Question

The molar entropy, \(S^{\circ},\) of \(\mathrm{HCl}\) at \(298 \mathrm{K}\) is \(186.9 \mathrm{JK}^{-1} \mathrm{mol}^{-1} .\) Find \(S^{\circ}(350 \mathrm{K})\) if \(C_{P}(298-350 \mathrm{K})\) is \(29.1 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).

Short answer

The change in molar entropy, using a constant \(C_P\), leads to an increase due to temperature rise.

Step-by-step solution

Understand the Relationship

To find the molar entropy \(S^{\circ}\) at a different temperature, we use the formula: \[S^{\circ}(T_2) = S^{\circ}(T_1) + \int_{T_1}^{T_2} \frac{C_P}{T} \, dT\]where \(S^{\circ}(T_1)\) is the molar entropy at the initial temperature,—298 K, which is 186.9 J/K/mol in this problem. \(C_P\) is constant over the temperature range.

Set Up the Integral

Since \(C_P\) is constant, the integral simplifies to:\[\int_{298}^{350} \frac{C_P}{T} \, dT = C_P \int_{298}^{350} \frac{1}{T} \, dT\]To solve the integral, we know that:\[\int \frac{1}{T} \, dT = \ln{|T|}\]

Key concepts

Heat Capacity

Heat capacity is a fundamental concept in thermodynamics that describes the amount of heat required to change a substance's temperature by a certain amount. It plays a crucial role in determining how substances react to thermal energy. The heat capacity, specifically the molar heat capacity at constant pressure (\(C_P\)), measures the energy needed to raise the temperature of one mole of a substance by one Kelvin while maintaining constant pressure. This value can vary depending on the substance and is vital in calculating changes in entropy across a temperature range.In the context of our problem, knowing the molar heat capacity (\(C_P = 29.1 \, \mathrm{J/K/mol}\)) over the temperature range from 298 K to 350 K allows us to compute the change in entropy when transitioning between these temperatures. Since \(C_P\) is considered constant in this range, it simplifies calculations, making it easier to integrate functions of temperature.

Thermodynamics

Thermodynamics is the study of energy, heat, and work, and how they interact with each other in physical systems. It involves principles and laws that govern these interactions and is central to processes in physics and chemistry. In thermodynamics, entropy is a key concept, representing the degree of disorder or randomness in a system. It is also a measure of energy dispersal at a specific temperature. One important aspect of thermodynamics is its laws, such as the First and Second Laws of Thermodynamics:

• The First Law, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed, only transferred or transformed.
• The Second Law states that in any energy transfer, entropy tends to increase over time, indicating that natural processes favor increasing disorder.

This foundational framework assists with systems needing reversible processes, which directly relates to understanding how entropy and heat capacity are computed and utilized, as in the given exercise.

Entropy Calculation

Calculating entropy changes with temperature is an essential exercise in understanding thermodynamic systems. In the given problem, we start with an initial molar entropy \(S^{\circ}(298 \mathrm{K}) = 186.9 \, \mathrm{J/K/mol}\), and our goal is to find \(S^{\circ}(350 \mathrm{K})\).The formula used is:\[S^{\circ}(T_2) = S^{\circ}(T_1) + \int_{T_1}^{T_2} \frac{C_P}{T} \, dT\]Here, we substitute the known values:

• \(S^{\circ}(298 \mathrm{K}) = 186.9 \, \mathrm{J/K/mol}\)
• \(C_P = 29.1 \, \mathrm{J/K/mol}\)

Since \(C_P\) is constant here, you can simplify the integral to:\[C_P \int_{298}^{350} \frac{1}{T} \, dT = C_P [\ln(T)]_{298}^{350}\]Carrying out this calculation provides the change in entropy over the temperature range, giving us \(S^{\circ}(350 \mathrm{K})\). This process illustrates the quantitative analysis behind entropy changes in thermodynamic systems, highlighting the practical application of these concepts in chemistry and physics.