Question

The normal freezing point of \(n\)-octane \(\left(\mathrm{C}_{1} \mathrm{H}_{2 n}\right)\) 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

The freezing of n-octane is an exothermic process. The freezing of n-octane is spontaneous at temperatures below 216.15 K and nonspontaneous at temperatures above 216.15 K. At the freezing point (216.15 K or -57°C), liquid and solid n-octane coexist, and the process is in equilibrium.

Step-by-step solution

(a) Identifying the Freezing Process as Endothermic or Exothermic

Freezing is the process of a liquid transforming into a solid state. This occurs when a substance loses energy (in the form of heat) since the molecules slow down their movement and form a more stable structure. An endothermic process is one in which heat is absorbed by the system, while an exothermic process is one in which heat is released by the system. Freezing is generally considered an exothermic process since heat is released during the formation of a solid structure. Therefore, the freezing of n-octane is an exothermic process.

(b) Temperature Range for Spontaneous Freezing

For a process to be spontaneous, the change in free energy (ΔG) should be negative. Considering that freezing is an exothermic process (ΔH < 0) and that the change in entropy (ΔS) would be negative as the liquid (higher entropy) changes state to a solid (lower entropy), we can write the expression for free energy change as: ΔG = ΔH - TΔS, where T is the temperature in Kelvin. Since TΔS is always positive for the freezing process (both T and ΔS are positive), and ΔH is negative, we have to find the temperature range where this condition holds true: ΔG < 0 ⇒ ΔH - TΔS < 0 ⇒ TΔS > ΔH. The freezing point given is -57°C. Converting this to Kelvin, we have: T = -57°C + 273.15 = 216.15 K. Since the entropy change (ΔS) is always positive for the freezing of n-octane (as the entropy of liquid is greater than that of the solid), the freezing process would be spontaneous at temperatures below 216.15 K.

(c) Temperature Range for Nonspontaneous Freezing

For the freezing of n-octane to be nonspontaneous, the change in free energy (ΔG) should be positive. Since we're still looking at an exothermic process (ΔH < 0) and the change in entropy (ΔS) remains negative, we have to find the temperature range where: ΔG > 0 ⇒ ΔH - TΔS > 0 ⇒ TΔS < ΔH. Using the freezing point in Kelvin (216.15 K) and given that ΔS is always positive, the freezing process would be nonspontaneous at temperatures above 216.15 K.

(d) Equilibrium Temperature

At the equilibrium temperature, liquid n-octane and solid n-octane coexist, and the change in free energy (ΔG) is equal to zero: ΔG = 0 ⇒ ΔH - TΔS = 0 ⇒ TΔS = ΔH. Since we already know that the freezing point of n-octane is 216.15 K (-57°C), at this temperature, both liquid and solid n-octane coexist, and the process is in equilibrium.

Key concepts

Endothermic and Exothermic Processes

Understanding the difference between endothermic and exothermic processes is essential when studying thermodynamics and its implications in chemical reactions or phase changes. An endothermic process occurs when a system absorbs heat from its surroundings, leading to a temperature decrease in the environment. Conversely, an exothermic process releases heat, increasing the environmental temperature.

For example, when a substance like n-octane freezes, it transforms from a higher-energy liquid state to a lower-energy solid state, and in the process, it releases energy in the form of heat, making freezing an exothermic process. This can be thought of as the molecules 'settling down' into a more stable and less energetic structure. The key takeaway is that whether a process is endothermic or exothermic depends on the direction of heat flow between the system and surroundings.

Spontaneous and Nonspontaneous Processes

A process is classified as spontaneous if it can proceed on its own without any external input of energy, whereas a nonspontaneous process requires external energy to occur. Spontaneity is not about how fast a process occurs but whether it can happen under the given conditions.

The spontaneity of a process is largely determined by changes in free energy, typically denoted as \( \Delta G \). A negative value of \( \Delta G \) implies a spontaneous process, whereas a positive \( \Delta G \) indicates a nonspontaneous one. For instance, the freezing of n-octane is spontaneous at temperatures below its freezing point since energy is released (exothermic) and it proceeds without external help.

Free Energy and Entropy

The concepts of free energy (G) and entropy (S) provide deep insight into the spontaneity of chemical reactions and phase changes. Free energy is a thermodynamic function representing the maximum amount of work that can be performed by a system at a constant temperature and pressure. It incorporates both enthalpy (\( H \)), a measure of heat content, and entropy, which is a measure of disorder or randomness.

In the equation \( \Delta G = \Delta H - T\Delta S \), a change in free energy (\( \Delta G \)) is related to the change in enthalpy (\( \Delta H \)), temperature (\( T \)), and change in entropy (\( \Delta S \)). Decreased free energy indicates a system's tendency to do work or move towards stability, while entropy gives us essential information about the dispersal of energy in a system during a process. As such, when n-octane freezes, the decrease in entropy (due to the formation of a more ordered solid from a disordered liquid) is accompanied by a release of heat, affecting the free energy change.

Phase Equilibrium

The state of phase equilibrium is reached when two phases of a substance, such as liquid and solid n-octane, coexist at a specific condition without changing over time. At this point, the free energy of the system is minimized, and there is a balance between the opposing processes of freezing and melting.

At the freezing point of n-octane, which is \(216.15\ K\), the free energy change (\( \Delta G \) is zero, signifying that the substance is at phase equilibrium. This indicates that there is no net change in either the amount of liquid or solid n-octane, and the two phases can coexist indefinitely, as long as external conditions such as temperature and pressure remain constant.