Problem 47 - Chemistry The Molecular Science

Question

Hydrogen burns in air with considerable heat transfer to the surroundings. Consider the decomposition of water to gaseous hydrogen and oxygen. Without doing any calculations, and basing your prediction on the enthalpy change and the entropy change, is this reaction product-favored at \(25^{\circ} \mathrm{C} ?\) Explain your answer briefly.

Step-by-Step Solution

Answer

The reaction is not product-favored at \(25^{\circ} \mathrm{C}\) due to positive \(\Delta H\).

Step 1: Understanding the Reaction

The decomposition of water into gaseous hydrogen and oxygen can be represented by the equation: \(2\text{H}_2\text{O}(l) \rightarrow 2\text{H}_2(g) + \text{O}_2(g)\). This reaction involves breaking the strong O-H bonds in water to form gaseous hydrogen and oxygen molecules.

Step 2: Analyzing Enthalpy Change

In the reaction, energy is required to break the bonds of water molecules. Therefore, the enthalpy change \((\Delta H)\) for the decomposition of water is positive, indicating an endothermic process. This means the reaction absorbs heat, which generally works against it being product-favored.

Step 3: Analyzing Entropy Change

Entropy \((\Delta S)\) measures the disorder or randomness of a system. The decomposition increases the number of gas molecules (from liquid water to gaseous hydrogen and oxygen), leading to increased disorder. Thus, the entropy change is positive, which favors the products.

Step 4: Applying Gibbs Free Energy Concept

The spontaneity of a reaction at constant temperature and pressure can be determined using Gibbs free energy change \((\Delta G)\), calculated as \(\Delta G = \Delta H - T\Delta S\). For a reaction to be spontaneous (product-favored), \(\Delta G\) must be negative.

Step 5: Prediction Based on Enthalpy and Entropy

While \(\Delta H\) is positive, making the reaction less favorable, \(\Delta S\) is also positive, which helps in favoring products. However, since \(\Delta H\) is typically large due to the endothermic nature of the reaction, the temperature \(T\Delta S\) term would need to be very large to make \(\Delta G\) negative. At \(25^{\circ}\mathrm{C}\), it is unlikely that \(\Delta G\) will be negative.

Key Concepts

Enthalpy Change

In thermodynamics, enthalpy change (\(\Delta H\)) represents the heat absorbed or released during a chemical reaction at constant pressure. It helps us understand whether a reaction requires or releases energy. When the decomposition of water into hydrogen and oxygen occurs, the reaction absorbs energy to break the O-H bonds, making \(\Delta H\) positive. This is known as an endothermic process.

Endothermic reactions absorb heat from the surroundings, and they tend to be less favorable because they need an external energy source. In contrast, exothermic reactions release energy, often making them more spontaneous and product-favored. The positive enthalpy change in this reaction implies the system requires a significant amount of energy input, working against it being favorable. Understanding whether a reaction is endothermic or exothermic helps predict its likelihood of proceeding without external intervention.

Entropy Change

Entropy (\(\Delta S\)) measures the degree of disorder or randomness in a system. The more disordered the system, the higher the entropy. In the decomposition of water, the entropy change is positive because the reaction goes from fewer molecules in a liquid state to more molecules in a gaseous state.

In this transformation:

• Liquid water (\(2\text{H}_2\text{O}(l)\)) has a relatively low entropy due to the ordered structure of the liquid.
• The formation of gaseous hydrogen \((2\text{H}_2(g))\) and oxygen \((\text{O}_2(g))\) increases randomness as gases are much more disordered than liquids.

A positive entropy change usually favors the formation of products, as systems naturally evolve towards greater disorder. This tendency aligns with the concept that increasing randomness leads to more energetically favorable conditions.

Gibbs Free Energy

Gibbs Free Energy (\(\Delta G\)) is a key thermodynamic function used to predict the spontaneity of a reaction. It combines both enthalpy and entropy changes according to the equation:
\[ \Delta G = \Delta H - T\Delta S \]
Here, \(T\) represents the temperature in Kelvin.

For a reaction to be spontaneous or product-favored, \(\Delta G\) must be negative. If \(\Delta G\) is positive, the reaction is not spontaneous under the given conditions. In the decomposition of water:

• The positive \(\Delta H\) (endothermic) makes it less favorable.
• The positive \(\Delta S\) (increased disorder) helps favor the product.

At standard conditions, such as \(25^{\circ}\mathrm{C}\), the temperature term \(T\Delta S\) is typically not enough to overcome the large \(\Delta H\). Thus, \(\Delta G\) remains positive, indicating a non-spontaneous reaction without additional energy input. Understanding \(\Delta G\) allows chemists to determine whether a reaction will occur naturally or if it requires special conditions.