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 \(127^{\circ} \mathrm{C}\).

Step by step solution

01

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.

02

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.

03

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.

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