Question

The following data are given for the two solid forms of \(\mathrm{HgI}_{2}\) at \(298 \mathrm{K}\). $$\begin{array}{llll} \hline & \Delta H_{f}^{\circ} & \Delta G_{f,}^{\circ} & S^{\circ} \\ & \text { kJ mol }^{-1} & \text {kJ mol }^{-1} & \text {J mol }^{-1} \text {K }^{-1} \\ \hline \mathrm{HgI}_{2} \text { (red) } & -105.4 & -101.7 & 180 \\ \mathrm{Hg} \mathrm{I}_{2} \text { (yellow) } & -102.9 & (?) & (?) \\ \hline \end{array}$$ Estimate values for the two missing entries. To do this, assume that for the transition \(\mathrm{HgI}_{2}(\mathrm{red}) \longrightarrow\) \(\mathrm{HgI}_{2}(\text { yellow }),\) the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(25^{\circ} \mathrm{C}\) have the same values that they do at the equilibrium temperature of \(127^{\circ} \mathrm{C}\).

Short answer

\(\Delta G_{f}^{\circ}\) for the yellow form of \(\mathrm{HgI}_{2}\) is -155.94 kJ/mol and \(\Delta S^{\circ}\) for the yellow form of \(\mathrm{HgI}_{2}\) is 180.97 J/mol·K

Step-by-step solution

Understanding The Thermodynamic Identity

A fundamental equation in thermodynamics is the Gibbs-Helmholtz equation, which states that \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\). This equation connects the Gibbs free energy change, the enthalpy change, the absolute temperature and the entropy change. It shows that the Gibbs free energy change is the enthalpy change minus the product of absolute temperature and the entropy change.

Calculating \(\Delta G_{f}^{\circ}\) for the Yellow Form

We can calculate \(\Delta G_{f}^{\circ}\) for the yellow form using the Gibbs-Helmholtz equation as follows: \(\Delta G_{f}^{\circ} (\text {yellow}) = \Delta H_{f}^{\circ} (\text {yellow}) - T \Delta S^{\circ} (\text {red}) \). Given that \(\Delta H_{f}^{\circ} (\text {yellow}) = -102.9\) kJ/mol, \(T = 298\) K and \(\Delta S^{\circ} (\text {red}) = 180 \) J/mol·K, we have \(\Delta G_{f}^{\circ} (\text {yellow}) = -102.9 \times 10^3 - 298 \times 180 = -155.94 \) kJ/mol.

Calculating \(\Delta S^{\circ}\) for the Yellow Form

We can calculate the \(\Delta S^{\circ}\) for the yellow form, using the equation: \(\Delta S^{\circ} (\text {yellow}) = \Delta S^{\circ} (\text {red}) - \frac{\Delta H_{f}^{\circ} (\text {yellow}) - \Delta H_{f}^{\circ} (\text {red})}{T} \). Given that \(\Delta H_{f}^{\circ} (\text {yellow}) = -102.9\) kJ/mol, \(\Delta H_{f}^{\circ} (\text {red}) = -105.4\) kJ/mol and \(T = 298\) K, we have \(\Delta S^{\circ} (\text {yellow}) = 180 + \frac{(-102.9+105.4) \times 10^3}{298} = 180.97\) J/mol·K.

Key concepts

Enthalpy Change

Enthalpy change, denoted by \( \Delta H \), is a measure of the total heat content in a system. It's vital in understanding how energy is absorbed or released during a chemical reaction. When a system undergoes a reaction at constant pressure, the enthalpy change reflects the difference in the heat content between the products and reactants.

In the context of the given exercise, we are looking at the enthalpy changes for two forms of mercury(II) iodide, red and yellow. The provided values are \(-105.4\, \text{kJ/mol}\) for the red form and \(-102.9\, \text{kJ/mol}\) for the yellow form.

The negative values indicate that the formation of these compounds from their elements releases energy, making the reaction exothermic. The difference in enthalpy change between the red and yellow forms can also suggest variations in their stability or bonding environments. The red form has a more negative enthalpy value, implying it might be more stable or have stronger bonds in comparison to the yellow form.

• Enthalpy Change shows heat absorbed or released.
• Negative \(\Delta H\) indicates an exothermic process.
• Affects understanding of compound stability.

Entropy Change

Entropy change, represented as \( \Delta S \), is another crucial thermodynamic quantity, describing the disorder or randomness in a system. Systems tend to move towards higher entropy or disorder over time, which is one of the fundamental principles of thermodynamics.

For the red and yellow forms of mercury(II) iodide, \( \Delta S^{\circ} \) of the red form is given as \(180\, \text{J/mol}\cdot\text{K}\), while for the yellow form we calculated it to be approximately \(180.97\, \text{J/mol}\cdot\text{K}\).

This small difference in entropy suggests a slight increase in disorder as the compound transitions from red to yellow. Such differences are often linked to changes in the crystalline structure or molecular arrangement of the compounds.

• Entropy Change measures system disorder.
• Higher entropy means more disorder.
• Small entropy differences indicate minor structural changes.

Gibbs Free Energy

Gibbs free energy change \( \Delta G \) is a key concept used to predict the spontaneity of a process at constant temperature and pressure. The Gibbs-Helmholtz equation, \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \), elegantly links enthalpy, entropy, and temperature.

This equation implies that a negative \( \Delta G \) indicates a spontaneous reaction. Conversely, a positive \( \Delta G \) suggests that the reaction is non-spontaneous under the prevailing conditions.

From the exercise, we've calculated \( \Delta G_{f}^{\circ} \) for the yellow form as \(-155.94\, \text{kJ/mol}\). This negative value shows that the formation of the yellow form from its elements is spontaneous under standard conditions. The balance between enthalpy and entropy factors ensures that the reaction naturally proceeds in this direction under specific temperatures and pressures.

• Gibbs Free Energy predicts spontaneity.
• Negative \(\Delta G\) means spontaneous reaction.
• Integrates enthalpy and entropy influences.