Question

For the gaseous reaction of carbon monoxide and chlorine to form phosgene \(\left(\mathrm{COCl}_{2}\right)\) (a) Calculate \(\Delta S^{\circ}\) at \(298 \mathrm{~K}\left(\Delta H^{\circ}=-220 . \mathrm{kJ} / \mathrm{mol}\right.\) and \(\Delta G^{\circ}=\) \(-206 \mathrm{~kJ} / \mathrm{mol})\) (b) Assuming that \(\Delta S^{\circ}\) and \(\Delta H^{\circ}\) change little with temperature, calculate \(\Delta G^{\circ}\) at \(450 . \mathrm{K}\).

Short answer

a) \(\Delta S^{\text{\degree}} = -47 \text{ J/(mol \cdot K)}\) b) \(\Delta G^{\text{\degree}} = -198.85 \text{ kJ/mol}\)

Step-by-step solution

Write down the given values

Let's identify and write down the given values for the reaction. We have: \[\begin{equation}\begin{aligned}&\text{Temperature } T = 298 \text{ K} \ &\text{Enthalpy change } \Delta H^{\text{\degree}} = -220 \text{ kJ/mol} \ &\text{Gibbs free energy change } \Delta G^{\text{\degree}} = -206 \text{ kJ/mol}\end{aligned}\end{equation}\]

Calculate \(\Delta S^{\text{\circ}}\)

To find \(\Delta S^{\text{\circ}}\) at 298 K, use the Gibbs free energy equation: \[\begin{equation}\Delta G^{\text{\degree}} = \Delta H^{\text{\degree}} - T \Delta S^{\text{\degree}}\end{equation}\]Rearrange to solve for \(\Delta S^{\text{\degree}}\): \[\begin{equation}\Delta S^{\text{\degree}} = \frac{(\Delta H^{\text{\degree}} - \Delta G^{\text{\degree}})}{T}\end{equation}\]Substitute the known values:\[\begin{equation}\Delta S^{\text{\degree}} = \frac{(-220 \text{ kJ/mol} + 206 \text{ kJ/mol})}{298 \text{ K}}\end{equation}\]Note: Since the energies are in kJ/mol, the result will be in kJ/(mol \cdot K)\[\begin{equation}\Delta S^{\text{\degree}} = -0.047 \text{ kJ/(mol \cdot K)}\end{equation}\]Convert to J/(mol \cdot K) (1 kJ = 1000 J):\[\begin{equation}\Delta S^{\text{\degree}} = -47 \text{ J/(mol \cdot K)}\end{equation}\]

Calculate \(\Delta G^{\text{\circ}}\) at 450 K

Using the Gibbs free energy equation at 450 K:\[\begin{equation}\Delta G^{\text{\degree}} = \Delta H^{\text{\degree}} - T \Delta S^{\text{\degree}}\end{equation}\]Substitute the known values for \(\Delta H^{\text{\degree}}\), T, and \(\Delta S^{\text{\degree}}\):\[\begin{equation}\Delta G^{\text{\degree}} = -220 \text{ kJ/mol} - (450 \text{ K} \times -0.047 \text{ kJ/(mol \cdot K)})\end{equation}\]Perform the multiplication and addition:\[\begin{equation}\Delta G^{\text{\degree}} = -220 \text{ kJ/mol} + 21.15 \text{ kJ/mol} = -198.85 \text{ kJ/mol}\end{equation}\]

Key concepts

enthalpy change

Enthalpy change, denoted as \(\text{\Delta H}\), measures the heat absorbed or released in a chemical reaction at constant pressure. It is an important quantity in thermodynamics because it helps us understand energy exchanges. In the given exercise, the enthalpy change for the reaction forming phosgene is \(-220 \text{ kJ/mol}\). This value is negative, indicating that the reaction releases energy, making it exothermic.
When calculating enthalpy change, it is important to keep track of the sign. A negative value signals that the system is losing energy to the surroundings, while a positive value indicates energy absorption.
Understanding enthalpy change is essential for predicting reaction behavior and determining if a process is exothermic or endothermic.

entropy change

Entropy change, symbolized as \(\text{\Delta S}\), is a measure of the disorder or randomness in a system. Entropy increases when the system becomes more disorganized. In the exercise, we calculated \(\text{\Delta S}\) for the phosgene formation reaction: \(\text{\Delta S \degree} = -47 \text{ J/(mol ⋅ K)}\). A negative \(\text{\Delta S}\) value indicates the system is becoming more ordered, possibly due to the formation of more complex molecules from simpler reactants.
The entropy change influences the direction and feasibility of a reaction. When combined with enthalpy change, entropy helps determine the Gibbs free energy, indicating whether a reaction occurs spontaneously.
In real-life applications, entropy change helps predict the efficiency of chemical reactions, energy transformations, and the behavior of substances at different temperatures.

temperature dependence of Gibbs free energy

The Gibbs free energy change, denoted as \(\text{\Delta G}\), is temperature-dependent. It is calculated using the equation \[ \Delta G^{\text{\degree}} = \Delta H^{\text{\degree}} - T \Delta S^{\text{\degree}} \]. This relation shows how temperature can influence the spontaneity of a reaction.
From the exercise, the calculation at 450 K demonstrated the temperature's effect on \(\text{\Delta G}\). Since \(\text{\Delta H}\) and \(\text{\Delta S}\) are assumed constant with temperature, Gibbs free energy varies directly with temperature. At higher temperatures, entropy contributions can significantly affect \(\text{\Delta G}\).
This interdependence is crucial in chemical engineering and reaction kinetics, where temperature control is vital for driving reactions towards desired products and achieving maximum efficiency.

chemical thermodynamics

Chemical thermodynamics studies the principles governing energy and matter transformations in chemical processes. Key concepts include enthalpy, entropy, and Gibbs free energy. These parameters help predict reaction direction and feasibility.
For instance, the reaction forming phosgene relies on enthalpy and entropy changes to determine Gibbs free energy. By understanding these variables, we can assess if the reaction occurs spontaneously and its energy efficiency.
In practical applications, chemical thermodynamics assists in optimizing reaction conditions in industries. It aids in designing processes that ensure favorable energy changes, improved reaction yields, and minimal environmental impact.