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) \(218 \mathrm{~J} \mathrm{~K}^{-\mathrm{t}}\) (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 correct answer is (a) 218 J/K.

Step 1: Understanding the variables

We are given the mass of the water as 36 g and the temperature as 373 K. The molar enthalpy of evaporation, \( \Delta H \), is given as 40.63 kJ/mol. We need to find the change in entropy, \( \Delta S \).

Step 2: Convert mass to moles

First, convert the mass of water to moles since \( \Delta H \) is given per mole. The molecular weight of water is 18 g/mol. Thus, the number of moles of water is: \[ n = \frac{36 \, \text{g}}{18 \, \text{g/mol}} = 2 \, \text{mol}. \]

Step 3: Calculate the entropy change

The change in entropy, \( \Delta S \), is calculated using the formula \( \Delta S = \frac{\Delta H}{T} \) where \( \Delta H \) is the enthalpy change and \( T \) is the temperature in Kelvin. Since \( \Delta H \) is given in kJ/mol, convert it to J/mol: \( 40.63 \, \text{kJ/mol} = 40630 \, \text{J/mol}. \) We have: \[ \Delta S = \frac{40630 \, \text{J/mol}}{373 \, \text{K}} \times 2 \, \text{mol} = \frac{40630 \, \times \, 2}{373} = 218 \, \text{J/K} \].

Step 4: Select the correct option

From the calculation, the entropy change is 218 J/K for the evaporation of 36 g of water at 373 K. Therefore, the correct answer is option (a) 218 J/K.

Key Concepts

Enthalpy of Evaporation

The enthalpy of evaporation, often denoted as \( \Delta H \), describes the energy required to convert a substance from a liquid to a gas at constant pressure. It essentially tells us how much heat energy is needed to break the intermolecular forces keeping the molecules in a liquid state, allowing them to transition into a gaseous state. This process requires heat, making it an endothermic reaction, which means that the enthalpy change **\( \Delta H \)** is always positive.

In the given exercise, the enthalpy of evaporation for water is given as **40.63 kJ/mol**. This value indicates how much energy is needed to evaporate one mole of water at a constant pressure.

In practical terms, knowing the enthalpy of evaporation helps in determining the change in energy and subsequently the entropy change, which is linked to the efficiency and spontaneity of the evaporation process.

Molar Mass of Water

To convert grams to moles, it is vital to understand the concept of molar mass, which is the mass of one mole of a substance. The molar mass of water \( \text{H}_2\text{O} \) can be calculated by summing the atomic masses of its constituent elements: hydrogen (H) and oxygen (O).

• Hydrogen: 1 g/mol (there are 2 hydrogens in a water molecule, so 2 g/mol)
• Oxygen: 16 g/mol

Together, this gives water a molar mass of **18 g/mol**.

When dealing with a problem that involves converting mass to moles, simply divide the mass given by the molar mass. In the exercise, we are given **36 g** of water. By dividing this by **18 g/mol**, we find that it equates to **2 moles** of water, which is essential for the further calculation of the entropy change.

Entropy Calculation

Entropy, symbolized as \( \Delta S \), is a measure of disorder or randomness in a system. The change in entropy during the phase transition from liquid to gas can be determined using the formula:
\[ \Delta S = \frac{\Delta H}{T} \]
where \( \Delta H \) is the enthalpy change and **T** is the absolute temperature (in Kelvin).

In our example, we know:

• \( \Delta H = 40.63 \, \text{kJ/mol} = 40630 \, \text{J/mol} \)
• Temperature **T** is given as **373 K**.

By substituting into the equation, and knowing we have 2 moles, we perform:
\[ \Delta S = \frac{40630 \, \text{J/mol} \times 2}{373 \, \text{K}} = 218 \, \text{J/K} \]
This process effectively determines the level of disorder introduced when the phase change occurs, confirming the entropy change of **218 J/K**.

Temperature Conversion

In thermodynamics, particularly when working with entropy, converting temperatures to Kelvin is essential since Kelvin is the standard unit of thermodynamic temperature measurement. This scale begins at absolute zero, and unlike Celsius or Fahrenheit, it ensures calculations remain accurate across various conditions.

For temperature conversion from Celsius to Kelvin, simply add **273.15** to the Celsius value. However, in this exercise, the temperature is already provided in Kelvin, **373 K**, indicating **100°C**, the boiling point of water at standard atmospheric pressure.

Having the temperature in Kelvin is crucial for direct usage in entropy and enthalpy calculations, as all these thermodynamic equations utilize this absolute scale to maintain consistency and precision. Always ensure temperatures are in Kelvin when using them in formulas related to thermodynamics to avoid errors in calculations.