Problem 84 - A Complete Resource Book in Chemistry for JEE Main

Question

The entropy change when \(36 \mathrm{~g}\) of water evaporates at \(373 \mathrm{~K}\) is \(\left(\Delta \mathrm{H}=40.63 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)\) (a) \(218 \mathrm{~J} \mathrm{~K}^{-1}\) (b) \(150 \mathrm{~J} \mathrm{~K}^{-1}\) (c) \(118 \mathrm{~J} \mathrm{~K}^{-1}\) (d) \(200 \mathrm{~J} \mathrm{~K}^{-1}\)

Step-by-Step Solution

Answer

The entropy change is approximately 218 J K^{-1}.

Step 1: Identify the Formula for Entropy Change

The formula for the change in entropy during a phase change is given by \( \Delta S = \frac{\Delta H}{T} \), where \( \Delta S \) is the entropy change, \( \Delta H \) is the enthalpy change per mole, and \( T \) is the temperature in Kelvin at which the change occurs.

Step 2: Determine the Number of Moles of Water

First, convert the mass of water from grams to moles using the molar mass of water (\(18 \mathrm{~g/mol}\)).\[\text{Moles of water} = \frac{36 \mathrm{~g}}{18 \mathrm{~g/mol}} = 2 \text{ moles}\]

Step 3: Calculate Total Enthalpy Change for Given Mass

The total enthalpy change is calculated by multiplying the number of moles by the enthalpy change per mole:\[\Delta H_{\text{total}} = 2 \text{ moles} \times 40.63 \mathrm{~kJ/mol} = 81.26 \mathrm{~kJ}\]. Convert \(\Delta H_{\text{total}}\) to \( \mathrm{J} \) (\(1 \mathrm{~kJ} = 1000 \mathrm{~J}\)):\[81.26 \mathrm{~kJ} = 81260 \mathrm{~J}\]

Step 4: Calculate Entropy Change

Use the formula from Step 1 to find \( \Delta S \):\[ \Delta S = \frac{\Delta H_{\text{total}}}{T} = \frac{81260 \mathrm{~J}}{373 \mathrm{~K}} \approx 217.85 \mathrm{~J} \mathrm{~K}^{-1} \]

Step 5: Select the Most Appropriate Answer

The calculated entropy change is approximately \(217.85 \mathrm{~J} \mathrm{~K}^{-1}\), which rounds to \(218 \mathrm{~J} \mathrm{~K}^{-1}\), so the correct answer is (a) \(218 \mathrm{~J} \mathrm{~K}^{-1}\).

Key Concepts

Phase Change

Phase change, also known as a phase transition, describes the process where a substance changes from one state of matter to another, such as from liquid to gas. In the problem, we are dealing with the phase change of water evaporating at 373 K. Here, water changes from liquid to vapor. This process requires an absorption of energy since the molecules move from a densely packed liquid state into a more separated gaseous state. This energy absorbed during the phase change significantly affects the entropy, or the measure of disorder, of the system. Evaporation increases the number of possible states for molecules, thereby increasing the system's entropy.

• Evaporation: This is the specific phase change from liquid to gas as seen in the exercise.
• Energy Requirement: The energy needed to overcome intermolecular forces in the liquid.
• Entropy Increase: Represents the disorder increase resulting from more spread-out gaseous molecules.

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \, \), is a measure of the heat absorbed or released during a chemical reaction under constant pressure. In this exercise, the enthalpy change describes the amount of energy needed to convert water from liquid to vapor at a specific temperature. This value is given as 40.63 kJ/mol. For water undergoing evaporation, the enthalpy change is positive, indicating that the process absorbs heat from the surroundings.To determine the total enthalpy change for the entire mass of water, you need to calculate it based on the number of moles of water under consideration. Multiplying the number of moles by the given enthalpy change per mole gives the total enthalpy change, crucial for finding the entropy change during the phase transition.

• Energy Input: Required to break the hydrogen bonds present in liquid water.
• Units: Typically expressed in joules or kilojoules per mole.
• Relation to Entropy: More energy input generally leads to an increase in entropy in such phase changes.

Moles of Water

To solve the problem effectively, determining the number of moles of water is a vital step. In chemical processes, we often convert mass to moles to use thermodynamic equations. The molar mass of water, \( 18 \text{ g/mol} \, \), serves as the conversion factor between grams and moles.For instance, when given 36 g of water, dividing by the molar mass results in 2 moles of water, as shown in the exercise. Using moles instead of grams provides a basis for calculating other properties, like enthalpy or entropy changes, that depend on the amount of substance on a molecular scale.

• Conversion: Mass to moles conversion helps standardize calculations.
• Molar Mass of Water: Essential for calculating moles (18 g/mol).
• Role in Thermodynamics: Foundational for accurately applying entropic and enthalpy calculations.