Problem 54 - Chemistry The Molecular Science

Question

Use a mathematical equation to show how the statement leads to the conclusion cited: If \(\Delta_{\mathrm{r}} H\) and \(\Delta_{\mathrm{r}} S\) have the same sign, then the magnitude of \(T\) determines whether \(\Delta_{\mathrm{r}} G\) is negative and whether the reaction is product-favored.

Step-by-Step Solution

Answer

Temperature (\( T \)) dictates if \( \Delta_{\mathrm{r}} G \) is negative when \( \Delta_{\mathrm{r}} H \) and \( \Delta_{\mathrm{r}} S \) have the same sign.

Step 1: Understand the Relationship

The relationship between enthalpy change (\( \Delta_{\mathrm{r}} H \)), entropy change (\( \Delta_{\mathrm{r}} S \)), and Gibbs free energy change (\( \Delta_{\mathrm{r}} G \)) is given by the equation: \[\Delta_{\mathrm{r}} G = \Delta_{\mathrm{r}} H - T \Delta_{\mathrm{r}} S\]This equation shows how the Gibbs free energy depends on both enthalpy, entropy, and temperature (\( T \)).

Step 2: Determine Same Sign Implications

When \( \Delta_{\mathrm{r}} H \) and \( \Delta_{\mathrm{r}} S \) have the same sign, there are two possibilities:1. Both are positive: This means \( \Delta_{\mathrm{r}} G \) becomes more negative as \( T \) increases (since \(-T\Delta_{\mathrm{r}} S\) becomes more negative), making the reaction more product-favored at high temperatures.2. Both are negative: \( \Delta_{\mathrm{r}} G \) becomes less negative as \( T \) increases (since \(-T\Delta_{\mathrm{r}} S\) becomes less negative), making the reaction more product-favored at low temperatures.

Step 3: Analyzing Temperature's Role

The sign of \( \Delta_{\mathrm{r}} G \) determines if the reaction is spontaneous (product-favored). For product-favored reactions, \( \Delta_{\mathrm{r}} G \) must be negative. Therefore, at a lower temperature, \( \Delta_{\mathrm{r}} G \) may be negative when \( \Delta_{\mathrm{r}} H \) and \( \Delta_{\mathrm{r}} S \) are negative, while at higher temperatures, it may be negative when both are positive.

Key Concepts

Enthalpy Change

Enthalpy change, often symbolized as \( \Delta_{\mathrm{r}} H \), represents the heat absorbed or released during a reaction at constant pressure. It plays a crucial role in determining the favorability of a reaction. If the enthalpy change is negative, the reaction releases heat, indicating an exothermic reaction. On the other hand, if \( \Delta_{\mathrm{r}} H \) is positive, the reaction absorbs heat, making it endothermic.

In the context of Gibbs free energy, understanding enthalpy helps us see how heat interactions impact whether a reaction proceeds spontaneously. When \( \Delta_{\mathrm{r}} H \) is negative, it can contribute to making \( \Delta_{\mathrm{r}} G \) negative, thus favoring the formation of products. Conversely, a positive \( \Delta_{\mathrm{r}} H \) can make it harder for \( \Delta_{\mathrm{r}} G \) to be negative unless other factors, such as entropy, compensate.

Entropy Change

Entropy change \( \Delta_{\mathrm{r}} S \) is a measure of the disorder or randomness introduced during a chemical reaction. A positive entropy change indicates an increase in disorder, while a negative \( \Delta_{\mathrm{r}} S \) suggests a decrease in randomness. In terms of Gibbs free energy, entropy change is key as it factors in how surroundings and molecules are impacted by a reaction.

Because entropy reflects molecular dynamics, its effect varies based on temperature. For instance, if both \( \Delta_{\mathrm{r}} H \) and \( \Delta_{\mathrm{r}} S \) are positive, higher temperatures will amplify the influence of entropy (\(-T\Delta_{\mathrm{r}} S\)), potentially leading to negative \( \Delta_{\mathrm{r}} G \) and making the reaction favorable. Thus, understanding entropy change is vital for predicting when reactions will succeed based on inherent molecular changes.

Temperature Effects on Reactions

Temperature is a fundamental factor that affects the spontaneity of chemical reactions through its impact on Gibbs free energy \( \Delta_{\mathrm{r}} G \). The equation \( \Delta_{\mathrm{r}} G = \Delta_{\mathrm{r}} H - T \Delta_{\mathrm{r}} S \) shows that temperature not only shifts the balance between enthalpy and entropy but can determine if a process is product-favored.

When \( \Delta_{\mathrm{r}} H \) and \( \Delta_{\mathrm{r}} S \) have the same sign:

• If they are both positive, increasing temperature makes \(-T\Delta_{\mathrm{r}} S\) more negative, which can make \( \Delta_{\mathrm{r}} G \) negative—favoring the reaction at high temperatures.
• If they are both negative, lowering the temperature helps \( \Delta_{\mathrm{r}} G \) become negative, favoring the reaction at low temperatures.

This temperature-dependent behavior helps chemists control and optimize reactions for desired outcomes. Studying these effects provides vital insight into energy management and reaction predictability across different conditions.