Question

The element cesium (Cs) freezes at \(28.4^{\circ} \mathrm{C}\), and its molar enthalpy of fusion is \(\Delta H_{\text {fus }}=2.09 \mathrm{~kJ} / \mathrm{mol}\). (a) When molten cesium solidifies to \(\mathrm{Cs}(\mathrm{s})\) at its normal melting point, is \(\Delta S\) positive or negative? (b) Calculate the value of \(\Delta S\) when \(15.0 \mathrm{~g}\) of \(\mathrm{Cs}(l)\) solidifies at \(28.4^{\circ} \mathrm{C}\).

Short answer

When cesium freezes, the change in entropy (ΔS) is negative as the particles become more ordered in the solid state. The value of ΔS when 15.0 g of cesium solidifies at 28.4°C is approximately 0.782 J/K.

Step-by-step solution

Part (a): Determine the sign of ΔS

When a substance changes from liquid to solid, its entropy decreases because the particles become more ordered in a solid state. Therefore, the change in entropy (ΔS) will be negative.

Part (b): Calculate ΔS

To calculate ΔS, we will use the formula: ΔS = - (ΔH_fus) / T where ΔH_fus is the molar enthalpy of fusion, and T is the temperature in Kelvin. Step 1: Convert the temperature to Kelvin T(K) = T(°C) + 273.15 T = 28.4 + 273.15 = 301.55 K Step 2: Calculate the number of moles of cesium Molar mass of cesium (Cs) = 132.90 g/mol n = mass / molar mass n = 15.0 g / 132.90 g/mol = 0.1129 mol Step 3: Calculate ΔS ΔS = - (ΔH_fus) / T ΔS = - (-2.09 kJ/mol) / 301.55 K We need to convert kJ to J: ΔS = - (-2090 J/mol) / 301.55 K ΔS = 6.93 J/(mol·K) Step 4: Calculate ΔS for 15.0 g of Cs ΔS_total = n * ΔS ΔS_total = 0.1129 mol * 6.93 J/(mol·K) ΔS_total = 0.782 J/K The value of ΔS when 15.0 g of Cs solidifies at 28.4°C is approximately 0.782 J/K.

Key concepts

Molar Enthalpy of Fusion

The molar enthalpy of fusion, denoted as \( \Delta H_{\text{fus}} \), is the amount of energy needed to change 1 mole of a substance from solid to liquid at its melting point. This concept is crucial in understanding phase changes as it represents the energy required to overcome intermolecular forces keeping the solid structure intact. For cesium, this value is given as 2.09 kJ/mol.
The process involves heat absorption to break the orderly crystal lattice of a solid, turning it into a less ordered liquid. During solidification, the reverse process occurs, and energy equivalent to the molar enthalpy of fusion is released as the atoms settle into a more ordered crystalline structure. Remember, even though cesium absorbs or releases heat during fusion or solidification, its temperature remains constant at the melting/freezing point until the phase transition is complete.
Enticingly, for most substances, including cesium, the act of solidifying releases this energy back and is a process that can be quantified, giving us deeper insight into the energetics of phase changes.

Phase Transition

Phase transitions refer to the process of changing a substance from one state of matter to another, such as from solid to liquid or liquid to gas. In the case of cesium, transitioning from a liquid to a solid state is a typical example of a phase change known as freezing. These transitions are typically governed by temperature and pressure conditions.
At cesium’s melting point of 28.4°C, the energy provided by the surroundings is just enough to maintain the cesium atoms in a liquid state. When this energy is withdrawn, or the liquid cesium cools, it solidifies into a regular crystalline structure, reducing the movement of its atoms and decreasing its entropy.
This phase transition is critical in thermodynamics because it involves significant energy exchange. The molar enthalpy of fusion dictates the energy change throughout the transition, while entropy change provides insights into molecular order adjustments. Understood in these terms, phase transitions not only highlight the physical changes in state but also embody larger principles of energy redistribution and molecular dynamics at play.

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. Entropy and enthalpy are central concepts within this field, especially when discussing phase transitions like the freezing of cesium.In thermodynamics, phase transitions are framed by the laws that dictate energy conservation and interactions. The first law, essentially an expression of energy conservation, tells us that energy in a closed system can neither be created nor destroyed, only transformed. Cesium absorbing or releasing heat during phase changes exemplifies this as internal energy fluctuations dictate the system's state or phase change.
The change in entropy, \( \Delta S \), echoes the second law of thermodynamics, where processes in an isolated system will tend toward a state of maximum entropy. However, in controlled processes like freezing, where cesium turns from liquid to a more ordered solid, we see a local decrease in entropy, balanced by the energy exchange with the environment.
These thermodynamic principles help us predict and quantify the effects of temperature and energy changes on various phases, giving us a comprehensive look at how energy transformations define the properties and behaviors of substances like cesium during phase transitions.

Calculating ΔS

The calculation of entropy change, \( \Delta S \), is essential to understanding how ordered or disordered a system becomes during phase transitions. For cesium, when 15.0 g of the element transitions from liquid to solid, we calculate \( \Delta S \) using the formula: \[ \Delta S = - \frac{\Delta H_{\text{fus}}}{T} \]where \( \Delta H_{\text{fus}} \) is the molar enthalpy of fusion (in J/mol) and \( T \) is the temperature in Kelvin.First, convert the given temperature from Celsius to Kelvin, ensuring accuracy in the thermodynamic equations. The temperature at which cesium freeze is 28.4°C, which translates into 301.55 K.
Next, determine the number of moles of cesium using its molar mass (132.90 g/mol): \[ n = \frac{15.0 \text{ g}}{132.90 \text{ g/mol}} = 0.1129 \text{ mol} \]Apply the previously mentioned formula by inserting \( \Delta H_{\text{fus}} \) as 2.09 kJ/mol (converted to 2090 J/mol) and \( T \) as 301.55 K to calculate the entropy change per mole. Finally, multiply the result by the moles of cesium to find the total entropy change: \[ \Delta S_{\text{total}} = 0.1129 \text{ mol} \times 6.93 \text{ J/(mol·K)} = 0.782 \text{ J/K} \]By calculating \( \Delta S \), we gain insights into the entropy, representing a shift in order within the cesium during freezing, highlighting a critical aspect of phase transitions.