Question

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

(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

(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.

(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.

(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.

(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.

Key concepts

Freezing Point

The freezing point is the temperature at which a liquid turns into a solid. For a substance like n-octane, this occurs at -57°C. At this specific temperature, the molecules in the liquid slow down sufficiently to form solid structures.
The freezing point is unique to each substance and is affected by the substance's intrinsic properties.

• It marks the transition between the liquid and solid phase under standard atmospheric pressure.
• For n-octane, below -57°C, the solid form is favored, while above it, the liquid form prevails.

Understanding the freezing point helps us predict the conditions under which a substance will change its state.

Exothermic Process

An exothermic process is one that releases heat to the surroundings. During freezing, like in n-octane, the substance goes from a liquid to a solid state.
In this process, the energy that was keeping the molecules in a disordered state (liquid) is released.

• This release of energy occurs because the molecules need less energy to maintain the solid state.
• Thus, the process is exothermic because heat is a byproduct.

This is crucial in determining how substances interact with their environment during phase changes.

Spontaneous Process

A spontaneous process occurs naturally under the given conditions without any external influence. For freezing, the conditions for spontaneity rely on temperature and energy considerations.
For n-octane, freezing is spontaneous below -57°C. Here, both entropy and enthalpy changes favor the process:

• The enthalpy change ( ∆H < 0 ) indicates heat release is favorable.
• The entropy of the surrounding increases (∆S > 0), making the overall process conducive.

Therefore, n-octane will spontaneously freeze at temperatures lower than its normal freezing point.

Phase Equilibrium

Phase equilibrium refers to the state where the rates of phase change are equal. At -57°C, n-octane achieves this equilibrium.
In this state, the rate of liquid turning to solid matches the rate of solid turning to liquid.

• This balance is essential for maintaining consistent properties over time.
• During equilibrium, there's no net change in phase, only transitions within the established balance.

Understanding phase equilibrium helps in determining conditions suitable for certain applications and storage requirements of substances like n-octane.