Question

The normal freezing point of 1 -propanol \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}\right)\) is \(-127{ }^{\circ} \mathrm{C}\). (a) Is the freezing of 1 -propanol an endothermic or exothermic process? (b) In what temperature range is the freezing of 1 -propanol a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) Is there any temperature at which liquid and solid 1-propanol are in equilibrium? Explain.

Short answer

(a) The freezing of 1-propanol is an exothermic process. (b) The freezing of 1-propanol is a spontaneous process at temperatures lower than its normal freezing point (-127°C). (c) The freezing of 1-propanol is a nonspontaneous process at temperatures higher than its normal freezing point (-127°C). (d) Liquid and solid 1-propanol are in equilibrium at -127°C.

Step-by-step solution

Identify if the freezing process is endothermic or exothermic

The freezing process is a phase transition in which a substance changes from a liquid to a solid state. The particles in a liquid have more energy than those in a solid state. When a substance, such as 1-propanol, freezes, it releases energy in the form of heat. This makes the freezing process an exothermic process.

Determine the temperature range for spontaneous freezing

During the freezing process, the enthalpy change is negative (since it is an exothermic process). To determine the temperature range when the freezing process is spontaneous, we consider the following: - The change in the Gibbs free energy, ∆G, has to be negative for the process to be spontaneous. - The relationship between Gibbs free energy, enthalpy change, and entropy change is given by the equation: ∆G = ∆H - T∆S As the ΔH is negative (exothermic process) and the ΔS is negative as well (due to a decrease in disorder when going from liquid to solid), for ∆G to be negative, T has to be such that the term TΔS is smaller than ΔH, meaning that the temperature needs to be low enough. The spontaneous freezing of 1-propanol occurs at temperatures lower than its normal freezing point, which is -127°C.

Determine the temperature range for nonspontaneous freezing

For the freezing process to be nonspontaneous, the change in Gibbs free energy, ∆G, must be positive (or zero). Using the same equation, ∆G = ∆H - T∆S, because ΔH and ΔS are negative, T has to be high enough so that the term - TΔS is greater than ΔH. Hence, the nonspontaneous freezing of 1-propanol occurs at temperatures higher than its normal freezing point, which is -127°C.

Determine the temperature where liquid and solid are in equilibrium

At the equilibrium between the liquid and solid phase, the change in Gibbs free energy, ∆G, is zero, meaning there is no net driving force for any phase change. Using the equation ∆G = ∆H - T∆S, and knowing that ΔH and ΔS are negative, the term - TΔS must be equal to ΔH. This is the case at the normal freezing point, -127°C. So, liquid and solid 1-propanol are in equilibrium at -127°C.

Key concepts

Freezing Process

The freezing process is a fascinating phase transition where a substance shifts from a liquid state to a solid state. In this process, the molecules within a liquid lose energy, resulting in them coming closer together and forming a more ordered structure—that of a solid. This transition from liquid to solid involves the release of energy, as the liquid molecules are giving up some of their kinetic energy as heat to the surroundings. Thus, freezing is recognized as an exothermic process. This release of energy happens because the system no longer requires the energy to maintain the less ordered liquid state.

Understanding this process is crucial in thermodynamics, especially when studying how and why matter changes its state under different conditions.

Gibbs Free Energy

Gibbs Free Energy is a central concept in understanding spontaneity in thermodynamic processes. It is defined by the equation \[ \Delta G = \Delta H - T\Delta S \] where

• \( \Delta G \) is the change in Gibbs Free Energy,
• \( \Delta H \) is the change in enthalpy, and
• \( T\Delta S \) is the temperature multiplied by the change in entropy.

For a process to be spontaneous, \( \Delta G \) must be negative. This reflects that the process can occur naturally without needing external energy input. In the case of freezing, as it releases energy \( (\Delta H < 0) \) and decreases disorder \( (\Delta S < 0) \), the temperature plays a crucial role in determining spontaneity. The energy balance expressed by \( \Delta G \) helps us predict whether the freezing will occur spontaneously based on the surrounding temperature relative to the freezing point.

Phase Transition

Phase transitions, such as freezing, involve changing a substance's state, often resulting in alterations in physical properties such as density, structure, and energy. Freezing specifically deals with transitioning from a less ordered liquid state to a more ordered solid state. During freezing, the system releases heat, evidencing an exothermic process.

These transitions are fundamental in thermodynamics, as they involve distinct enthalpy (heat transfer) and entropy (disorder) changes. Recognizing how these properties shift helps in understanding why certain processes occur spontaneously under specific conditions—highlighting the intricate balance orchestrated by Gibbs Free Energy during these transitions.

Enthalpy Change

Enthalpy change \( (\Delta H) \) is yet another crucial ingredient in the thermodynamics of freezing. It represents the heat absorbed or released during a process at constant pressure.

In the freezing of 1-propanol, the enthalpy change is negative, denoting heat release. This negative \( \Delta H \) reflects the exothermic nature of freezing, as the system sheds energy to transition into a solid state. Understanding this enthalpy change is vital, as it is a primary driver in determining the spontaneity of the phase change, especially when coupled with entropy in the Gibbs Free Energy equation.

Entropy Change

Entropy change \( (\Delta S) \) is a measure of the disorder or randomness in a system. In the context of freezing, \( \Delta S \) is negative, indicating a reduction in disorder as the system moves from a liquid to a solid state. The molecules become more orderly and packed in a regular structure in the solid, leading to this decrease in entropy.

This decrease is vital to understand, as it must be counterbalanced by the enthalpy change in the Gibbs Free Energy equation to determine if a process like freezing is spontaneous. Despite the negative entropy change, freezing may still be spontaneous if the temperature is low enough to allow the overall Gibbs Free Energy change to be negative. By comprehending how entropy interacts with other factors, we gain deeper insights into the spontaneity and feasibility of different thermodynamic processes.