Phosgene, \(\mathrm{COCl}_{2}\), can be formed by the reaction of chloroform, \(\mathrm{CHCl}_{3}(l)\), with oxygen: $$ \begin{gathered} 2 \mathrm{CHCl}_{3}(l)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{COCl}_{2}(g)+2 \mathrm{HCl}(g) \\ \Delta H^{\circ}=-353,2 \mathrm{~kJ} ; \Delta G^{\circ}=-452.4 \mathrm{~kJ} \text { at } 25^{\circ} \mathrm{C} \end{gathered} $$ (a) Calculate \(\Delta S^{\circ}\) for the reaction. Is the sign reasonable? (b) Calculate \(S^{\circ}\) for phosgene. (c) Calculate \(\Delta H_{f}^{\circ}\) for phosgene.

Given the standard Gibbs free energy change (\(\Delta G^{\circ}\)) and the standard enthalpy change (\(\Delta H^{\circ}\)) for the reaction, we can use the Gibbs-Helmholtz equation to find the standard entropy change (\(\Delta S^{\circ}\)) for the reaction at 25°C: $$ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} $$ where \(T\) is the temperature in kelvins. First, we convert the given temperature (25°C) to kelvins: $$ T = 25 + 273.15 = 298.15\,\text{K}. $$ Now we can solve for \(\Delta S^{\circ}\): $$ \Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T} $$

We now plug in the given values for \(\Delta H^{\circ}\), \(\Delta G^{\circ}\), and \(T\), and find \(\Delta S^{\circ}\): $$ \Delta S^{\circ} = \frac{-353.2\,\text{kJ} - (-452.4\,\text{kJ})}{298.15\,\text{K}} = \frac{99.2\,\text{kJ}}{298.15\,\text{K}} $$ $$ \Delta S^{\circ} = 332.866\,\text{J/K} $$ The positive sign of \(\Delta S^{\circ}\) is reasonable because it suggests an increase in entropy in the reaction due to the conversion of a liquid and a gas into two gases.

To calculate the standard entropy (\(S^{\circ}\)) for phosgene, we must first determine the standard entropy change for each reactant and product. We cannot calculate the standard entropy for phosgene directly from the given data, however it can be found using standard entropy values from a thermodynamic data table. Using standard entropy values from a thermodynamic data table for each reactant and product: $$ S^{\circ}_{\mathrm{CHCl}_{3}} = 214.3 \,\text{J/K}\cdot\text{mol} \text{ (liquid)}\\ S^{\circ}_{\mathrm{O}_{2}} = 205.0 \,\text{J/K}\cdot\text{mol} \text{ (gas)}\\ S^{\circ}_{\mathrm{COCl}_{2}}= \text{? (gas)}\\ S^{\circ}_{\mathrm{HCl}} = 186.9 \,\text{J/K}\cdot\text{mol} \text{ (gas)} $$ Based on the balanced chemical equation, the standard entropy change (\(\Delta S^{\circ}\)) for the reaction can also be written as: $$ \Delta S^{\circ} = 2S^{\circ}_{\mathrm{COCl}_{2}} + 2S^{\circ}_{\mathrm{HCl}} - 2S^{\circ}_{\mathrm{CHCl}_{3}} - S^{\circ}_{\mathrm{O}_{2}} \\ $$ Next, plug in the values for \(\Delta S^{\circ}\), \(S^{\circ}_{\mathrm{CHCl}_{3}}\), \(S^{\circ}_{\mathrm{O}_{2}}\), and \(S^{\circ}_{\mathrm{HCl}}\), we can solve for \(S^{\circ}_{\mathrm{COCl}_{2}}\): $$ 332.866 = 2 S^{\circ}_{\mathrm{COCl}_{2}} + 2(186.9) - 2(214.3) - 205 $$ Solve for \(S^{\circ}_{\mathrm{COCl}_{2}}\): $$ S^{\circ}_{\mathrm{COCl}_{2}} = \frac{332.866 + 2(214.3) - 410}{2} = 325.733\,\text{J/K}\cdot\text{mol} $$

To calculate the standard enthalpy of formation (\(\Delta H^{\circ}_{f}\)) for phosgene, we can use the balanced chemical equation in combination with the given standard enthalpy change for the reaction: $$ 2\Delta H^{\circ}_{f}(\mathrm{CHCl}_{3}) + \Delta H^{\circ}_{f}(\mathrm{O}_{2}) = 2\Delta H^{\circ}_{f}(\mathrm{COCl}_{2}) + 2\Delta H^{\circ}_{f}(\mathrm{HCl}) \\ $$ Since the standard enthalpy of formation for elemental oxygen is zero (\(\Delta H^{\circ}_{f}(\mathrm{O}_{2}) = 0\)), we can simplify the equation: $$ 2\Delta H^{\circ}_{f}(\mathrm{CHCl}_{3}) = 2\Delta H^{\circ}_{f}(\mathrm{COCl}_{2}) + 2\Delta H^{\circ}_{f}(\mathrm{HCl}) \\ $$ From a thermodynamic data table, we obtain the standard enthalpies of formation for each reactant and product: $$ \Delta H^{\circ}_{f}(\mathrm{CHCl}_{3}) = 104.061\,\text{kJ/mol} \\ \Delta H^{\circ}_{f}(\mathrm{HCl}) = -92.31\,\text{kJ/mol} \\ $$ Now, using the values from the table we can solve for \(\Delta H^{\circ}_{f}(\mathrm{COCl}_{2})\): $$ 2(104.061) = 2\Delta H^{\circ}_{f}(\mathrm{COCl}_{2}) + 2(-92.31) \\ $$ $$ \Delta H^{\circ}_{f}(\mathrm{COCl}_{2}) = \frac{2(104.061) + 2(92.31)}{2} = 98.1855\,\text{kJ/mol} $$ Thus, the standard enthalpy of formation for phosgene is \(\Delta H^{\circ}_{f}(\mathrm{COCl}_{2}) = 98.1855\,\text{kJ/mol}\).