12.11 Using values for $\Delta_{ ext {fus }} H^{\circ}$ and $\Delta_{ ext {fus }} S^{\circ}$ at 298 kelvins, estimate the standard boiling point of mercury, $\Delta_{\mathrm{f}} H^{\circ}=$ $61.4 \mathrm{~kJ} / \mathrm{mol}$ and $S^{\circ}=175.0 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}$ for $\mathrm{Hg}(\mathrm{g})$. Determine the vapor pressure of mercury at $25^{\circ} \mathrm{C}$.

Question

12.11 Using values for $\Delta_{	ext {fus }} H^{\circ}$ and $\Delta_{	ext {fus }} S^{\circ}$ at 298 kelvins, estimate the standard boiling point of mercury, $\Delta_{\mathrm{f}} H^{\circ}=$ $61.4 \mathrm{~kJ} / \mathrm{mol}$ and $S^{\circ}=175.0 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}$ for $\mathrm{Hg}(\mathrm{g})$. Determine the vapor pressure of mercury at $25^{\circ} \mathrm{C}$.

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Identify the given values**: The problem provides the enthalpy of fusion ($\Delta_{ ext{fus}} H^{\circ}$) and the entropy of fusion ($\Delta_{ ext{fus}} S^{\circ}$) for mercury at 298 K. These values are $\Delta_{ ext{fus}} H^{\circ} = 61.4 \, ext{kJ/mol}$ and $\Delta_{ ext{fus}} S^{\circ} = 175.0 \, ext{J/mol} \cdot ext{K}$. ‖ ‖ 2 ⋮ **Calculate the standard boiling point**: The standard boiling point can be estimated using the formula $T_b = \frac{\Delta_{ ext{fus}} H^{\circ}}{\Delta_{ ext{fus}} S^{\circ}}$. Converting $\Delta_{ ext{fus}} H^{\circ}$ to J/mol: $61.4 \, ext{kJ/mol} = 61400 \, ext{J/mol}$. Then, $T_b = \frac{61400 \, ext{J/mol}}{175.0 \, ext{J/mol} \cdot ext{K}} = 350.29 \, ext{K}$. ‖ ‖ 3 ⋮ **Determine the vapor pressure at 25°C**: Use the Clausius-Clapeyron equation: $\ln \left( \frac{P_2}{P_1} ight) = \frac{\Delta H_{ ext{vap}}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} ight)$. Assuming $P_1$ is the vapor pressure at the boiling point (1 atm), $T_1 = 350.29 \, ext{K}$, $T_2 = 298 \, ext{K}$, and $\Delta H_{ ext{vap}} \approx \Delta_{ ext{fus}} H^{\circ} = 61400 \, ext{J/mol}$. Solving for $P_2$ at $25^{\circ} ext{C}$: $\ln \left( \frac{P_2}{1 \, ext{atm}} ight) = \frac{61400}{8.314} \left( \frac{1}{350.29} - \frac{1}{298} ight)$. ‖ ∻Answer∻ ⚹ The standard boiling point of mercury is approximately 350.29 K. The vapor pressure of mercury at 25°C can be calculated using the Clausius-Clapeyron equation. ⚹ ∻Key Concept∻ ⚹ The Clausius-Clapeyron equation relates the vapor pressure of a substance to its temperature and enthalpy of vaporization. ⚹ ∻Explanation∻ ⚹ By using the enthalpy and entropy of fusion, we can estimate the boiling point. The Clausius-Clapeyron equation then allows us to determine the vapor pressure at a different temperature. ⚹◊What is the relationship between the standard boiling point of a substance and its vapor pressure at a given temperature?⍭ Generate me a similar question◊

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