Chapter 18

Consider the following facts: Water freezes spontaneously at \(-5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm},\) and ice has a more ordered structure than liquid water. Explain how a spontaneous process can lead to a decrease in entropy.

Ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)\) dissolves spontaneously and endothermically in water. What can you deduce about the sign of \(\Delta S\) for the solution process?

For reactions carried out under standard-state conditions, Equation (18.10) takes the form \(\Delta G^{\circ}=\Delta H^{\circ}-\) \(T \Delta S^{\circ} .\) (a) Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature, derive the equation $$\ln \frac{K_{2}}{K_{1}}=\frac{\Delta H^{\circ}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$$ where \(K_{1}\) and \(K_{2}\) are the equilibrium constants at \(T_{1}\) and \(T_{2}\), respectively. (b) Given that at \(25^{\circ} \mathrm{C} K_{\mathrm{c}}\) is \(4.63 \times 10^{-3}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol} $$ calculate the equilibrium constant at \(65^{\circ} \mathrm{C}\)

Under what conditions does a substance have a standard entropy of zero? Can a substance ever have a negative standard entropy?

Consider the following Bronstead acid-base reaction at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{HF}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{HCl}(a q)+\mathrm{F}^{-}(a q) $$. (a) Predict whether \(K\) will be greater or smaller than unity, (b) Does \(\Delta S^{\circ}\) or \(\Delta H^{\circ}\) make a greater contribution to \(\Delta G^{\circ} ?\) (c) Is \(\Delta H^{\circ}\) likely to be positive or negative?

Crystallization of sodium acetate from a supersaturated solution occurs spontaneously (see p. 426). What can you deduce about the signs of \(\Delta S\) and \(\Delta H ?\)

Predict whether the entropy change is positive or negative for each of these reactions: (a) \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{O}(g)+\mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

The reaction \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) proceeds spontaneously at \(25^{\circ} \mathrm{C}\) even though there is a decrease in the number of microstates of the system (gases are converted to a solid). Explain.

Use the following data to determine the normal boiling point, in kelvins, of mercury. What assumptions must you make in order to do the calculation? $$ \begin{aligned} \mathrm{Hg}(l): & \Delta H_{\mathrm{f}}^{\circ} &=0 \text { (by definition) } \\\ & S^{\circ} &=77.4 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} \\ \mathrm{Hg}(g): & \Delta H_{\mathrm{f}}^{\circ} &=60.78 \mathrm{~kJ} / \mathrm{mol} \\ & S^{\circ} &=174.7 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} \end{aligned} $$

The molar heat of vaporization of ethanol is \(39.3 \mathrm{~kJ} / \mathrm{mol}\) and the boiling point of ethanol is \(78.3^{\circ} \mathrm{C}\) Calculate \(\Delta S\) for the vaporization of \(0.50 \mathrm{~mol}\) ethanol.