Problem 67 - Chemistry The Central Science

Question

A particular constant-pressure reaction is spontaneous at \(390 \mathrm{~K}\). The enthalpy change for the reaction is \(+23.7 \mathrm{~kJ}\). What can you conclude about the sign and magnitude of \(\Delta S\) for the reaction?

Step-by-Step Solution

Answer

Since the given reaction is spontaneous at the given temperature of 390 K, we can conclude that the entropy change, \(\Delta S\), is positive and has a magnitude greater than \(60.77 \frac{\mathrm{J}}{\mathrm{K}}\).

Step 1: Identify the given values and the Gibbs free energy equation

For this problem, we are given the following: - Temperature, \(T = 390 \mathrm{~K}\) - Enthalpy change, \(\Delta H = +23.7 \mathrm{~ kJ}\) The Gibbs free energy equation is: \(\Delta G = \Delta H - T\Delta S\)

Step 2: Rewrite the equation to solve for entropy change, \(\Delta S\)

We know that the reaction is spontaneous, so the Gibbs free energy must be negative \(\Delta G < 0\). We can rewrite the equation to solve for \(\Delta S\): \(\Delta S = \frac{\Delta H - \Delta G}{T}\)

Step 3: Evaluate the inequality for \(\Delta S\)

Since \(\Delta G < 0\), we have the following inequality: \(\Delta S > \frac{\Delta H}{T}\)

Step 4: Substitute the given values and solve

Now substitute the given values of \(\Delta H\) and \(T\) into the inequality: \(\Delta S > \frac{23.7 \mathrm{~kJ}}{390 \mathrm{~K}}\) To make units consistent, we convert kJ to J: \(23.7 \mathrm{~kJ} = 23700 \mathrm{~J}\) \(\Delta S > \frac{23700 \mathrm{~J}}{390 \mathrm{~K}} \) Now, divide 23700 by 390: \(\Delta S > 60.77 \frac{\mathrm{J}}{\mathrm{K}}\) #Conclusion#Since the given reaction is spontaneous at the given temperature of 390 K, we can conclude that the entropy change, \(\Delta S\), is positive and has a magnitude greater than \(60.77 \frac{\mathrm{J}}{\mathrm{K}}\).

Key Concepts

Spontaneity of Reactions

In chemistry, spontaneous reactions are processes that occur naturally without the need for external energy. Whether a reaction is spontaneous can be determined using Gibbs free energy (\( \Delta G \)).
A negative \( \Delta G \) indicates a spontaneous reaction, whereas a positive \( \Delta G \) means the reaction is non-spontaneous. At equilibrium, \( \Delta G = 0 \).
For a reaction, the Gibbs free energy is calculated using the formula:

• \( \Delta G = \Delta H - T\Delta S \)

Here:

• \( \Delta H \) is the enthalpy change, representing the heat absorbed or released.
• \( T \) is the temperature in Kelvin.
• \( \Delta S \) is the entropy change, which measures the disorder or randomness.

When considering spontaneity:

• If \( \Delta H \) is positive and \( \Delta S \) is also positive, the reaction can be spontaneous at high temperatures.
• If \( \Delta H \) is negative and \( \Delta S \) is positive, the reaction is spontaneous at all temperatures.
• If \( \Delta H \) is negative and \( \Delta S \) is negative, the reaction is spontaneous at low temperatures.

This interplay of energy, temperature, and disorder enables us to predict and understand chemical reactions' behavior.

Entropy Change

Entropy change (\( \Delta S \)) is a critical factor in determining the spontaneity of a reaction. Entropy, often described as the measure of disorder, signifies how disorganized or spread out energy is in a system. An increase in entropy (\( \Delta S > 0 \)) generally implies a more spontaneous process in nature.
For spontaneous reactions, the entropy change plays a vital role. Even if a reaction absorbs heat (\( \Delta H > 0 \)) and may not seem favorable, a large positive \( \Delta S \) can drive the reaction to be spontaneous. This principle arises in processes like the melting of ice, where the solid state is more structured than the liquid state, leading to an increase in disorder.
Understanding \( \Delta S \) involves recognizing that nature tends to favor processes that result in increased randomness. In practical terms:

• A system tends to move from an ordered to a disordered state.
• Both the system and its surroundings' entropy changes are pivotal for overall spontaneity.
• Entropy often increases in reactions where gases are produced.

In calculations, we link entropy to temperature and enthalpy through Gibbs free energy, considering all these variables gives a comprehensive understanding of reaction behaviors.

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), refers to the total energy change in a reaction when it occurs at constant pressure. It is represented as the heat absorbed or released during a reaction.
Enthalpy change is crucial for understanding reactions as it directly impacts the Gibbs free energy, which dictates spontaneity:

• The equation for Gibbs free energy, \( \Delta G = \Delta H - T\Delta S \), clearly shows the pivotal role of enthalpy.
• Positive \( \Delta H \), found in endothermic reactions, indicates the system absorbs heat from the surroundings.
• Negative \( \Delta H \), found in exothermic reactions, means the system releases heat.

For our problem, the reaction had a \( \Delta H = +23.7 \text{ kJ} \). This positive value hints the reaction needs heat. However, if the reaction is spontaneous, a significant positive \( \Delta S \) can offset this energy need and favor the reaction at given conditions.
Distinct variations in \( \Delta H \) define types of reactions:

• Endothermic (\( \Delta H > 0 \)) processes, like melting, usually increase system entropy, requiring heat input.
• Exothermic (\( \Delta H < 0 \)) processes, like combustion, release heat, often increasing disorder as heat disperses.

Thus, by understanding enthalpy changes, we comprehend how energy transforms and influences spontaneity in reactions.