Problem 37 - Chemistry: Principles and Reactions

Question

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.

Step-by-Step Solution

Answer

(a) The standard entropy change (ΔS°) for the reaction is calculated to be 332.866 J/K. The positive sign is reasonable because it suggests an increase in entropy due to the conversion of a liquid and a gas into two gases. (b) The standard entropy (S°) for phosgene is found to be 325.733 J/K⋅mol. (c) The standard enthalpy of formation (ΔHf°) for phosgene is calculated to be 98.1855 kJ/mol.

Step 1: Use the Gibbs-Helmholtz equation to find ∆S

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 \(\Delta S^{\circ}\): $$ \Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T} $$

Step 2: Calculate ∆S for the reaction

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.

Step 3: Calculate S° for phosgene

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 \(S^{\circ}_{\mathrm{COCl}_{2}}\): $$ S^{\circ}_{\mathrm{COCl}_{2}} = \frac{332.866 + 2(214.3) - 410}{2} = 325.733\,\text{J/K}\cdot\text{mol} $$

Step 4: Calculate ∆H° for phosgene

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 \(\Delta H^{\circ}_{f}(\mathrm{COCl}_{2}) = 98.1855\,\text{kJ/mol}\).

Key Concepts

Gibbs Free Energy

Understanding Gibbs Free Energy is crucial when studying chemical reactions. It's denoted as \( G \) and provides essential info on whether a process will occur spontaneously under constant temperature and pressure. A reaction with a negative \( \Delta G^\circ \) suggests it's thermodynamically favored; it can occur without added energy. Conversely, a positive \( \Delta G^\circ \) means the reaction is non-spontaneous.
In our exercise, the Gibbs free energy change for the formation of phosgene is \( \Delta G^\circ = -452.4 \text{ kJ} \), a negative value, hinting at spontaneity. This measurement, combined with enthalpy and temperature, allows us to determine the reaction's entropy change using the Gibbs-Helmholtz equation: \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \).

Enthalpy Change

Enthalpy change, expressed as \( \Delta H \), represents the heat absorbed or released in a reaction at constant pressure, indicating if the process is endothermic (absorbing heat) or exothermic (releasing heat). An exothermic reaction, like our example where \( \Delta H^\circ = -353.2 \text{ kJ/mol} \), signals that energy is being emitted to the surroundings.
The calculation and interpretation of \( \Delta H \) play a key role in understanding the energy aspects of reactions. Knowing the \( \Delta H \), along with \( \Delta G \), allows us to compute entropy changes and thus understand the disorder introduced by the chemical process.

Entropy Change

Entropy is a measure of disorder or randomness in a system. The change in entropy, \( \Delta S \), informs us about the degree of disorder resulting from a reaction. An increase in entropy, which parallels an increase in randomness, is typical when a reaction results in a greater number of gas molecules or when solids or liquids transform into gases.
A positive entropy change, found in our exercise as \( \Delta S^\circ = 332.866 \text{ J/K} \), suggests the reaction leads to a more disordered system, here due to the formation of gaseous products from a liquid and a gas. This aligns with the second law of thermodynamics, which states that for a process to be spontaneous, the total entropy of the universe should increase.

Standard Enthalpy of Formation

The standard enthalpy of formation \( \Delta H^\circ_{f} \) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. It's a fundamental principle in thermochemistry, helping chemists understand the energy changes during the formation of compounds from their elemental forms.
In our scenario, we seek the \( \Delta H^\circ_{f} \) of phosgene, which involves calculating the energy change when phosgene is created from its constituent elements. The given exercise provided us with the necessary information to deduce that the standard enthalpy of formation for phosgene is \( \Delta H^\circ_{f}(\mathrm{COCl}_{2}) = 98.1855 \text{ kJ/mol} \), using the known enthalpies of formation for the reactants and accounting for the enthalpy change of the reaction.