Question

What is the entropy change (in \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) ) when \(1 \mathrm{~mol}\) of ice is converted into water at \(0^{\circ} \mathrm{C} ?\) (The enthalpy change for the conversion of ice to liquid water is \(6.0\) \(\mathrm{kJ} \mathrm{mol}^{-1}\) at \(0^{\circ} \mathrm{C}\) ) (a) \(2.198\) (b) \(21.98\) (c) \(20.13\) (d) \(2.013\)

Short answer

The entropy change is approximately \( 21.98 \ \mathrm{JK}^{-1} \ \mathrm{mol}^{-1} \), so the correct option is (b).

Step-by-step solution

Understanding the Concept

The change in entropy, denoted as \( \Delta S \), can be found using the formula \( \Delta S = \frac{\Delta H}{T} \), where \( \Delta H \) is the change in enthalpy and \( T \) is the temperature in Kelvin.

Conversion of Temperature

Convert the given temperature from Celsius to Kelvin. The formula to convert Celsius to Kelvin is \( T_K = T_{\degree C} + 273.15 \). Therefore, the temperature at \( 0^{\circ} \mathrm{C} \) in Kelvin is \( 273.15 \) K.

Applying the Formula to Find Entropy Change

Use the entropy formula \( \Delta S = \frac{\Delta H}{T} \). Here, \( \Delta H \) is given as \( 6.0 \ \mathrm{kJ} \ \mathrm{mol}^{-1} \), which is \( 6000 \ \mathrm{J} \ \mathrm{mol}^{-1} \), and \( T \) is \( 273.15 \ \mathrm{K} \). So, the entropy change is \( \Delta S = \frac{6000}{273.15} \ \mathrm{JK}^{-1} \ \mathrm{mol}^{-1} \).

Calculation

Calculate \( \Delta S \) using the values: \( \Delta S = \frac{6000}{273.15} \approx 21.98 \ \mathrm{JK}^{-1} \ \mathrm{mol}^{-1} \).

Key concepts

Enthalpy Change

Enthalpy change, represented by \( \Delta H \), is a measure of the total energy change in a system when a chemical reaction occurs or a phase change takes place, such as when ice melts into water.
It is a crucial concept in thermodynamics, helping to understand how energy is absorbed or released.
In our example, the enthalpy change for converting 1 mole of ice to liquid water at \( 0^{\circ} \mathrm{C} \) is \( 6.0 \ \mathrm{kJ} \, \mathrm{mol}^{-1} \).
This implies that 6 kJ of energy is required to transform 1 mole of ice to water, indicating the phase change needs energy input.
When dealing with enthalpy changes:

• Energy absorbed is characterized by positive \( \Delta H \).
• Energy released is characterized by negative \( \Delta H \).

In general, understanding enthalpy is foundational for analyzing reactions and predicting their energetic behavior.

Temperature Conversion

Converting temperature scales is an essential skill in thermodynamics. Often, temperatures are provided in degrees Celsius (\( ^{\circ} \mathrm{C} \)), but calculations usually require temperatures in Kelvin (K), the absolute temperature scale.
To convert Celsius to Kelvin, add 273.15 to the Celsius temperature.
For example, \( 0^{\circ} \mathrm{C} \) becomes:

• \( T_{K} = 0 + 273.15 \)

Thus, \( T_{K} = 273.15 \ \mathrm{K} \).
In scientific equations, Kelvin is often used because it starts at absolute zero, the lowest possible temperature where particles have minimal thermal motion.
Remember, always convert to Kelvin for accurate thermodynamic calculations.

Entropy Formula

The entropy formula is crucial in calculating changes in entropy, denoted as \( \Delta S \).
The formula is \( \Delta S = \frac{\Delta H}{T} \), where:

• \( \Delta H \) is the enthalpy change, often given in joules per mole.
• \( T \) is the temperature in Kelvin.

Entropy reflects the degree of disorder or randomness in a system—a higher entropy suggests greater disorder.
In the melting of ice example, \( \Delta H \) is \( 6000 \, \mathrm{J} \, \mathrm{mol}^{-1} \) (converted from \( 6.0 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \)) and \( T \) is \( 273.15 \, \mathrm{K} \).
Thus, the entropy change is calculated as follows:

• \( \Delta S = \frac{6000}{273.15} = 21.98 \, \mathrm{JK}^{-1} \, \mathrm{mol}^{-1} \)

This indicates the amount of disorder added to the system when ice melts to water.
Understanding this formula helps in many areas of science and engineering, especially when evaluating system changes.