Problem 79 - Chemistry and Chemical Reactivity

Question

Iodine, I \(_{2},\) dissolves readily in carbon tetrachloride. For this process, \(\Delta H^{\circ}=0 \mathrm{kJ} / \mathrm{mol}.\) $$\mathrm{I}_{2}(\mathrm{s}) \rightarrow \mathrm{I}_{2}\left(\text { in } \mathrm{CCl}_{4} \text { solution }\right)$$ What is the sign of \(\Delta_{r} G^{\circ} ?\) Is the dissolving process entropy-driven or enthalpy-driven? Explain briefly.

Step-by-Step Solution

Answer

The sign of \(\Delta_{r} G^{\circ}\) is negative; the process is entropy-driven.

Step 1: Understanding the Problem and Given Data

We're provided with the reaction notation: \(\mathrm{I}_{2}(\mathrm{s}) \rightarrow \mathrm{I}_{2}\left(\text { in } \mathrm{CCl}_{4} \text { solution }\right)\). The standard enthalpy change for this process is given as \(\Delta H^{\circ}=0 \mathrm{kJ}/\mathrm{mol}.\) We need to determine the sign of the standard Gibbs free energy change \(\Delta_{r} G^{\circ}\) and whether the process is entropy-driven or enthalpy-driven.

Step 2: Applying the Gibbs Free Energy Equation

The relationship between Gibbs free energy change (\(\Delta G^{\circ}\)), enthalpy change (\(\Delta H^{\circ}\)), and entropy change (\(\Delta S^{\circ}\)) is given by the equation: \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ},\)where \(T\) is the temperature in Kelvin.

Step 3: Analyzing the Enthalpy Change

It's given that \(\Delta H^{\circ} = 0 \mathrm{kJ}/\mathrm{mol}.\) This suggests there is no change in enthalpy when iodine dissolves in carbon tetrachloride, meaning the dissolving is neither endothermic nor exothermic.

Step 4: Determining the Gibbs Free Energy Change

Since \(\Delta H^{\circ} = 0\), the equation simplifies to: \(\Delta G^{\circ} = - T\Delta S^{\circ}.\) This implies the sign of \(\Delta G^{\circ}\) depends entirely on the term \(- T\Delta S^{\circ}\). For a spontaneous dissolving process, \(\Delta G^{\circ}\) should be negative, suggesting \(\Delta S^{\circ}\) must be positive, indicating an increase in disorder or randomness.

Step 5: Identifying Whether the Process is Entropy-Driven

Since \(\Delta H^{\circ} = 0\) and the sign of \(\Delta G^{\circ}\) depends on \(- T\Delta S^{\circ}\), the process is driven by entropy changes. A positive \(\Delta S^{\circ}\) leads to a negative \(\Delta G^{\circ}\), making the process spontaneous.

Key Concepts

Iodine dissolution

The process of iodine dissolution refers to the ability of solid iodine (\( \text{I}_2 \text{(s)} \)) to dissolve in another substance, like carbon tetrachloride (\( \text{CCl}_4 \)). During this occurrence, iodine transitions from a solid state into a solute within the carbon tetrachloride solution. This transformation doesn't entail any chemical change; rather, it's a physical process. Iodine atoms tend to dissolve easily in non-polar solvents such as \( \text{CCl}_4 \).- Iodine has weak intermolecular forces in its solid state, allowing it to dissolve when a suitable solvent is present.- During dissolution, the iodine molecules disperse evenly in the solvent, leading to a homogenous mixture.Although the dissolution involves only physical changes, it plays a critical role in various chemical processes. It's worth noting that the dissolution of iodine in \( \text{CCl}_4 \) doesn't involve any change in enthalpy (\( \Delta H^\circ = 0 \text{kJ/mol} \)), which makes understanding its underlying driving forces even more interesting.

Entropy-driven process

An entropy-driven process is characterized by predominance of entropy change over enthalpy change. Entropy (\( \Delta S \)) measures the disorder or randomness of a system. In iodine dissolution into carbon tetrachloride, entropy change (\( \Delta S^\circ \)) governs the process since there is no enthalpy change (\( \Delta H^\circ = 0 \)).- Entropy-driven processes often involve an increase in randomness or disorder, such as the dispersal of iodine molecules within the solvent.- The equation \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \) illustrates how Gibbs free energy (\( \Delta G^\circ \)) is influenced by entropy changes.For a spontaneous process like iodine dissolution, \( \Delta G^\circ \) must be negative. This necessitates \( \Delta S^\circ \) being positive, indicating an increase in disorder. Thus, the dissolution becomes favorable due to the entropy-related changes, making the process primarily driven by entropy.

Enthalpy change

Enthalpy change (\( \Delta H \)) represents the heat absorbed or released in a reaction at constant pressure. However, in the case of iodine dissolving in carbon tetrachloride, the enthalpy change (\( \Delta H^\circ \)) is zero.- A zero enthalpy change indicates there's neither absorption nor release of heat during the dissolution process.- This means that the amount of energy required to break intermolecular forces of solid iodine is perfectly offset by the energy released when iodine interacts with \( \text{CCl}_4 \) molecules.This balance between energy absorbed and released suggests that no net energy is involved in the process. In such scenarios, other factors like entropy take center stage in determining the spontaneity of the reaction. Here, since \( \Delta H^\circ = 0 \), the changes in Gibbs free energy, making the process spontaneous, are wholly attributed to variations in entropy.