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) \(118 \mathrm{JK}^{-1}\) (d) \(200 \mathrm{~J} \mathrm{~K}^{-1}\)

Step-by-Step Solution

Answer

(a) 218 J/K

Step 1: Calculate Moles of Water

Determine the number of moles of water using the formula \( n = \frac{m}{M} \), where \( m \) is the mass of water and \( M \) is the molar mass of water. For water, \( M = 18 \mathrm{~g/mol} \). We have \( n = \frac{36 \mathrm{~g}}{18 \mathrm{~g/mol}} = 2 \mathrm{~mol} \).

Step 2: Use the Entropy Change Formula

The entropy change when a substance changes phase at constant temperature is given by the formula \( \Delta S = \frac{\Delta H}{T} \), where \( \Delta H \) is the enthalpy change and \( T \) is the temperature in Kelvin.

Step 3: Plug in the Values

Using the given \( \Delta H = 40.63 \mathrm{~kJ/mol} \) and converting it to \( \mathrm{J/mol} \) by multiplying by 1000, we get \( \Delta H = 40630 \mathrm{~J/mol} \). \( T = 373 \mathrm{~K} \). The change in entropy for 2 moles of water is \( \Delta S = \frac{40630 \mathrm{~J/mol} \times 2 \mathrm{~mol}}{373 \mathrm{~K}} = 217.99 \mathrm{~J/K} \).

Step 4: Choose the Closest Correct Answer

The calculated entropy change \( 217.99 \mathrm{~J/K} \) is closest to option (a) \( 218 \mathrm{~J/K} \). Therefore, option (a) is the correct answer.

Key Concepts

Enthalpy of Vaporization

When a substance changes from a liquid to a gas, it requires energy, and this energy is known as the enthalpy of vaporization. It refers to the amount of heat needed to vaporize one mole of a liquid at constant pressure. When water evaporates, it needs energy to overcome the attraction between its molecules.
This is why water requires heat to turn into vapor. The enthalpy of vaporization for water at its boiling point (373 K) is given as 40.63 kJ/mol. This means that each mole of water needs 40.63 kJ of energy to transform into a gas. In calculations, it is often necessary to convert this energy into Joules by multiplying by 1000, since 1 kJ = 1000 J.
When calculating entropy changes, having the enthalpy of vaporization in Joules per mole makes the math consistent.

Molar Mass of Water

The molar mass of a substance is the weight of one mole of its particles. For water, this is a crucial value in calculations involving phase changes. The molar mass of water is 18 g/mol, which is calculated by adding together the atomic masses of its constituent elements: hydrogen (approximately 1 g/mol) and oxygen (approximately 16 g/mol). So, - H: 2 x 1 g/mol = 2 g/mol- O: 1 x 16 g/mol = 16 g/molTogether, they sum up to the 18 g/mol total.Understanding the molar mass is essential when determining how many moles of water are present in a given mass. For instance, if you have 36 grams of water, you can calculate the number of moles by using the relationship:\[ n = \frac{m}{M} \]where \( n \) is the number of moles, \( m \) is the mass in grams, and \( M \) is the molar mass. Thus, for 36 grams of water with a molar mass of 18 g/mol, there are 2 moles of water present.

Temperature in Kelvin

When dealing with thermodynamic processes, such as phase transitions, it is crucial to express temperature in Kelvin rather than Celsius or Fahrenheit. This is because the Kelvin scale starts at absolute zero, making it ideal for scientific calculations that involve change or energy.To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. For water's boiling point:\[ \text{Temperature in Celsius} = 100^\circ \text{C} \]\[ \text{Temperature in Kelvin} = 100 + 273.15 = 373.15 \text{ K} \]However, in many calculations, this is simplified to 373 K for easier math.
Using Kelvin allows us to apply formulas like the entropy change formula directly, ensuring consistent and accurate results. Entropy, denoted as \( \Delta S \), depends on temperature being in Kelvin for precise measurements when using the formula:\[ \Delta S = \frac{\Delta H}{T} \] Here, \( \Delta H \) is the enthalpy change in Joules, ensuring the units align correctly with the temperature in Kelvin.