Question

The enthalpy of vaporization of a liquid is \(30 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and entropy of vaporization is \(5 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}\). The boiling point of the liquid at 1 atm is: (a) \(250 \mathrm{~K}\) (b) \(400 \mathrm{~K}\) (c) \(450 \mathrm{~K}\) (d) \(600 \mathrm{~K}\)

Short answer

The boiling point of the liquid is \(600 \mathrm{~K}\) (option d).

Step-by-step solution

Understand the Relationship

To find the boiling point, we need to understand that at the boiling point, the Gibbs free energy change \( \Delta G \) is zero during phase transition. The Gibbs free energy change is given by \( \Delta G = \Delta H - T \Delta S \), where \( \Delta H \) is the enthalpy change, \( \Delta S \) is the entropy change, and \( T \) is the temperature in Kelvin.

Applying the Condition for Boiling Point

At boiling point, \( \Delta G = 0 \). Therefore, the equation \( \Delta H - T \Delta S = 0 \) simplifies to \( T = \frac{\Delta H}{\Delta S} \). This equation relates the enthalpy and entropy of vaporization to the boiling point temperature.

Plug in the Given Values

Given \( \Delta H = 30 \mathrm{~kJ} \mathrm{~mol}^{-1} = 30000 \mathrm{~J} \mathrm{~mol}^{-1} \) (converted to Joules) and \( \Delta S = 5 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \). Substitute these values into the equation for temperature: \[ T = \frac{30000}{5} \]

Calculate the Temperature

Perform the division: \[ T = \frac{30000}{5} = 600 \mathrm{~K} \]. This is the temperature at which the liquid boils at 1 atm pressure.

Key concepts

Enthalpy of Vaporization

The enthalpy of vaporization is a crucial concept when discussing phase changes, especially when a liquid turns into a gas. It represents the amount of energy required to vaporize one mole of a liquid at its boiling point under standard pressure. Think of it as the energy needed to break the intermolecular forces that hold the molecules together in the liquid state.
For example, in our given problem, the enthalpy of vaporization is specified as 30 kJ/mol. This means that to convert one mole of the liquid completely into its gaseous form, 30,000 Joules of energy is needed. This energy input overcomes the attractive forces between molecules in the liquid, allowing them to disperse freely as a gas.
The enthalpy of vaporization is always a positive value because it requires an input of energy to transition a substance from liquid to gas. Understanding this concept helps us make predictions about a substance's boiling point and comprehend the heat exchange during phase changes.

Entropy of Vaporization

Entropy, in simple terms, is a measure of disorder or randomness within a system. When a liquid turns into a gas, the entropy increases because the gas molecules move more freely and occupy more space than in the liquid state.
In the example problem, the entropy of vaporization is 5 J/mol·K. This means that for every mole of the substance that is vaporized, there is an increase in disorder equivalent to 5 Joules per Kelvin.
The concept of entropy is crucial when dealing with phase changes as it gives insight into the spontaneity of the process. Higher entropy signifies greater disorder, which is a natural tendency in thermodynamics. During the vaporization, the system's entropy increases, reflecting the change from a more orderly state (liquid) to a less orderly state (vapor). This concept, along with enthalpy, ties into the calculation of Gibbs free energy to determine boiling points.

Gibbs Free Energy

Gibbs free energy is an essential concept used to determine whether a process at constant temperature and pressure will occur spontaneously. For phase transitions, like vaporization, it is particularly relevant because it relates enthalpy and entropy changes of the system to predict the equilibrium state.
The formula to calculate Gibbs free energy is given by \( \Delta G = \Delta H - T \Delta S \). Here, \( \Delta G \) represents the change in Gibbs free energy, \( \Delta H \) is the change in enthalpy, \( \Delta S \) is the change in entropy, and \( T \) is the temperature in Kelvin.
At the boiling point, the liquid-gas equilibrium means \( \Delta G = 0 \). This implies that there is no net energy change and the system is in balance, making vaporization possible without additional energy input. This relationship allows us to calculate the boiling point as seen in our problem. By setting \( \Delta H - T \Delta S = 0 \), and rearranging, we can solve for \( T \) to find the boiling point using the given enthalpy and entropy. This illustrates how crucial Gibbs free energy is in understanding and predicting the conditions under which phase changes occur.