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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 at temperatures lower than -57°C. (c) Freezing of n-octane is non-spontaneous at temperatures above -57°C. (d) At -57°C, liquid and solid n-octane are in equilibrium.

Step by step solution

01

(a) Determine if the freezing process is endothermic or exothermic.

Freezing is the process in which a substance changes from its liquid state to its solid state. When a substance freezes, it releases heat energy into its surroundings. Since energy is being released during the process, freezing is an exothermic process. Thus, freezing of n-octane is an exothermic process.
02

(b) Identify the temperature range for spontaneous freezing.

For the freezing of n-octane to be spontaneous, it should be favorable in terms of entropy change (∆S) and enthalpy change (∆H). An exothermic process (∆H < 0) and an increase in the entropy of the surroundings (∆S > 0) make spontaneous process. Since freezing is an exothermic process, we only need to consider the temperature range where the entropy change is positive. Considering that the normal freezing point of n-octane is -57°C, spontaneous freezing will occur at temperatures below -57°C for the process to be favorable in terms of both ∆H and ∆S. Therefore, n-octane freezing is spontaneous at temperatures lower than -57°C.
03

(c) Identify the temperature range for non-spontaneous freezing.

When a process is non-spontaneous, it means either the enthalpy change (∆H) or entropy change (∆S) are unfavorable for the process to occur on its own. In the case of n-octane freezing, we know that it is an exothermic process and favorable in terms of ΔH. Therefore, ΔS must be the unfavorable factor. A process is non-spontaneous when its entropy change is negative (∆S < 0). As discussed in part (b), n-octane freezing is spontaneous at temperatures lower than -57°C. So, it is non-spontaneous at temperatures above -57°C.
04

(d) Determine if there's a temperature at which liquid and solid n-octane are in equilibrium.

At the normal freezing point of a substance, the solid and liquid phases are in equilibrium. It means that the rate of freezing (solidification) is equal to the rate of melting (liquefaction). For n-octane, this equilibrium occurs at its normal freezing point, -57°C. At this temperature, both liquid and solid n-octane coexist in equilibrium.

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Most popular questions from this chapter

(a) What do you expect for the sign of \(\Delta S\) in a chemical reaction in which 2 mol of gaseous reactants are converted to 3 mol of gaseous products? (b) For which of the processes in Exercise 19.11 does the entropy of the system increase?

Reactions in which a substance decomposes by losing CO are called decarbonylation reactions. The decarbonylation of acetic acid proceeds according to: $$ \mathrm{CH}_{3} \mathrm{COOH}(l) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{CO}(g) $$ By using data from Appendix \(\mathrm{C}\) , calculate the minimum temperature at which this process will be spontaneous under standard conditions. Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not vary with temperature.

Predict the sign of the entropy change of the system for each of the following reactions: $$\begin{array}{l}{\text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)} \\ {\text { (b) } \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)} \\ {\text { (c) } 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g)} \\ {\text { (d) } \mathrm{Al}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Al}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)}\end{array}$$

Consider the reaction 2 \(\mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) .(\mathbf{a})\) 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.

The reaction $$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons 3 \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is the basis of a suggested method for removal of SO \(_{2}\) from power-plant stack gases. The standard free energy of each substance is given in Appendix C. (a) What is the equilibrium constant for the reaction at 298 \(\mathrm{K} ?\) (b) In principle, is this reaction a feasible method of removing \(S O_{2} ?\) (c) If \(P_{S O_{2}}=P_{H_{2} S}\) and the vapor pressure of water is 25 torr, calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at 298 K. (d) Would you expect the process to be more or less effective at higher temperatures?
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