Problem 66 - A Complete Resource Book in Chemistry for JEE Main

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) \(600 \mathrm{~K}\)

Step-by-Step Solution

Answer

The boiling point of the liquid at 1 atm is 600 K (option d).

Step 1: Understanding the Relationship

At the boiling point, the change in Gibbs free energy for the vaporization process is zero, i.e., \( \Delta G = \Delta H - T_b \Delta S = 0 \). This equation implies \( T_b \Delta S = \Delta H \), where \( T_b \) is the boiling point.

Step 2: Solve for Boiling Point

Rearrange the equation \( T_b \Delta S = \Delta H \) to solve for the boiling point: \( T_b = \frac{\Delta H}{\Delta S} \).

Step 3: Substitute Known Values

Substitute \( \Delta H = 30,000 \, \text{J/mol} \) and \( \Delta S = 5 \, \text{J/mol K} \) into the equation: \[ T_b = \frac{30,000}{5} \]

Step 4: Calculate Boiling Point

Perform the division: \[ T_b = 600 \, \text{K} \]

Step 5: Choose the Correct Option

The calculated boiling point is \( 600 \, \text{K} \), which corresponds to option (d).

Key Concepts

Enthalpy of Vaporization

The enthalpy of vaporization, often symbolized as \( \Delta H_{\text{vap}} \), is a crucial concept in thermodynamics. It represents the amount of energy needed to convert a liquid into a gas at its boiling point. This process involves overcoming intermolecular forces keeping the molecules together in the liquid state.
Understanding the enthalpy of vaporization involves recognizing that energy is required to break these forces, allowing molecules to enter the gas phase. The stronger the forces, the higher \( \Delta H_{\text{vap}} \).

• The units of enthalpy of vaporization are typically joules per mole (J/mol).
• It is key to energy changes during phase transitions.
• A larger \( \Delta H \) values indicate more energy required.

The provided exercise mentions an enthalpy of vaporization of \( 30 \text{kJ/mol} \). Converting this into joules, it's \( 30,000 \text{J/mol} \). This value helps determine the boiling point of the liquid when combined with entropy information using the formula \( T_b = \frac{\Delta H}{\Delta S} \).

Entropy of Vaporization

The entropy of vaporization, denoted as \( \Delta S_{\text{vap}} \), measures the change in disorder or randomness when a liquid becomes a vapor. This concept stems from the Second Law of Thermodynamics, which states that systems naturally progress toward higher entropy.

• Signifies the increase in randomness from liquid to gas.
• Measured in joules per mole per kelvin (J/mol·K).
• Higher values indicate a larger increase in disorder.

In the example, \( \Delta S \) is given as \( 5 \, \text{J/mol K} \). When a liquid changes to a gas, the molecules become more dispersed, reflecting an increase in entropy. This change is vital to understanding how temperature affects phase transitions, especially boiling. Using the formula \( T_b = \frac{\Delta H}{\Delta S} \), the entropy value allows one to solve for the boiling temperature of the substance.

Boiling Point Calculation

Calculating the boiling point involves understanding the relationship between enthalpy and entropy of vaporization. The boiling point, \( T_b \), is the temperature at which a liquid's vapor pressure equals atmospheric pressure, and it can be found when the Gibbs free energy change, \( \Delta G \), equals zero.
To calculate the boiling point, use:\[ T_b = \frac{\Delta H}{\Delta S} \]

• \( \Delta G = 0 \) at boiling point, as vapor and liquid phases are in equilibrium.
• Adjust units so both \( \Delta H \) and \( \Delta S \) are compatible.
• Rearrange and solve for \( T_b \).

For example, substituting the enthalpy \( 30,000 \, \text{J/mol} \) and entropy \( 5 \, \text{J/mol K} \) into the equation results in \( T_b = 600 \text{K} \). This calculation aligns with the Kelvin scale, indicating at this temperature, the liquid will convert to gas under 1 atm pressure. Understanding this formula is crucial in thermodynamics as it links energy changes to physical states.