Titanium metal is extensively used in aerospace industry because the metal imparts strength to structures but does not unduly add to their masses. The metal is produced by the reduction of `TiCl_(4(l))` which in turn is produced from mineral rutile `[TiO_(2(s))]`. can the following reaction for production of `TiCl_(4(l))` be carried out at `25^(@)C`?
`TiO_(2(s))+2Cl_(2(g))toTiCl_(4(l))+O_(2(g))`
Given that `H_(f)^(@)` for `TiO_(2(s)),TiCl_(4(l)),Cl_(2(g))` and `O_(2(g))` are `-944.7, -804.2,0.0,0.0 kJ mol^(-1)`. also `S^(@)` for `TiO_(2(g)),TiCl_(4(l)),Cl_(2(g))` and `O_(2(g))` are `50.3,252.3,233.0,205.1 J mol^(-1) K^(-1)` respectively.

Answer

To determine whether the reaction

\[ \text{TiO}_2(s) + 2\text{Cl}_2(g) \rightarrow \text{TiCl}_4(l) + \text{O}_2(g) \]

can be carried out at \(25^\circ C\), we need to calculate the standard Gibbs free energy change (\(\Delta G^\circ\)) for the reaction. The reaction is feasible at this temperature if \(\Delta G^\circ < 0\).

### Step 1: Calculate \(\Delta H^\circ\)

The standard enthalpy change (\(\Delta H^\circ\)) for the reaction can be calculated using the standard heats of formation (\(H_f^\circ\)) of the reactants and products:

\[ \Delta H^\circ = \sum H_f^\circ \text{(products)} - \sum H_f^\circ \text{(reactants)} \]

Given:
- \(H_f^\circ (\text{TiO}_2(s)) = -944.7 \, \text{kJ/mol}\)
- \(H_f^\circ (\text{TiCl}_4(l)) = -804.2 \, \text{kJ/mol}\)
- \(H_f^\circ (\text{Cl}_2(g)) = 0.0 \, \text{kJ/mol}\) (element in standard state)
- \(H_f^\circ (\text{O}_2(g)) = 0.0 \, \text{kJ/mol}\) (element in standard state)

Calculating \(\Delta H^\circ\):

\[ \Delta H^\circ = [H_f^\circ (\text{TiCl}_4(l)) + H_f^\circ (\text{O}_2(g))] - [H_f^\circ (\text{TiO}_2(s)) + 2 \cdot H_f^\circ (\text{Cl}_2(g))] \]

\[ \Delta H^\circ = [-804.2 + 0] - [-944.7 + 2 \cdot 0] \]

\[ \Delta H^\circ = -804.2 + 944.7 = 140.5 \, \text{kJ/mol} \]

### Step 2: Calculate \(\Delta S^\circ\)

The standard entropy change (\(\Delta S^\circ\)) for the reaction can be calculated using the standard entropies (\(S^\circ\)) of the reactants and products:

\[ \Delta S^\circ = \sum S^\circ \text{(products)} - \sum S^\circ \text{(reactants)} \]

Given:
- \(S^\circ (\text{TiO}_2(s)) = 50.3 \, \text{J/mol K}\)
- \(S^\circ (\text{TiCl}_4(l)) = 252.3 \, \text{J/mol K}\)
- \(S^\circ (\text{Cl}_2(g)) = 233.0 \, \text{J/mol K}\)
- \(S^\circ (\text{O}_2(g)) = 205.1 \, \text{J/mol K}\)

Calculating \(\Delta S^\circ\):

\[ \Delta S^\circ = [S^\circ (\text{TiCl}_4(l)) + S^\circ (\text{O}_2(g))] - [S^\circ (\text{TiO}_2(s)) + 2 \cdot S^\circ (\text{Cl}_2(g))] \]

\[ \Delta S^\circ = [252.3 + 205.1] - [50.3 + 2 \cdot 233.0] \]

\[ \Delta S^\circ = 457.4 - [50.3 + 466.0] = 457.4 - 516.3 = -58.9 \, \text{J/mol K} \]

### Step 3: Calculate \(\Delta G^\circ\)

Now, we can calculate the standard Gibbs free energy change (\(\Delta G^\circ\)) at \(25^\circ C\) (or \(298 \, K\)) using the formula:

\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]

Convert \(\Delta H^\circ\) to J/mol:

\[ \Delta H^\circ = 140.5 \, \text{kJ/mol} = 140500 \, \text{J/mol} \]

Now substituting the values:

\[ \Delta G^\circ = 140500 \, \text{J/mol} - 298 \cdot (-58.9 \, \text{J/mol K}) \]

\[ \Delta G^\circ = 140500 + 17548.2 = 158048.2 \, \text{J/mol} = 158.05 \, \text{kJ/mol} \]

### Conclusion

Since \(\Delta G^\circ = 158.05 \, \text{kJ/mol} > 0\), the reaction cannot be carried out at \(25^\circ C\).