Problem 50 - A Complete Resource Book in Chemistry for JEE Main

Question

The entropy change in the fusion of 1 mol of a solid melting at \(27^{\circ} \mathrm{C}\) (Latent heat of fusion, \(2930 \mathrm{~J} \mathrm{~mol}^{-1}\) ) is (a) \(9.77 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (b) \(10.73 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (c) \(2930 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (d) \(108.5 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The correct answer is (a) \(9.77 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\).

Step 1: Understanding the Problem

We need to calculate the change in entropy when 1 mole of a solid melts at its melting point temperature, given the latent heat of fusion. The formula to use is: \[ \Delta S = \frac{\Delta H}{T} \] where \(\Delta S\) is the change in entropy, \(\Delta H\) is the latent heat of fusion, and \(T\) is the absolute temperature in Kelvin.

Step 2: Convert Temperature to Kelvin

The melting temperature is given as \(27^{\circ} \mathrm{C}\). We convert this to Kelvin using the formula: \[ T(\mathrm{K}) = T(^{\circ} \mathrm{C}) + 273.15 \] \[ T = 27 + 273.15 = 300.15 \mathrm{K} \]

Step 3: Apply the Entropy Change Formula

Using the formula \(\Delta S = \frac{\Delta H}{T}\), we substitute the given values: \(\Delta H = 2930 \mathrm{~J} \mathrm{~mol}^{-1}\) and \(T = 300.15 \mathrm{K}\). Thus, \[ \Delta S = \frac{2930}{300.15} \approx 9.764 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \] Rounding to two decimal places, \(\Delta S \approx 9.76 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\).

Step 4: Checking Against Given Choices

Compare the calculated entropy change, \(9.76 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), with the given options. We see that option (a) \(9.77 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) is very close to our computed value, likely due to rounding differences.

Key Concepts

Latent Heat of Fusion

When a solid changes into a liquid at its melting point, it absorbs energy without a change in temperature. This specific amount of heat required for this phase change is known as the Latent Heat of Fusion. It is an essential concept in thermodynamics.

For example, consider ice melting into water. Even though its temperature remains constant during the process, energy, quantified as latent heat, is needed to break the molecular bonds holding the solid structure.

Key points to remember:

• The latent heat of fusion is usually given in units like Joules per mole (J/mol).
• In the exercise example, this value is given as 2930 J/mol.
• This value indicates how much energy is needed to melt 1 mole of the solid substance.

Understanding this concept helps in calculating the entropy change during a phase transition.

Temperature Conversion to Kelvin

Temperature scales are crucial in scientific calculations. Kelvin is the SI unit for temperature and is widely used in thermodynamics because it starts from absolute zero — the point where all kinetic motion in particles theoretically ceases.

To convert temperature from Celsius to Kelvin, add 273.15 to the Celsius value. This is a straightforward and essential conversion in calculations involving energy and entropy.

• For instance, in the problem, the melting point temperature of 27°C was converted to Kelvin.
• Using: \[ T(K) = T(°C) + 273.15 \]
• The result is 300.15 K.

Converting to Kelvin ensures that temperature differences and calculations remain consistent with the laws of thermodynamics.

Entropy Formula

Entropy is a measure of disorder or randomness in a system, often associated with energy dispersal. When heat is transferred into or out of a system, the entropy changes. To calculate this change during a phase transition, the formula used is:\[ \Delta S = \frac{\Delta H}{T} \]

This ties together the concepts of latent heat and temperature. Here, \( \Delta S \) represents entropy change, \( \Delta H \) is the latent heat of fusion, and \( T \) is the temperature in Kelvin.

• The given latent heat of fusion \( 2930 \, \text{J/mol} \) and the converted temperature \( 300.15 \, \text{K} \) were used in this formula.
• After calculation, the change in entropy is approximately \( 9.76 \, \text{JK}^{-1} \text{mol}^{-1} \).

Understanding this formula allows for the determination of how much energy is disorderly spread within a system during transitions like melting. This quantification is crucial in both academic studies and practical applications involving energy systems.