Question

Titanium is obtained by the reduction of \(\mathrm{TiCl}_{4}(1)\) which in turn is produced from the mineral rutile \(\left(\mathrm{TiO}_{2}\right)\) (a) With data from Appendix D, determine \(\Delta G^{\circ}\) at 298 K for this reaction. $$\mathrm{TiO}_{2}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{TiCl}_{4}(1)+\mathrm{O}_{2}(\mathrm{g})$$ (b) Show that the conversion of \(\mathrm{TiO}_{2}(\mathrm{s})\) to \(\mathrm{TiCl}_{4}(1)\) with reactants and products in their standard states, is spontaneous at \(298 \mathrm{K}\) if the reaction in (a) is coupled with the reaction $$2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{g})$$

Short answer

Calculate \(\Delta G^{\circ}\) using the formula \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\) for both reactions, and then add. If the sum is negative, it proves that the reaction is spontaneous when these two reactions are coupled, otherwise it is not spontaneous at \(298 K\).

Step-by-step solution

Calculation of Gibbs Free Energy for the First Reaction

Using the formula for Gibbs free energy change for a reaction, \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\), plug in the values given in Appendix D for the enthalpy change \(\Delta H^{\circ}\) and the entropy change \(\Delta S^{\circ}\) for the conversion of \(\mathrm{TiO}_{2}(\mathrm{s})\) to \(\mathrm{TiCl}_{4}(1)\) and solve to find \(\Delta G^{\circ}\) at \(298 K\).

Calculation of Gibbs Free Energy for the Second Reaction

Similar to Step 1, calculate \(\Delta G^{\circ}\) for the reaction \(2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{g})\) using the enthalpy change \(\Delta H^{\circ}\) and the entropy change \(\Delta S^{\circ}\) values for this reaction from Appendix D.

Coupling of Reactions

Add both reactions and their respective \(\Delta G^{\circ}\) values. If the resulting \(\Delta G^{\circ}\) is negative, the reaction is spontaneous, otherwise the reaction is not spontaneous at \(298 K\).

Key concepts

Chemical Thermodynamics

Chemical thermodynamics deals with the study of energy and work relations in chemical reactions. It is fundamental in predicting whether a chemical reaction will occur spontaneously and in understanding the energy changes that accompany reactions.

One of the key concepts in thermodynamics is 'Gibbs Free Energy' (G), a quantity used to measure the maximum amount of work that a chemical reaction can perform. The Gibbs Free Energy equation is expressed as:

Spontaneous Reactions

Spontaneous reactions are processes that occur naturally without external intervention once they've been initiated. For a reaction to be spontaneous, it must result in a decrease in the system's Gibbs Free Energy, meaning \(\Delta G < 0\). To determine spontaneity, we not only consider the energy changes but also the disorder in the system, known as entropy.

To assess spontaneity at a constant temperature and pressure, we use the equation \(\Delta G = \Delta H - T\Delta S\), where \(\Delta H\) represents the change in enthalpy (heat energy), \(\Delta S\) the change in entropy, and T the temperature in Kelvin. If \(\Delta G\) is negative, the reaction tends to proceed without external energy; if positive, the reaction is non-spontaneous under the stated conditions; and if zero, the system is at equilibrium.

Enthalpy and Entropy Calculations

Enthalpy (\(\Delta H\)) and entropy (\(\Delta S\)) calculations are pivotal in evaluating thermochemical equations and predicting whether a reaction will occur spontaneously. Enthalpy is a measure of the heat change during a reaction at constant pressure. If a reaction releases heat, it is exothermic (\(\Delta H < 0\)), and if it absorbs heat, it is endothermic (\(\Delta H > 0\)).

Entropy, on the other hand, quantifies the disorder or randomness within a system. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. Consequently, \(\Delta S\) is positive when randomness increases and negative when the system becomes more ordered.

For the given exercise involving titanium extraction, we calculate \(\Delta H\) and \(\Delta S\) from provided data and use them to find the \(\Delta G\) for each step. By combining these enthalpy and entropy values, we can determine if the overall process is energetically favorable.