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Chapter 17: Problem 34

For mercury, the enthalpy of vaporization is \(58.51 \mathrm{~kJ} / \mathrm{mol}\) and the entropy of vaporization is \(92.92 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). What is the normal boiling point of mercury?

Short Answer

Expert verified
The normal boiling point of mercury is approximately \(629.6 \ \text{K}\).

Step by step solution

01

Simplify the Clausius-Clapeyron equation

Since \(\ln{1} = 0\), we can simplify the Clausius-Clapeyron equation as follows: \[0 = \frac{-\Delta H_\text{vap}}{R} (\frac{1}{T_\text{b}} - \frac{1}{T_1})\]
02

Use \(\Delta S_\text{vap}\) to find the relationship between \(\Delta H_\text{vap}\) and \(T_\text{b}\)

We have the relationship: \[\Delta S_\text{vap} = \frac{\Delta H_\text{vap}}{T_\text{b}}\] Rearranging for \(T_\text{b}\): \[T_\text{b} = \frac{\Delta H_\text{vap}}{\Delta S_\text{vap}}\]
03

Plug in the values of enthalpy and entropy of vaporization for mercury

We are given the values: \[\Delta H_\text{vap} = 58.51 \ \text{kJ/mol}\] \[\Delta S_\text{vap} = 92.92 \ \text{J/K} \cdot \text{mol}\] First, let's convert the enthalpy of vaporization to J/mol: \[\Delta H_\text{vap} = 58.51 \ \text{kJ/mol} \times \frac{1000 \ \text{J}}{1 \ \text{kJ}} = 58510 \ \text{J/mol}\] Now we can plug the values into the equation: \[T_\text{b} = \frac{58510 \ \text{J/mol}}{92.92 \ \text{J/K} \cdot \text{mol}}\]
04

Calculate the normal boiling point of mercury

Dividing the enthalpy by the entropy: \[T_\text{b} = \frac{58510 \ \text{J/mol}}{92.92 \ \text{J/K} \cdot \text{mol}} = 629.6 \ \text{K}\] Therefore, the normal boiling point of mercury is approximately \(629.6 \ \text{K}\).

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Most popular questions from this chapter

List three different ways to calculate the standard free energy change, \(\Delta G^{\circ}\), for a reaction at \(25^{\circ} \mathrm{C}\). How is \(\Delta G^{\circ}\) estimated at temperatures other than \(25^{\circ} \mathrm{C}\) ? What assumptions are made?

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2}\). Consider the following reactions and approximate standard free energy changes: $$\begin{array}{clr}\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array}$$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\) using the following data: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6}\) at \(600 . \mathrm{K}\) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37}\) at 600. \(\mathrm{K}\)

Hydrogen cyanide is produced industrially by the following exothermic reaction: $$2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \stackrel{\mathrm{low}^{\circ} \mathrm{c}}{\mathrm{Pt} \cdot \mathrm{Rh}} 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ Is the high temperature needed for thermodynamic or kinetic reasons?

Given the following data: $$\begin{array}{lr}2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta G^{\circ}=-6399 \mathrm{~kJ} \\\\\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-394 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta G^{\circ}=-237 \mathrm{~kJ} \end{array}$$ calculate \(\Delta G^{\circ}\) for the reaction $$6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$$
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