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Chapter 19: Problem 16

The normal freezing point of \(n\) -octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is \(-57{ }^{\circ} \mathrm{C}\). (a) Is the freezing of \(n\) -octane an endothermic or exothermic process? (b) In what temperature range is the freezing of \(n\) -octane a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) Is there any temperature at which liquid \(n\) -octane and solid \(n\) -octane are in equilibrium? Explain.

Short Answer

Expert verified
(a) The freezing of n-octane is an exothermic process. (b) Freezing of n-octane is spontaneous when the temperature is below its normal freezing point of -57°C. (c) Freezing becomes nonspontaneous when the temperature is above -57°C. (d) Liquid and solid n-octane are in equilibrium at the normal freezing point, -57°C.

Step by step solution

01

(a) Endothermic or Exothermic Process

To determine if the freezing of n-octane is endothermic or exothermic, we need to recall that an endothermic process absorbs heat, while an exothermic process releases heat. Now, we know that freezing is a phase transition from liquid to solid, and in this process, energy is released as the particles slow down and form a more ordered, solid structure. Hence, the freezing of n-octane is an exothermic process as energy is released in the form of heat.
02

(b) Temperature Range for Spontaneous Freezing

The temperature range for the spontaneous freezing of n-octane can be determined by thinking about where the energy released during the phase change can dissipate into the surroundings. As the temperature of the surroundings decreases, the process of freezing becomes more spontaneous because the energy released during freezing can be more easily absorbed by the surrounding environment. So, in general, the spontaneous freezing of n-octane occurs when the temperature is below its normal freezing point, which is -57°C.
03

(c) Temperature Range for Nonspontaneous Freezing

As mentioned earlier, the freezing process becomes more spontaneous as the temperature decreases. Therefore, it implies that freezing becomes less spontaneous or nonspontaneous when the temperature is higher. Hence, the nonspontaneous freezing of n-octane occurs when the temperature is above its normal freezing point, that is, greater than -57°C.
04

(d) Equilibrium Temperature for Liquid and Solid n-Octane

The temperature at which liquid n-octane and solid n-octane are in equilibrium is known as the normal freezing point. At equilibrium, the rate of freezing equals the rate of melting, and both phases can coexist without changing the overall composition. In this case, the equilibrium temperature is given: -57°C. At this temperature, both liquid and solid n-octane coexist in equilibrium.

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Most popular questions from this chapter

Consider the following reaction between oxides of nitrogen: $$ \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g) $$ (a) Use data in Appendix \(\mathrm{C}\) to predict how \(\Delta \mathrm{G}^{\circ}\) for the reaction varies with increasing temperature. (b) Calculate \(\Delta G^{\circ}\) at \(800 \mathrm{~K}\), assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with temperature. Under standard conditions is the reaction spontaneous at \(800 \mathrm{~K} ?\) (c) Calculate \(\Delta G^{\circ}\) at \(1000 \mathrm{~K}\). Is the reaction spontaneous under standard conditions at this temperature?

The \(K_{b}\) for methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{b} .\) (b) By using the value of \(K_{b},\) calculate \(\Delta G^{\circ}\) for the equilibrium in part (a). (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{H}^{+}\right]=6.7 \times 10^{-9} \mathrm{M},\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]=2.4 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=0.098 \mathrm{M} ?\)

For a particular reaction, \(\Delta H=-32 \mathrm{~kJ}\) and \(\Delta S=-98 \mathrm{~J} / \mathrm{K}\). Assume that \(\Delta H\) and \(\Delta S\) do not vary with temperature. (a) At what temperature will the reaction have \(\Delta G=0 ?(\mathbf{b})\) If \(T\) is increased from that in part (a), will the reaction be spontaneous or nonspontaneous?

The element gallium (Ga) freezes at \(29.8^{\circ} \mathrm{C},\) and its molar enthalpy of fusion is \(\Delta H_{\text {fus }}=5.59 \mathrm{~kJ} / \mathrm{mol}\). (a) When molten gallium solidifies to \(\mathrm{Ga}(s)\) at its normal melting point, is \(\Delta S\) positive or negative? (b) Calculate the value of \(\Delta S\) when \(60.0 \mathrm{~g}\) of \(\mathrm{Ga}(l)\) solidifies at \(29.8^{\circ} \mathrm{C}\)

Consider the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\). (a) Using data from Appendix \(\mathrm{C},\) calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are 0.40 atm and 1.60 atm, respectively.
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