Problem 97
Question
Which of these reactions would you expect to be spontaneous at relatively high temperatures? At relatively low temperatures? Explain a. \(C H_{3} O H(1) \rightarrow C H_{3} O H(g)\) b. \(C H_{3} O H(g) \rightarrow C H_{3} O H(1)\) c. \(C H_{3} O H(s) \rightarrow C H_{3} O H(1)\)
Step-by-Step Solution
Verified Answer
Reaction a, where methanol changes from liquid to gas (CH3OH(l) → CH3OH(g)), is spontaneous at relatively high temperatures due to the positive ΔH and ΔS values. Reaction b, where methanol changes from gas to liquid (CH3OH(g) → CH3OH(l)), is spontaneous at relatively low temperatures as energy is released through the negative ΔH value and the negative ΔS value. Reaction c, where methanol changes from solid to liquid (CH3OH(s) → CH3OH(l)), is spontaneous at relatively high temperatures since ΔH and ΔS values are positive for melting.
1Step 1: Reaction a: CH3OH(l) → CH3OH(g)
For this reaction, we are converting the liquid methanol to its gaseous state. The enthalpy change (ΔH) would be positive, as energy is required to break the intermolecular forces in the liquid phase. The entropy change (ΔS) would also be positive, since the gaseous state has more freedom of movement and more randomness.
As T increases, TΔS will increase and dominate the term ΔH - TΔS, therefore making ΔG more negative and the reaction more spontaneous. Hence, this reaction is spontaneous at relatively high temperatures.
2Step 2: Reaction b: CH3OH(g) → CH3OH(l)
For this reaction, we are converting the gaseous methanol to its liquid state. The enthalpy change (ΔH) would be negative here, as energy is released when intermolecular forces are formed during the transition to the more ordered liquid phase. The entropy change (ΔS) would be negative, since the more ordered liquid state has less freedom of movement and less randomness than the gaseous state.
As T increases, TΔS will become more negative and less dominant compared to ΔH in the term ΔH - TΔS, making ΔG less negative or even positive and the reaction non-spontaneous. However, at low temperatures, TΔS will be a less negative term, and the reaction will be spontaneous. Therefore, this reaction is spontaneous at relatively low temperatures.
3Step 3: Reaction c: CH3OH(s) → CH3OH(l)
This reaction involves converting solid methanol to its liquid state, which is melting. The enthalpy change (ΔH) is positive, as energy is required to break the intermolecular forces in the solid phase to form the liquid phase. The entropy change (ΔS) is also positive, since the liquid state has more freedom of movement and more randomness compared to the solid state.
Similar to reaction a, as T increases, the term TΔS will increase and dominate the term ΔH - TΔS, making ΔG more negative and the reaction more spontaneous. So, this reaction is spontaneous at relatively high temperatures.
Key Concepts
Spontaneity of ReactionsTemperature DependencePhase ChangesEnthalpy and Entropy
Spontaneity of Reactions
Gibbs Free Energy (\( \Delta G \)) is the key to understanding whether a chemical reaction is spontaneous or not. A reaction is considered spontaneous if it occurs naturally under given conditions without needing extra energy. We can predict spontaneity by calculating Gibbs Free Energy using the equation:
Each reaction has a different balance of enthalpy and entropy changes, affecting its spontaneity. Higher entropy (more disorder) and lower enthalpy (less energy requirement) contribute to a negative \( \Delta G \), making the reaction naturally favorable.
- \( \Delta G = \Delta H - T\Delta S \)
Each reaction has a different balance of enthalpy and entropy changes, affecting its spontaneity. Higher entropy (more disorder) and lower enthalpy (less energy requirement) contribute to a negative \( \Delta G \), making the reaction naturally favorable.
Temperature Dependence
The temperature plays a crucial role in determining the spontaneity of a reaction. As evident from the Gibbs Free Energy equation, the term \( T\Delta S \) influences the spontaneity greatly depending on the temperature value.
- At high temperatures: Entropy (\( \Delta S \)) plays a more significant role as \( T\Delta S \) outweighs \( \Delta H \) resulting in a more negative \( \Delta G \)
- At low temperatures: Enthalpy (\( \Delta H \)) becomes more dominant, often resulting in a positive \( \Delta G \)
Phase Changes
In a phase change, the substance transitions from one state of matter to another, such as from solid to liquid or liquid to gas. Phase changes are common in reactions and are crucial for understanding the behavior of materials under different conditions.
- Melting and boiling are processes where solid melts into liquid or liquid turns into gas, typically occurring at higher temperatures. These processes often involve positive \( \Delta H \) and \( \Delta S \) values.
- Freezing and condensation, moving from gas to liquid or liquid to solid, happen with negative \( \Delta H \) and \( \Delta S \) values since they transition to more ordered states.
Enthalpy and Entropy
Enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) are thermodynamic properties crucial to understanding reactions.
**Enthalpy** is the heat content of a system. \( \Delta H \) indicates whether a reaction absorbs (endothermic) or releases (exothermic) heat.
**Enthalpy** is the heat content of a system. \( \Delta H \) indicates whether a reaction absorbs (endothermic) or releases (exothermic) heat.
- Positive \( \Delta H \): Endothermic reaction, absorbs heat. Example: ice melting.
- Negative \( \Delta H \): Exothermic reaction, releases heat. Example: fuel burning.
- Positive \( \Delta S \): More disorder, typical in melting or boiling.
- Negative \( \Delta S \): Less disorder, common in freezing or condensation.
Other exercises in this chapter
Problem 94
Use Hess’s law and the changes in enthalpy for the following two generic reactions to calculate ?H for the reaction \(2 \mathrm{A}+\mathrm{B}_{2} \mathrm{C}_{3}
View solution Problem 96
Predict how the entropy of the system changes for the reaction \(\mathrm{CaCO}_{3}(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})
View solution Problem 98
Explain how an exothermic reaction changes the entropy of the surroundings. Does the enthalpy change for such a reaction increase or decrease \(\Delta G_{\text
View solution Problem 99
Calculate \(\Delta G_{\text { system for each process, and state whether }}\) the process is spontaneous or nonspontaneous. a) \(\Delta H_{\text { system }}=145
View solution