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Chapter 17: Problem 38

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 629.44 K.

Step by step solution

01

Choose appropriate values for \(P_1\) and \(T_1\)

Since we are looking for the boiling point, we are interested in the temperature at which the vapor pressure is equal to the atmospheric pressure. So, we can choose \(P_1 = 0\) and \(T_1 = 0~K\) as the reference state.
02

Substitute the given values into the Clausius-Clapeyron equation

From the given data, we know: \(\Delta H_{vap} = 58.51~kJ/mol = 58510~J/mol\) \(\Delta S_{vap} = 92.92 \frac{J}{K\cdot mol}\) From the definition of entropy, we have: \(\Delta H_{vap} = T_2 \Delta S_{vap}\), Where \(T_2\) is the boiling point we are looking for.
03

Solve for \(T_2\)

Divide both sides of the equation by \(\Delta S_{vap}\) to get: \[T_2 = \frac{\Delta H_{vap}}{\Delta S_{vap}}\] Now, substitute the given values and solve for \(T_2\): \[T_2 = \frac{58510 \mathrm{~J/mol}}{92.92 \mathrm{~J/K\cdot mol}}\] \[T_2 = 629.44 \mathrm{~K}\] The normal boiling point of mercury is approximately 629.44 K.

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Most popular questions from this chapter

Consider the reaction: $$ \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{~S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ for which \(\Delta H\) is \(-233 \mathrm{~kJ}\) and \(\Delta S\) is \(-424 \mathrm{~J} / \mathrm{K}\). a. Calculate the free energy change for the reaction \((\Delta G)\) at \(393 \mathrm{~K}\). b. Assuming \(\Delta H\) and \(\Delta S\) do not depend on temperature, at what temperatures is this reaction spontaneous?

Given the following data: $$ \begin{aligned} 2 \mathrm{H}_{2}(g)+\mathrm{C}(s) & \longrightarrow \mathrm{CH}_{4}(g) & & \Delta G^{\circ}=-51 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & \Delta G^{\circ}=-474 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(g) & & \Delta G^{\circ}=-394 \mathrm{~kJ} \end{aligned} $$ Calculate \(\Delta G^{\circ}\) for \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(l)\).

The synthesis of glucose directly from \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) and the synthesis of proteins directly from amino acids are both nonspontaneous processes under standard conditions. Yet it is necessary for these to occur for life to exist. In light of the second law of thermodynamics, how can life exist?

Given the values of \(\Delta H\) and \(\Delta S\), which of the following changes will be spontaneous at constant \(T\) and \(P\) ? a. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=300 . \mathrm{K}\) b. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) c. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=298 \mathrm{~K}\) d. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}\)

The equilibrium constant \(K\) for the reaction $$ 2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g) $$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus \(1 / T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{~K}\) and a \(y\) -intercept of \(-14.51\). Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise 79
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