Problem 21 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The enthalpy of vapourization of liquid is \(30 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and entropy of vapourization is \(75 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}\). The boiling point of the liquid at \(1 \mathrm{~atm}\) is [2004S] (a) \(250 \mathrm{~K}\) (b) \(400 \mathrm{~K}\) (c) \(450 \mathrm{~K}\) (d) \(600 \mathrm{~K}\)

Step-by-Step Solution

Answer

The boiling point of the liquid is 400 K, option (b).

Step 1: Understand the Given Values

We have the enthalpy of vaporization \( \Delta H_{\text{vap}} = 30 \text{ kJ mol}^{-1} = 30000 \text{ J mol}^{-1} \) and the entropy of vaporization \( \Delta S_{\text{vap}} = 75 \text{ J mol}^{-1} \text{ K}^{-1} \). We need to find the boiling point temperature \( T_b \) where the vapour pressure equals atmospheric pressure \( 1 \text{ atm} \).

Step 2: Apply Gibbs Free Energy Relationship

At the boiling point, the change in Gibbs Free Energy \( \Delta G \) is zero for phase transitions. Thus, we can use the formula \( \Delta G = \Delta H - T \Delta S \). At vaporization temperature, \( \Delta G = 0 \), so \( 0 = \Delta H_{\text{vap}} - T_b \Delta S_{\text{vap}} \).

Step 3: Solve for Boiling Point Temperature

Rearrange the equation \( 0 = \Delta H_{\text{vap}} - T_b \Delta S_{\text{vap}} \) to find \( T_b \): \[ T_b = \frac{\Delta H_{\text{vap}}}{\Delta S_{\text{vap}}} \]. Substitute in the given values: \( T_b = \frac{30000 \text{ J mol}^{-1}}{75 \text{ J mol}^{-1} \text{ K}^{-1}} = 400 \text{ K} \).

Step 4: Conclude with the Correct Option

The calculated boiling point is \( 400 \text{ K} \), which corresponds to option (b).

Key Concepts

Enthalpy of Vaporization

Enthalpy of vaporization is a key concept when discussing phase changes, particularly from liquid to gas. It represents the amount of heat energy required to convert one mole of a liquid into a gas at constant pressure. This energy is needed to overcome intermolecular forces holding the liquid together, allowing the molecules to escape into the gaseous phase.

In our problem, the enthalpy of vaporization, denoted as \( \Delta H_{\text{vap}} \), is given as \( 30 \text{ kJ mol}^{-1} \), which can also be expressed as \( 30000 \text{ J mol}^{-1} \). This value gives us an idea of how much energy is needed to vaporize the substance under study.

Understanding this concept is crucial because it helps predict boiling points and understand the energy requirements for phase transitions. If a substance has a high enthalpy of vaporization, it indicates strong intermolecular attractions and thus a higher boiling point.

Entropy of Vaporization

Entropy of vaporization, denoted as \( \Delta S_{\text{vap}} \), measures the degree of disorder or randomness introduced when a liquid becomes a gas. During vaporization, molecules move from a more ordered state (liquid) to a less ordered state (gas), increasing the system's entropy.

In the provided exercise, the entropy of vaporization is \( 75 \text{ J mol}^{-1} \text{ K}^{-1} \). This figure indicates how much disorder is added to the system per mole of liquid vaporized. Generally, the greater the entropy of vaporization, the higher the temperature needed to achieve boiling, assuming entropy and enthalpy are in balance.

Understanding entropy helps in analyzing the spontaneity of phase changes. Systems tend to evolve towards higher entropy, a principle behind the natural tendency of liquids to vaporize under suitable conditions.

Gibbs Free Energy

Gibbs Free Energy (\( \Delta G \)) is a fundamental property in thermodynamics that predicts the spontaneity of a reaction or process. At constant temperature and pressure, a process is considered spontaneous if \( \Delta G \) is negative. For phase transitions like vaporization, \( \Delta G \) becomes zero at the boiling point because the two phases are in equilibrium.

Our exercise uses the formula for Gibbs Free Energy \( \Delta G = \Delta H - T \Delta S \), where \( \Delta H \) is the enthalpy change and \( \Delta S \) is the entropy change. At the boiling point, we set \( \Delta G = 0 \) because this signifies that the liquid and gas phases can coexist. From our problem, rearranging the formula provides \( T_b = \frac{\Delta H_{\text{vap}}}{\Delta S_{\text{vap}}} \). Using the given values, the boiling point is calculated to be \( 400 \text{ K} \).

This understanding of Gibbs Free Energy emphasizes its role in determining phase transition points and highlights why spontaneous processes like boiling occur at certain temperatures.