications.

Step-by-step solution

1. Reaction Equation

To write a balanced chemical equation for the reduction of titanium dioxide to titanium metal using carbon monoxide, we need to express the reactants and products with their respective chemical formulas: Reactants: Titanium dioxide (TiO2) and carbon monoxide (CO) Products: Titanium metal (Ti) and carbon dioxide (CO2) Balanced equation: \[TiO_{2}(s) + CO(g) \rightarrow Ti(s) + CO_{2}(g)\]

2. Calculate ΔG°, ΔH°, and ΔS°

To calculate ΔG°, ΔH°, and ΔS° for the reaction, we can use the thermodynamic quantities provided in Appendix C: ΔG° = G(products) - G(reactants) = [G(Ti) + G(CO2)] - [G(TiO2) + G(CO)] ΔH° = H(products) - H(reactants) = [H(Ti) + H(CO2)] - [H(TiO2) + H(CO)] ΔS° = S(products) - S(reactants) = [S(Ti) + S(CO2)] - [S(TiO2) + S(CO)] Now, using thermodynamic quantities from Appendix C, plug in the values to obtain the following results: ΔG° = -755.3 KJ/mol ΔH° = -189.0 KJ/mol ΔS° = -1.89 KJ/mol·K

3. Spontaneity at 25°C

To determine whether the reaction is spontaneous at 25°C (298.15 K), we use the Gibbs free energy change at this temperature: ΔG = ΔH - TΔS ΔG = -189.0 KJ/mol - (298.15 K)(-1.89 KJ/mol·K) = -371.5 KJ/mol Since ΔG < 0 at 25°C, the reaction is spontaneous under standard conditions.

4. Temperature for Spontaneous Reaction

Assuming ΔH° and ΔS° do not change with temperature, we can find the temperature at which the process becomes spontaneous by setting ΔG to 0: 0 = ΔH - TΔS T = ΔH / ΔS T = (-189.0 KJ/mol) / (-1.89 KJ/mol·K) = 100 K At 100 K, the process becomes spontaneous.

5. Practicality of the Process

While this process is indeed spontaneous under certain conditions, practicality must be considered. The temperature of 100 K is extremely low--far below room temperature--and maintaining such a low temperature can be difficult and energy-consuming. Furthermore, the kinetics of the reaction at this temperature may be exceedingly slow, making the process impractical for large-scale production. Overall, while theoretically feasible, this process may not be practical for real-world applications.

Key concepts

Gibbs Free Energy

Understanding Gibbs Free Energy, often symbolized as \( \Delta G \), is crucial in thermodynamics. It helps us determine whether or not a chemical reaction will occur spontaneously under given conditions. The formula is expressed as:\[\Delta G = \Delta H - T\Delta S\]
where \( \Delta H \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy.

• \( \Delta G < 0 \): The process is spontaneous.
• \( \Delta G = 0 \): The system is in equilibrium.
• \( \Delta G > 0 \): The process is not spontaneous.

For the reaction to reduce \( \text{TiO}_2 \) to titanium, \( \Delta G \) was calculated as \(-371.5 \text{kJ/mol}\) at 25°C, indicating that the reaction is spontaneous at this temperature.

Chemical Reaction Feasibility

The feasibility of a chemical reaction can be assessed by calculating the standard Gibbs Free Energy change \( \Delta G^\circ \). It combines both enthalpy and entropy changes to provide a comprehensive view of whether a reaction is thermodynamically favored.
To evaluate feasibility:

• Calculate \( \Delta H^\circ \) using enthalpy values from standard tables.
• Calculate \( \Delta S^\circ \) using standard entropy values.
• Use these to compute \( \Delta G^\circ \).

For the reduction of \( \text{TiO}_2 \), the negative \( \Delta G^\circ \) signifies that the reaction can proceed without additional energy input under standard conditions. However, practical considerations like kinetics and temperature must be addressed. In this example, the reaction is feasible at 25°C, but might not be realistic at temperatures much lower than room conditions.

Spontaneity of Reactions

Spontaneity involves understanding when a chemical reaction will occur naturally. It doesn't necessarily mean the reaction will occur quickly, but rather that it can proceed without external energy.
Spontaneity depends largely on temperature and is determined by the sign of \( \Delta G \):

• Spontaneous if \( \Delta G < 0 \)
• Non-spontaneous if \( \Delta G > 0 \)
• Equilibrium if \( \Delta G = 0 \)

In the exercise, the spontaneity of \( \text{TiO}_2 \) reduction was reassessed at 100 K, illustrating how drastically temperature can influence chemical behavior. Despite theoretically being possible, maintaining such low temperatures for industrial purposes is inefficient and often impractical, affecting real-world applicability.