Problem 46
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) \(2.013\)
Step-by-Step Solution
Verified Answer
The entropy change is approximately \( 21.98 \ \mathrm{JK}^{-1} \ \mathrm{mol}^{-1} \), so the correct option is (b).
1Step 1: 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.
2Step 2: 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.
3Step 3: 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 \( \Delta S = \frac{6000}{273.15} \ \mathrm{JK}^{-1} \ \mathrm{mol}^{-1} \).
4Step 4: 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 ChangeTemperature ConversionEntropy Formula
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:
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 \( \Delta H \).
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:
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.
To convert Celsius to Kelvin, add 273.15 to the Celsius temperature.
For example, \( 0^{\circ} \mathrm{C} \) becomes:
- \( T_{K} = 0 + 273.15 \)
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:
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:
Understanding this formula helps in many areas of science and engineering, especially when evaluating system changes.
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.
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:
- \( \Delta S = \frac{6000}{273.15} = 21.98 \, \mathrm{JK}^{-1} \, \mathrm{mol}^{-1} \)
Understanding this formula helps in many areas of science and engineering, especially when evaluating system changes.
Other exercises in this chapter
Problem 44
Standard enthalpy and standard entropy changes for the oxidation of ammonia at \(298 \mathrm{~K}\) are \(-382.64 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) and \(-145
View solution Problem 45
The molar heat capacity of water at constant pressure, \(\mathrm{C}\), is \(75 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\). When \(1.0 \mathrm{~kJ}\) of heat is suppl
View solution Problem 47
For the reaction, \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \ma
View solution Problem 48
2 mole of an ideal gas at \(27^{\circ} \mathrm{C}\) temperature is expanded reversibly from \(2 \mathrm{~L}\) to \(20 \mathrm{~L}\). Find entropy change in cal.
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