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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

For a liquid, which would you expect to be larger, \(\Delta S_{\text {fusion }}\) or \(\Delta S_{\text {evaporation }}\) ? Why?

Consider two perfectly insulated vessels. Vessel 1 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and water at \(0^{\circ} \mathrm{C}\). Vessel 2 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and a saltwater solution at \(0^{\circ} \mathrm{C}\). Consider the process \(\mathrm{H}_{2} \mathrm{O}(s) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\) a. Determine the sign of \(\Delta S, \Delta S_{\text {surr }}\), and \(\Delta S_{\text {univ }}\) for the process in vessel 1 . b. Determine the sign of \(\Delta S, \Delta S_{\text {surt }}\), and \(\Delta S_{\text {univ }}\) for the process in vessel 2 . (Hint: Think about the effect that a salt has on the freezing point of a solvent.)

Using data from Appendix 4, calculate \(\Delta H^{\circ}, \Delta G^{\circ}\), and \(K\) (at 298 K) for the production of ozone from oxygen: $$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$ At \(30 \mathrm{~km}\) above the surface of the earth, the temperature is about 230\. \(\mathrm{K}\) and the partial pressure of oxygen is about \(1.0 \times 10^{-3}\) atm. Estimate the partial pressure of ozone in equilibrium with oxygen at \(30 \mathrm{~km}\) above the earth's surface. Is it reasonable to assume that the equilibrium between oxygen and ozone is maintained under these conditions? Explain.

At what temperatures will the following processes be spontaneous? a. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) b. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\) c. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) d. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)

Consider the following reaction at \(25.0^{\circ} \mathrm{C}\) : $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ} / \mathrm{mol}\) and \(-176.6 \mathrm{~J} / \mathrm{K}\). mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\). Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\).
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