Question

The entropy change in the fusion of \(1 \mathrm{~mol}\) 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) \(108.5 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

Short answer

The entropy change is approximately \(9.77 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), which matches option (a).

Step-by-step solution

Understanding the problem

The problem asks for the entropy change, \(\Delta S\), during the fusion (melting) of 1 mole of a solid at \(27^{\circ} \mathrm{C}\) with a given latent heat of fusion, \(2930 \mathrm{~J/mol}\). Use the relation for entropy change during phase transitions.

Formula for entropy change

For a phase change, the formula for entropy change is given by:\[\Delta S = \frac{q_{\text{rev}}}{T}\]where \(q_{\text{rev}}\) is the reversible heat exchange (latent heat of fusion) and \(T\) is the temperature in Kelvin.

Convert temperature to Kelvin

Convert the given temperature from Celsius to Kelvin using the formula:\[T (\text{K}) = T(\text{{in °C}}) + 273.15\]Thus, \(27^{\circ} \mathrm{C} = 27 + 273.15 = 300.15 \mathrm{K}\).

Calculate the entropy change

Substitute \(q_{\text{rev}} = 2930 \mathrm{~J/mol}\) and \(T = 300.15 \mathrm{K}\) into the entropy formula:\[\Delta S = \frac{2930 \mathrm{~J/mol}}{300.15 \mathrm{~K}}\]Perform the division to find \(\Delta S\).

Perform the calculation

Calculate \(\Delta S = \frac{2930}{300.15} \approx 9.76 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\).

Match with options

Compare the calculated entropy change with the provided options. The closest match to \(9.76 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) is option (a) \(9.77 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\).

Key concepts

Latent Heat of Fusion

Latent heat of fusion refers to the amount of energy needed to change a substance from a solid to a liquid at its melting point. This energy doesn't increase the substance's temperature but enables the molecules to overcome their fixed positions in the lattice of the solid. For instance, when you heat an ice cube, its temperature rises until it reaches 0°C, its melting point. From here, the extra energy added doesn't make the liquid water hotter. Instead, it's utilized to break the bonds between ice particles, transforming them into water.

• It is measured in joules per mole (J/mol).
• This property varies with different substances; for ice, it's approximately 334 J/g.

Understanding this concept is crucial, particularly because it is involved in calculating phase transition phenomena like changes in entropy.

Temperature Conversion to Kelvin

Converting a temperature from Celsius to Kelvin is essential in scientific calculations, including those dealing with entropy changes. The Kelvin scale, unlike the Celsius scale, starts at absolute zero, which is the lowest limit of temperature. Converting temperatures to Kelvin is straightforward and involves adding 273.15 to the Celsius measurement. Therefore, to convert:

• Measure the given temperature in degrees Celsius.
• Add 273.15 to the Celsius temperature.

For example, if you have a temperature of 27°C, you convert it as follows:\[T (\text{K}) = 27 + 273.15 = 300.15 \text{ K}\]Using the Kelvin scale is particularly handy when you are inserting it into formulas, such as the entropy formula which requires temperature in Kelvin. This ensures consistency and accuracy across scientific calculations.

Entropy Formula

Entropy refers to the degree of disorder or randomness in a system. In the context of a phase transition, the entropy change measures how much disorder is introduced when a substance melts or solidifies. The formula itself is relatively simple:\[\Delta S = \frac{q_{\text{rev}}}{T}\]Where:

• \(\Delta S\) is the change in entropy.
• \(q_{\text{rev}}\) is the reversible heat exchange, or the latent heat of fusion.
• \(T\) is the temperature in Kelvin at which the phase transition occurs.

To compute the entropy change during the melting of a solid, insert the values of latent heat and temperature in the above formula. For example, with a latent heat of 2930 J/mol and a temperature of 300.15 K, you calculate it as:\[\Delta S = \frac{2930}{300.15} \approx 9.76 \text{ JK}^{-1} \text{ mol}^{-1}\]This calculated value reflects the entropy change resulting from the melting process, providing insights into the energy dynamics involved in phase changes.