Entropy Calculations: Videos & Practice Problems

Understanding the entropy of a system and its surroundings is crucial in predicting the spontaneity of a chemical reaction. Entropy, a measure of disorder or randomness, is denoted in units of joules per Kelvin (J/K). The total entropy change in the universe (ΔS_universe) is the sum of the entropy change of the system (ΔS_system) and the surroundings (ΔS_surroundings). A positive ΔS_universe indicates a spontaneous process, aligning with the second law of thermodynamics, which states that the entropy of the universe tends to increase. Conversely, a negative value suggests a non-spontaneous process, while a value of zero indicates equilibrium.

The entropy change of the surroundings can be calculated using the formula:

ΔS_surroundings=-ΔH_reaction/T where ΔH_reaction is in kilojoules and T is the temperature in Kelvin, assuming constant pressure and temperature conditions.

To calculate the entropy change of a system, the standard molar entropy (S°) of each substance is used. These values, which are always greater than zero, are provided due to the vast number of substances and their unique entropies. The standard entropy change of a reaction (ΔS°_reaction)

Entropy of the Universe

When calculating entropy, it is essential to consider the total entropy change in the universe, which can be expressed with the following equation:

\[ \Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} \]

In this equation, \(\Delta S_{universe}\) represents the total change in entropy, while \(\Delta S_{system}\) refers to the change in entropy of the system (or reaction), and \(\Delta S_{surroundings}\) denotes the change in entropy of the surroundings. The units for entropy change are typically measured in joules per Kelvin (J/K).

The total change in entropy can yield three possible outcomes: positive, negative, or zero. A positive value indicates that the entropy of the universe is increasing, aligning with the second law of thermodynamics, which states that the entropy of the universe tends to increase over time. This increase signifies a spontaneous process. Conversely, a negative value suggests a decrease in entropy, indicating a non-spontaneous process. When the total change in entropy equals zero, the system is at equilibrium, meaning it is neither spontaneous nor non-spontaneous.

Understanding these conditions is crucial for predicting the spontaneity of reactions and the overall behavior of systems in thermodynamics.

Total Entropy Example

To determine the spontaneity of a reaction, we calculate the total entropy change, denoted as ΔS_total. This is done by summing the entropy change of the surroundings (ΔS_surroundings) and the standard entropy change of the reaction (ΔS_reaction). In this case, the change in entropy of the surroundings is given as 2.7 joules per Kelvin, while the standard entropy change of the reaction is -450 kilojoules per Kelvin.

First, we need to convert the standard entropy change from kilojoules to joules. Since 1 kilojoule equals 1000 joules, we convert -450 kilojoules to:

ΔS_reaction = -450,000 joules per Kelvin.

Now, we can calculate the total entropy change:

ΔS_total = ΔS_surroundings + ΔS_reaction = 2.7 joules per Kelvin - 450,000 joules per Kelvin.

This results in:

ΔS_total = -449,997.3 joules per Kelvin.

Since the total entropy change is negative, this indicates that the reaction is non-spontaneous. In thermodynamics, a negative total entropy change suggests that the reaction does not favor the formation of products under standard conditions.

Entropy of Surroundings Formula

Understanding the entropy of the surroundings is crucial in thermodynamics, particularly when analyzing chemical reactions. The change in the entropy of the surroundings can be calculated using the formula:

$\Delta S_{surroundings} = -\frac{\Delta H_{reaction}}{T}$

In this equation, $\Delta S_{surroundings}$ represents the change in entropy of the surroundings, while $\Delta H_{reaction}$ denotes the change in enthalpy of the reaction, typically measured in kilojoules (kJ). The temperature (T) must be expressed in Kelvin (K) to ensure consistency in units.

This relationship highlights that the entropy change of the surroundings is inversely related to the temperature at which the reaction occurs. As the enthalpy change of the reaction increases (indicating a more exothermic reaction), the entropy of the surroundings also increases, reflecting the dispersal of energy into the environment. Conversely, at higher temperatures, the impact of the enthalpy change on the entropy of the surroundings diminishes, illustrating the importance of temperature in thermodynamic calculations.

Entropy of Surroundings Example

To determine the change in the entropy of the universe for a given chemical reaction at 32 degrees Celsius, we start with the provided values: the change in enthalpy (ΔH) of the reaction is -140 kJ, and the change in entropy (ΔS) of the reaction is 3.6 J/K. The total change in entropy of the universe (ΔS_universe) can be expressed as the sum of the change in entropy of the surroundings (ΔS_surroundings) and the change in entropy of the reaction (ΔS_reaction).

The formula for the change in entropy of the surroundings is given by:

ΔS_surroundings = -ΔH / T

where ΔH must be in joules and T is the temperature in Kelvin.

First, we convert the enthalpy from kilojoules to joules:

ΔH = -140 kJ × 1000 = -140,000 J

Next, we convert the temperature from degrees Celsius to Kelvin:

T = 32 °C + 273.15 = 305.15 K

Now, we can calculate the change in entropy of the surroundings:

ΔS_surroundings = -(-140,000 J) / 305.15 K = 458.79 J/K

With both components known, we can now find the total change in entropy of the universe:

ΔS_universe = ΔS_surroundings + ΔS_reaction

ΔS_universe = 458.79 J/K + 3.6 J/K = 462.39 J/K

Thus, the change in the entropy of the universe for the reaction at 32 degrees Celsius is 462.39 J/K.

Entropy of the System

Entropy is a crucial concept in thermodynamics, particularly when analyzing chemical reactions. Each substance involved in a reaction has a standard molar entropy, denoted as \( S^\circ \), which is a measure of the disorder or randomness of the system. These values are typically provided in reference materials, as memorizing them for all compounds is impractical. Unlike standard molar enthalpies, represented as \( \Delta H^\circ \), the standard molar entropy of substances in their natural state is always greater than zero.

The change in standard entropy for a reaction can be calculated using the following formula:

\[\Delta S^\circ_{\text{reaction}} = \Sigma n S^\circ_{\text{products}} - \Sigma m S^\circ_{\text{reactants}}\]

In this equation, \( \Delta S^\circ_{\text{reaction}} \) is expressed in joules per Kelvin (J/K). The symbol \( \Sigma \) represents the summation of the standard molar entropies of the products and reactants, where \( n \) indicates the number of moles of each product and \( m \) indicates the number of moles of each reactant. The standard molar entropy values are given in joules per mole per Kelvin (J/(mol·K)).

When calculating entropy changes, it is essential to consider standard conditions, which are defined as a temperature of 25 degrees Celsius (298 K) and a pressure of 1 atmosphere. By applying the entropy change formula, one can effectively determine the entropy of the system for various chemical reactions, enhancing the understanding of the spontaneity and feasibility of these processes.

Entropy of Reaction Example

To calculate the change in standard entropy (\(\Delta S^\circ\)) for the reaction of nitrogen monoxide (NO) and oxygen (O₂) producing nitrogen dioxide (NO₂), we start by identifying the standard molar entropies of each substance involved. The reaction can be represented as:

2 NO(g) + O₂(g) → 2 NO₂(g)

Given the standard molar entropies:

• NO: 210.8 J/(mol·K)
• O₂: 205.2 J/(mol·K)
• NO₂: 240.1 J/(mol·K)

To find the change in standard entropy, we apply the formula:

\[\Delta S^\circ = S^\circ_{\text{products}} - S^\circ_{\text{reactants}}\]

Calculating the entropy for the products:

\[S^\circ_{\text{products}} = 2 \times S^\circ_{\text{NO}_2} = 2 \times 240.1 \, \text{J/(mol·K)} = 480.2 \, \text{J/K}\]

Calculating the entropy for the reactants:

\[S^\circ_{\text{reactants}} = 2 \times S^\circ_{\text{NO}} + 1 \times S^\circ_{\text{O}_2} = 2 \times 210.8 \, \text{J/(mol·K)} + 1 \times 205.2 \, \text{J/(mol·K)} = 421.6 \, \text{J/K} + 205.2 \, \text{J/K} = 626.8 \, \text{J/K}\]

Now, substituting these values into the entropy change formula:

\[\Delta S^\circ = 480.2 \, \text{J/K} - 626.8 \, \text{J/K} = -146.6 \, \text{J/K}\]

This negative value indicates a decrease in entropy for the reaction, which is consistent with the formation of fewer gas molecules from the reactants. The enthalpy of the reaction, given as -114.14 kJ, is not necessary for this calculation, as the change in entropy is determined solely by the standard molar entropies of the reactants and products.

For the following reaction at 27 °C, calculate ∆S°_rxn, ∆S_surr, and ∆S_tot. Determine if reaction is favorable.
Fe₂O₃ (s) + 3 H₂ (g) → 2 Fe (s) + 3 H₂O (g) ∆H_rxn = 98.8 kJ

• A. ∆S°_rxn = 0.1415 J/K, ∆S_surr = 0.3292 J/K, ∆S_tot = −0.1877 J/K, not favorable
• B. ∆S°_rxn = 141.5 J/K, ∆S_surr = −329.17 J/K, ∆S_tot = −187.67 J/K, not favorable
• C. ∆S°_rxn = 70.75 J/K, ∆S_surr = 164.58 J/K, ∆S_tot = −93.84 J/K, favorable
• D. ∆S°_rxn = 283.0 J/K, ∆S_surr = 658.34 J/K, ∆S_tot = −375.34 J/K, favorable

Answer: (B)

Here’s what students ask on this topic:

Which of the following has the largest standard molar entropy, S° (298.15 K)?

To determine which substance has the largest standard molar entropy (S°) at 298.15 K, you would typically compare the S° values of the substances in question. Entropy is a measure of the disorder or randomness in a system, and in general, the standard molar entropy increases with the complexity of the molecule, the number of atoms in the molecule, and the number of possible microstates (ways the energy can be distributed within the molecule).

Here are some general trends that can help you determine which substance might have the largest S°:

1. Gases have higher entropy than liquids, which in turn have higher entropy than solids.
2. Among substances in the same state of matter, those with larger, more complex molecules tend to have higher entropy.
3. Substances with more atoms or a greater variety of atoms generally have higher entropy.
4. For isomers, the more branched isomers tend to have higher entropy than linear ones.

Without specific substances to compare, these guidelines can help predict which might have the highest standard molar entropy. If you have a list of specific substances, you would look up their standard molar entropy values in a reference source such as a textbook or a chemical data database to make a direct comparison.

Which of the following have a standard molar entropy, $s^{°}$, of zero?

The standard molar entropy ($S^{°}$) of zero is a unique value assigned to a perfectly ordered crystalline substance at absolute zero temperature (0 Kelvin) according to the third law of thermodynamics. At this temperature, the entropy, which is a measure of disorder or randomness in a system, is considered to be at its minimum because the particles in the substance have the least amount of motion and are in their most ordered state.

No real substances actually have a standard molar entropy of zero at temperatures above absolute zero because there is always some degree of thermal motion contributing to entropy. However, for the sake of calculations and conventions in thermodynamics, the standard molar entropy of an element in its most stable form at 1 bar of pressure and a specified standard temperature (usually 298.15 K) is often compared to this theoretical zero point.

It's important to note that while the third law provides a reference point, actual standard molar entropy values for substances at standard temperature and pressure are always positive.

Which process is necessarily driven by an increase in the entropy of the surroundings?

The process that is necessarily driven by an increase in the entropy of the surroundings is a spontaneous process. In thermodynamics, a process is considered spontaneous if it occurs without needing to be driven by an external force. The second law of thermodynamics states that for a process to be spontaneous, the total entropy change of the system and its surroundings must increase. This means that while the entropy of a system (for example, the substances involved in a chemical reaction) might decrease, the entropy of the surroundings must increase by a greater amount, leading to an overall increase in the total entropy of the universe. This net increase in entropy is what drives spontaneous processes. Examples of such processes include the melting of ice at room temperature and the diffusion of a perfume from one part of a room to another.

Which of the following would lead to the highest gain in entropy in the surroundings?

The gain in entropy in the surroundings is highest when a process involves the transfer of a large amount of heat to the surroundings at a low temperature. According to the second law of thermodynamics, entropy, which is a measure of disorder or randomness, increases with the transfer of heat.

When a process occurs at a lower temperature, the same amount of heat transfer results in a larger change in entropy than it would at a higher temperature. This is because entropy change (ΔS) is calculated by the formula:

$\Delta S=\frac{q}{T}$

where q is the heat transferred and T is the temperature in Kelvin. As T is in the denominator, a lower T results in a higher ΔS for the same amount of heat q.

Therefore, a process that transfers a large amount of heat to the surroundings at the lowest possible temperature would lead to the highest gain in entropy in the surroundings.